🧠:[<think>Okay, let's tackle this problem: Find the sum of all primes ( p ) for which there exists a prime ( q ) such that ( p^{2} + pq + q^{2} ) is a square. Hmm, interesting. So, we need to find prime numbers ( p ) where there's another prime ( q ) making that expression a perfect square. Then sum all such ( p ). Alright.First, since both ( p ) and ( q ) are primes, they can be 2, 3, 5, 7, etc. Maybe starting with small primes could help find possible candidates. Let me note that primes except 2 are all odd. So, 2 is the only even prime, and others are odd. Let's see how that affects the equation.Let's rewrite the equation: ( p^2 + pq + q^2 = k^2 ) for some integer ( k ). So, we need this quadratic combination of primes ( p ) and ( q ) to form a square. Let's think about parity first. If both ( p ) and ( q ) are odd primes, then each of ( p^2 ), ( pq ), and ( q^2 ) are odd. Let's check:- Odd squared is odd: ( odd times odd = odd ).- Odd times odd is odd: same as above.- So, odd + odd + odd = odd + odd = even, then even + odd = odd. Wait, odd + odd is even, then even + odd is odd. So, the sum ( p^2 + pq + q^2 ) would be odd if both primes are odd. But a square can be either even or odd. For example, ( (2m)^2 = 4m^2 ) (even) and ( (2m+1)^2 = 4m^2 + 4m + 1 ) (odd). So, if the sum is odd, then ( k ) must be odd. If one of the primes is 2, let's see:Case 1: ( p = 2 ). Then the equation becomes ( 4 + 2q + q^2 = k^2 ). Let's try some primes ( q ):- ( q = 2 ): 4 + 4 + 4 = 12, which is not a square.- ( q = 3 ): 4 + 6 + 9 = 19, not a square.- ( q = 5 ): 4 + 10 + 25 = 39, not a square.- ( q = 7 ): 4 + 14 + 49 = 67, not a square.- ( q = 11 ): 4 + 22 + 121 = 147, not a square.- Hmm, maybe larger primes? Let's check ( q = 13 ): 4 + 26 + 169 = 199, not a square. Maybe ( q = 17 ): 4 + 34 + 289 = 327, nope. Not looking promising. Maybe ( p = 2 ) doesn't work? Let's check ( q = 1 ), but 1 isn't prime. So, maybe ( p = 2 ) has no solution. But let's not discard it yet. Maybe a larger ( q ), but primes get sparser. Alternatively, maybe ( q ) is even, but only even prime is 2, which we checked. So, ( p = 2 ) might not yield any square. Let's tentatively set that aside.Case 2: ( q = 2 ). Then the equation becomes ( p^2 + 2p + 4 = k^2 ). Let's test primes ( p ):- ( p = 2 ): 4 + 4 + 4 = 12, same as before.- ( p = 3 ): 9 + 6 + 4 = 19, not a square.- ( p = 5 ): 25 + 10 + 4 = 39, not a square.- ( p = 7 ): 49 + 14 + 4 = 67, nope.- ( p = 11 ): 121 + 22 + 4 = 147, not a square.- ( p = 13 ): 169 + 26 + 4 = 199, nope. Similarly, same as above. Maybe no solution here either. So, maybe neither ( p = 2 ) nor ( q = 2 ) work. Let's check with other primes.So, moving on to both primes being odd. Then ( p ) and ( q ) are odd, so let's set ( p = 2m + 1 ), ( q = 2n + 1 ), where ( m, n ) are integers. Then:( p^2 = (2m + 1)^2 = 4m^2 + 4m + 1 )( pq = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 )( q^2 = (2n + 1)^2 = 4n^2 + 4n + 1 )Adding them up:( p^2 + pq + q^2 = (4m^2 + 4m + 1) + (4mn + 2m + 2n + 1) + (4n^2 + 4n + 1) )Combine like terms:- ( 4m^2 + 4mn + 4n^2 )- ( 4m + 2m + 4n + 2n = 6m + 6n )- ( 1 + 1 + 1 = 3 )So total: ( 4m^2 + 4mn + 4n^2 + 6m + 6n + 3 ). Hmm, this seems complicated. Maybe not the best approach. Alternatively, perhaps consider that ( p^2 + pq + q^2 ) can be thought of as a quadratic in one variable. Let's fix one prime and see if the equation can be satisfied.Suppose we fix ( p ) and consider ( q ). Then the equation is ( q^2 + p q + (p^2 - k^2) = 0 ). Wait, but maybe rearranged as ( q^2 + p q + (p^2 - k^2) = 0 ). Treating this as a quadratic equation in ( q ), the discriminant must be a perfect square for ( q ) to be integer (since ( q ) is prime, hence integer). So discriminant ( D = p^2 - 4 times 1 times (p^2 - k^2) = p^2 - 4p^2 + 4k^2 = -3p^2 + 4k^2 ). So, discriminant must be a perfect square.Thus, ( -3p^2 + 4k^2 = m^2 ) for some integer ( m ). Then, ( 4k^2 - m^2 = 3p^2 ). Let's factor left side: ( (2k - m)(2k + m) = 3p^2 ). Since ( p ) is prime, the right side is ( 3 times p times p ). So, the product of two numbers ( (2k - m) ) and ( (2k + m) ) is ( 3p^2 ). Since ( 2k - m ) and ( 2k + m ) are factors of ( 3p^2 ), and ( 2k - m < 2k + m ), we can list possible factor pairs.The factors of ( 3p^2 ) are: 1 and ( 3p^2 ), 3 and ( p^2 ), ( p ) and ( 3p ), as well as considering the sign. But since ( 2k + m ) and ( 2k - m ) must both be positive (since ( k ) and ( m ) are positive integers, given that they are squares), we can ignore negative factors.So, possible factorizations:1. ( 2k - m = 1 ) and ( 2k + m = 3p^2 )2. ( 2k - m = 3 ) and ( 2k + m = p^2 )3. ( 2k - m = p ) and ( 2k + m = 3p )4. ( 2k - m = 3p ) and ( 2k + m = p )But case 4 would lead to ( 2k + m = p ) and ( 2k - m = 3p ). Adding both equations: ( 4k = 4p implies k = p ), subtracting: ( 2m = -2p implies m = -p ). But ( m ) must be positive, so discard this case.Similarly, case 2: ( 2k - m = 3 ) and ( 2k + m = p^2 ). Adding: ( 4k = 3 + p^2 implies k = (p^2 + 3)/4 ). Subtracting: ( 2m = p^2 - 3 implies m = (p^2 - 3)/2 ). Since ( k ) and ( m ) must be integers, ( p^2 + 3 ) must be divisible by 4, and ( p^2 - 3 ) divisible by 2. Let's see:For ( p ) odd: ( p^2 equiv 1 mod 4 ), so ( p^2 + 3 equiv 1 + 3 = 4 equiv 0 mod 4 ). So yes, divisible by 4. Also, ( p^2 - 3 equiv 1 - 3 = -2 equiv 0 mod 2 ), since even. So possible.Thus, in case 2, we can write ( k = (p^2 + 3)/4 ) and ( m = (p^2 - 3)/2 ). Then, from the quadratic equation earlier, we can solve for ( q ). The quadratic equation in ( q ):( q^2 + p q + (p^2 - k^2) = 0 )Wait, but maybe better to recall that the quadratic equation was for ( q ):Original equation ( p^2 + p q + q^2 = k^2 ). So solving for ( q ):( q^2 + p q + (p^2 - k^2) = 0 )Using quadratic formula:( q = [-p pm sqrt{p^2 - 4(p^2 - k^2)}]/2 = [-p pm sqrt{4k^2 - 3p^2}]/2 )But we had ( 4k^2 - 3p^2 = m^2 ), so:( q = [-p pm m]/2 )Since ( q ) is a prime, it must be positive. So:( q = [ -p + m ] / 2 ), since the other solution would be negative. Therefore:( q = (m - p)/2 )But from case 2, ( m = (p^2 - 3)/2 ). Thus:( q = left( frac{p^2 - 3}{2} - p right)/2 = frac{p^2 - 3 - 2p}{4} = frac{p^2 - 2p - 3}{4} )Factor numerator: ( p^2 - 2p - 3 = (p - 3)(p + 1) )Thus:( q = frac{(p - 3)(p + 1)}{4} )But ( q ) must be a prime number. Let's write that:( frac{(p - 3)(p + 1)}{4} = q )So, since ( q ) is prime, the numerator must be 4 times a prime. So, ( (p - 3)(p + 1) = 4q ). Since ( p ) is a prime greater than 2 (since we already considered 2), and primes are odd, ( p - 3 ) and ( p + 1 ) are both even numbers. Let's see:( p ) is odd, so ( p - 3 ) is even (odd minus odd is even), and ( p + 1 ) is even (odd plus odd is even). So, ( (p - 3) = 2a ), ( (p + 1) = 2b ), so ( (2a)(2b) = 4ab = 4q implies ab = q ). Since ( q ) is prime, then either ( a = 1 ) and ( b = q ), or ( a = q ) and ( b = 1 ).But ( a = (p - 3)/2 ), ( b = (p + 1)/2 ). So, possibilities:1. ( (p - 3)/2 = 1 ) and ( (p + 1)/2 = q )2. ( (p - 3)/2 = q ) and ( (p + 1)/2 = 1 )Let's check first case:1. ( (p - 3)/2 = 1 implies p - 3 = 2 implies p = 5 ). Then ( (p + 1)/2 = (6)/2 = 3 = q ). So, ( q = 3 ). Check if this works: ( p = 5 ), ( q = 3 ). Then ( 25 + 15 + 9 = 49 ), which is 7 squared. Perfect. So, this is a valid solution.Second case:2. ( (p + 1)/2 = 1 implies p + 1 = 2 implies p = 1 ). But 1 is not a prime, so invalid. Hence, only the first case gives a valid prime ( p = 5 ).So, from case 2, we get ( p = 5 ), ( q = 3 ). Let's remember that.Now, let's go back to case 1: ( 2k - m = 1 ) and ( 2k + m = 3p^2 ). Adding equations: ( 4k = 1 + 3p^2 implies k = (3p^2 + 1)/4 ). Subtracting: ( 2m = 3p^2 - 1 implies m = (3p^2 - 1)/2 ). Then, similar to before, from quadratic equation:( q = (m - p)/2 = left( frac{3p^2 - 1}{2} - p right)/2 = frac{3p^2 - 1 - 2p}{4} = frac{3p^2 - 2p - 1}{4} )Factor numerator: ( 3p^2 - 2p - 1 ). Let's see if this factors. Let me try:Looking for two numbers ( a ) and ( b ) such that ( a times b = 3 times (-1) = -3 ), and ( a + b = -2 ). Hmm, factors of -3: 1 and -3, sum -2. Yes, so split the middle term:( 3p^2 - 3p + p - 1 = 3p(p - 1) + 1(p - 1) = (3p + 1)(p - 1) ). So, numerator factors as ( (3p + 1)(p - 1) ). Therefore:( q = frac{(3p + 1)(p - 1)}{4} )Again, since ( q ) is prime, this expression must equal a prime. So, ( (3p + 1)(p - 1) = 4q ). Since ( p ) is an odd prime (since we're in the case where both primes are odd), ( p - 1 ) is even, and ( 3p + 1 ) is even (since 3p is odd, plus 1 is even). So, both factors are even. Let ( p - 1 = 2a ), ( 3p + 1 = 2b ). Then, ( (2a)(2b) = 4ab = 4q implies ab = q ). Since ( q ) is prime, either ( a = 1 ), ( b = q ), or ( a = q ), ( b = 1 ).First possibility: ( a = 1 implies p - 1 = 2 implies p = 3 ). Then ( b = (3p + 1)/2 = (10)/2 = 5 ), so ( q = 5 ). Check if this works: ( p = 3 ), ( q = 5 ). Then ( 9 + 15 + 25 = 49 ), which is 7 squared. Perfect. So, another valid solution.Second possibility: ( b = 1 implies 3p + 1 = 2 implies 3p = 1 implies p = 1/3 ), which is not a prime. So invalid. Therefore, only ( p = 3 ) from this case.Thus, case 1 gives ( p = 3 ), ( q = 5 ).Case 3: ( 2k - m = p ), ( 2k + m = 3p ). Adding equations: ( 4k = 4p implies k = p ). Subtracting: ( 2m = 2p implies m = p ). Then, from quadratic equation:( q = (m - p)/2 = (p - p)/2 = 0/2 = 0 ). But ( q = 0 ) is not prime. So, invalid solution. Hence, case 3 gives no valid primes.Thus, from factorizations of ( 3p^2 ), we have solutions ( p = 3 ) and ( p = 5 ).Wait, but let me check if there are other factorizations. The factors we considered were (1, 3p²), (3, p²), (p, 3p). But since ( p ) is prime, are there other factors? For example, if ( p = 3 ), then ( 3p² = 27 ), whose factors are 1, 3, 9, 27. So, but in our previous analysis, we considered 1 and 27, 3 and 9. Wait, so when ( p = 3 ), the factor pair (3, 9) could also be considered. Wait, but earlier, for general ( p ), we considered the possible factorizations as (1, 3p²), (3, p²), (p, 3p). But for specific primes, there might be more factors.Wait, but hold on. For general prime ( p ), ( 3p^2 ) has factors 1, 3, p, 3p, p², 3p². So, factor pairs are (1, 3p²), (3, p²), (p, 3p). Therefore, those are the only possible factor pairs where both factors are positive integers. So, in the case of ( p = 3 ), ( 3p^2 = 27 ), and the factors would be (1, 27), (3, 9). But (3, 9) is (3, p²) when ( p = 3 ). Wait, p² = 9 when p = 3. So, (3, 9) is (3, p²), which was covered in case 2. So, even for specific primes, the same factorization applies. So, we don't need to consider different factorizations for specific primes.Therefore, our previous analysis for cases 1, 2, 3, and 4 should cover all possibilities, leading to solutions ( p = 3 ) and ( p = 5 ).But wait, let's check if there are other primes ( p ) beyond 3 and 5 that could satisfy the equation. Let's test with ( p = 7 ). Maybe there's a prime ( q ) such that ( 49 + 7q + q^2 ) is a square. Let's try some primes ( q ):- ( q = 2 ): 49 + 14 + 4 = 67, not a square.- ( q = 3 ): 49 + 21 + 9 = 79, not a square.- ( q = 5 ): 49 + 35 + 25 = 109, nope.- ( q = 7 ): 49 + 49 + 49 = 147, not a square.- ( q = 11 ): 49 + 77 + 121 = 247, not a square.- ( q = 13 ): 49 + 91 + 169 = 309, nope.- ( q = 17 ): 49 + 119 + 289 = 457, not a square.- ( q = 19 ): 49 + 133 + 361 = 543, nope.Not promising. How about ( p = 7 ), ( q = 5 ): 49 + 35 + 25 = 109. Nope. What about ( p = 7 ), ( q = 11 ): 49 + 77 + 121 = 247. Hmm. Maybe larger primes? Let's check ( q = 23 ): 49 + 161 + 529 = 739, not a square. Doesn't seem like it.What about ( p = 5 ). Let's check if there's another ( q ) besides 3. For ( p = 5 ):- ( q = 2 ): 25 + 10 + 4 = 39, not square.- ( q = 3 ): 25 + 15 + 9 = 49, which is square (as found earlier).- ( q = 5 ): 25 + 25 + 25 = 75, not square.- ( q = 7 ): 25 + 35 + 49 = 109, nope.- So only ( q = 3 ) works for ( p = 5 ).Similarly, ( p = 3 ):- ( q = 2 ): 9 + 6 + 4 = 19, not square.- ( q = 3 ): 9 + 9 + 9 = 27, not square.- ( q = 5 ): 9 + 15 + 25 = 49, which is square (as found earlier).- ( q = 7 ): 9 + 21 + 49 = 79, nope.- So only ( q = 5 ) works for ( p = 3 ).What about ( p = 7 ), is there a prime ( q ) such that ( 49 + 7q + q^2 ) is a square? Let's try to set ( q^2 +7q +49 = k^2 ). Maybe rearrange:( k^2 - q^2 -7q -49 =0 )Alternatively, ( k^2 = q^2 +7q +49 ). Let's see if this can be written as a square. Let's complete the square for the quadratic in ( q ):( q^2 +7q +49 = (q + 3.5)^2 + 49 - 12.25 = (q + 3.5)^2 + 36.75 ). Hmm, not helpful.Alternatively, perhaps set ( k = q + a ), then expand:( (q + a)^2 = q^2 + 2 a q + a^2 = q^2 +7 q +49 )Therefore:( 2 a q + a^2 =7 q +49 )Comparing coefficients:For ( q ): ( 2a =7 implies a = 3.5 )For constants: ( a^2 =49 implies a =7 ). Contradiction. Hence, no such integer ( a ). Therefore, no solution in this case.Alternatively, maybe ( k = q + t ), where ( t ) is some integer greater than 1. Then:( (q + t)^2 = q^2 + 2 t q + t^2 = q^2 +7 q +49 )Thus:( (2 t -7) q =49 - t^2 )Then:( q = frac{49 - t^2}{2 t -7} )Since ( q ) is a positive prime, the numerator and denominator must both be integers. Let's try some integer values of ( t ):- ( t =4 ): denominator ( 8 -7=1 ), numerator ( 49 -16=33 implies q=33 ). 33 is not prime.- ( t=5 ): denominator 10-7=3, numerator 49-25=24. 24/3=8. 8 not prime.- ( t=6 ): denominator 12-7=5, numerator 49-36=13. 13/5=2.6, not integer.- ( t=7 ): denominator 14-7=7, numerator 49-49=0. q=0, invalid.- ( t=3 ): denominator 6-7=-1, numerator 49-9=40. q= -40, invalid.- ( t=8 ): denominator 16-7=9, numerator 49-64= -15. Negative, invalid.- ( t=2 ): denominator 4-7=-3, numerator 49 -4=45. q= -15, invalid.- ( t=1 ): denominator 2-7=-5, numerator 49 -1=48. q= -48/5, invalid.- ( t=0 ): denominator -7, numerator 49, q= -7, invalid.- ( t=9 ): denominator 18-7=11, numerator 49 -81= -32, invalid.No valid primes here. So, ( p =7 ) has no solution. How about ( p =11 )?Let’s check ( p =11 ). Then equation ( 121 +11 q +q^2 =k^2 ). Let’s try some primes ( q ):- ( q=2 ):121 +22 +4=147, not square.- ( q=3 ):121 +33 +9=163, nope.- ( q=5 ):121 +55 +25=201, nope.- ( q=7 ):121 +77 +49=247, not a square.- ( q=11 ):121 +121 +121=363, nope.- ( q=13 ):121 +143 +169=433, not a square.- ( q=17 ):121 +187 +289=597, nope.- ( q=19 ):121 +209 +361=691, not a square.Not looking good. Alternatively, perhaps using the same approach as above, set ( k = q + t ):( (q + t)^2 = q^2 +11 q +121 )Which gives:( 2 t q + t^2 =11 q +121 )Rearranged:( (2 t -11) q =121 -t^2 )Hence, ( q = (121 - t^2)/(2 t -11) ). Let’s try integer ( t ):- ( t=6 ): denominator 12 -11=1, numerator 121 -36=85. q=85, not prime.- ( t=7 ): denominator 14 -11=3, numerator 121 -49=72. 72/3=24, not prime.- ( t=8 ): denominator 16 -11=5, numerator 121 -64=57. 57/5=11.4, not integer.- ( t=9 ): denominator 18 -11=7, numerator 121 -81=40. 40/7≈5.71, not integer.- ( t=10 ): denominator 20 -11=9, numerator 121 -100=21. 21/9≈2.33, nope.- ( t=11 ): denominator 22 -11=11, numerator 121 -121=0. q=0, invalid.- ( t=5 ): denominator 10 -11=-1, numerator 121 -25=96. q= -96, invalid.- Etc. No valid primes here either. So, ( p =11 ) doesn't work.So far, only ( p =3 ) and ( p =5 ) have solutions. Let's check ( p =7 ) again with another approach. Wait, maybe I missed a prime ( q ). Let me try ( q = 15 ) for ( p =7 ). But 15 is not prime. ( q = 11 ): we did that. 247, not square. Maybe ( q =13 ): 49 + 91 +169 = 309. 17² = 289, 18²=324. So between 17² and 18². Not a square.Alternatively, let's check ( p =7 ), maybe with larger ( q ). ( q = 23 ): 49 + 161 +529 = 739. 27²=729, 28²=784. Nope. So no luck.What about ( p = q )? If ( p = q ), then the equation becomes ( p² + p² + p² = 3p² ). So, ( 3p² =k² implies k = p sqrt{3} ). Which is irrational unless ( p =0 ), which is not prime. So, no solution when ( p = q ).Wait, unless ( p =0 ), which isn't prime. So, no solutions there.So, maybe only ( p =3 ) and ( p =5 ).But let's check another prime, say ( p =7 ), but different approach. Let's set ( q = r ), then ( p^2 + p q + q^2 =k^2 ). Let's think of this as an equation in integers. Maybe parametrization? For the equation ( x^2 + xy + y^2 = z^2 ), perhaps there's a parametric solution. Let me recall that equations of the form ( x^2 + xy + y^2 = z^2 ) are related to triangles with angles of 60 degrees, using the Law of Cosines. Maybe parametrization exists.Alternatively, note that ( x^2 + xy + y^2 = left(x + frac{y}{2}right)^2 + frac{3 y^2}{4} ). So, similar to an ellipse equation. Not sure if that helps.Alternatively, divide both sides by ( y^2 ):( left(frac{x}{y}right)^2 + left(frac{x}{y}right) + 1 = left( frac{z}{y} right)^2 )Let ( t = x/y ), then equation is ( t² + t + 1 = (z/y)² ). So, looking for rational solutions ( t ) such that ( t² + t +1 ) is a square. This is similar to finding rational points on an elliptic curve. But since we need integer solutions where ( x ) and ( y ) are primes, this might be restrictive.Alternatively, recall that in our previous cases, the solutions were ( (3,5) ) and ( (5,3) ), giving 49. Let's check if there are other small primes where flipping ( p ) and ( q ) gives a square. For example, ( p =5 ), ( q =3 ), which we did. Similarly, ( p =3 ), ( q =5 ). Are there others?Let me check ( p =7 ), ( q = something ). Wait, but we tried that. Alternatively, let's try ( p =7 ), ( q = 16 ). But 16 isn't prime. Not helpful.Wait, perhaps looking back at the factorization approach. We found solutions when factoring led to ( p =3 ) and ( p =5 ). Maybe these are the only solutions. To confirm, let's check ( p =7 ) in the cases from earlier. For case 1 and case 2.From case 2, which gave us ( p =5 ):We had ( q = frac{(p - 3)(p + 1)}{4} ). For ( p =7 ):( q = frac{(4)(8)}{4} = 8 ). Not prime. Similarly, for ( p =11 ):( q = frac{(8)(12)}{4} =24 ), not prime. For ( p =13 ):( q = frac{(10)(14)}{4} =35 ), not prime. So, no primes here.Similarly, case 1: ( q = frac{(3p +1)(p -1)}{4} ). For ( p =7 ):( q = frac{(22)(6)}{4} = frac{132}{4}=33 ), not prime. ( p =11 ):( q = frac{(34)(10)}{4}= frac{340}{4}=85 ), not prime. So, no good.Thus, only ( p =3 ) and ( p =5 ) satisfy the conditions where ( q ) is prime. So, the primes are 3 and 5. Their sum is 3 +5=8. But wait, wait. Wait, let's check if ( p =2 ) could have a solution we missed.Earlier, when ( p =2 ), we tried small primes ( q ) and didn't find any squares. Let's see if ( q ) could be a larger prime. Let’s suppose ( p =2 ), so equation is (4 + 2q + q² =k²). Let's rearrange:( q² +2q +4 =k² ). Let’s complete the square for ( q ):( q² +2q +1 = (q +1)^2 ), so ( (q +1)^2 +3 =k² implies k² - (q +1)^2 =3 implies (k - q -1)(k + q +1)=3 ). The factors of 3 are (1,3) or (-1,-3). Since ( k ) and ( q ) are positive, we take positive factors:( k - q -1 =1 )( k + q +1 =3 )Adding equations:( 2k =4 implies k=2 )Subtracting equations:( 2q +2=2 implies q=0 ). Not prime. Alternatively, negative factors:( k - q -1=-1 )( k + q +1=-3 )But this would lead to negative ( k ), which isn't possible. Hence, no solution for ( p=2 ).Therefore, only primes ( p=3 ) and ( p=5 ) work. Their sum is 8.But wait, wait a second. Let me confirm once again. The problem says "the sum of all primes ( p )" such that there exists a prime ( q ). So, both 3 and 5. Sum is 8. But let me check if there are other primes beyond 3 and 5. Let's check ( p =7 ), ( q = something ). Wait, perhaps ( q ) is not prime. Wait, no, ( q ) must be prime.Wait, but in the equation ( p^2 + pq + q^2 =k^2 ), both ( p ) and ( q ) must be primes, but ( k ) can be any integer. So, for example, if ( p =7 ), maybe ( q ) is a larger prime such that the sum is a square. Let's try with ( p =7 ), ( q = 11 ): 49 +77 +121=247. Not a square. ( q =13 ):49 +91 +169=309. Not a square. ( q =17 ):49 +119 +289=457. Not a square. ( q =19 ):49 +133 +361=543. Not a square. ( q =23 ):49 +161 +529=739. Not a square. ( q =29 ):49 +203 +841=1093. Not a square.How about ( p =7 ), ( q =43 ):49 +301 +1849=2199. Not a square. 46²=2116, 47²=2209. So, 2199 is between them. Not a square. Hmm.Alternatively, think of the equation as ( p^2 + pq + q^2 =k^2 ). Let me note that for ( p ) and ( q ) both odd primes, the sum is odd + odd + odd = odd, so ( k ) must be odd. Let’s let ( k =2m +1 ), then ( k²=4m² +4m +1 ). So, equation:( p² +pq + q² =4m² +4m +1 ). Hmm, not sure if helpful.Alternatively, let’s consider symmetry. The equation is symmetric in ( p ) and ( q ), so if ( (p, q) ) is a solution, so is ( (q, p) ). We found ( (3,5) ) and ( (5,3) ), which are distinct. But since the problem asks for primes ( p ), regardless of ( q ), so both 3 and 5 are included, but since we are summing all such primes ( p ), even if ( p ) and ( q ) are swapped, each ( p ) is counted once if it has at least one prime ( q ). So, even though 3 and 5 can be swapped, we only include each once in the sum.But wait, in our analysis, we only found 3 and 5 as possible primes. Let's check another angle. Suppose someone claims ( p =7 ) and ( q = 15 ), but 15 isn't prime. Or another composite. But no, ( q ) must be prime.Alternatively, perhaps using Pell’s equation. Wait, not sure. Let’s think again.Suppose we rearrange the equation as ( k^2 = p^2 + pq + q^2 ). Let’s divide both sides by ( q^2 ):( left( frac{k}{q} right)^2 = left( frac{p}{q} right)^2 + frac{p}{q} +1 )Let ( x = frac{p}{q} ), then:( left( frac{k}{q} right)^2 = x^2 +x +1 )Let ( y = frac{k}{q} ), so ( y^2 =x^2 +x +1 ). Multiply both sides by 4:(4y^2 =4x^2 +4x +4)(4y^2 = (2x +1)^2 +3)So, (4y^2 - (2x +1)^2 =3). Let’s set ( u =2y ), ( v =2x +1 ). Then:( u^2 -v^2 =3 implies (u -v)(u +v)=3 ). Since ( u ) and ( v ) are positive integers (since ( x = frac{p}{q} >0 ), ( y = frac{k}{q} >0 )), we have:( u -v =1 )( u +v =3 )Solving:Adding equations: (2u=4 implies u=2). Then (v=1). Hence,( u=2 implies y=1 )( v=1 implies 2x +1=1 implies x=0 ). But ( x= p/q=0 implies p=0 ), which is invalid. So, only solution here is trivial. Therefore, the only solution in integers for Pell-type equation is trivial. Hence, no non-trivial solutions. Therefore, this suggests that the only solutions are those found earlier.Alternatively, maybe the substitution is not the best. Let me check. Wait, but we already found solutions with ( p =3 ), ( q=5 ) and vice versa. So, perhaps there's a mistake in assuming ( x ) and ( y ) are integers. They are rationals, since ( p ) and ( q ) are integers. But our Pell equation approach led to only trivial solutions, but we know there are non-trivial solutions. Hence, maybe this approach is missing something.Alternatively, note that when ( p =3 ), ( q=5 ), then ( x =3/5 ), ( y =7/5 ). Then,( (7/5)^2 = (3/5)^2 + (3/5) +1 implies 49/25 =9/25 +15/25 +25/25=49/25 ). Correct. So, but these are rational solutions, not integers. So, perhaps the Pell equation approach is not helpful here.Given that, and since we've checked primes up to 11 and beyond without finding any new solutions, it's likely that only ( p =3 ) and ( p =5 ) satisfy the condition.Therefore, the sum of all such primes ( p ) is ( 3 +5 =8 ).Final Answerboxed{8}</think>To find the sum of all primes ( p ) for which there exists a prime ( q ) such that ( p^2 + pq + q^2 ) is a square, we start by analyzing the equation ( p^2 + pq + q^2 = k^2 ) for primes ( p ) and ( q ).1. Case Analysis for Small Primes: - Testing small primes ( p ) and ( q ) (both odd and even) reveals that neither ( p = 2 ) nor ( q = 2 ) yield a square result. - Considering both primes as odd, we rewrite the equation and analyze possible factorizations and discriminants.2. Quadratic Equation Analysis: - The equation ( p^2 + pq + q^2 = k^2 ) is treated as a quadratic in ( q ). The discriminant must be a perfect square, leading to factorizations of ( 3p^2 ). - Factor pairs of ( 3p^2 ) lead to solving for ( q ) and checking if it results in a prime.3. Key Solutions: - For ( p = 3 ), solving the quadratic equation yields ( q = 5 ), and checking ( 3^2 + 3 cdot 5 + 5^2 = 49 ) (a square). - For ( p = 5 ), solving the quadratic equation yields ( q = 3 ), and checking ( 5^2 + 5 cdot 3 + 3^2 = 49 ) (a square).4. Verification for Larger Primes: - Testing larger primes (e.g., 7, 11) and various values of ( q ) shows no valid solutions where the result is a square.5. Conclusion: - The only primes ( p ) that satisfy the condition are 3 and 5. Their sum is ( 3 + 5 = 8 ).[boxed{8}],<think>Okay, so I need to find all prime numbers p such that there exists another prime q where p² + pq + q² is a perfect square. Then, sum all those primes p. Hmm, primes... quadratic expressions... squares. Let me start by understanding the problem step by step.First, let me rewrite the given equation: p² + pq + q² = k², where k is some integer. Both p and q are primes. My goal is to find all such primes p (with corresponding primes q) that satisfy this equation and then sum them up.Since both p and q are primes, maybe I can try small primes first and see if I can find any patterns or possible candidates. Let's list out the first few primes: 2, 3, 5, 7, 11, 13, 17, etc. Let's start testing these.Case 1: p = 2. Then, we need to find a prime q such that 2² + 2*q + q² is a square. Let's compute this expression for different q's.For q = 2: 4 + 4 + 4 = 12, which is not a square.q = 3: 4 + 6 + 9 = 19, not a square.q = 5: 4 + 10 + 25 = 39, not a square.q = 7: 4 + 14 + 49 = 67, not a square.q = 11: 4 + 22 + 121 = 147, not a square.q = 13: 4 + 26 + 169 = 199, not a square.Hmm, maybe p = 2 doesn't work. Let's try p = 3.Case 2: p = 3.Find prime q such that 9 + 3q + q² is a square.q = 2: 9 + 6 + 4 = 19, not a square.q = 3: 9 + 9 + 9 = 27, not a square.q = 5: 9 + 15 + 25 = 49, which is 7². Oh! This works. So when p = 3 and q = 5, we get 49, which is a square. So 3 is a candidate.Let me check q = 7: 9 + 21 + 49 = 79, not a square.q = 11: 9 + 33 + 121 = 163, not a square.So p = 3 works with q = 5.Case 3: p = 5.Check if there's a prime q where 25 + 5q + q² is a square.q = 2: 25 + 10 + 4 = 39, not a square.q = 3: 25 + 15 + 9 = 49, which is 7². Wait, so when p = 5 and q = 3, we also get 49. But q has to be prime, which 3 is. So this also works. Therefore, p = 5 is another candidate.Wait, so p and q can be swapped here? If p = 3 and q = 5 works, then p = 5 and q = 3 also works because the equation is symmetric in p and q. So both p = 3 and p = 5 will be included. Let me confirm.Yes, because swapping p and q gives the same expression. Therefore, if (p, q) is a solution, then (q, p) is also a solution if q is prime. So if we have p = 3 and q = 5, then p = 5 and q = 3 is also a solution. Therefore, both 3 and 5 should be included.But wait, let's check for p = 5 with other q's.q = 5: 25 + 25 + 25 = 75, not a square.q = 7: 25 + 35 + 49 = 109, not a square.q = 11: 25 + 55 + 121 = 201, not a square.So only q = 3 works for p = 5.Case 4: p = 7.Check if there exists a prime q where 49 + 7q + q² is a square.q = 2: 49 + 14 + 4 = 67, not a square.q = 3: 49 + 21 + 9 = 79, not a square.q = 5: 49 + 35 + 25 = 109, not a square.q = 7: 49 + 49 + 49 = 147, not a square.q = 11: 49 + 77 + 121 = 247, not a square.q = 13: 49 + 91 + 169 = 309, not a square.q = 17: 49 + 119 + 289 = 457, not a square.Not seeing anything here. Maybe p = 7 is not a solution.Case 5: p = 11.Check prime q's:q = 2: 121 + 22 + 4 = 147, not a square.q = 3: 121 + 33 + 9 = 163, not a square.q = 5: 121 + 55 + 25 = 201, not a square.q = 7: 121 + 77 + 49 = 247, not a square.q = 11: 121 + 121 + 121 = 363, not a square.Hmm, seems like p = 11 isn't working. Let's try p = 13.Wait, maybe before going higher, let's check if there are other possible p and q pairs. So far, we have p = 3 and p = 5. Let me check p = 2 again, maybe with a larger q.Wait, p = 2, q = 7: 4 + 14 + 49 = 67. Not a square. q = 11: 4 + 22 + 121 = 147. Not a square. q = 13: 4 + 26 + 169 = 199. Still not a square. Maybe p = 2 doesn't work. Let's check p = 7 again. Maybe with a larger q?q = 11: 49 + 77 + 121 = 247, which is 13² + 78? Wait 13² is 169, 14² is 196, 15² is 225, 16² is 256. 247 is between 15² and 16². Not a square.Alternatively, perhaps there's a different approach here instead of checking case by case, which might take too long. Let me think algebraically.Given that p and q are primes, and p² + pq + q² is a square. Let's denote k² = p² + pq + q². Since p and q are primes, maybe there's a way to factor this expression or find some relationship between p and q.The expression p² + pq + q² resembles the expansion for (p + q)², but (p + q)² = p² + 2pq + q². So our expression is (p + q)² - pq. So k² = (p + q)² - pq.Hmm, so (p + q)² - k² = pq. Which factors as [(p + q) - k][(p + q) + k] = pq.Since p and q are primes, pq has only factors 1, p, q, pq. So the product [(p + q - k)][(p + q + k)] = pq. Let me denote A = p + q - k and B = p + q + k, so that AB = pq. Since A and B are positive integers (assuming k is less than p + q, which it must be because k² = p² + pq + q² < (p + q)² when pq > 0), and A < B. Also, A and B must be factors of pq. Since p and q are primes, the factors are 1, p, q, pq.Therefore, the possible pairs (A, B) are (1, pq), (p, q). But since A and B must satisfy A < B and A * B = pq.But also, since A = p + q - k and B = p + q + k, then B - A = 2k. So B - A must be even, which it is because B and A are both integers.Additionally, A + B = 2(p + q). So, given that AB = pq and A + B = 2(p + q). Let's write equations based on possible factor pairs.First possible pair: A = 1, B = pq. Then, A + B = 1 + pq = 2(p + q). So 1 + pq = 2p + 2q. Rearranged: pq - 2p - 2q + 1 = 0. Let's add 4 to both sides: pq - 2p - 2q + 1 + 4 = 4. So pq - 2p - 2q + 5 = 4? Wait, maybe that's not helpful. Alternatively, factor:pq - 2p - 2q + 1 = 0Let me add 4 to both sides: pq - 2p - 2q + 1 + 4 = 4Then pq - 2p - 2q + 5 = 4. Hmm, not helpful.Alternatively, rearrange:pq - 2p - 2q = -1Add 4 to both sides: pq - 2p - 2q + 4 = 3Factor: (p - 2)(q - 2) = 3Ah! That's clever. Let's verify:Left-hand side: (p - 2)(q - 2) = pq - 2p - 2q + 4So yes, pq - 2p - 2q + 4 = (p - 2)(q - 2) = 3.So (p - 2)(q - 2) = 3. Since p and q are primes greater than or equal to 2, so (p - 2) and (q - 2) are integers. 3 can be factored as 1*3 or 3*1. Since p and q are primes, let's see possible values.Case 1: (p - 2) = 1 and (q - 2) = 3. Then p = 3 and q = 5. Both primes.Case 2: (p - 2) = 3 and (q - 2) = 1. Then p = 5 and q = 3. Both primes.Alternatively, could (p - 2) = -1 and (q - 2) = -3, but since primes are at least 2, p - 2 and q - 2 can't be negative. So only positive factors. So the only solutions here are (p, q) = (3, 5) and (5, 3). Which matches the earlier examples we found.Therefore, in the case where (A, B) = (1, pq), we get these solutions. So that's two possible primes p: 3 and 5.Now check the other factor pair: (A, B) = (p, q). But since A * B = pq, and A = p, B = q. Then A + B = p + q = 2(p + q). Wait, no. Wait, if (A, B) = (p, q), then A + B = p + q. But from earlier, A + B = 2(p + q). Therefore, p + q = 2(p + q). Which implies p + q = 0. But since p and q are primes (at least 2), this is impossible. So this factor pair doesn't yield any solutions.Similarly, if we take (A, B) = (q, p), same problem. So the only valid factor pair is (1, pq), leading to the equation (p - 2)(q - 2) = 3, which gives p = 3, q = 5 and p = 5, q = 3.Are there any other factor pairs? The factors of pq are 1, p, q, pq. So possible (A, B) pairs are (1, pq), (p, q). We checked both. So unless there's a case where A and B are not co-prime, but since pq is the product of two primes, the factors are as above.Therefore, the only solutions are p = 3 and p = 5. Hence, the sum is 3 + 5 = 8.Wait, but let me check if there might be other solutions where A and B are not co-prime. Wait, if p = q, then the expression becomes p² + p² + p² = 3p², which is 3 times a square. But 3p² is a square only if 3 is a square, which it isn't. So p cannot equal q. Therefore, p and q must be distinct primes. So the only solutions are p = 3 and p = 5.But just to be thorough, let me check p = 7 again with a different approach. Suppose there exists a prime q such that 49 + 7q + q² is a square. Let's write this as q² + 7q + 49 = k². Let's rearrange: k² - q² - 7q = 49. Maybe factor this as (k - q)(k + q) - 7q = 49. Not sure. Alternatively, complete the square for q.q² + 7q + 49 = k².Complete the square for q:q² + 7q + (49/4) = k² + (49/4) - 49.Wait, q² + 7q + 49 = (q + 3.5)^2 + 49 - 12.25 = (q + 3.5)^2 + 36.75. Hmm, not helpful since we need integers.Alternatively, consider that q² + 7q + 49 = k². Let's denote q as variable, and check if for primes q, this expression is a square.For q = 2: 4 + 14 + 49 = 67, not square.q = 3: 9 + 21 + 49 = 79, not square.q = 5: 25 + 35 + 49 = 109, not square.q = 7: 49 + 49 + 49 = 147, not square.q = 11: 121 + 77 + 49 = 247, not square.q = 13: 169 + 91 + 49 = 309, not square.So no, p = 7 doesn't work.How about p = 7 and q = 19? Let's see: 49 + 133 + 361 = 543, which is not a square. 23² = 529, 24² = 576. So no.Alternatively, maybe there's a case where p and q are larger primes. Let me try p = 7 and q = 19. Wait, I just did that, doesn't work. Maybe p = 11 and q = something.Wait, but according to the earlier algebraic approach, the only solutions are when (p - 2)(q - 2) = 3, so p = 3 and 5. So perhaps those are the only solutions. Let me verify if there are other factorizations possible.Suppose, for some reason, we considered the equation AB = pq where A and B are not necessarily (1, pq) or (p, q). But since pq is a product of two primes, the only possible factorizations are 1 * pq, p * q. So unless p or q is 2, but even then, for example, if p = 2, then pq = 2q, so factors would be 1, 2, q, 2q. Let's see if p = 2 allows another factorization.Wait, let's revisit the equation (p + q - k)(p + q + k) = pq. If p = 2, then pq = 2q. So factors of 2q are 1, 2, q, 2q. So possible factor pairs (A, B):(1, 2q): Then, A = 1, B = 2q. Then, A + B = 1 + 2q = 2(p + q) = 2(2 + q) = 4 + 2q. So 1 + 2q = 4 + 2q => 1 = 4. Contradiction.(2, q): Then A = 2, B = q. Then, A + B = 2 + q = 2(p + q) = 4 + 2q. So 2 + q = 4 + 2q => -2 = q. Impossible since q is a prime.Similarly, if p = 2, the other factor pairs (q, 2) but q is prime, so same as above.Therefore, for p = 2, there's no solution. Which aligns with our earlier testing. So even when p = 2, the only possible factor pairs don't work. Therefore, p = 2 is excluded.Therefore, based on the algebraic approach, the only solutions are p = 3 and p = 5. Hence, their sum is 8.But let me just confirm once more with another perspective. Suppose we consider the equation p² + pq + q² = k². Let's divide both sides by q² to make it a quadratic in (p/q):(p/q)² + (p/q) + 1 = (k/q)².Let x = p/q, which is a rational number since p and q are primes (unless q divides p, but since they are primes, x is either 1 or a fraction). Then, x² + x + 1 must be a square of a rational number (k/q). Let's suppose x is a rational number a/b where a and b are integers with no common factors.Then, (a/b)² + (a/b) + 1 = (c/b)², where c is an integer. Multiply both sides by b²:a² + ab + b² = c².So this is similar to the original equation but with integers a and b instead of primes. Perhaps there are solutions where a and b are integers, which might give rise to solutions where a and b are primes. But maybe we can parametrize all solutions of a² + ab + b² = c².This is a Diophantine equation. Let me see if I can find a parametrization for this.Looking up parametrization for a² + ab + b² = c². Alternatively, think of this as a quadratic form. Maybe using substitution variables.Let me set variables substitution. Let’s denote u = a + b, v = a - b. Wait, not sure. Alternatively, express in terms of variables to complete the square.Alternatively, note that a² + ab + b² = c² can be rewritten as (2a + b)² + 3b² = 4c². Let me check:Expand (2a + b)²: 4a² + 4ab + b². Then, (2a + b)² + 3b² = 4a² + 4ab + b² + 3b² = 4a² + 4ab + 4b² = 4(a² + ab + b²) = 4c². Therefore, the equation becomes (2a + b)² + 3b² = 4c². Hmm, which resembles an equation of the form x² + 3y² = 4z².Let me divide both sides by 4: ( (2a + b)/2 )² + 3*(b/2)^2 = z². Let me set m = (2a + b)/2 and n = b/2. Then the equation becomes m² + 3n² = z². So we have m² + 3n² = z². Which is a well-known Diophantine equation. The solutions to this can be parametrized.Alternatively, perhaps parametrizing m, n, z. For example, solutions can be found using identity:(m + n√3)(m - n√3) = z².But maybe it's easier to look for small solutions and find a pattern.Alternatively, look for integer solutions to m² + 3n² = z².Assume that m, n, z are coprime in pairs. Suppose they are not, then factor out the common divisor.Suppose there exists a solution where m, n, z are coprime. Then, m² ≡ z² mod 3. Therefore, either m ≡ z ≡ 0 mod 3 or m ≡ ±z mod 3.If m ≡ 0 mod 3, then z must also be ≡ 0 mod 3. Let m = 3k, z = 3l. Then the equation becomes 9k² + 3n² = 9l² → 3k² + n² = 3l² → n² = 3(l² - k²). Therefore, 3 divides n², so 3 divides n. Let n = 3m. Then, 9m² = 3(l² - k²) → 3m² = l² - k² → (l - k)(l + k) = 3m².Since l and k are integers, factors of 3m². Since l - k and l + k must both be even or both be odd. But (l - k) + (l + k) = 2l, which is even, so both factors are even or both odd. However, their product is 3m². If m is even, then 3m² is even. If m is odd, 3m² is odd. So if m is odd, then l - k and l + k are both odd. If m is even, they are both even.But let's suppose that m is coprime with 3. Then, the factors (l - k) and (l + k) must be 3 and m², but considering their product is 3m². However, this may not hold. It's getting complicated.Alternatively, maybe this approach isn't the best. Let's think back to our original equation. We found solutions at p=3 and p=5, maybe those are the only ones. Let's check p=7 again with q=?Wait, but according to the earlier algebraic approach where AB = pq, the only possible solutions come from (p - 2)(q - 2) = 3, so p=3,5. Therefore, these are the only solutions.But just to make sure, let's consider p=7 and q=?Suppose there exists a prime q where 49 + 7q + q² is a square. Let me rearrange the equation:q² + 7q + 49 = k².Let me compute discriminant for q as variable: 7² - 4*1*(49 - k²) = 49 - 196 + 4k² = 4k² - 147.But for q to be integer, discriminant must be a perfect square. So 4k² - 147 = m² for some integer m. Therefore, 4k² - m² = 147. Which can be written as (2k - m)(2k + m) = 147.Factor pairs of 147: 1*147, 3*49, 7*21. Since 2k - m < 2k + m and both positive.Case 1: 2k - m = 1, 2k + m = 147. Then adding: 4k = 148 → k=37, m=2k -1=73.Case 2: 2k - m = 3, 2k + m =49. Adding: 4k=52→k=13, m=2k -3=23.Case3: 2k -m=7, 2k +m=21. Adding:4k=28→k=7, m=2k -7=7.So possible k=37,13,7.Now, for each k, solve for q.Original equation: q² +7q +49 =k².For k=37: q² +7q +49=1369 → q² +7q -1320=0. Discriminant=49 +5280=5329=73². So q=(-7±73)/2. Positive solution: (66)/2=33. Not prime (33=3*11).For k=13: q² +7q +49=169→q² +7q -120=0. Discriminant=49 +480=529=23². Solutions: (-7±23)/2. Positive solution:16/2=8. Not prime.For k=7: q² +7q +49=49→q² +7q=0→q(q +7)=0. Solutions q=0 or q=-7. Not primes.Thus, even though there are integer solutions for k, q is not prime in any case. Hence, p=7 doesn't work.Similarly, let's check p=11. So equation q² +11q +121=k². Then discriminant for q: 121 -4*(121 -k²)=4k² -363. So 4k² -363 must be a perfect square. Let me set 4k² - m²=363. So (2k -m)(2k +m)=363. Factors of 363:1*363, 3*121, 11*33.Case1: 2k -m=1, 2k +m=363. Adding:4k=364→k=91, m=2k -1=181.Then equation q² +11q +121=91²=8281. So q² +11q -8160=0. Discriminant=121 +32640=32761. Square root: 181. So q=(-11±181)/2. Positive solution:170/2=85. Not prime.Case2: 2k -m=3, 2k +m=121. Adding:4k=124→k=31, m=2*31 -3=59.Equation q² +11q +121=961→q² +11q -840=0. Discriminant=121 +3360=3481=59². Solutions: (-11±59)/2. Positive:48/2=24. Not prime.Case3:2k -m=11, 2k +m=33. Adding:4k=44→k=11, m=22 -11=11.Equation q² +11q +121=121→q² +11q=0→q(q +11)=0. Solutions q=0 or -11. Not primes.Thus, no solutions for p=11. So p=3 and p=5 are the only ones.Therefore, the sum of all such primes p is 3 + 5 = 8.Final AnswerThe sum of all primes ( p ) is boxed{8}.</think>To find the sum of all primes ( p ) such that there exists a prime ( q ) where ( p^2 + pq + q^2 ) is a square, we start by analyzing the equation ( p^2 + pq + q^2 = k^2 ) for primes ( p ) and ( q ).First, we test small primes:- For ( p = 2 ), no prime ( q ) makes the expression a square.- For ( p = 3 ), we find ( q = 5 ) works: ( 3^2 + 3 cdot 5 + 5^2 = 49 = 7^2 ).- For ( p = 5 ), we find ( q = 3 ) works: ( 5^2 + 5 cdot 3 + 3^2 = 49 = 7^2 ).Next, we consider the algebraic approach. Rewriting the equation as ( (p + q)^2 - pq = k^2 ), we factorize it as ( (p + q - k)(p + q + k) = pq ). Given ( p ) and ( q ) are primes, the factors must be ( 1 ) and ( pq ). Solving ( (p - 2)(q - 2) = 3 ), we find the solutions ( (p, q) = (3, 5) ) and ( (5, 3) ).Testing other primes (7, 11, etc.) shows no solutions. The algebraic approach confirms only ( p = 3 ) and ( p = 5 ) work.Thus, the sum of all such primes ( p ) is ( 3 + 5 = 8 ).[boxed{8}]]

🔑:Answer: 83 and 5 both work, because 3^{2}+3 cdot 5+5^{2}=49. Now, say p^{2}+p q+q^{2}=k^{2}, for a positive integer k. Then (p+q)^{2}-k^{2}=p q, or:(p+q+k)(p+q-k)=p qSince p+q+k is a divisor of p q, and it is greater than p and q, p+q+k=p q. Then p+q-k=1. So:2 p+2 q=p q+1 Leftrightarrow p q-2 p-2 q+4=3 Leftrightarrow(p-2)(q-2)=3This shows that one of p and q is 3 and the other is 5 .

🧠:[<think>Okay, so I need to figure out how many three-digit numbers are reentrant numbers. From the problem statement, a reentrant number is one where the tens digit is less than both the hundreds digit and the units digit. Examples given are 201, 636, 734, etc. Let me think about how to approach this.First, let's break down what a three-digit number is. A three-digit number can be represented as ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit. Each of these digits has certain constraints:- A can be from 1 to 9 (since it's a three-digit number, the hundreds digit can't be 0)- B can be from 0 to 9- C can be from 0 to 9But for a reentrant number, the tens digit B must be less than both A and C. So, for each possible value of B, we need to count how many combinations of A and C satisfy A > B and C > B. Then sum that over all possible B.Let me formalize this. For each B from 0 to 9, the number of possible A's is (9 - B) because A has to be greater than B and A ranges from 1 to 9. Similarly, the number of possible C's is (9 - B) because C has to be greater than B and C ranges from 0 to 9. Wait, but hold on. For C, since it can be 0, if B is 0, then C has to be greater than 0. But actually, if B is 0, C can be from 1 to 9, which is 9 digits, which would be (9 - B) where B is 0. So yes, that formula works.But wait, let's check with an example. Take B = 0. Then A must be greater than 0, so A can be 1-9 (9 choices). C must be greater than 0, so C can be 1-9 (9 choices). So total numbers for B=0 would be 9*9=81. If B=1, then A must be 2-9 (8 choices) and C must be 2-9 (8 choices), so 8*8=64. Continuing this way, for each B, the number of possible A and C digits would be (9 - B) * (10 - B -1)? Wait, no. Wait, if B is the tens digit, then A must be in 1-9 and greater than B, and C must be in 0-9 and greater than B.Wait, so for A, which is from 1 to 9, if B is given, the number of possible A is (9 - B) since A must be greater than B. For example, if B=3, then A can be 4-9, which is 6 numbers, so 9 - 3 = 6.For C, which is from 0 to 9, if B is given, the number of possible C is (9 - B) as well. Wait, because if B is 3, then C has to be 4-9, which is 6 numbers. But if B=0, then C has to be 1-9, which is 9 numbers. So (9 - B) works here because when B=0, 9 - 0 = 9, which is correct. When B=9, the number of possible C's would be (9 - 9) = 0, which is correct because C can't be greater than 9. Similarly, if B=8, then C can be 9, which is (9 - 8)=1.Wait, hold on. Let's formalize:For C (units digit), which can be from 0 to 9. If the tens digit B must be less than C, then C must be from (B + 1) to 9. So the number of possible C's is 9 - B. For example, B=3: 4-9 is 6 numbers, which is 9 - 3 = 6. Similarly, B=0: 1-9 is 9 numbers, which is 9 - 0 = 9. So that's correct.For A (hundreds digit), which must be from 1 to 9 and greater than B. So A must be from (B + 1) to 9. Therefore, the number of possible A's is 9 - B. For example, B=3: A can be 4-9, which is 6 numbers. B=0: A can be 1-9, which is 9 numbers. So yes, same as C.Therefore, for each B from 0 to 8 (since B=9 would result in 0 possibilities for both A and C), the number of reentrant numbers is (9 - B) * (9 - B). So the total number of reentrant numbers is the sum from B=0 to B=8 of (9 - B)^2.Wait, let me confirm. If B=0: 9*9=81B=1: 8*8=64B=2:7*7=49...B=8:1*1=1So the total is 81 + 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1Let me compute that:81 + 64 = 145145 + 49 = 194194 + 36 = 230230 + 25 = 255255 + 16 = 271271 + 9 = 280280 + 4 = 284284 + 1 = 285So total number is 285. Wait, is that correct?Wait, but let me check the formula again. For each B from 0 to 8 inclusive, since B=9 would have (9 - 9)=0 for both A and C. So we have B=0 to 8, which is 9 terms. So sum_{k=1 to 9} k^2, but in reverse? Because when B=0, (9 - 0)^2=9^2=81, when B=1, (9 -1)^2=8^2=64, ..., B=8, (9 -8)^2=1^2=1. So yes, it's the sum of squares from 1^2 to 9^2, but in reverse order. The sum of squares from 1^2 to n^2 is n(n + 1)(2n + 1)/6. Therefore, sum_{k=1}^9 k^2 = 9*10*19/6 = (9*10*19)/6. Let's compute that: 9*10=90, 90*19=1710, 1710/6=285. So that's correct. Therefore, the total number is 285.Therefore, the answer should be 285. Let me just check with an example given: 201. The tens digit is 0, which is less than 2 and 1. Wait, wait, hold on, in 201, the tens digit is 0, which is less than the hundreds digit 2, but the units digit is 1. Wait, 0 < 1? Wait, no. Wait, in the problem statement, the example given is 201. But in 201, the tens digit is 0, which is less than the hundreds digit 2 and less than the units digit 1. Wait, 0 is less than 1? Yes, 0 < 1. So 0 is less than both 2 and 1. Wait, but 0 is less than 2 and 0 is less than 1, so that satisfies the condition. So 201 is a reentrant number.Similarly, 636: tens digit is 3, which is less than 6 and 6. So 3 < 6 and 3 < 6. Correct.734: tens digit is 3, less than 7 and 4. 3 < 7 and 3 < 4. Yes.Wait, but in 734, the units digit is 4, which is greater than 3. So that's okay. So the tens digit must be less than both the hundreds and the units digit. So the units digit can be equal to the hundreds digit or different, as long as it's greater than the tens digit.So the initial logic holds. For each B, the number of possible A and C digits where A > B and C > B. So the count is (9 - B) * (9 - B). Summing from B=0 to B=8 gives 285. Therefore, the total number of reentrant numbers is 285.But let me just confirm with another example. Let's take B=5. Then A can be 6,7,8,9 (4 choices) and C can be 6,7,8,9 (4 choices). So 4*4=16 numbers. Then for B=5, there are 16 numbers. Then when we sum all of them, we get 285. Let me check for B=5, the count is 16. Then in the total sum, 81 (B=0) +64 (B=1)=145, +49 (B=2)=194, +36 (B=3)=230, +25 (B=4)=255, +16 (B=5)=271, +9 (B=6)=280, +4 (B=7)=284, +1 (B=8)=285. Yes, that adds up. So seems correct.Therefore, the answer is 285. So the total number of reentrant numbers is boxed{285}.Final Answerboxed{285}</think>To determine the number of reentrant numbers, we consider a three-digit number (ABC) where the tens digit (B) is less than both the hundreds digit (A) and the units digit (C). For each possible value of (B) (from 0 to 8), we calculate the number of valid combinations of (A) and (C):- The hundreds digit (A) must be greater than (B) and can range from (B + 1) to 9, giving (9 - B) choices.- The units digit (C) must be greater than (B) and can range from (B + 1) to 9, giving (9 - B) choices.Thus, for each (B), the number of valid combinations is ((9 - B) times (9 - B)). Summing this over all possible values of (B) from 0 to 8 gives the total number of reentrant numbers:[sum_{B=0}^{8} (9 - B)^2]This sum is equivalent to the sum of the squares of the first 9 positive integers:[sum_{k=1}^{9} k^2 = frac{9 times 10 times 19}{6} = 285]Therefore, the total number of reentrant numbers is (boxed{285}).,<think>Okay, so I need to figure out how many three-digit numbers are "reentrant numbers." The definition given is that the tens digit has to be less than both the hundreds digit and the units digit. For example, 201 is a reentrant number because the tens digit is 0, which is less than 2 (the hundreds digit) and 1 (the units digit). Similarly, 636 has a tens digit of 3, which is less than 6 and 6. First, let me confirm what the problem is asking. A three-digit number has digits in the hundreds, tens, and units places. The tens digit must be strictly less than both the hundreds and the units digits. So, for a number like 734, the tens digit is 3, which is less than 7 and 4. Wait, but 3 is less than 7 and 4? Yes, 3 < 7 and 3 < 4. So that's correct.Now, my goal is to count all such three-digit numbers. Let me think about how to approach this systematically.First, let's recall that three-digit numbers range from 100 to 999. So the hundreds digit (let's call it H) can be from 1 to 9, the tens digit (T) from 0 to 9, and the units digit (U) from 0 to 9. The condition is that T < H and T < U. So, the problem reduces to counting all possible combinations of H, T, U where H is 1-9, T is 0-9, U is 0-9, and T is less than both H and U.To solve this, perhaps I can fix the tens digit T and then count the number of possible H and U digits that are greater than T. Let's try this approach. Let's iterate over each possible value of T (from 0 to 8, since T must be less than both H and U, which can be at most 9). For each T, determine how many possible H and U digits there are such that H > T and U > T. Then, the total number for each T is the product of the number of possible H's and the number of possible U's. Then sum over all T's.Yes, that seems like a good plan.So, let's formalize this:For a given T (tens digit), the hundreds digit H must be in {T+1, T+2, ..., 9}, so there are 9 - T possible choices for H. Similarly, the units digit U must be in {T+1, T+2, ..., 9}, so there are also 9 - T possible choices for U. Therefore, for each T, the number of three-digit numbers with that T being the tens digit and satisfying the reentrant condition is (9 - T) * (9 - T) = (9 - T)^2.But wait, hold on. H can be from 1 to 9, but if T is 0, then H must be greater than 0, which is the same as H being from 1 to 9. However, when T is 1, H must be greater than 1, so H is from 2 to 9, which is 8 choices. Similarly, for T=2, H from 3 to 9, 7 choices, etc. So actually, the number of possible H for each T is (9 - T) but starting from H=1 when T=0.Wait, but the original consideration was that H is at least 1. So when T is 0, H can be 1-9, so 9 choices. When T is 1, H must be 2-9, so 8 choices. Similarly, for T=2, H is 3-9, 7 choices, etc., down to T=8, where H must be 9, which is 1 choice. For T=9, but T can't be 9 because H and U have to be greater than T, which would require H and U to be at least 10, which isn't possible. So T can only go up to 8.Similarly, for U, when T is 0, U must be 1-9, 9 choices. When T is 1, U must be 2-9, 8 choices, and so on. So, for each T, the number of possible U's is (9 - T). Therefore, for each T from 0 to 8, the number of reentrant numbers with that T is (9 - T) * (9 - T) = (9 - T)^2.Therefore, the total number of reentrant numbers is the sum from T=0 to T=8 of (9 - T)^2.Alternatively, since T is from 0 to 8, let's let k = 9 - T. When T=0, k=9; when T=8, k=1. So, the sum becomes k^2 from k=1 to k=9.But the sum of squares from 1 to n is given by the formula n(n + 1)(2n + 1)/6. So, if n=9, the sum is 9*10*19/6. Let me compute that.First, compute 9*10=90, then 90*19=1710. Then divide by 6: 1710 / 6 = 285. So the total number of reentrant numbers would be 285. Hmm.Wait, but hold on. Let me check this reasoning again. Because for each T from 0 to 8, the count is (9 - T)^2. So the total is sum_{T=0}^{8} (9 - T)^2. Let's make substitution k = 9 - T. Then when T=0, k=9; when T=8, k=1. So the sum becomes sum_{k=1}^9 k^2. Which is 285. So that's correct.But let me verify this with a concrete example. For T=0, number of numbers is 9*9=81. For T=1, 8*8=64. Then T=2, 7*7=49. T=3, 6*6=36. T=4, 5*5=25. T=5, 4*4=16. T=6, 3*3=9. T=7, 2*2=4. T=8, 1*1=1. Now, adding these up: 81 + 64 = 145; 145 + 49 = 194; 194 + 36 = 230; 230 +25=255; 255 +16=271; 271 +9=280; 280 +4=284; 284 +1=285. Yes, that adds up to 285. So that seems correct.But wait, let me check the initial examples given in the problem statement. For example, 201: H=2, T=0, U=1. So, T=0, which is allowed, and 0 < 2 and 0 < 1. So that's included in the count. Similarly, 636: H=6, T=3, U=6. So T=3 < 6 and 3 < 6. 734: H=7, T=3, U=4. 3 < 7 and 3 < 4. All these are included.But let me check another example. What about a number like 121? H=1, T=2, U=1. Wait, T=2 is not less than H=1 or U=1, so it's not a reentrant number. So that wouldn't be counted. What about 312? T=1, which is less than H=3 and U=2. Yes, T=1 < 3 and 1 < 2. So 312 should be a reentrant number. Let's check if it's counted in the total. For T=1, the number of possibilities is 8*8=64. So yes, 312 is one of them. So seems correct.Wait, but in the example given in the problem statement, 201 is given, which is H=2, T=0, U=1. Then 636 is H=6, T=3, U=6. Wait, in 636, U=6 is equal to H=6. The problem statement says the tens digit is less than both the hundreds and the units digits. So, in this case, T=3 < 6 and T=3 < 6. So equality is allowed for H and U? The problem says "less than both the hundreds and the units digits." So if H or U is equal to T, then it doesn't satisfy the condition. Wait, but in 636, the units digit is equal to the hundreds digit, which is 6, but T=3 is less than both. So 636 is acceptable. Wait, no, the problem states "the tens digit is less than both the hundreds and the units digits." So as long as T is less than both H and U, regardless of whether H and U are equal or not. So 636 is acceptable because 3 < 6 and 3 < 6.Similarly, if we have a number like 212, T=1, which is less than H=2 and U=2. That's acceptable. So the hundreds and units digits can be equal as long as they are both greater than the tens digit.Therefore, in our counting, when we fix T, H can be from T+1 to 9, and U can be from T+1 to 9. So H and U can be the same, different, etc., as long as they are both greater than T. So the count (9 - T) * (9 - T) is correct, as each of H and U can independently be chosen from T+1 to 9.Therefore, the total number is 285. Hmm. But let me just verify this with another approach to make sure.Alternative approach: For each hundreds digit H from 1 to 9, and each units digit U from 0 to 9, how many tens digits T satisfy T < H and T < U. Then sum over all H and U.But this might be more complicated, but let's try.So for given H and U, the number of T's that are less than both H and U is the number of T's less than min(H, U). Since T is a digit from 0 to 9, the number of T's less than min(H, U) is min(H, U). For example, if H=5 and U=7, then min(H, U)=5, so T can be 0-4, which is 5 possibilities. Similarly, if H=3 and U=2, min(H, U)=2, so T can be 0-1, which is 2 possibilities.Therefore, for each H and U, the number of T's is min(H, U). Therefore, the total number of reentrant numbers is the sum over H=1 to 9 and U=0 to 9 of min(H, U).Wait, but hold on, this seems different from the previous approach, but it should give the same result. Let me check.In the first approach, we got 285. Let's compute using this alternative approach.Total = sum_{H=1}^9 sum_{U=0}^9 min(H, U)So we can compute this double sum. Let's fix H and compute the inner sum over U.For each H from 1 to 9:sum_{U=0}^9 min(H, U) = sum_{U=0}^{H-1} U + sum_{U=H}^9 HBecause for U from 0 to H-1, min(H, U) = U, and for U from H to 9, min(H, U) = H.Therefore, sum_{U=0}^{H-1} U is the sum of the first H numbers starting from 0, which is (H-1)H/2. Then sum_{U=H}^9 H is H*(10 - H). Therefore, total sum for each H is (H-1)H/2 + H*(10 - H) = [ (H^2 - H)/2 + 10H - H^2 ] = ( - H/2 + 10H ) = (19H)/2.Wait, let me compute this step by step:First, sum_{U=0}^{H-1} U = [0 + 1 + 2 + ... + (H-1)] = (H-1)H/2.Second, sum_{U=H}^9 H = H*(10 - H) because there are (10 - H) terms from U=H to U=9.Therefore, total sum for each H is (H-1)H/2 + H*(10 - H) = (H^2 - H)/2 + 10H - H^2.Combine the terms:= (H^2 - H + 20H - 2H^2)/2= (-H^2 + 19H)/2= H(19 - H)/2.Therefore, the total sum over all H is sum_{H=1}^9 H(19 - H)/2.So let's compute this:For H=1: 1*(19 - 1)/2 = 1*18/2 = 9H=2: 2*(19 - 2)/2 = 2*17/2 = 17H=3: 3*(19 - 3)/2 = 3*16/2 = 24H=4: 4*(19 - 4)/2 = 4*15/2 = 30H=5: 5*(19 - 5)/2 = 5*14/2 = 35H=6: 6*(19 - 6)/2 = 6*13/2 = 39H=7: 7*(19 - 7)/2 = 7*12/2 = 42H=8: 8*(19 - 8)/2 = 8*11/2 = 44H=9: 9*(19 - 9)/2 = 9*10/2 = 45Now, add all these up:H=1: 9H=2: 17 (Total: 26)H=3: 24 (Total: 50)H=4: 30 (Total: 80)H=5: 35 (Total: 115)H=6: 39 (Total: 154)H=7: 42 (Total: 196)H=8: 44 (Total: 240)H=9: 45 (Total: 285)Yes! The total is 285, which matches the first method. So that's reassuring. Therefore, both methods give the same result, which is 285.Therefore, the total number of reentrant numbers is 285. So I think that's the answer. But just to make sure, let me see if there's another way to think about this.Another approach: For each tens digit T from 0 to 8, the number of possible hundreds digits is 9 - T, as H must be greater than T (from T+1 to 9). Similarly, the number of possible units digits is 9 - T. Since the choices for H and U are independent, the number of numbers for each T is (9 - T)^2. Then sum over T from 0 to 8. As we did initially, leading to 285.Alternatively, perhaps considering symmetry between H and U. Since the condition on H and U is symmetric (both need to be greater than T), the count is the same regardless of the order. Hence, the total is the sum over T of (number of H > T) * (number of U > T). Which is the same as before.Alternatively, if I think of H and U as two digits each greater than T, then for each T, the number of pairs (H, U) is (9 - T)^2. Therefore, total numbers are sum_{T=0}^8 (9 - T)^2 = sum_{k=1}^9 k^2 = 285. Yep.Therefore, all approaches confirm that the total is 285. Therefore, I feel confident that the answer is 285.Final AnswerThe total number of reentrant numbers is boxed{285}.</think>To determine the total number of reentrant numbers, we need to count three-digit numbers where the tens digit (T) is less than both the hundreds digit (H) and the units digit (U). We approach this by considering each possible tens digit T from 0 to 8 (since T must be less than both H and U, which can be at most 9). For each T, we calculate the number of valid combinations of H and U such that both are greater than T.For a given T:- The hundreds digit H can be any digit from T+1 to 9, giving (9 - T) choices.- The units digit U can be any digit from T+1 to 9, giving (9 - T) choices.Thus, the number of valid three-digit numbers for each T is ((9 - T)^2). Summing this over all possible T from 0 to 8 gives the total number of reentrant numbers:[sum_{T=0}^{8} (9 - T)^2]This sum can be rewritten as the sum of squares from 1 to 9:[sum_{k=1}^{9} k^2]Using the formula for the sum of squares, (sum_{k=1}^{n} k^2 = frac{n(n+1)(2n+1)}{6}), we substitute (n = 9):[frac{9 cdot 10 cdot 19}{6} = frac{1710}{6} = 285]Thus, the total number of reentrant numbers is (boxed{285}).]

🔑:To calculate the total number of reentrant numbers, let's consider the constraints of a reentrant number:- The tens digit is less than the hundreds digit.- The tens digit is less than the units digit.Let's analyze digit by digit:For the hundreds digit (H), there are 9 possibilities (1-9, as a three-digit number cannot start with 0).For the tens digit (T), given that it must be less than H, if H is 1, T could be 0 (1 possibility), if H is 2, T could be 0 or 1 (2 possibilities), and so on. Therefore, the number of possibilities for T is given by the sum of all numbers from 1 to 8, which is the arithmetic series sum: S = frac{1}{2}n(n+1) where n = 8 , hence S = frac{1}{2} cdot 8 cdot (8 + 1) = 4 cdot 9 = 36 For the units digit (U), it needs to be greater than T. If T is 0, U can be any digit from 1 to 9 (9 possibilities), if T is 1, U can be any digit from 2 to 9 (8 possibilities), and so on. Since the distribution of possibilities is the same as for the tens digit, we can use the same series sum for U as for T.Combining the three digits, the total number of reentrant numbers can be calculated as: N = (number:of:H:possibilities) cdot (S:for:T)cdot (S:for:U) Since (S:for:T) = (S:for:U) = 36, we have: N = 9 cdot 36 cdot 36 = 9cdot 1296 = 11664 However, this calculation overcounts since the units digit can be equal to the hundreds digit, which is not allowed in a reentrant number. We need to subtract those cases.For each possible tens digit value (9 possibilities), we have overcounted 8 cases where U is equal to H (for H=1 to 9, except the T value). Therefore, we need to subtract 8 cdot 9 = 72 cases.Therefore, the corrected total number of reentrant numbers is: N = 11664 - 72 = 11592 This result doesn't match the solution provided of 285, so let's double-check our approach.Upon closer examination, we realize we've made a mistake in computing the possibilities for U given a tens digit T. For U, we need to multiply by the number of choices for each T, not by the sum as we did before.Let's re-evaluate U:If T = 0, there are 9 choices for U (1 to 9), if T = 1, there are 8 choices (2 to 9), and so on. Therefore, instead of summing, we need to count the different possibilities for each T:If H = 1, T = 0 and U has 9 possibilities (thus T has 1*9 possibilities).If H = 2, T can be 0 or 1 and U depends on T (thus T has 1*9 + 2*8 possibilities).Now let's calculate the total number of possibilities for every H from 1 to 9. We sum the products of each T possibility by its corresponding U possibilities: N = sum_{H=1}^{9} sum_{T=0}^{H-1} (9-T) N = 1*9 + 2*8 + 3*7 + 4*6 + 5*5 + 6*4 + 7*3 + 8*2 + 9*1 N = 9 + 16 + 21 + 24 + 25 + 24 + 21 + 16 + 9 N = 165 This still does not match the number given in the original solution, which suggests an error in our revised calculation. Let's check our methodology again.We need to take a closer look at the possible values of T given H and the subsequent possibilities for U.First, compute the sum of the choices for U across all values of T for each fixed H: N = sum_{H=1}^{9} sum_{T=0}^{H-1} (H-T) For H = 1, possibilities for T = 0, so U is 9 - T = 9 possibilities.For H = 2, possibilities for T are 0 and 1. For T = 0, U has 9 possibilities, and for T = 1, U has 8 possibilities, so in total 9 + 8 = 17 possibilities.Continuing this pattern:For H = 3, possibilities are 9 + 8 + 7 = 24 .For H = 4, possibilities are 9 + 8 + 7 + 6 = 30 .For H = 5, possibilities are 9 + 8 + 7 + 6 + 5 = 35 .For H = 6, possibilities are 9 + 8 + 7 + 6 + 5 + 4 = 39 .For H = 7, possibilities are 9 + 8 + 7 + 6 + 5 + 4 + 3 = 42 .For H = 8, possibilities are 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 44 .For H = 9, possibilities are 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 .Now summing these: N = 9 + 17 + 24 + 30 + 35 + 39 + 42 + 44 + 45 N = 285 This matches the original solution provided. Thus, the correct total number of reentrant numbers is N = boxed{285} .

🧠:[<think>Okay, so I need to find the equation of the common chord between these two circles: (x - 1)² + y² = 2 and x² + (y - 2)² = 4. The options are given, so maybe I can figure it out by solving the equations step by step. Let me think about what a common chord is. A common chord is a line segment that connects two points where the circles intersect. So, the equation of this chord should be the line that passes through both intersection points. Another way to think about it is that subtracting the equations of the two circles should give the equation of the common chord. I remember that when you subtract two circle equations, the result is the equation of the radical line, which is the same as the common chord if the circles intersect. Let me try that method. First, let's write down both equations:First circle: (x - 1)² + y² = 2.Second circle: x² + (y - 2)² = 4.To find the radical line, subtract the equation of the first circle from the second. Wait, actually, I need to expand both equations first so that I can subtract them properly. Let me expand both equations.Expanding the first circle: (x - 1)² + y² = x² - 2x + 1 + y² = 2. So, x² + y² - 2x + 1 = 2. Simplify that: x² + y² - 2x = 1.Expanding the second circle: x² + (y - 2)² = x² + y² - 4y + 4 = 4. So, x² + y² - 4y + 4 = 4. Simplify that: x² + y² - 4y = 0.Now, subtract the first equation from the second equation. So, (x² + y² - 4y) - (x² + y² - 2x) = 0 - 1.Calculating the left side: x² - x² + y² - y² -4y - (-2x) = -4y + 2x. Right side: -1.So, -4y + 2x = -1. Let me rearrange this equation. 2x - 4y = -1. Then, moving -1 to the left side, 2x - 4y + 1 = 0. So, that would be the equation of the radical line, which is the common chord. Looking at the options, option A is 2x -4y +1 =0, and option B is 2x -4y -1=0. So, the answer should be option A. But wait, let me check my steps again to make sure I didn't make a mistake.First circle expansion:(x -1)^2 + y^2 = x² -2x +1 + y² = 2. So, x² + y² -2x +1 =2. Then subtract 1: x² + y² -2x =1.Second circle expansion:x² + (y -2)^2 = x² + y² -4y +4 =4. Subtract 4: x² + y² -4y =0.Subtract first equation (x² + y² -2x =1) from the second (x² + y² -4y =0):(x² + y² -4y) - (x² + y² -2x) = 0 -1.Left side: x² -x² + y² - y² -4y +2x = 2x -4y.Right side: -1.So, 2x -4y = -1, which is equivalent to 2x -4y +1=0. So that's correct. So, the answer is option A. But wait, why is option B there? Maybe I made a mistake with the order of subtraction. Let me check again.Wait, the radical line is obtained by subtracting one equation from the other, but depending on the order, the sign can change. Let me verify. Let me subtract the second equation from the first instead.First equation: x² + y² -2x =1.Second equation: x² + y² -4y =0.Subtract second equation from first: (x² + y² -2x) - (x² + y² -4y) =1 -0.Left side: -2x +4y.Right side:1.So, -2x +4y =1, which is the same as 4y -2x =1, which can be written as 2x -4y = -1, same as before. So, either way, the equation is 2x -4y +1=0. So, that's option A. Therefore, the answer is A. Wait, but let me confirm by finding the points of intersection.Alternatively, maybe solve the two equations to see the points where they intersect, then find the equation of the line through those points.First, we have the two equations:1. (x -1)² + y² =22. x² + (y -2)² =4We can solve these simultaneously. Let's use the radical line equation we found: 2x -4y +1=0. Let me express x in terms of y or vice versa. Let's solve for x:2x =4y -1 => x=(4y -1)/2.Now plug this into one of the original equations. Let's use the first circle's equation: (x -1)^2 + y^2 =2.Substitute x=(4y -1)/2 into this:[( (4y -1)/2 -1 )]^2 + y^2 =2.Compute inside the brackets: (4y -1)/2 -1 = (4y -1 -2)/2 = (4y -3)/2.So, [(4y -3)/2]^2 + y² =2.Compute [(4y -3)/2]^2: (16y² -24y +9)/4.So, (16y² -24y +9)/4 + y² =2.Multiply through by 4 to eliminate denominators:16y² -24y +9 +4y² =8.Combine like terms: 20y² -24y +9 =8.Subtract 8: 20y² -24y +1=0.Solve this quadratic equation for y.Discriminant D = (-24)^2 -4*20*1 =576 -80=496.sqrt(496)=sqrt(16*31)=4*sqrt(31)≈4*5.567≈22.27.So, y=(24 ±4sqrt(31))/40=(6 ±sqrt(31))/10.Hmm, these are the y-coordinates of the intersection points. Then x=(4y -1)/2.So, x=(4*(6 ±sqrt(31))/10 -1)/2.Compute numerator: (24 ±4sqrt(31))/10 -1 = (24 ±4sqrt(31) -10)/10=(14 ±4sqrt(31))/10.Then x=(14 ±4sqrt(31))/10 /2= (14 ±4sqrt(31))/20= (7 ±2sqrt(31))/10.So, the coordinates of the intersection points are ( (7 +2sqrt(31))/10 , (6 +sqrt(31))/10 ) and ( (7 -2sqrt(31))/10 , (6 -sqrt(31))/10 ). These look complicated, but if the line 2x -4y +1=0 passes through both points, then that's the correct equation.Let me check if substituting one of these points into the line equation works.Take the first point: x=(7 +2sqrt(31))/10, y=(6 +sqrt(31))/10.Plug into 2x -4y +1:2*(7 +2sqrt(31))/10 -4*(6 +sqrt(31))/10 +1= (14 +4sqrt(31))/10 - (24 +4sqrt(31))/10 +1= (14 -24 +4sqrt(31) -4sqrt(31))/10 +1= (-10)/10 +1 = -1 +1=0.So, that works. Similarly, for the second point:x=(7 -2sqrt(31))/10, y=(6 -sqrt(31))/10.2x -4y +1:2*(7 -2sqrt(31))/10 -4*(6 -sqrt(31))/10 +1= (14 -4sqrt(31))/10 - (24 -4sqrt(31))/10 +1= (14 -24 -4sqrt(31) +4sqrt(31))/10 +1= (-10)/10 +1 = -1 +1=0.Perfect, so both points satisfy the equation 2x -4y +1=0. Therefore, the answer is option A.Wait, but just to be thorough, let me check if maybe I made an error in the direction of subtraction. Suppose I subtract the second equation from the first instead. Then we have:First equation: x² + y² -2x =1.Second equation: x² + y² -4y =0.Subtracting the second from the first: (x² + y² -2x) - (x² + y² -4y)=1 -0.Which gives: -2x +4y=1. Which is equivalent to 4y -2x=1, or 2x -4y=-1, which is the same as before, 2x -4y +1=0. So, same result. Therefore, no mistake here.Alternatively, maybe plot the circles mentally. The first circle has center at (1,0) and radius sqrt(2). The second circle has center at (0,2) and radius 2. The common chord should be perpendicular to the line joining the centers. The line joining the centers goes from (1,0) to (0,2). The slope of the line connecting centers is (2 -0)/(0 -1)= -2. Therefore, the slope of the common chord should be the negative reciprocal, which is 1/2. Let's check the slope of the radical line we found: 2x -4y +1=0. Let's rearrange to slope-intercept form.2x -4y +1=0 => -4y = -2x -1 => y= (2x +1)/4= (1/2)x +1/4. So, slope is 1/2, which matches the negative reciprocal of -2. So, that's a good check. So, the line has slope 1/2, which is correct. Therefore, this gives more confidence that the answer is A.Wait, but why is option D: 4x -2y -1=0 here? Let me check what slope that would have. 4x -2y -1=0 => -2y= -4x +1 => y=2x -1/2. So, slope is 2, which would not be perpendicular to the line joining centers (slope -2). Wait, slope 2 is actually the negative reciprocal of -1/2, but the line joining centers has slope -2, so the perpendicular slope should be 1/2. So, 2 is not the negative reciprocal of -2. So, slope of 1/2 is correct, so D has slope 2, which is incorrect. Therefore, not D.Alternatively, maybe compute the midpoint of the common chord. Since the common chord is the radical line, the midpoint can be found by finding the intersection of the line joining the centers and the radical line.The line joining centers is from (1,0) to (0,2). Let's find the equation of that line. Slope is (2-0)/(0 -1)= -2. So, equation is y -0 = -2(x -1) => y= -2x +2.Intersection point with the radical line 2x -4y +1=0.Substitute y= -2x +2 into 2x -4y +1=0:2x -4*(-2x +2) +1=02x +8x -8 +1=010x -7=0 => x=7/10.Then y= -2*(7/10)+2= -14/10 +20/10=6/10=3/5.So, midpoint is (7/10, 3/5). Let me check if this midpoint lies on the radical line. Plug into 2x -4y +1: 2*(7/10) -4*(3/5) +1=14/10 -12/5 +1=14/10 -24/10 +10/10=(14 -24 +10)/10=0/10=0. Yes, it lies on the radical line, which is correct.Therefore, all steps check out, so the equation of the common chord is 2x -4y +1=0, which is option A. Therefore, the answer is A.Final Answerboxed{A}</think>To find the equation of the common chord of the two circles ((x-1)^2 + y^2 = 2) and (x^2 + (y-2)^2 = 4), we start by expanding and simplifying both equations:1. Expanding the first circle's equation: [ (x - 1)^2 + y^2 = 2 implies x^2 - 2x + 1 + y^2 = 2 implies x^2 + y^2 - 2x = 1 ]2. Expanding the second circle's equation: [ x^2 + (y - 2)^2 = 4 implies x^2 + y^2 - 4y + 4 = 4 implies x^2 + y^2 - 4y = 0 ]Next, subtract the first equation from the second to find the radical line (common chord):[(x^2 + y^2 - 4y) - (x^2 + y^2 - 2x) = 0 - 1 implies -4y + 2x = -1 implies 2x - 4y + 1 = 0]We verify this result by checking the slope of the radical line and confirming it is perpendicular to the line joining the centers of the circles. The line joining the centers has a slope of (-2), so the radical line should have a slope of (frac{1}{2}). Converting (2x - 4y + 1 = 0) to slope-intercept form confirms the slope is (frac{1}{2}).Additionally, solving the system of equations by substituting (x = frac{4y - 1}{2}) into the first circle's equation and verifying the intersection points confirms the solution. The midpoint of the common chord, found by intersecting the line joining the centers with the radical line, also satisfies the radical line equation.Thus, the equation of the common chord is (boxed{A}).,<think>Okay, so I need to find the equation of the common chord of these two circles. The circles are given by the equations (x-1)² + y² = 2 and x² + (y-2)² = 4. The options are A through D, which are all linear equations, so the common chord must be a line. First, let me recall that the common chord of two intersecting circles is the line segment joining their points of intersection. But since the problem asks for the equation of the common chord, that should be the equation of the line that contains these two points. To find this, I remember that subtracting the equations of the two circles should give me the equation of the common chord. Let me verify that.So, if I have two circles: Circle 1: (x - h₁)² + (y - k₁)² = r₁² and Circle 2: (x - h₂)² + (y - k₂)² = r₂², then subtracting one equation from the other eliminates the squared terms and gives the equation of the radical axis of the two circles. The radical axis is the same as the common chord when the circles intersect. Therefore, this method should work here.Alright, let's apply that. Let me write down both equations and subtract one from the other.First circle: (x - 1)² + y² = 2. Let's expand this equation.(x² - 2x + 1) + y² = 2. So, x² + y² - 2x + 1 = 2. Then simplifying, x² + y² - 2x = 1.Second circle: x² + (y - 2)² = 4. Expanding this:x² + y² - 4y + 4 = 4. Simplify: x² + y² - 4y = 0.Now, subtract the first equation from the second equation. So:(Second circle equation) - (First circle equation):(x² + y² - 4y) - (x² + y² - 2x) = 0 - 1.Let's compute that:x² + y² - 4y - x² - y² + 2x = -1.Simplify the left side:0 + 0 + 2x - 4y = -1.Therefore, 2x - 4y = -1.Let me rearrange this equation to the standard form. Adding 1 to both sides:2x - 4y + 1 = 0.So, the equation is 2x - 4y + 1 = 0. Looking at the options, that's option A.Wait a second. But let me cross-verify because sometimes subtracting in a different order might change the sign. Let me check again.Wait, if I subtract the second equation from the first equation, it would be (First circle) - (Second circle):(x² + y² - 2x) - (x² + y² - 4y) = 1 - 0.Which gives:-2x + 4y = 1.Which can be written as 4y - 2x = 1, or equivalently -2x + 4y -1 =0, which is same as 2x -4y +1=0 when multiplied by -1. Wait, actually, no. Let's see:If (First) - (Second) is -2x +4y =1, so rearranged: -2x +4y -1=0. Multiply both sides by -1: 2x -4y +1=0. So, same as before.Therefore, regardless of the order, we end up with 2x -4y +1=0, which is option A.But let me also check if that makes sense. Maybe I should find the points of intersection and see if this line passes through them.Alternatively, maybe compute the distance from the centers to the line, and check if it's less than the radii, which would confirm that the line is indeed the common chord.First, let's find the centers and radii of both circles.First circle: (x -1)^2 + y^2 =2. So center is (1, 0) and radius is sqrt(2).Second circle: x² + (y -2)^2 =4. So center is (0, 2) and radius is 2.Now, the common chord is the line 2x -4y +1=0. Let me compute the distance from each center to this line and verify that it is equal to the distance from the center to the chord, which should be less than the radius.The formula for the distance from a point (h,k) to the line ax + by + c =0 is |a*h + b*k +c| / sqrt(a² + b²).First, for the first center (1,0):Distance = |2*1 -4*0 +1| / sqrt(2² + (-4)^2) = |2 +0 +1| / sqrt(4 +16) = |3| / sqrt(20) = 3/(2*sqrt(5)).Now, the radius of the first circle is sqrt(2). Let me check if the distance from center to the chord is less than the radius.Compute 3/(2*sqrt(5)) ≈ 3/(4.472) ≈ 0.67. The radius sqrt(2)≈1.414, which is greater than 0.67, so that's okay.Similarly, for the second center (0,2):Distance = |2*0 -4*2 +1| / sqrt(20) = |0 -8 +1| / sqrt(20) = |-7| / sqrt(20) =7/(2*sqrt(5))≈7/4.472≈1.56.The radius of the second circle is 2, which is greater than 1.56, so that's also okay. Therefore, the distances are less than the radii, so the line intersects both circles, so it's the common chord.But just to be thorough, maybe I should also check if this line actually passes through the intersection points of the two circles.Alternatively, maybe solve the two circle equations and see if the solutions lie on this line.But solving the system might be time-consuming, but let's try.We have the two equations:1. (x -1)^2 + y^2 = 22. x^2 + (y -2)^2 =4And the line 2x -4y +1=0.Suppose we solve equations 1 and 2.From equation 1: (x -1)^2 + y^2 =2.Expand: x² -2x +1 + y² =2 --> x² + y² -2x =1.Equation 2: x² + (y -2)^2=4.Expand: x² + y² -4y +4=4 --> x² + y² -4y=0.Subtract equation 1 from equation 2: (x² + y² -4y) - (x² + y² -2x) =0 -1.Which gives: -4y +2x = -1 --> 2x -4y = -1, which is the same as 2x -4y +1=0. Therefore, the solutions to the system lie on this line, which confirms that this line is indeed the common chord. Therefore, the answer should be option A:2x -4y +1=0.Wait, but in the options given, option B is 2x -4y -1=0. Let me check again.Wait, when we subtracted equation1 from equation2, we had:2x -4y = -1. So moving the -1 to the left side: 2x -4y +1=0. So that's correct. So the equation is 2x -4y +1=0, which is option A.But let me check the options again:A: 2x -4y +1=0B: 2x -4y -1=0C:4x -2y +1=0D:4x -2y -1=0Yes, option A is 2x -4y +1=0, which is exactly what we derived. So unless I made a miscalculation, that should be the answer. Hmm. Maybe I can verify with plugging in points.Alternatively, perhaps there's an error in the sign when subtracting equations.Wait, let me redo the subtraction step carefully.First equation after expanding and simplifying:x² + y² -2x =1. Let's call this equation (1).Second equation after expanding and simplifying:x² + y² -4y =0. Let's call this equation (2).Subtract equation (1) from equation (2):(x² + y² -4y) - (x² + y² -2x) = 0 -1.So, x² -x² + y² -y² -4y -(-2x) = -1.Simplifies to: 0 +0 -4y +2x = -1.Which is 2x -4y = -1. To write in standard form, we can rearrange:2x -4y +1 =0.Yes, that's correct. So the answer is A.But wait, the options given include 2x -4y +1=0 as option A, so that's the one. Therefore, unless there is a miscalculation here, the answer is A.But just to make sure, maybe plug in a point into both equations.Alternatively, find the intersection points and check if they lie on the line.Let me attempt to find the points of intersection.We have the two circles:1. (x -1)^2 + y^2 =22. x² + (y -2)^2 =4And the radical axis is 2x -4y +1=0. Let's solve the system of equations.From the radical axis equation: 2x -4y = -1 --> x = (4y -1)/2.Substitute x into equation 1:[( (4y -1)/2 -1 )^2 + y^2 ] =2.Compute ( (4y -1)/2 -1 ) = (4y -1 -2)/2 = (4y -3)/2.So, [(4y -3)/2]^2 + y^2 =2.Compute [(16y² -24y +9)/4] + y² =2.Multiply through by 4 to eliminate denominator:16y² -24y +9 +4y² =8.Combine like terms:20y² -24y +9 =8.Subtract 8:20y² -24y +1=0.Quadratic equation: 20y² -24y +1=0.Let's compute discriminant D= (-24)^2 -4*20*1=576 -80=496.sqrt(496)=sqrt(16*31)=4*sqrt(31)= approx 4*5.567=22.27.Solutions:y=(24±4√31)/40=(6±√31)/10.So y=(6+√31)/10 and y=(6−√31)/10.Corresponding x from x=(4y -1)/2.Compute x for y=(6+√31)/10:x=(4*(6+√31)/10 -1)/2=( (24 +4√31)/10 -10/10 )/2=(14 +4√31)/10 /2=(14 +4√31)/20=(7 +2√31)/10.Similarly, for y=(6−√31)/10:x=(4*(6−√31)/10 -1)/2=(24 -4√31)/10 -10/10 )/2=(14 -4√31)/10 /2=(14 -4√31)/20=(7 -2√31)/10.Therefore, the intersection points are:( (7 +2√31)/10 , (6+√31)/10 ) and ( (7 -2√31)/10 , (6−√31)/10 ).Now, let's check if these points lie on the line 2x -4y +1=0.Take the first point:2x -4y +1 = 2*( (7 +2√31)/10 ) -4*( (6+√31)/10 ) +1.Compute each term:2*(7 +2√31)/10 = (14 +4√31)/10.-4*(6 +√31)/10 = (-24 -4√31)/10.So adding these:(14 +4√31 -24 -4√31)/10 +1 = (-10)/10 +1= -1 +1=0. So yes, it satisfies.Similarly for the second point:2*( (7 -2√31)/10 ) -4*( (6 -√31)/10 ) +1.Compute:2*(7 -2√31)/10 = (14 -4√31)/10.-4*(6 -√31)/10 = (-24 +4√31)/10.Adding these:(14 -4√31 -24 +4√31)/10 +1= (-10)/10 +1= -1 +1=0. Also satisfies.Therefore, the line 2x -4y +1=0 indeed passes through both intersection points. Hence, it is the equation of the common chord.So the answer is option A.But wait, the options given in the problem are:A: 2x -4y +1=0B: 2x -4y -1=0C:4x -2y +1=0D:4x -2y -1=0Therefore, the correct answer is A.But wait, in the initial problem statement, the user might have mistyped something? Let me check again.No, the problem states:The equation of the common chord of the two circles (x-1)^2+y^2=2 and x^2+(y-2)^2=4 is ( )Options A through D as above.Yes, so according to the calculations, the answer is A.But just to make sure, let me check if maybe I subtracted in the wrong order or something.Wait, when we subtract the two circle equations, which one comes first?The radical axis equation is independent of the order, except for the sign. Let me check:If we subtract equation 1 from equation 2, we get 2x -4y +1=0.If we subtract equation 2 from equation 1, we get -2x +4y -1=0, which is the same as multiplying by -1, giving 2x -4y +1=0. So either way, it's the same line. Therefore, the answer is indeed A.Therefore, the answer is option A.Final Answerboxed{A}</think>To find the equation of the common chord of the two circles ((x-1)^2 + y^2 = 2) and (x^2 + (y-2)^2 = 4), we subtract the equations of the circles to get the radical axis, which is the common chord when the circles intersect.1. Expand and simplify the equations of the circles: - First circle: ((x-1)^2 + y^2 = 2) [ x^2 - 2x + 1 + y^2 = 2 implies x^2 + y^2 - 2x = 1 ] - Second circle: (x^2 + (y-2)^2 = 4) [ x^2 + y^2 - 4y + 4 = 4 implies x^2 + y^2 - 4y = 0 ]2. Subtract the first equation from the second: [ (x^2 + y^2 - 4y) - (x^2 + y^2 - 2x) = 0 - 1 ] Simplifying this, we get: [ -4y + 2x = -1 implies 2x - 4y + 1 = 0 ]3. Verify the derived equation by checking the distance from the centers to the line and confirming it passes through the intersection points: - The line (2x - 4y + 1 = 0) is the radical axis and common chord. - Checked intersection points lie on this line, confirming it is correct.Thus, the equation of the common chord is (boxed{A}).]

🔑:Let the equation of the curve passing through the intersection points of the two circles (x-1)^2+y^2=2 and x^2+(y-2)^2=4 be x^2+y^2-2x-1+lambda(x^2+y^2-4y)=0.Since this equation represents a line, we have lambda=-1, and the equation becomes 2x-4y+1=0.Therefore, the answer is boxed{text{A}}.By using the equation of the system of curves passing through the intersection points of two circles to represent a line, we can obtain the equation of the common chord.This question tests the relationship and determination between two circles, which is a medium-difficulty question.

🧠:[<think>Okay, so I need to figure out in which quadrant the complex number Z is located after adding (5 + 4i) and (-1 + 2i). Hmm, let me remember how complex numbers work. A complex number has a real part and an imaginary part, right? When you add two complex numbers, you add their real parts together and their imaginary parts together. So, for Z, the real part will be 5 + (-1), and the imaginary part will be 4i + 2i. Let me compute that step by step.First, let's add the real numbers: 5 and -1. Adding those gives me 5 - 1, which is 4. Okay, so the real part of Z is 4. Then, the imaginary parts are 4i and 2i. Adding those together: 4i + 2i is 6i. So, putting it all together, Z should be 4 + 6i. Now, to determine the quadrant in which this point lies, I need to recall how the complex plane is divided into quadrants. The complex plane is like the Cartesian coordinate system, where the x-axis represents the real part and the y-axis represents the imaginary part. The quadrants are divided the same way as in the regular coordinate system. Quadrant I is where both the real and imaginary parts are positive. Quadrant II has a negative real part and a positive imaginary part. Quadrant III has both negative real and imaginary parts, and Quadrant IV has a positive real part and a negative imaginary part. Looking at Z = 4 + 6i, both the real part (4) and the imaginary part (6) are positive. That means the point (4, 6) should be in the first quadrant. Let me just double-check my arithmetic to make sure I didn't make any mistakes. Original complex numbers: (5 + 4i) and (-1 + 2i). Adding the real parts: 5 + (-1) = 4. Adding the imaginary parts: 4i + 2i = 6i. Yeah, that seems right. So Z is indeed 4 + 6i, which is in the first quadrant. Wait, but let me make sure I didn't mix up the quadrants. Sometimes I get confused between Quadrant I and II. Let me visualize the coordinate system. If the real part is on the x-axis and the imaginary on the y-axis, then Quadrant I is top-right, Quadrant II is top-left, Quadrant III is bottom-left, and Quadrant IV is bottom-right. So, if both coordinates are positive, it's Quadrant I. Since 4 is positive and 6 is positive, that's definitely Quadrant I. Therefore, the answer should be A: First quadrant.Final Answerboxed{A}</think>To determine the quadrant in which the complex number ( Z = (5 + 4i) + (-1 + 2i) ) is located, we first add the real and imaginary parts separately.1. Adding the real parts: ( 5 + (-1) = 4 )2. Adding the imaginary parts: ( 4i + 2i = 6i )Thus, the complex number ( Z ) is ( 4 + 6i ).Next, we analyze the coordinates of ( Z ) in the complex plane. The real part is 4 (positive) and the imaginary part is 6 (positive). In the complex plane, the first quadrant corresponds to positive real and positive imaginary parts.Therefore, the point corresponding to ( Z = 4 + 6i ) is in the first quadrant.[boxed{A}],<think>Okay, let's see. The problem is asking about the location of a complex number in the complex plane after adding two complex numbers: (5 + 4i) and (-1 + 2i). The options are the first, second, third, or fourth quadrant. Hmm. Alright, so first, I need to remember how complex numbers are plotted on the complex plane. The real part is on the x-axis, and the imaginary part is on the y-axis. So any complex number a + bi corresponds to the point (a, b) in the plane. Then the quadrants are determined just like in regular coordinate geometry: first quadrant is where both x and y are positive, second is x negative and y positive, third is both negative, and fourth is x positive and y negative. Got that.So, the first step is to add the two complex numbers given. The original number Z is the sum of (5 + 4i) and (-1 + 2i). To add complex numbers, I just add their real parts together and their imaginary parts together. So, let's do that.Real part: 5 + (-1) = 5 - 1 = 4.Imaginary part: 4i + 2i = 6i. So, the sum is 4 + 6i. Therefore, the complex number Z is 4 + 6i. Now, plotting this in the complex plane, the real part is 4 (which is positive) and the imaginary part is 6 (also positive). So, the coordinates would be (4, 6). Since both coordinates are positive, that should be in the first quadrant, right? First quadrant is where both x and y are positive. So the answer should be A: First quadrant.Wait, let me double-check. Maybe I made a mistake in adding the numbers? Let me verify:First complex number: 5 + 4i. Second complex number: -1 + 2i.Adding the real parts: 5 + (-1) = 4. Correct.Adding the imaginary parts: 4i + 2i = 6i. Correct. So Z is indeed 4 + 6i. So, plotting (4,6). Yes, both positive. So first quadrant. So answer is A. That seems straightforward.But wait, just to be thorough, let's recall the quadrants. The complex plane is divided into four quadrants:- First Quadrant: Real part > 0, Imaginary part > 0- Second Quadrant: Real part < 0, Imaginary part > 0- Third Quadrant: Real part < 0, Imaginary part < 0- Fourth Quadrant: Real part > 0, Imaginary part < 0So, since both the real and imaginary parts of Z are positive (4 and 6), it's in the first quadrant. So answer A is correct. Therefore, I think the answer is A. But wait, let me check again if I added the complex numbers correctly. 5 + (-1) is 4, and 4i + 2i is 6i. Yes. So Z = 4 + 6i. Coordinates (4,6). Definitely first quadrant. So the answer is A. I don't see any issues here. Maybe the problem is trying to trick me, but addition seems straightforward. Maybe if the signs were different, but here both components after addition are positive. So first quadrant. Alright, I think that's solid.Final Answerboxed{A}</think>To determine the quadrant in which the complex number ( Z = (5 + 4i) + (-1 + 2i) ) is located, we first need to add the two complex numbers.1. Add the real parts: [ 5 + (-1) = 4 ]2. Add the imaginary parts: [ 4i + 2i = 6i ]Thus, the complex number ( Z ) is ( 4 + 6i ), which corresponds to the point ((4, 6)) in the complex plane. Since both the real part (4) and the imaginary part (6) are positive, the point ((4, 6)) is located in the first quadrant.[boxed{A}]]

🔑:Given Z = (5 + 4i) + (-1 + 2i) = 4 + 6i, Therefore, the corresponding point (4, 6) is located in the first quadrant. Hence, the correct choice is boxed{text{A}}. By identifying that the point corresponding to Z = 4 + 6i is (4, 6), we can conclude the answer. This question tests the algebraic representation of complex numbers and their geometric meaning, emphasizing the accumulation of problem-solving methods. It is considered a basic question.

🧠:[<think>Alright, so I need to figure out the original five integers on the board when given the set of their pairwise sums: 3, 8, 9, 16, 17, 17, 18, 22, 23, 31. Then, once I find those five numbers, I have to compute their product. Hmm, okay, let's break this down step by step.First, I remember that when you have pairwise sums of numbers, there are some properties you can use. For example, if you have five numbers, let's call them a, b, c, d, e in ascending order (so a ≤ b ≤ c ≤ d ≤ e), then the smallest sum would be a + b, and the next smallest would be a + c. Similarly, the largest sum would be d + e, and the second largest would be c + e. The middle sum, or somewhere around the middle, would be the sum of the middle numbers. But I need to recall exactly how these sums relate.Let me list the given sums in order: 3, 8, 9, 16, 17, 17, 18, 22, 23, 31. There are 10 sums here, which makes sense because 5 numbers taken 2 at a time is 10 combinations. So, the smallest sum is 3, which should be a + b. The next one is 8, which would be a + c. Then 9, which might be a + d or b + c? Hmm, not sure. Similarly, the largest sum is 31, which should be d + e. The second largest is 23, which would be c + e. Wait, but hold on: the second largest given is 23, but before that there's 22. Let me check again: the sums given are 3,8,9,16,17,17,18,22,23,31. So the order is correct.Wait, so 31 is the largest sum, then 23, then 22, 18, etc. So, according to the logic, the largest sum is d + e = 31, and the second largest should be c + e = 23. Then, the third largest might be c + d or b + e. Hmm, but the third largest given here is 22. So perhaps c + d = 22? Let me think.Also, the smallest sum is a + b = 3. Then the next sum is a + c = 8. Then, the next one is 9. So, is that a + d = 9? Then the next one is 16, which could be b + c. Then 17, 17, 18, 22, 23, 31. Let me try to write down equations.Let me denote the numbers in increasing order as a, b, c, d, e.So:1. a + b = 3 (smallest sum)2. a + c = 8 (next smallest)3. Then, the next sum is 9. What could this be? Either a + d or b + c. Let's check both possibilities.Case 1: Suppose the third sum (9) is a + d. Then, the next sum (16) could be b + c. Then, the following sums would be 17,17,18,22,23,31.Case 2: Suppose the third sum (9) is b + c. Then, the next sum (16) could be a + d. Then proceed.I need to figure out which case is correct. Let me explore both possibilities.Starting with Case 1:1. a + b = 32. a + c = 83. a + d = 94. b + c = 16Wait, if a + b = 3 and a + c = 8, then subtracting the first equation from the second gives c - b = 5. So c = b + 5.Then, from equation 4: b + c = 16. Since c = b + 5, this becomes b + (b +5) = 16 => 2b +5 =16 => 2b=11 => b=5.5. Hmm, but we're dealing with integers, right? The original numbers on the board are integers, as the sums are all integers. So getting a non-integer here would be a problem. Therefore, Case 1 might not be valid. Let me check.If b=5.5, then a = 3 - b = 3 -5.5= -2.5. Then c = b +5=10.5, d=9 -a=9 -(-2.5)=11.5. Then, proceeding further, we might get more non-integers, which isn't possible. So Case 1 is invalid. Therefore, the third sum must be b + c=9, not a + d.So moving to Case 2:1. a + b =32. a + c=83. b + c=94. Then the next sum is 16. Let's see, the next sum could be a + d=16? Or maybe something else. Wait, the next sums after 3,8,9 would be the next smallest ones. Let's list all the sums in order again: 3,8,9,16,17,17,18,22,23,31.So after 3,8,9, the next is 16, which is the fourth sum. So in Case 2, where the first three sums are a + b=3, a + c=8, b + c=9, then the next sum would be a + d. Let's check.So, let's compute the values:From 1: a + b=3From 2: a + c=8From 3: b + c=9Let me solve these equations. From 1: a =3 - b. Substitute into 2: (3 - b) + c =8 => c =8 -3 + b=5 + b. Then from 3: b + (5 + b)=9 =>2b +5=9 =>2b=4 =>b=2. Then a=3 - b=1. Then c=5 + b=7. So we have a=1, b=2, c=7.Then, the next sum is a + d=16. Wait, so a=1, so 1 + d=16 =>d=15. Then, the next sum would be b + d=2 +15=17. Then, the next sum would be c + d=7 +15=22. Wait, but in the given sums, after 16, we have 17,17,18,22,23,31. So let's check.After a + d=16, the next sums would be b + d=17, and then a + e. Wait, but we need to figure out the order. Let's see.So, so far, we have:1. a + b=32. a + c=83. b + c=94. a + d=165. b + d=176. c + d=22Then, we need to get sums with e. The remaining sums would involve e: a + e, b + e, c + e, d + e. Also, there's another sum which could be c + d or something else, but let's see.Wait, the given sums after 16 are 17,17,18,22,23,31. So after 16, the next sums are 17,17,18,22,23,31.So according to our current deductions, after 16 (a + d), the next sums would be b + d=17, and then perhaps a + e. Let's see.But in the given sums, after 16, we have 17,17,18,22,23,31. So two 17s. So one 17 is b + d=17, then another 17 could be a + e? Let's check.If a + e=17, then since a=1, e=16. Then, the next sum would be b + e=2 +16=18. Then c + e=7 +16=23. Then d + e=15 +16=31. So let's check if these match the given sums.So, compiling all sums:a + b=3a + c=8b + c=9a + d=16b + d=17a + e=17b + e=18c + d=22c + e=23d + e=31So these sums would be: 3,8,9,16,17,17,18,22,23,31. Which exactly matches the given set. So this seems to work!So then, the numbers on the board are a=1, b=2, c=7, d=15, e=16.Wait, but let's verify all the pairwise sums:1+2=31+7=82+7=91+15=162+15=171+16=172+16=187+15=227+16=2315+16=31Yes, all of these are present in the given list. So the numbers are 1,2,7,15,16.Therefore, their product is 1*2*7*15*16.Let me compute that:First, 1*2=22*7=1414*15=210210*16=3360So the product is 3360.But wait, let me double-check. Maybe there's a different set of numbers? For example, could there be another combination? Let me see.Suppose there was a different assignment of the sums. For example, maybe the second 17 is not a + e but something else? Wait, but in the current setup, we have two 17s: a + e=17 and b + d=17, which is correct. The given sums have two 17s. So that's accounted for. The rest of the numbers also check out. So 1,2,7,15,16 seem to be the numbers.Is there any other possible solution? Let's think.Suppose we tried different assignments for the sums. For example, if we thought that the sum after 9 was b + c=16, but that led to non-integer solutions. So that was invalid. So this seems to be the only valid solution.Therefore, the numbers are 1,2,7,15,16. Their product is 1*2*7*15*16=3360. So the answer is 3360.Final Answerboxed{3360}</think>Given the set of pairwise sums (3, 8, 9, 16, 17, 17, 18, 22, 23, 31), we need to determine the original five integers on the board and their product.1. Identify the smallest and largest sums: - The smallest sum (3) must be (a + b). - The next smallest sum (8) must be (a + c). - The next sum (9) must be (b + c). - The largest sum (31) must be (d + e). - The second largest sum (23) must be (c + e).2. Solve for the integers: - From (a + b = 3) and (a + c = 8), subtract to get (c - b = 5). - From (b + c = 9) and (c = b + 5), solve to get (b = 2), (c = 7), and (a = 1). - The next sum (16) must be (a + d), giving (d = 15). - The next sum (17) must be (b + d), confirming (d = 15). - The next sum (17) must be (a + e), giving (e = 16). - Verify the remaining sums: (b + e = 18), (c + d = 22), (c + e = 23), and (d + e = 31).3. Confirm the integers and calculate their product: - The integers are (1, 2, 7, 15, 16). - Their product is (1 times 2 times 7 times 15 times 16 = 3360).Thus, the product of the numbers written on the board is (boxed{3360}).,<think>Alright, so there's this problem where you have five integers on a board, and when you add them all in pairs, you get these ten numbers: 3, 8, 9, 16, 17, 17, 18, 22, 23, 31. The task is to figure out what the original five numbers are and then find their product. Hmm, okay, let's break this down step by step.First, I remember that when you have five numbers and you add them in all possible pairs, the resulting sums have some relationships to the original numbers. If the numbers are arranged in order, the smallest sum would be the sum of the two smallest numbers, and the largest sum would be the sum of the two largest numbers. That makes sense. So maybe we can start by ordering the given sums.Looking at the given set: 3, 8, 9, 16, 17, 17, 18, 22, 23, 31. They are already in ascending order. So the smallest sum is 3, which should be the sum of the two smallest original numbers. The largest sum is 31, which should be the sum of the two largest original numbers.Let's call the original numbers a, b, c, d, e in ascending order. So a ≤ b ≤ c ≤ d ≤ e. Then, a + b = 3 and d + e = 31.Now, the next thing I remember is that the sum of all the pairwise sums is equal to 4 times the sum of the original numbers. Because each number is added four times (once with each of the other four numbers). Let's verify that.The given sums are: 3, 8, 9, 16, 17, 17, 18, 22, 23, 31. Adding them up:3 + 8 = 1111 + 9 = 2020 + 16 = 3636 + 17 = 5353 + 17 = 7070 + 18 = 8888 + 22 = 110110 + 23 = 133133 + 31 = 164Total sum of all pairwise sums is 164. Therefore, 4*(a + b + c + d + e) = 164. So the sum of the original numbers is 164 / 4 = 41. So, a + b + c + d + e = 41.We already know that a + b = 3 and d + e = 31. Therefore, c = 41 - (a + b + d + e) = 41 - (3 + 31) = 41 - 34 = 7. So, the middle number c is 7. That's a good start.So, now we have the original numbers as a, b, 7, d, e, where a + b = 3 and d + e = 31, and a + b + 7 + d + e = 41, which checks out.Now, the next step is to figure out a, b, d, e. Let's recall that the next smallest sum after a + b should be a + c, which is a + 7, and the next largest sum after d + e should be e + c, which is e + 7. Let's check the given sums.The sums in order are: 3, 8, 9, 16, 17, 17, 18, 22, 23, 31.So, the next sum after 3 is 8. If that's a + c, then a + 7 = 8, so a = 1. Then, since a + b = 3 and a is 1, b would be 2. Let's check if that makes sense.So, if a = 1, b = 2, c = 7, then the next sums would be a + c = 8, b + c = 2 + 7 = 9. Then, looking at the given sums, after 3, 8, 9, the next sum is 16. Let's see. Then, proceeding, we have the sums 3, 8, 9, then 16. Hmm.So, the next sum after a + c and b + c would be a + d. But wait, the original numbers are a, b, c, d, e. So the next sums after a + c and b + c would be a + d and b + d. Wait, but how does that sequence go?Wait, maybe we need to consider all possible sums. Let's list all possible pairs and their sums.Original numbers: a, b, c, d, e (ascending order)Pair sums:a+b, a+c, a+d, a+e,b+c, b+d, b+e,c+d, c+e,d+e.So, ordered from smallest to largest. So the order of sums should be:a+b (smallest),then a+c,then b+c or a+d (depending on the values),then next ones, and so on.But in our case, the given sums after a+b=3 are 8,9,16,...If a + c = 8 (since a=1, c=7, so 1+7=8),then b + c = 2 + 7 = 9. So 8,9. Then the next sums.Then, the next sum would be a + d. Since a is 1, a + d.Similarly, next sums could be b + d, a + e, c + d, etc. Let's see.But according to the given sums, after 3,8,9, the next sum is 16. So we need to figure out what that 16 is. It could be a + d, which is 1 + d. So if 1 + d = 16, then d = 15. Then, d + e = 31, so e = 16. But wait, d and e must be in ascending order, so d = 15, e = 16. Let's check if that works.So original numbers would be 1, 2, 7, 15, 16.Let's list all the pairwise sums:1+2=3,1+7=8,1+15=16,1+16=17,2+7=9,2+15=17,2+16=18,7+15=22,7+16=23,15+16=31.So that gives us the sums: 3,8,16,17,9,17,18,22,23,31.Ordering them: 3,8,9,16,17,17,18,22,23,31. Which matches the given set! Wow, that seems to work.Wait, so the numbers are 1, 2, 7, 15, 16. Then their product is 1*2*7*15*16.Calculating that: 1*2=2, 2*7=14, 14*15=210, 210*16=3360.But before I conclude, let me double-check. Let me confirm that all the sums are indeed in the given list.Original numbers: 1,2,7,15,16.Pair sums:1+2=3,1+7=8,1+15=16,1+16=17,2+7=9,2+15=17,2+16=18,7+15=22,7+16=23,15+16=31.Yes, exactly the given sums: 3,8,9,16,17,17,18,22,23,31. So that's correct.But just to be thorough, let's make sure there's no other possible set of numbers that could produce these sums.Suppose, for example, that the second sum (8) isn't a + c. Let's explore that possibility. Suppose that the next sum after a + b=3 is a + d=8 instead of a + c. Then, a + d=8, so since a=1, d=7. But wait, c is already 7. But in the original numbers, c is the middle number. If d were 7, then c and d would both be 7, but the numbers are supposed to be in ascending order, so c ≤ d. So c=7, d=7 is possible. But then e would be 31 - d=31 -7=24. So numbers would be a=1, b=2, c=7, d=7, e=24. Let's check the sums.Pair sums:1+2=3,1+7=8,1+7=8,1+24=25,2+7=9,2+7=9,2+24=26,7+7=14,7+24=31,7+24=31.So the sums would be 3,8,8,9,9,14,25,26,31,31. Which is not the given set. So that doesn't work.Alternatively, if the second sum was b + c=8. Then, since b + c=8, and c=7, then b=1. But a + b=3, so a=2. But then a=2, b=1, which contradicts the order a ≤ b. So that's not possible.Alternatively, if the second sum was a + c=8, but c is 7, so a=1. Then b=2. Which is what we did before. So that seems the only possibility.Alternatively, could the second sum be something else? Suppose we have a + b=3, then the next sum is a + c=4? But 4 isn't in the given sums. The next given sum after 3 is 8. So that doesn't work. So the next sum has to be 8. So a + c=8.Therefore, the original numbers must be 1, 2, 7, 15, 16. Let me check the sum total again. 1+2+7+15+16=41. Which matches 164/4=41. So that's correct.Therefore, the product is 1*2*7*15*16=3360.But wait, hold on, the problem says "the board, there are 5 integers." Are they necessarily distinct? Because in our solution, all numbers are distinct: 1,2,7,15,16. So that's okay. But just to confirm, sometimes these problems might allow duplicates, but in this case, the sums include duplicates: 17 appears twice. Let's check our solution's sums:1+16=17,2+15=17.So two 17s. Which is exactly in the given set. So that's okay.Is there any other possible set of numbers? For example, if c wasn't 7? Let's check.Suppose c was something else. Wait, but we calculated c as 41 - (3 + 31) = 7. So unless our initial assumption is wrong. Wait, how did we get c?The total sum of the original numbers is 41, which is 164/4. That's because each number is added four times in the total sum. Let me confirm that. If you have five numbers, and you do all pairwise sums, there are 10 pairs. Each number is paired with four others, so each number appears four times in the total sum. So total sum of all pairs is 4*(a + b + c + d + e). Therefore, (a + b + c + d + e)=164/4=41. So that's correct.Therefore, c must be 7. So that's fixed.Hence, the numbers must be 1,2,7,15,16. Therefore, their product is 1*2*7*15*16.Calculating step by step:1*2=22*7=1414*15=210210*16=3360So the product is 3360. Therefore, the answer should be 3360. But just to make sure, let me see if there's another approach or possible numbers.Suppose, for example, that a + c is not 8. Wait, but after a + b=3, the next smallest sum has to be a + c or b + c, but since a ≤ b ≤ c, a + c ≤ b + c. Since a ≤ b. So the next sum after a + b=3 should be a + c. If a + c is not 8, then it must be something else, but in the given sums, the next one is 8. So unless a + c=8 or b + a=8? Wait, no. Since a + b=3, which is the smallest sum, the next one has to be a + c. Since a is the smallest, and c is the next after b. Wait, but c is the middle number. So original numbers: a, b, c, d, e. So after a and b, c is the third number.So, the possible sums in order would be:a + b,a + c,a + d,a + e,b + c,b + d,b + e,c + d,c + e,d + e.But depending on the values, the order can change. For example, a + d might be larger than b + c, depending on how spread out the numbers are.But in our case, after a + b=3, the next two sums are 8 and 9. If a + c=8, then b + c would be 9. Since a=1, c=7, then b=2, and indeed b + c=9. So that's consistent.If we suppose that the next two sums are a + c=8 and b + c=9, then the next sum would be a + d. In the given list, after 3,8,9 comes 16. So a + d=16. Since a=1, d=15. Then, since d + e=31, e=16. So that gives us d=15, e=16.So the numbers are 1,2,7,15,16. Let's check all sums again:1+2=3,1+7=8,1+15=16,1+16=17,2+7=9,2+15=17,2+16=18,7+15=22,7+16=23,15+16=31.Yes, that gives exactly the given sums. So that's correct.Is there another possible configuration? Let's think. Suppose that instead of a + d=16, the next sum after 9 is b + d=16. Then, since b=2, 2 + d=16 ⇒ d=14. Then d + e=31 ⇒ e=17. Then, the numbers would be 1,2,7,14,17. Let's check the sums:1+2=3,1+7=8,1+14=15,1+17=18,2+7=9,2+14=16,2+17=19,7+14=21,7+17=24,14+17=31.The sums would be 3,8,15,18,9,16,19,21,24,31. Which when ordered are 3,8,9,15,16,18,19,21,24,31. This doesn't match the given sums, which have 16,17,17,18,22,23. So that's different. Therefore, this configuration doesn't work.Alternatively, if after 3,8,9, the next sum is a + e=16. Then a=1, so e=15. But then d + e=31 ⇒ d=16. So numbers would be 1,2,7,16,15. But then we need to sort them: 1,2,7,15,16. Wait, that's the same as the previous set. Just d and e swapped. But since we sorted them in ascending order, it's the same as before. So that doesn't change anything.Alternatively, if the next sum after 9 is c + d=16. Then c=7, so 7 + d=16 ⇒ d=9. Then d + e=31 ⇒ e=22. Then the numbers are 1,2,7,9,22. Let's check the sums:1+2=3,1+7=8,1+9=10,1+22=23,2+7=9,2+9=11,2+22=24,7+9=16,7+22=29,9+22=31.The sums here would be 3,8,10,23,9,11,24,16,29,31. Ordered: 3,8,9,10,11,16,23,24,29,31. Which is different from the given set, so that's not correct.Therefore, the only configuration that works is 1,2,7,15,16. Their product is 1*2*7*15*16=3360.I think that's solid. Let me check once more if all the sums are accounted for.Original numbers: 1,2,7,15,16.Pair sums:1+2=3,1+7=8,1+15=16,1+16=17,2+7=9,2+15=17,2+16=18,7+15=22,7+16=23,15+16=31.List of sums: 3,8,16,17,9,17,18,22,23,31.Order them: 3,8,9,16,17,17,18,22,23,31. Which matches exactly the given problem statement. Therefore, the numbers are correct. The product is indeed 1×2×7×15×16=3360.Final AnswerThe product of the numbers on the board is boxed{3360}.</think>Given five integers on a board, we know their pairwise sums are: (3, 8, 9, 16, 17, 17, 18, 22, 23, 31). We need to determine the original numbers and their product.1. Identify the smallest and largest sums: - The smallest sum (3) must be the sum of the two smallest numbers. - The largest sum (31) must be the sum of the two largest numbers.2. Calculate the total sum of the original numbers: - The sum of all pairwise sums is (164). - Each number is added four times in the total sum, so the sum of the original numbers is (164 / 4 = 41).3. Determine the middle number: - Let the original numbers be (a, b, c, d, e) in ascending order. - Given (a + b = 3) and (d + e = 31), the middle number (c = 41 - (3 + 31) = 7).4. Find the remaining numbers: - The next smallest sums after (3) are (8) and (9), which must be (a + c) and (b + c). Thus, (a = 1) and (b = 2). - The next largest sum after (31) is (16), which must be (a + d). Thus, (d = 15) and (e = 16) (since (d + e = 31)).5. Verify all pairwise sums: - Original numbers: (1, 2, 7, 15, 16). - Pairwise sums: (1+2=3), (1+7=8), (1+15=16), (1+16=17), (2+7=9), (2+15=17), (2+16=18), (7+15=22), (7+16=23), (15+16=31). - These sums match the given set exactly.6. Calculate the product: - The product of the numbers (1, 2, 7, 15, 16) is (1 times 2 times 7 times 15 times 16 = 3360).The product of the numbers on the board is (boxed{3360}).]

🔑:Answer: 3360 (numbers on the board: 1,2,7,15,16).

🧠:[<think>Okay, let's see. The problem says that each digit of the natural number N is strictly greater than the one to its left. So, for example, a number like 12345 or 13579 would fit because each digit is increasing from left to right. Then we need to find the sum of the digits of the number 9*N. Hmm, interesting.First, I need to recall some properties of numbers with strictly increasing digits. These are numbers where each digit is larger than the previous one. For instance, 1234, 1357, 24689, etc. Such numbers are also known as strictly increasing numbers. One important thing about these numbers is that they don't have any repeating digits, and each digit is unique and in ascending order.Now, the problem is asking for the sum of the digits of 9*N. Let me think. Maybe there's a pattern when multiplying such numbers by 9. Since 9 is a multiple of 9, multiplying any number by 9 will give a number whose digits sum to a multiple of 9. But here, the question is specifically about the sum of the digits of 9*N. Maybe there's a specific sum that always occurs, regardless of N? The problem doesn't specify N, so it must be a general result.Wait, let me check with an example. Let's take a simple N where digits are increasing. For instance, let N = 12. Then 9*12 = 108. The sum of the digits is 1 + 0 + 8 = 9. Another example: N = 123. 9*123 = 1107. Sum of digits: 1 + 1 + 0 + 7 = 9. Hmm, interesting. Let me try another one. N = 135. 9*135 = 1215. Sum: 1 + 2 + 1 + 5 = 9. Hmm, same result. Wait, maybe the sum is always 9?Wait, but let's check with a longer number. N = 1234. 9*1234 = 11106. Sum: 1 + 1 + 1 + 0 + 6 = 9. Again 9. Another one: N = 12345. 9*12345 = 111105. Sum: 1+1+1+1+0+5=9. Wait, so is this a coincidence? Let me try a non-consecutive increasing number. N=1345. 9*1345=12105. Sum:1+2+1+0+5=9. Still 9. Hmm, so is this a general result? For any N with strictly increasing digits, 9*N has digit sum 9? That seems to be the case here.But wait, maybe test a larger N. Let's take N=123456789. That's the largest such number with all digits increasing. 9*123456789=1111111101. Wait, let me calculate that. 123456789 *9: 123456789*10=1234567890, subtract 123456789: 1234567890 -123456789=1111111101. Sum of digits: 1+1+1+1+1+1+1+1+0+1= 9. Again 9. Wow. So, regardless of N, as long as digits are strictly increasing, multiplying by 9 gives a number whose digits sum to 9. Therefore, the answer is 9. So the sum is always 9?But let me check a different case. What if N is a single-digit number? For example, N=5. Then 9*5=45. Sum of digits:4+5=9. Still 9. So even single-digit numbers, which trivially satisfy the "each digit strictly greater than the one to its left" because there's only one digit, also work. Another single-digit: N=9. 9*9=81. Sum:8+1=9. Still 9.Another edge case: N=1. 9*1=9. Sum is 9. Hmm, same. So it seems that regardless of the number of digits in N, as long as it's strictly increasing, multiplying by 9 gives a digit sum of 9.Therefore, the answer should be 9. So the sum of the digits of 9*N is always 9. Therefore, the answer is 9.But to make sure I didn't miss any exceptions, let me think. Let's take N=19. Wait, 19 is a two-digit number where 9 >1, so digits are increasing. 9*19=171. Sum:1+7+1=9. Still 9. Another example: N=29. 9*29=261. Sum:2+6+1=9. Hmm. If N=39: 9*39=351. Sum:3+5+1=9. Still 9. So even when the number has 9 as the last digit, the sum remains 9.Wait, but if N has a 9 in it, like N=19. 9*19=171. Sum 9. If N=199? Wait, but N can't have two 9s because digits must be strictly increasing. So N can't have duplicate digits. So the digits are unique and increasing. Therefore, N can have at most 9 digits (from 1 to 9). So, all possible N here have unique digits in increasing order. Therefore, even if N contains a 9, when multiplied by 9, the digit sum is 9.Therefore, I think the answer is 9. So regardless of N, as long as it's a strictly increasing digit number, multiplying by 9 gives a number with digit sum 9. Therefore, the answer is 9.Final AnswerThe sum of the digits of the number (9N) is boxed{9}.</think>Given a natural number ( N ) where each digit is strictly greater than the one to its left, we need to find the sum of the digits of the number ( 9N ).1. Understanding the structure of ( N ): - ( N ) is a number with strictly increasing digits, such as 123, 13579, etc. - These numbers have unique digits in ascending order.2. Testing with examples: - For ( N = 12 ): ( 9 times 12 = 108 ). Sum of digits: ( 1 + 0 + 8 = 9 ). - For ( N = 123 ): ( 9 times 123 = 1107 ). Sum of digits: ( 1 + 1 + 0 + 7 = 9 ). - For ( N = 135 ): ( 9 times 135 = 1215 ). Sum of digits: ( 1 + 2 + 1 + 5 = 9 ). - For ( N = 1234 ): ( 9 times 1234 = 11106 ). Sum of digits: ( 1 + 1 + 1 + 0 + 6 = 9 ). - For ( N = 123456789 ): ( 9 times 123456789 = 1111111101 ). Sum of digits: ( 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1 = 9 ).3. Edge cases: - Single-digit numbers: ( N = 5 ): ( 9 times 5 = 45 ). Sum of digits: ( 4 + 5 = 9 ). - Numbers containing 9: ( N = 19 ): ( 9 times 19 = 171 ). Sum of digits: ( 1 + 7 + 1 = 9 ).4. Conclusion: - Regardless of the specific strictly increasing number ( N ), multiplying by 9 results in a number whose digits sum to 9.Thus, the sum of the digits of the number ( 9N ) is (boxed{9}).,<think>Alright, let's tackle this problem step by step. So, we have a natural number N where each digit is strictly greater than the one to its left. That means N is a number with digits in increasing order, right? For example, something like 1234 or 13579. Now, the question is asking for the sum of the digits of the number 9*N. Hmm, interesting. Let me think about how multiplying such a number by 9 affects its digits and their sum.First, let's recall that when you multiply a number by 9, there are some patterns. For example, 9 times a number with digits all 1s (like 11, 111, etc.) results in numbers like 99, 999, which have digit sums of 18, 27, etc. But in our case, the digits of N are increasing, so it's not all 1s. Let me try with a small example first to see if I can spot a pattern.Take N = 123. Then 9*123 = 1107. The digits of 1107 are 1, 1, 0, 7. The sum is 1 + 1 + 0 + 7 = 9. Wait, that's 9. Let me try another example. Suppose N = 12. Then 9*12 = 108. The digits are 1, 0, 8. Sum is 1 + 0 + 8 = 9. Hmm, same result. Let me try a longer number. N = 1234. Then 9*1234 = 11106. The digits sum to 1+1+1+0+6 = 9. Again 9! Wait, is this a coincidence?Let me test another one. Let's take N = 13579. Multiply by 9: 13579*9. Let's calculate that. 13579*10 is 135790, subtract 13579 gives 135790 - 13579 = 122211. Now, the sum of the digits: 1 + 2 + 2 + 2 + 1 + 1 = 9. Again 9! Wow. So is the sum of the digits of 9*N always 9 when N has strictly increasing digits?But wait, let's check another example. Let me take N = 19. Wait, 19 has digits 1 and 9. 9*19 = 171. The digits sum to 1 + 7 + 1 = 9. Still 9. How about N = 29. 9*29 = 261. Sum: 2 + 6 + 1 = 9. Hmm, same result.Wait a second, maybe this is a general property? Let me see. If N is a number with digits in strictly increasing order, then multiplying by 9 gives a number whose digits sum to 9. If that's the case, then the answer would be 9. But why does this happen?Let me try to understand why. First, note that numbers with strictly increasing digits are numbers like 123...k, where each digit is one more than the previous. But actually, strictly increasing digits could be any digits as long as each digit is greater than the one before. So numbers like 125, 136, 1479, etc. So maybe there's a relationship between these numbers and multiples of 9?Wait, 9 is a special number because it's one less than 10, the base of our numeral system. So multiplying by 9 can lead to patterns where digits carry over in a particular way. Let me think about how multiplication by 9 affects such numbers.Let’s take a number N with digits d1, d2, ..., dk where d1 < d2 < ... < dk. So each digit is strictly increasing. When we multiply this number by 9, we can analyze the digit-by-digit multiplication with carries.Alternatively, perhaps there's a clever way to note that such numbers N are actually numbers that can be written as combinations of digits without repetition. For example, 123 is 1, 2, 3; 135 is 1, 3, 5; etc. But how does that relate to multiplying by 9?Alternatively, maybe we can consider that N is a number whose digits are all distinct and increasing, so it's a combination of digits from 1 to 9 (since digits are strictly increasing, the digits can't repeat and must be in ascending order). Therefore, such numbers are subsets of {1,2,...,9} arranged in order. Therefore, maybe N is a number like 123456789, or a subset of those digits.But how does multiplying such a number by 9 affect the digits? Let me try with the largest such number: 123456789. Let's compute 9*123456789. Let me compute that.123456789 * 9: Let's break it down. 123456789 * 10 = 1234567890, subtract 123456789 gives 1234567890 - 123456789 = 111111101. Wait, 1234567890 - 123456789 is 1111111101? Wait, let me check that subtraction:1234567890-123456789-----------1111111101Wait, let me verify:Starting from the rightmost digit:0 - 9: can't do that, borrow 1 from the next digit (which is 9, becomes 8). So 10 - 9 = 1.Next digit: 8 - 8 (since we borrowed 1 earlier) = 0.Next: 8 - 7 = 1.Next: 7 - 6 = 1.Next: 6 - 5 = 1.Next: 5 - 4 = 1.Next: 4 - 3 = 1.Next: 3 - 2 = 1.Next: 2 - 1 = 1.Next: 1 - 0 (since there are no more digits in the subtrahend) = 1.So the result is 1111111101. The digits are seven 1s, followed by 0, then 1. Wait, sum of digits: 1*7 + 0 + 1 = 8. Wait, that's 8, not 9. Hmm, that contradicts my earlier examples. Did I compute something wrong?Wait, 123456789 * 9: Let me compute it another way. Maybe the subtraction method isn't correct here.123456789 * 9:Let's compute digit by digit:Starting from the right:9 * 9 = 81, write down 1, carryover 8.Next digit: 8 * 9 = 72 + carryover 8 = 80, write down 0, carryover 8.Next digit: 7 * 9 = 63 + carryover 8 = 71, write down 1, carryover 7.Next digit: 6 * 9 = 54 + carryover 7 = 61, write down 1, carryover 6.Next digit: 5 * 9 = 45 + carryover 6 = 51, write down 1, carryover 5.Next digit: 4 * 9 = 36 + carryover 5 = 41, write down 1, carryover 4.Next digit: 3 * 9 = 27 + carryover 4 = 31, write down 1, carryover 3.Next digit: 2 * 9 = 18 + carryover 3 = 21, write down 1, carryover 2.Next digit: 1 * 9 = 9 + carryover 2 = 11, write down 11.So putting it all together: 11 1 1 1 1 1 1 0 1. Wait, that would be 111111101. Wait, the first digit is 11? No, actually, when you write down the leftmost digit, you have to handle the carryover. Let me reconstruct the number step by step:Starting from the rightmost digit:1 (from 81)Then 0 (from 80)Then 1 (from 71)Then 1 (from 61)Then 1 (from 51)Then 1 (from 41)Then 1 (from 31)Then 1 (from 21)Then 1 (from 11) with the carryover 1.Wait, actually, the leftmost multiplication was 1*9 + carryover 2 = 9 + 2 = 11. So we have to write down 1 and carryover 1. But since there are no more digits, we add the carryover 1. So the leftmost digits become 11. So the number is 1111111101. Wait, but this is 10 digits. Let's count:Digits: 1,1,1,1,1,1,1,1,0,1. That's 10 digits. The sum is 1*8 + 0 +1 = 9. Wait, but earlier when I computed the subtraction, I thought the sum was 8. Wait, now with the digit-by-digit multiplication, the sum is 9. Let me verify with another method.Compute 123456789 * 9:123456789 * 9.123456789 * 9 is equal to 123456789 * (10 - 1) = 1234567890 - 123456789.Compute 1234567890 minus 123456789.1234567890-123456789=1111111101.Wait, that's 10 digits: 1,1,1,1,1,1,1,1,0,1. So sum is 8*1 + 0 +1 = 9. Wait, so my previous calculation was correct. So even for this large number, the digit sum is 9. Then why did I initially miscalculate? Probably a miscount. So the sum is indeed 9.So this suggests that regardless of N, as long as N is a number with strictly increasing digits, multiplying by 9 gives a number whose digits sum to 9. But why? What's the underlying reason?Let me think about the properties of such numbers. Since each digit is strictly increasing, the digits are all unique and range from 1 to 9 (since 0 can't be part of a number with strictly increasing digits unless it's the first digit, which can't be). So N is composed of digits 1 through 9 in some increasing order.But how does multiplying such a number by 9 lead to a digit sum of 9?Wait, perhaps it's related to the fact that such numbers are divisible by 9? Wait, no. For example, 123: 123 is not divisible by 9 (1+2+3=6, which is not a multiple of 9). But when we multiply by 9, 9*N would be divisible by 9, so the digit sum must be a multiple of 9. However, in our examples, the digit sum is exactly 9. So 9*N is not just any multiple of 9, but specifically one with digit sum 9. That's interesting.Is there a theorem or property that connects numbers with increasing digits multiplied by 9 to having digit sum 9? Maybe not directly, but perhaps we can look for a pattern.Let me try a different approach. Let's represent N as a sum of its digits multiplied by powers of 10. Suppose N = d1*10^{k-1} + d2*10^{k-2} + ... + dk*10^0, where d1 < d2 < ... < dk.Then 9*N = 9*d1*10^{k-1} + 9*d2*10^{k-2} + ... + 9*dk*10^0.However, when we compute this, each term 9*di*10^{i} will result in possible carries when written in base 10. The key here is that because the digits di are strictly increasing, the way the multiplication by 9 interacts with each digit and the subsequent carries might lead to a consistent digit sum.Alternatively, perhaps using the concept that the sum of digits modulo 9 is equal to the number modulo 9. Since 9*N is obviously divisible by 9, its digit sum must be a multiple of 9. But in our examples, the digit sum is exactly 9. So maybe for numbers N with strictly increasing digits, multiplying by 9 gives a number with digit sum 9. Therefore, the answer is 9.But wait, let's test with N = 12345. 9*12345 = 111105. Sum of digits: 1+1+1+1+0+5=9. Yes, 9 again. Another example: N = 89. 9*89=801. Sum:8+0+1=9.Wait, but what about N=1? 9*1=9. Sum is 9. Also works. N=2? 9*2=18. Sum 1+8=9. N=5? 9*5=45. 4+5=9. So even single-digit numbers work. So regardless of the number of digits, as long as N has strictly increasing digits (which for single-digit numbers is trivially true), multiplying by 9 gives a digit sum of 9.So perhaps this is a general result: For any natural number N with strictly increasing digits, the sum of the digits of 9*N is 9. Therefore, the answer is 9.But to be thorough, let's try a case where N has a larger digit that causes more carrying. For example, N = 19. As before, 9*19=171. Sum is 1+7+1=9. How about N=199? Wait, but N must have digits strictly increasing, so 199 has two 9s, which are not strictly increasing. So N can't be 199. Let's pick N=189. Wait, digits 1,8,9. They are increasing? 1 < 8 <9, yes. So 189*9. Let's compute that.189*9: 100*9=900, 80*9=720, 9*9=81. So 900 + 720 = 1620 + 81 = 1701. The digits are 1,7,0,1. Sum:1+7+0+1=9. Still 9. So even with a larger digit, it works.How about a number like N=123456789. As we saw earlier, sum is 9. So seems consistent.But why does this happen? Let's think about the structure of N. Since digits are strictly increasing, when we multiply by 9, each digit di is multiplied by 9, but because di < di+1, the carryover from each digit might not interfere destructively. Wait, maybe this is due to the fact that the digits are increasing, so when multiplied by 9, the carries propagate in a way that the digits in the product end up summing to 9.Alternatively, perhaps there's a mathematical proof for this. Let's consider that N is a number with digits d1 < d2 < ... < dk. Then 9*N is a number where each digit di is multiplied by 9, but considering the positional value (place value) of each digit. However, the exact digit-by-digit multiplication is complex due to carries. However, we can use the concept that the sum of the digits of a number is congruent to the number modulo 9. Since 9*N is obviously divisible by 9, its digit sum must be a multiple of 9. But in our examples, the sum is exactly 9. So maybe there's a property here that 9*N for such N can't have a digit sum higher than 9? But that seems unlikely, unless there's a specific constraint.Wait, let's consider that when N has digits in increasing order, the multiplication by 9 does not result in any digits beyond 9, and through the carrying process, the digits "collapse" into a sum of 9. But how?Wait, another approach: Since N has digits strictly increasing, then the difference between consecutive digits is at least 1. For example, if N is a 2-digit number ab with a < b, then N = 10a + b. Then 9N = 90a + 9b. Let's represent this as:If a and b are digits with a < b, then 90a + 9b = 9*(10a + b). Let's compute this:Take a=1, b=2: 90 + 18 = 108, digits sum to 1+0+8=9.Take a=1, b=3: 90 + 27 = 117, sum=1+1+7=9.Take a=1, b=5: 90 + 45 = 135, sum=1+3+5=9.Take a=2, b=3: 180 + 27 = 207, sum=2+0+7=9.So in each case, regardless of a and b, the sum is 9. For a two-digit number, the result holds.How about a three-digit number? Let's take N=123: 9*123=1107, sum=1+1+0+7=9.N=124: 9*124=1116, sum=1+1+1+6=9.N=135: 9*135=1215, sum=1+2+1+5=9.N=149: 9*149=1341, sum=1+3+4+1=9.So works here too.Wait, but why? Let's think of the three-digit number N = 100a + 10b + c, with a < b < c. Then 9N = 900a + 90b + 9c. Let's see:900a is going to contribute 9*a to the hundreds place, but since 900a is actually 9*a*100, so when written out, it's 9a in the hundreds place, but 9a could be two digits if a >= 2. For example, a=2: 900*2=1800. Wait, but 900a is 9a*100, so when added to 90b and 9c, the total is 9*(100a + 10b + c) = 9N.Wait, but let's compute 900a + 90b + 9c:= 9*(100a + 10b + c)But this is equal to 9N. But how does this relate to the digit sum?The digit sum of 9N is always 9. So even though the actual digits vary, their sum remains 9. So this must be due to the fact that when you distribute the multiplication over the digits, the carries that occur during the addition of 900a, 90b, and 9c interact in such a way that the total digit sum is preserved as 9.Alternatively, since N is a number with digits increasing, then when multiplied by 9, the digits in the product are arranged such that all the carries from each digit multiplication cancel out the higher digits, leading to a total digit sum of 9. But this is vague.Alternatively, perhaps using induction. Suppose that for any number with k digits in strictly increasing order, the digit sum of 9*N is 9. Then prove it for k+1 digits.But maybe that's not straightforward. Alternatively, notice that numbers with strictly increasing digits are exactly the numbers that do not have any carries when added to themselves 9 times. Wait, not sure.Wait, another idea: Since N has digits in increasing order, its digits are d1, d2, ..., dk with d1 < d2 < ... < dk. Then, the digits can be considered as a combination of digits from 1 to 9. When multiplied by 9, the resulting number's digits somehow sum to 9. But how?Alternatively, perhaps it's related to the fact that N is a multiple of a number whose digits sum to a certain value, and when multiplied by 9, it creates a number with digit sum 9.Alternatively, think of the multiplication by 9 as a kind of "complement" to a number with a digit sum that is a divisor. Wait, maybe not.Wait, let me think about the digital root. The digital root of a number is the iterative sum of its digits until a single digit is obtained. The digital root is equivalent to the number modulo 9. Since 9*N is a multiple of 9, its digital root is 9. However, the digital root being 9 does not necessarily mean that the sum of the digits is 9. For example, 18 has digital root 9 but digit sum 9; 99 has digital root 9 but digit sum 18. So in our case, the sum of the digits is exactly 9, not just a multiple of 9. So this is a stronger condition.But why does multiplying a number with strictly increasing digits by 9 result in a digit sum of exactly 9?Wait, perhaps when you multiply such a number by 9, you avoid multiple carries in the same digit place, leading to an overall sum of 9. For example, in the case of 1234 * 9 = 11106. The multiplication by 9 causes each digit to carry over into the next, but since the digits are increasing, each carry is "absorbed" by the next higher digit's multiplication, leading to a single carry-over chain that results in digits summing to 9.Wait, let's take 1234 * 9:Starting from the rightmost digit:4*9 = 36. Write down 6, carryover 3.Next digit: 3*9 + 3 = 27 + 3 = 30. Write down 0, carryover 3.Next digit: 2*9 + 3 = 18 + 3 = 21. Write down 1, carryover 2.Next digit: 1*9 + 2 = 9 + 2 = 11. Write down 1, carryover 1.Finally, the carryover 1 is written down.So the result is 11106. The sum is 1+1+1+0+6=9.Similarly, if we have a longer number like 12345:5*9=45, write 5, carryover 4.4*9 +4=40, write 0, carryover4.3*9+4=31, write1, carryover3.2*9+3=21, write1, carryover2.1*9+2=11, write1, carryover1.Carryover1 is written.Result:111105. Sum 1+1+1+1+0+5=9.So each multiplication step creates a carryover that affects the next digit's multiplication, but since the digits are increasing, each next digit is larger, so when we multiply and add the carryover, it's just enough to cause a carryover again, but the digits that are written down (except possibly the last one) end up being 1s and 0s. However, in the first example with N=123, we had digits 1,1,0,7. Wait, why did the last digit become 7?Wait, let's check N=123:3*9=27, write 7, carryover2.2*9 +2=20, write0, carryover2.1*9 +2=11, write1, carryover1.Carryover1 is written.Result: 1107. Sum 1+1+0+7=9.Ah, here the last digit before the final carryover is 7. So it's not all 1s and 0s. So why does the sum still add up to 9?It's because when the last digit multiplication (in this case, the leftmost digit) plus the carryover equals a number that, when written down, contributes a digit and a carryover. Then the final carryover adds another digit. So the total digits contributed from each step plus the final carryover all sum up in such a way that their total is 9.But how does that work in general?Wait, let's formalize this. Let's suppose we have a number N with digits d1, d2, ..., dk from left to right, with d1 < d2 < ... < dk. When we multiply N by 9, starting from the rightmost digit (dk), each digit di is multiplied by 9 and added to the carryover from the previous digit. Let's denote the carryover after multiplying di as ci.Starting with c0 = 0.For each digit di (from right to left):Multiply di by 9: 9*di.Add the carryover ci-1: total = 9*di + ci-1.The digit written down is total mod 10.The new carryover ci = floor(total / 10).After processing all digits, we add the final carryover ck as a new digit.Now, if we can show that the sum of all the digits written down (including the final carryover) is 9, then we have proven the result.Let’s denote the digits of 9*N as a1, a2, ..., am, where m is the number of digits. Then, sum_{i=1}^m ai = sum_{i=1}^k (9*di + ci-1 - 10*ci) + ck.Wait, because each digit ai = (9*di + ci-1) - 10*ci.Therefore, the total sum is sum_{i=1}^k (9*di + ci-1 - 10*ci) + ck.Let's expand this:= sum_{i=1}^k 9*di + sum_{i=1}^k ci-1 - 10*sum_{i=1}^k ci + ck= 9*sum_{i=1}^k di + sum_{i=0}^{k-1} ci - 10*sum_{i=1}^k ci + ckNote that sum_{i=0}^{k-1} ci = c0 + c1 + ... + ck-1.Similarly, 10*sum_{i=1}^k ci = 10*(c1 + c2 + ... + ck).Therefore, substituting:= 9*sum_{i=1}^k di + (c0 + c1 + ... + ck-1) - 10*(c1 + ... + ck) + ck= 9*sum_{i=1}^k di + c0 + c1 + ... + ck-1 - 10*c1 - ... - 10*ck + ck= 9*sum di + c0 -9*c1 -9*c2 - ... -9*ck= 9*sum di + c0 -9*(c1 + c2 + ... + ck)But c0 is the initial carryover, which is 0. So:= 9*sum di -9*(c1 + c2 + ... + ck)= 9*(sum di - sum_{i=1}^k ci)But the total sum of the digits is also equal to 9*(sum di - sum ci). However, we need to relate this to the actual digit sum. Wait, but the sum we computed earlier is equal to this expression. But how do we know that sum di - sum ci = 1?Wait, because the total digit sum is 9*(sum di - sum ci). But we also know from the examples that the digit sum is 9. Therefore:9*(sum di - sum ci) = 9Therefore, sum di - sum ci = 1.So if we can show that sum di - sum ci =1, then the digit sum is 9.Therefore, we need to show that for numbers N with strictly increasing digits, when multiplied by 9, the sum of the digits di minus the sum of the carryovers ci equals 1.Hmm, how can we prove that sum di - sum ci =1?Let’s try induction.Base case: N is a single-digit number. Then N = d1, and 9*N = 9*d1. The digit sum is 9*d1. Wait, but for single-digit numbers, 9*d1 is a two-digit number if d1 >=2. For example, d1=1: 9*1=9 (sum=9). d1=2: 18 (sum=9). d1=3: 27 (sum=9). So for single-digit numbers, the sum is 9, which is 9*(d1) - sum ci. Wait, in this case, sum di = d1, and sum ci is the carryover when multiplying 9*d1. For example, d1=2: 9*2=18. Here, during the multiplication, we have carryover c1=1. But the sum di - sum ci = 2 -1=1. Therefore, 9*(2 -1)=9. Similarly, for d1=1: 9*1=9. There is no carryover, sum ci=0. Therefore, 9*(1 -0)=9. So in the single-digit case, sum di - sum ci =1 (since for d1=1, 1 -0=1; for d1=2, 2 -1=1; etc.) Therefore, sum di - sum ci=1, hence digit sum is 9*(1)=9.Assume that for a k-digit number N with strictly increasing digits, the sum di - sum ci=1. Then, for a k+1-digit number, we need to show that the same holds.However, induction might be tricky here since adding a new digit affects the carryovers in a non-trivial way. Alternatively, maybe we can find a relation for the carryovers.Let’s consider the process of multiplying by 9. Each digit di is multiplied by 9, and the carryover propagates to the next digit. Let’s denote:Starting from the rightmost digit dk:9*dk = 10*ck-1 + akThen 9*dk-1 + ck-1 = 10*ck-2 + ak-1...Continuing up to the first digit:9*d1 + c1 = 10*c0 + a1But c0 =0 since there are no more digits.From these equations, we can write:For each i from 1 to k:9*di + ci = 10*ci-1 + aiSumming all these equations:sum_{i=1}^k (9*di + ci) = sum_{i=1}^k (10*ci-1 + ai)Left side: 9*sum di + sum ciRight side: 10*sum ci-1 + sum aiRearranging:9*sum di + sum ci = 10*sum ci-1 + sum aiBring sum ci to the right:9*sum di = 10*sum ci-1 + sum ai - sum ciNote that sum ci-1 (from i=1 to k) is sum_{j=0}^{k-1} cj, and sum ci (from i=1 to k) is sum_{j=1}^k cj. Therefore:sum ci-1 (from i=1 to k) = c0 + c1 + ... + ck-1sum ci (from i=1 to k) = c1 + c2 + ... + ckTherefore, 10*sum ci-1 - sum ci = 10*(c0 + c1 + ... + ck-1) - (c1 + ... + ck)=10*c0 + 10*c1 + ... +10*ck-1 - c1 - ... - ck=10*c0 +9*c1 +9*c2 + ... +9*ck-1 - ckBut c0=0, so:=9*c1 +9*c2 + ... +9*ck-1 - ckTherefore, substituting back into the equation:9*sum di = 9*c1 +9*c2 + ... +9*ck-1 - ck + sum ai - sum ciWait, this seems a bit convoluted. Maybe there's another way.From the equation:9*sum di + sum ci = 10*sum ci-1 + sum aiRearranged:9*sum di = 10*sum ci-1 - sum ci + sum aiBut sum ci-1 = c0 + c1 + ... + ck-1 = sum ci (from i=1 to k) - ck + c0Since c0=0, sum ci-1 = sum ci (from i=1 to k) - ckTherefore:9*sum di = 10*(sum ci - ck) - sum ci + sum ai=10*sum ci -10*ck - sum ci + sum ai=9*sum ci -10*ck + sum aiTherefore:9*sum di -9*sum ci = sum ai -10*ckBut from our previous result, sum ai = 9*(sum di - sum ci) +10*ckHmm, not sure if this helps.Alternatively, recall that we need to show sum di - sum ci =1.From the previous equation:sum ai =9*sum di -9*sum ci +10*ckBut sum ai is the digit sum of 9*N, which we want to be 9. Therefore:9*sum di -9*sum ci +10*ck=9Divide both sides by 9:sum di - sum ci + (10/9)*ck=1But ck is an integer, so (10/9)*ck must also be an integer, which implies that ck is a multiple of 9. However, ck is a carryover from the multiplication of d1*9 + c1. Since d1 is a digit (1-9), c1 can be at most floor((9*9 + c0)/10). Wait, but c0=0, so c1 <= floor(81/10)=8. Similarly, subsequent carryovers are also less than 10. Therefore, ck can't be a multiple of 9 unless ck=0 or 9. But ck is a carryover digit, which is less than 10, so ck=0 or 9. If ck=9, then (10/9)*9=10, so sum di - sum ci +10=1 => sum di - sum ci= -9, which is impossible since sum di >=1 and sum ci >=0. Therefore, ck must be 0, leading to:sum di - sum ci=1Therefore, proving that sum di - sum ci=1 would lead to sum ai=9.But how do we prove that sum di - sum ci=1?Given that ck=0 (from above), which requires that the final carryover after processing the leftmost digit is 0. Which would require that 9*d1 + c1 <10. However, when d1 is 1, and c1 is up to 8, this is possible. For example, let's take N=123:Processing digits from right to left:3: 9*3=27, carryover=2, digit=72: 9*2 +2=20, carryover=2, digit=01:9*1 +2=11, carryover=1, digit=1Final carryover=1. So sum di=1+2+3=6, sum ci=2+2+1=5. 6-5=1. Which matches sum ai=9*(1)=9.Similarly, for N=1234:4: 36, carryover3, digit63:27+3=30, carryover3, digit02:18+3=21, carryover2, digit11:9+2=11, carryover1, digit1Final carryover1. Sum di=1+2+3+4=10, sum ci=3+3+2+1=9. 10-9=1. So sum ai=9.Another example, N=12:2:18, carryover1, digit81:9+1=10, carryover1, digit0Final carryover1. Sum di=1+2=3, sum ci=1+1=2. 3-2=1.So in each case, sum di - sum ci=1.Therefore, if we can show that for any number with strictly increasing digits, sum di - sum ci=1, then the digit sum of 9*N is 9.But why does this hold?Because when we process each digit from right to left, each multiplication by 9 plus the previous carryover generates a new digit and a new carryover. The key insight is that the total "loss" in the sum di - sum ci is exactly 1, regardless of the number of digits.But why is the total loss 1?Maybe because when you multiply each digit by 9 and handle the carries, the process effectively subtracts 1 from the total sum of the digits. For example, each time you have a carryover, you're effectively moving a value of 10 from one digit to the next, which is a net loss of 9. But since the total sum is 9*sum di, the sum of the digits of 9*N is 9*(sum di) -9*(sum ci). Therefore, if 9*(sum di - sum ci) =9, then sum di -sum ci=1.But how does the process guarantee that sum di -sum ci=1?It seems to be a result of the fact that each digit is multiplied by 9, and when you add the carryovers, the total sum of digits reduces by 9*(sum ci), but we need this to result in a digit sum of 9. Therefore, 9*(sum di - sum ci)=9 => sum di - sum ci=1.But why does this difference always equal 1?Perhaps it's due to the fact that the initial multiplication by 9 of the rightmost digit produces a carryover, and each subsequent multiplication and carryover propagation effectively maintains this difference. However, the exact reason is still a bit elusive.Alternatively, considering the entire process as a kind of base-10 complement. Since multiplying by 9 is similar to 10-1, and numbers with strictly increasing digits have a certain structure that complements this multiplication. But I can't quite formalize this intuition.In any case, all empirical examples and the partial proof through the digit-by-digit multiplication analysis suggest that the sum of the digits of 9*N is always 9, regardless of the number of digits in N, as long as N has strictly increasing digits.Therefore, the answer should be 9.Final Answerboxed{9}</think>Given a natural number ( N ) where each digit is strictly greater than the one to its left, we need to find the sum of the digits of the number ( 9N ).1. Understanding the problem: - ( N ) is a number with strictly increasing digits (e.g., 123, 13579). - When multiplied by 9, the digits of ( 9N ) might change due to carries, but we need the sum of these digits.2. Testing with examples: - For ( N = 123 ), ( 9 times 123 = 1107 ), sum of digits is ( 1 + 1 + 0 + 7 = 9 ). - For ( N = 12 ), ( 9 times 12 = 108 ), sum of digits is ( 1 + 0 + 8 = 9 ). - For ( N = 1234 ), ( 9 times 1234 = 11106 ), sum of digits is ( 1 + 1 + 1 + 0 + 6 = 9 ). - For ( N = 13579 ), ( 9 times 13579 = 122211 ), sum of digits is ( 1 + 2 + 2 + 2 + 1 + 1 = 9 ).3. General observation: - The sum of the digits of ( 9N ) is consistently 9 across various examples. - This suggests a pattern where multiplying any number with strictly increasing digits by 9 results in a digit sum of 9.4. Mathematical reasoning: - Numbers with strictly increasing digits have unique digits from 1 to 9. - Multiplying by 9 results in a number whose digits sum to a multiple of 9 (since ( 9N ) is divisible by 9). - Empirical results and digit-by-digit multiplication analysis show that the digit sum is exactly 9 due to the specific carryover patterns and the structure of the original number.5. Conclusion: - The sum of the digits of ( 9N ) is always 9 for any natural number ( N ) with strictly increasing digits.[boxed{9}]]

🔑:Answer: 9.Solution: Note that 9 N=10 N-N. Let's perform this subtraction in a column. In the units place, there will be the difference between 10 and the last digit of the number N, in the tens place - the last and the second-to-last digit, decreased by 1. In all subsequent places, there will be the difference between two adjacent digits, as the smaller digit will always be subtracted from the larger one. Each digit will appear once as the minuend and once as the subtrahend. When finding the sum of the digits, all these minuends and subtrahends will cancel each other out. Only the aforementioned 10-1=9 will remain.Criteria: only the answer, the answer with verification on specific cases - 0 points, the scheme of proof by example -2 points, the transition 9 N=10 N-N-1 point. These points mathrm{NE} are not summed.## Solutions to the tasks of the first stage of the All-Siberian Open Olympiad for schoolchildren in mathematics 2014-15## 9th grade## Each task is worth 7 points