🧠:[<think>Okay, let's see. I need to find the value of a + b + c given that f(x) = x³ + a x² + b x + c, where a, b, c are distinct non-zero integers. Also, f(a) = a³ and f(b) = b³. Hmm, that's interesting. So, the function evaluated at a gives a³, which is like the cubic term without the other coefficients. Similarly for b. Let me write down what I know.First, f(a) = a³. Let's substitute x = a into the function:f(a) = a³ + a*(a²) + b*(a) + c = a³. Wait, so that simplifies to:a³ + a³ + a b + c = a³. So combining the terms:2a³ + a b + c = a³. Subtracting a³ from both sides:a³ + a b + c = 0. Similarly, f(b) = b³. Substitute x = b:f(b) = b³ + a*(b²) + b*(b) + c = b³. Simplifying:b³ + a b² + b² + c = b³. Wait, hold on, that middle term is a*b² + b*b = a b² + b² = b²(a + 1). So:b³ + b²(a + 1) + c = b³. Subtracting b³ from both sides:b²(a + 1) + c = 0. So now we have two equations:1. a³ + a b + c = 02. b²(a + 1) + c = 0Since both equal zero, maybe we can set them equal to each other? Let me subtract the second equation from the first or something.Wait, both equations have c. Let me subtract the second equation from the first equation to eliminate c:(a³ + a b + c) - (b²(a + 1) + c) = 0 - 0Which simplifies to:a³ + a b - b²(a + 1) = 0Let me factor that:a³ + a b - a b² - b² = 0Hmm. Let's rearrange terms:a³ - a b² + a b - b² = 0Factor terms where possible. Let's look for common factors in the first two terms and the last two terms.First two terms: a(a² - b²) which is a(a - b)(a + b)Last two terms: b(a - b)Wait, let me check:a³ - a b² = a(a² - b²) = a(a - b)(a + b)a b - b² = b(a - b)So combining these:a(a - b)(a + b) + b(a - b) = 0Factor out (a - b):(a - b)[a(a + b) + b] = 0So either (a - b) = 0 or the other factor is zero. But since a, b, c are distinct integers, a ≠ b. Therefore, the other factor must be zero:a(a + b) + b = 0Let's expand that:a² + a b + b = 0Factor out b from the last two terms:a² + b(a + 1) = 0So a² = -b(a + 1)Therefore, b = -a² / (a + 1)But since a and b are integers, and a ≠ -1 (because denominator is a + 1, and if a = -1, denominator is zero, but a is a non-zero integer). So a must be an integer not equal to -1, 0. So a ∈ integers {-1, 0}, same with b and c.So b must be equal to -a² divided by (a + 1), and since b is an integer, (a + 1) must divide -a². So (a + 1) divides -a². Let's write that as (a + 1) | a². Because if (a + 1) divides -a², then it also divides a². So (a + 1) is a divisor of a².Therefore, (a + 1) divides a². So, for integer a ≠ -1, a + 1 divides a². Let's note that.Let me consider that for integer a, a + 1 divides a². So, we can write a² = k(a + 1) for some integer k. Therefore, k = a² / (a + 1). Which must be integer. So, how can a + 1 divide a²?Perhaps we can express a² in terms of (a + 1):a² = (a + 1 - 1)² = (a + 1)² - 2(a + 1) + 1Therefore, a² = (a + 1)^2 - 2(a + 1) + 1. Then,a² = (a + 1)(a + 1 - 2) + 1 = (a + 1)(a - 1) + 1.So, a² divided by (a + 1) is equal to (a - 1) + 1/(a + 1). Therefore, in order for (a + 1) to divide a², the remainder when a² is divided by (a + 1) must be zero. The remainder is 1, as shown above. Therefore, 1/(a + 1) must be an integer. Which implies that (a + 1) divides 1. Therefore, a + 1 is ±1. Therefore, possible values of a + 1 are 1 or -1.Therefore, a + 1 = 1 ⇒ a = 0, but a is non-zero. Disregarded.Or a + 1 = -1 ⇒ a = -2. So a = -2 is the only possibility.Wait, that's a key point. If a + 1 divides 1, then a + 1 is ±1. But since a is non-zero, a + 1 can't be 1 (as that would make a = 0), so only possible is a + 1 = -1, hence a = -2.Therefore, a = -2.Then, b = -a² / (a + 1) = -(-2)^2 / (-2 + 1) = -4 / (-1) = 4.So b = 4.Now, we can find c from one of the equations. Let's use equation 2: b²(a + 1) + c = 0. Plugging in a = -2, b = 4:(4)^2(-2 + 1) + c = 0 ⇒ 16*(-1) + c = 0 ⇒ -16 + c = 0 ⇒ c = 16.Alternatively, check with equation 1: a³ + a b + c = 0. Plugging in a = -2, b = 4:(-2)^3 + (-2)(4) + c = 0 ⇒ -8 -8 + c = 0 ⇒ -16 + c = 0 ⇒ c = 16. Same result.So, a = -2, b = 4, c = 16.But wait, the problem states that a, b, c are distinct non-zero integers. Let's check:a = -2, b = 4, c = 16. All distinct, all non-zero. Good.Therefore, a + b + c = (-2) + 4 + 16 = 18.Wait, but let me check if these values satisfy the original conditions.First, f(a) = f(-2) should equal (-2)^3 = -8.Compute f(-2):(-2)^3 + (-2)*(-2)^2 + 4*(-2) + 16 = -8 + (-2)*(4) + (-8) + 16 = -8 -8 -8 +16 = (-24) +16 = -8. Correct.Then f(b) = f(4) should equal 4^3 = 64.Compute f(4):4^3 + (-2)*(4)^2 + 4*(4) + 16 = 64 + (-2)*16 + 16 + 16 = 64 -32 +16 +16 = 64 -32 is 32, 32 +16 is 48, 48 +16 is 64. Correct.Therefore, it satisfies both f(a) = a³ and f(b) = b³. Also, a, b, c are distinct and non-zero. So seems like the answer is 18.But let me check if there are other possible values. Wait, earlier reasoning led me to a must be -2 because (a + 1) must divide 1, leading to a + 1 = ±1, but since a can't be 0, only a = -2.Therefore, that's the only solution. So a + b + c = -2 +4 +16=18. So answer is 18.Wait, but just to make sure, let me see if there are other ways this could happen. Suppose that when I factored out (a - b), but since a ≠ b, we had that expression must be zero. So that gives us the equation leading to a = -2. So seems like that's the only solution.Alternatively, perhaps if there was a miscalculation in the division step. Let me check the division part again.We had that (a + 1) divides a², so we wrote a² = k(a +1). Then to find integer k, we can perform polynomial division or use the remainder theorem. The remainder when a² is divided by (a +1) is equal to (-1)^2 = 1. So remainder is 1, so as above. Therefore, (a +1) divides a² implies that (a +1) divides 1, hence a +1 = ±1, so a =0 or a = -2. But a ≠0, so a = -2. So yes, that's the only solution.Therefore, the answer is 18. So boxed, it's boxed{18}.Final Answerboxed{18}</think>Given the function ( f(x) = x^3 + ax^2 + bx + c ) where ( a, b, c ) are distinct non-zero integers, and the conditions ( f(a) = a^3 ) and ( f(b) = b^3 ), we need to find ( a + b + c ).First, substituting ( x = a ) into the function:[f(a) = a^3 + a cdot a^2 + b cdot a + c = a^3]Simplifying, we get:[2a^3 + ab + c = a^3 implies a^3 + ab + c = 0 quad text{(1)}]Next, substituting ( x = b ) into the function:[f(b) = b^3 + a cdot b^2 + b cdot b + c = b^3]Simplifying, we get:[b^3 + ab^2 + b^2 + c = b^3 implies b^2(a + 1) + c = 0 quad text{(2)}]Subtracting equation (2) from equation (1):[a^3 + ab - b^2(a + 1) = 0]Factoring out ( (a - b) ):[(a - b)[a(a + b) + b] = 0]Since ( a neq b ), we have:[a(a + b) + b = 0 implies a^2 + ab + b = 0 implies a^2 = -b(a + 1)]Thus, ( b = -frac{a^2}{a + 1} ). For ( b ) to be an integer, ( a + 1 ) must divide ( a^2 ). The remainder when ( a^2 ) is divided by ( a + 1 ) is 1, so ( a + 1 ) must be ±1. Since ( a neq 0 ), ( a + 1 = -1 ) leading to ( a = -2 ).Substituting ( a = -2 ) into ( b = -frac{a^2}{a + 1} ):[b = -frac{(-2)^2}{-2 + 1} = -frac{4}{-1} = 4]Using equation (2) to find ( c ):[b^2(a + 1) + c = 0 implies 16(-1) + c = 0 implies c = 16]Thus, ( a = -2 ), ( b = 4 ), and ( c = 16 ). Therefore, ( a + b + c = -2 + 4 + 16 = 18 ).The final answer is (boxed{18}).,<think>Okay, so I need to solve this problem where the function is given as f(x) = x³ + a x² + b x + c, and a, b, c are distinct non-zero integers. The conditions given are that f(a) = a³ and f(b) = b³. We need to find the value of a + b + c. Hmm, let's start by understanding what the problem is asking for.First, the function f(x) is a cubic polynomial. The coefficients a, b, c are all distinct non-zero integers. So, they can't be the same number, and none of them can be zero. Then, when we plug in x = a into the function, the result is a³. Similarly, plugging in x = b into the function gives b³. So, f(a) = a³ and f(b) = b³. But f(x) itself is x³ + a x² + b x + c. That means when we substitute x with a or b, the entire expression should simplify to just a³ or b³ respectively. Therefore, the additional terms a x² + b x + c must cancel out when x is a or b. Let me write down the equations based on the given conditions. For f(a) = a³:a³ + a(a)² + b(a) + c = a³Similarly, for f(b) = b³:b³ + a(b)² + b(b) + c = b³Let me simplify both equations. Starting with the first one:a³ + a(a²) + b(a) + c = a³Simplify the terms:a³ + a³ + a b + c = a³Combine like terms:2a³ + a b + c = a³Subtract a³ from both sides:a³ + a b + c = 0So, equation (1): a³ + a b + c = 0Similarly, for the second equation:b³ + a(b²) + b(b) + c = b³Simplify the terms:b³ + a b² + b² + c = b³Combine like terms:b³ + a b² + b² + c = b³Factor out b² from the middle terms:b³ + b²(a + 1) + c = b³Subtract b³ from both sides:b²(a + 1) + c = 0So, equation (2): b²(a + 1) + c = 0Now, we have two equations:1. a³ + a b + c = 02. b²(a + 1) + c = 0Since both equal to zero and both equal to c, we can set them equal to each other. Let me subtract the two equations or subtract one from the other to eliminate c. Let's subtract equation (2) from equation (1):(a³ + a b + c) - (b²(a + 1) + c) = 0 - 0Simplify:a³ + a b + c - b²(a + 1) - c = 0The c terms cancel out:a³ + a b - b²(a + 1) = 0Factor the terms:a³ + a b - a b² - b² = 0Let me rearrange the terms:a³ - a b² + a b - b² = 0Factor by grouping. Let's group the first two terms and the last two terms:a(a² - b²) + b(a - b) = 0Notice that a² - b² factors as (a - b)(a + b). So:a(a - b)(a + b) + b(a - b) = 0Factor out (a - b):(a - b)[a(a + b) + b] = 0So, either (a - b) = 0 or [a(a + b) + b] = 0.But since a, b, c are distinct integers, a ≠ b. Therefore, (a - b) ≠ 0. So, we must have:a(a + b) + b = 0Let me write this as:a² + a b + b = 0Factor out b from the last two terms:a² + b(a + 1) = 0So, equation (3): a² + b(a + 1) = 0This seems like a key equation. Let me note this equation. So, from here, we can express b in terms of a. Let me rearrange equation (3):b(a + 1) = -a²Therefore,b = -a² / (a + 1)Since a and b are integers, and a ≠ -1 (since a is non-zero, and a, b, c are distinct non-zero integers). So, (a + 1) must divide -a². So, (a + 1) divides -a². Let me note that (a + 1) divides -a². Therefore, (a + 1) is a divisor of a². Let me think about this. So, since a and a + 1 are consecutive integers, they are coprime. That is, gcd(a, a + 1) = 1. Therefore, a + 1 divides -a², but since a + 1 and a are coprime, then a + 1 must divide -1. Wait, is that right?Wait, if a + 1 divides -a², and since gcd(a + 1, a) = 1, then a + 1 divides the constant term of the polynomial. Let's see. Let's write a² as (a + 1 - 1)^2 = (a + 1)^2 - 2(a + 1) + 1. So, that's (a + 1)^2 - 2(a + 1) + 1. Then, when you divide a² by (a + 1), you get (a + 1) - 2 + 1/(a + 1). Therefore, for (a + 1) to divide a², the remainder must be zero. The remainder when a² is divided by (a + 1) is 1. Therefore, if (a + 1) divides a², then (a + 1) divides 1. Therefore, (a + 1) must be ±1. Therefore, a + 1 = 1 or a + 1 = -1. Therefore, a = 0 or a = -2. But a is a non-zero integer, so a = -2. So, a = -2.Wait, that seems like a key step. Let me check that again.Given that (a + 1) divides -a². Since gcd(a + 1, a) = 1, then (a + 1) divides -1. Therefore, a + 1 is either 1 or -1. Therefore, a = 0 or a = -2. But a is non-zero, so a = -2. Therefore, a = -2. Then, from equation (3):b = -a² / (a + 1) = -(-2)^2 / (-2 + 1) = -4 / (-1) = 4. So, b = 4.So, we have a = -2 and b = 4. Now, we can find c from either equation (1) or equation (2). Let's use equation (1):a³ + a b + c = 0Substitute a = -2 and b = 4:(-2)^3 + (-2)(4) + c = 0Compute:-8 - 8 + c = 0-16 + c = 0Therefore, c = 16Alternatively, check with equation (2):b²(a + 1) + c = 0Substitute b = 4, a = -2:4²(-2 + 1) + c = 016(-1) + c = 0-16 + c = 0Therefore, c = 16. Same result. So, c = 16.Now, check if a, b, c are distinct non-zero integers. a = -2, b = 4, c = 16. They are all distinct and non-zero. So that's good.Therefore, a + b + c = -2 + 4 + 16 = 18.Wait, but let's verify the original conditions to make sure. Let's compute f(a) when a = -2.f(a) = f(-2) = (-2)^3 + (-2)(-2)^2 + 4*(-2) + 16Compute step by step:(-2)^3 = -8(-2)(-2)^2 = (-2)(4) = -84*(-2) = -816 remains.So, total: -8 -8 -8 +16 = (-24) +16 = -8. But f(a) should equal a³ = (-2)^3 = -8. So that works.Similarly, check f(b) when b = 4:f(4) = 4^3 + (-2)(4)^2 + 4*(4) +16Compute:4^3 = 64(-2)(16) = -324*4 = 1616 remains.Total: 64 -32 +16 +16 = (64 -32) + (16 +16) = 32 +32 = 64. Which is equal to b³ = 4³ = 64. So that works.Therefore, all conditions are satisfied. So, the answer is a + b + c = -2 +4 +16 = 18.Wait, but let me check if there are other possible solutions. Because in the step where we concluded that (a + 1) divides -a², and then we thought that (a + 1) must divide 1, leading to a +1 = ±1. But is that the only possibility?Wait, let's re-examine that step. So, we had that (a + 1) divides -a², and since a and a +1 are coprime, then (a +1) divides -1. But why?Wait, let's suppose that d = a +1. Then, d divides -a². But since d and a are coprime (because d = a +1 and a are consecutive integers), then d divides -1. Because if d divides a² and gcd(d, a) =1, then d divides 1. Therefore, d must be ±1. Hence, a +1 = ±1, so a =0 or a = -2. Since a ≠0, then a = -2.Therefore, that conclusion is correct. So, there's no other possible a. So, the only possible a is -2, leading to b=4 and c=16. Therefore, the answer is 18. Therefore, the answer is boxed{18}.But wait, to be thorough, let me check if there are other possibilities if, for some reason, the assumption that gcd(a +1, a) =1 is incorrect? Wait, no. Since a and a +1 are consecutive integers, their gcd is always 1. Therefore, if d divides a² and d divides a +1, since gcd(d, a) =1, then d divides 1. Therefore, d must be ±1. Therefore, a +1 = ±1. Hence, a=0 or a=-2. Since a is non-zero, only a=-2. So, that's the only solution.Therefore, the answer is 18. So, I think that's it. Let me just check once again that all the conditions are satisfied.a, b, c: -2, 4, 16. All distinct, all non-zero. Check. f(a) =a³. Checked. f(b)=b³. Checked. So, all conditions are met. Therefore, the answer is 18.Final Answerboxed{18}</think>Given the function ( f(x) = x^3 + ax^2 + bx + c ) where ( a, b, c ) are distinct non-zero integers, and the conditions ( f(a) = a^3 ) and ( f(b) = b^3 ), we need to find ( a + b + c ).First, we substitute ( x = a ) into the function:[f(a) = a^3 + a cdot a^2 + b cdot a + c = a^3]Simplifying this, we get:[a^3 + a^3 + ab + c = a^3 implies 2a^3 + ab + c = a^3 implies a^3 + ab + c = 0 quad text{(1)}]Next, substituting ( x = b ) into the function:[f(b) = b^3 + a cdot b^2 + b cdot b + c = b^3]Simplifying this, we get:[b^3 + ab^2 + b^2 + c = b^3 implies ab^2 + b^2 + c = 0 implies b^2(a + 1) + c = 0 quad text{(2)}]Subtracting equation (2) from equation (1) to eliminate ( c ):[a^3 + ab - b^2(a + 1) = 0]Rearranging and factoring, we get:[a^3 + ab - ab^2 - b^2 = 0 implies a(a^2 - b^2) + b(a - b) = 0 implies (a - b)(a(a + b) + b) = 0]Since ( a neq b ), we have:[a(a + b) + b = 0 implies a^2 + ab + b = 0 implies a^2 + b(a + 1) = 0]Solving for ( b ):[b = -frac{a^2}{a + 1}]Since ( a ) and ( a + 1 ) are coprime, ( a + 1 ) must divide (-a^2). Given ( a + 1 ) divides (-1), we find ( a + 1 = pm 1 ). Since ( a neq 0 ), we get ( a = -2 ).Substituting ( a = -2 ):[b = -frac{(-2)^2}{-2 + 1} = -frac{4}{-1} = 4]Using equation (1) to find ( c ):[(-2)^3 + (-2)(4) + c = 0 implies -8 - 8 + c = 0 implies c = 16]Thus, ( a = -2 ), ( b = 4 ), and ( c = 16 ). Verifying, all conditions are satisfied. Therefore, ( a + b + c = -2 + 4 + 16 = 18 ).[boxed{18}]]

🔑:3. 18 .Let g(x)=f(x)-x^{3}=a x^{2}+b x+c.From the problem, we have g(a)=g(b)=0.Thus, g(x)=a(x-a)(x-b)Rightarrow b=-a(a+b), c=a^{2} bRightarrow b=-frac{a^{2}}{a+1}=1-a-frac{1}{a+1}.Since b is an integer, we have a+1= pm 1.Also, a neq 0, so a=-2, b=4, c=16.Therefore, a+b+c=18.

🧠:[<think>Okay, so I have this physics problem here. Let me try to work through it step by step. Let me read it again first."A mass m hangs from a massless spring connected to the roof of a box of mass M. When the box is held stationary, the mass-spring system oscillates vertically with angular frequency ω. If the box is dropped and falls freely under gravity, how will the angular frequency change? The options are A to E, with E saying it depends on M and m."Hmm. So the setup is a mass hanging from a spring inside a box. Normally, when the box is stationary, the mass oscillates with angular frequency ω. But if the box is in free fall, like when it's dropped, how does the angular frequency change?First, I need to recall how the angular frequency of a mass-spring system is determined. The angular frequency ω for a simple harmonic oscillator is given by ω = sqrt(k/m), where k is the spring constant and m is the mass. That's the formula when the spring is vertical or horizontal, right? Wait, but in the vertical case, when the box is stationary, gravity is acting on the mass, which causes the spring to stretch. So when the mass is hanging stationary, the spring is stretched by some amount Δx = mg/k. But when it oscillates, the restoring force is still governed by Hooke's law, and the effective "equilibrium" position is just shifted by gravity. But the key point is that the presence of gravity doesn't affect the angular frequency because when you set up the differential equation for the motion, the mg term cancels out when you consider displacement from the equilibrium position.So in the stationary box case, even though there's gravity, the angular frequency ω is still sqrt(k/m). So the formula remains the same. So gravity affects the equilibrium position but not the frequency of oscillation.Now, what happens when the entire box is in free fall? If the box is in free fall, then the entire system—box, spring, and mass—are all accelerating downward with acceleration g. So in the frame of the box, which is a non-inertial frame because it's accelerating, there is effectively zero gravity. Because in free fall, you're in a local inertial frame—Einstein's equivalence principle. So in the box's frame, it's like being in zero gravity. So the spring isn't stretched anymore? Wait, because if the box is in free fall, the mass is also in free fall. So in the box's frame, the spring would not be stretched? Because both the box and the mass are accelerating downward at g, so relative to the box, the mass isn't pulling the spring down. So the equilibrium position would be where the spring is unstretched.But if the spring is unstretched in free fall, then if you displace the mass from that equilibrium position, the restoring force would still be governed by Hooke's law. So if there's no effective gravity in the box's frame, then the oscillations would be similar to a horizontal spring-mass system. So the angular frequency would still be sqrt(k/m). Wait, but the original angular frequency when the box was stationary was sqrt(k/m). So does that mean the angular frequency remains the same? But option A says ω will be unchanged, so that would be the answer. But wait, wait. Let me think again.Wait, when the box is in free fall, the effective gravity inside the box is zero. So the spring isn't stretched due to gravity. But the mass is still attached to the spring. If you displace the mass, say, upward or downward, the spring will exert a force proportional to the displacement. So the equation of motion would be F = -kx, leading to the same angular frequency ω = sqrt(k/m). Therefore, the frequency remains the same. So the answer is A? But the options say D) Oscillations are impossible under these conditions. Wait, why would that be?Wait, maybe in free fall, the box and the mass are both accelerating downward at g. So if you consider the mass inside the box, from the box's perspective, there is no net force acting on the mass except the spring force. Wait, but when the box is in free fall, the only real force acting on the mass is gravity. But the box is also accelerating downward at g. So in the non-inertial frame of the box, there's a fictitious force upward equal to mg. But that fictitious force would exactly cancel the gravitational force. So in the box's frame, the net force on the mass is the spring force plus the fictitious force. Wait, but in free fall, the effective gravity is zero. So in that frame, the mass would behave as if it's in zero gravity. So if you have a spring in zero gravity, you can still have oscillations. For example, if you compress or extend the spring and let go, it will oscillate. So the angular frequency is still sqrt(k/m). Therefore, the frequency should remain the same. So why is option D there?Wait, maybe I'm missing something here. Let's think carefully. Suppose the box is in free fall. Then, the entire system is in free fall. So the spring is not stretched because both the box and the mass are accelerating at g. So the equilibrium position is when the spring is at its natural length. If you displace the mass from that equilibrium position, then the spring force will still act to restore it. So the mass can still oscillate around the new equilibrium position. Therefore, oscillations are possible, and the frequency is the same as before. So D is incorrect.But wait, maybe when the box is in free fall, the mass isn't able to oscillate because both the box and the mass are accelerating downward. So if the mass moves downward relative to the box, the box is moving downward as well. Hmm, but relative to the box, the mass can still move up and down. Because in the box's frame, which is in free fall, the effective gravity is zero. So the mass would oscillate as if in zero gravity, with the same frequency. So the angular frequency remains ω. Therefore, answer A.But maybe there is another factor. Let me think again. When the box is in free fall, the spring is not stretched initially. If you start with the spring in its natural length, then if you give the mass a displacement, it will oscillate. But wait, how do you start the oscillation? If the box is just dropped, then at the moment it is released, the spring is stretched by mg/k. Then, as the box starts to accelerate downward at g, the equilibrium point shifts. So maybe the system will oscillate around the new equilibrium.Wait, hold on. Let me consider the initial conditions. When the box is held stationary, the spring is stretched by Δx = mg/k. Then, when the box is released and starts to free fall, the box and the mass are both in free fall. In the frame of the box, the effective gravity is zero. So the equilibrium position shifts from Δx to zero. So the mass is now displaced from the new equilibrium (which is the natural length of the spring) by Δx. Therefore, the mass would start oscillating around the new equilibrium with amplitude Δx. So oscillations are possible. Therefore, the answer is A, ω remains the same. Because the angular frequency only depends on k and m.Alternatively, maybe when the box is in free fall, the mass and the box are both accelerating downward. So the spring is not stretched at all. But if the spring is not stretched, then the mass isn't experiencing any force from the spring. However, if the mass is displaced relative to the box, the spring will exert a force. So oscillations can still occur. For example, if you have a spring in space, zero gravity, you can still compress or stretch it and it will oscillate. So the frequency is still sqrt(k/m). Therefore, ω is the same. So answer A.Wait, but in the original case, when the box was stationary, the oscillation was vertical, affected by gravity. But in free fall, in the box's frame, there is no gravity. So the oscillation is like in zero-g. But the formula for ω in zero-g is still sqrt(k/m). Therefore, same ω. So answer A.But wait, why is option D even there? Maybe because when in free fall, the entire system is in free fall, so the mass can't oscillate? But that doesn't make sense. If you have a spring in free fall, you can still compress or extend it, and it will oscillate. Think of astronauts in the space station (which is in free fall) playing with springs. They can definitely oscillate. So in that case, the spring would oscillate with the same frequency as on Earth's surface, except in the station's frame, it's zero-g. So the frequency is the same. Therefore, answer A.Alternatively, maybe in the free-falling box, if the box is accelerating downward at g, then when the mass moves upward relative to the box, the box is moving downward, so the total acceleration of the mass relative to the ground is g minus the spring acceleration? Hmm, this seems more complicated. Let's analyze from the ground frame.In the ground frame, when the box is in free fall, both the box and the mass are accelerating downward at g. However, if the mass is displaced relative to the box, then the spring exerts a force on the mass. Let's model this.Let's denote the position of the box as Y(t). Since the box is in free fall, Y(t) = -½gt² + v₀t + y₀, but since it's dropped, initial velocity v₀=0, so Y(t) = -½gt² + y₀. The acceleration of the box is d²Y/dt² = -g.Let the position of the mass relative to the box be x(t). So the absolute position of the mass is Y(t) + x(t).The forces on the mass are gravity, mg downward, and the spring force. The spring force is -k x(t) upward, since if x is positive (mass is below the box's roof), the spring is stretched, so it pulls upward.So from Newton's second law for the mass:m * d²(Y + x)/dt² = -k x - m gBut d²Y/dt² = -g, so:m*(-g + d²x/dt²) = -k x - m gExpanding:- m g + m d²x/dt² = -k x - m gCancel the -m g terms on both sides:m d²x/dt² = -k xWhich simplifies to:d²x/dt² + (k/m) x = 0So the equation is the standard simple harmonic oscillator equation with angular frequency ω = sqrt(k/m). So the angular frequency is unchanged. Therefore, the answer is A.So in the ground frame, even though both the box and the mass are accelerating downward at g, the relative motion of the mass with respect to the box is still governed by the same differential equation, leading to the same angular frequency. Therefore, oscillations are possible and ω remains the same. So the answer is A.But why is option D there? Maybe someone might think that in free fall, since everything is accelerating, the spring can't exert a force? But that's not the case. The spring still follows Hooke's law, and the forces are still present. The key insight is that in the equation of motion, the gravitational acceleration cancels out when considering the relative motion between the mass and the box. Therefore, oscillations are possible, and the frequency is the same. So the correct answer is A.Final Answerboxed{A}</think>When the box is held stationary, the angular frequency (omega) of the mass-spring system is given by (omega = sqrt{frac{k}{m}}), where (k) is the spring constant and (m) is the mass. This frequency is determined by the balance between the spring force and the inertia of the mass, and gravity only affects the equilibrium position, not the frequency.When the box is dropped and falls freely under gravity, both the box and the mass are in free fall. In this scenario, the effective gravity within the box's frame of reference is zero. The key analysis involves considering the forces acting on the mass from both the ground frame and the box's frame:1. Ground Frame Analysis: - The box and the mass are accelerating downward with acceleration (g). - The forces on the mass are gravity ((mg) downward) and the spring force ((-kx) upward, where (x) is the displacement from the equilibrium position). - Using Newton's second law, the equation of motion for the mass relative to the box is derived as: [ m frac{d^2(Y + x)}{dt^2} = -kx - mg ] - Since the acceleration of the box (frac{d^2Y}{dt^2} = -g), substituting this into the equation simplifies to: [ m frac{d^2x}{dt^2} = -kx ] - This results in the standard simple harmonic oscillator equation (frac{d^2x}{dt^2} + frac{k}{m}x = 0), confirming the angular frequency remains (sqrt{frac{k}{m}}).2. Box's Frame Analysis: - In the box's non-inertial frame, the effective gravity is zero due to free fall. - The spring force is the only restoring force, leading to oscillations with the same angular frequency (sqrt{frac{k}{m}}).Both analyses confirm that the angular frequency (omega) remains unchanged. Therefore, the correct answer is:[boxed{A}],<think>Okay, let's try to figure this problem out. So, we have a mass m hanging from a spring inside a box of mass M. When the box is held stationary, the mass-spring system oscillates with an angular frequency ω. Now, the question is, if the box is dropped and falls freely under gravity, what happens to the angular frequency? The options are that it stays the same, increases, decreases, oscillations are impossible, or it depends on M and m.Hmm. I remember that the angular frequency for a mass-spring system is given by ω = sqrt(k/m), where k is the spring constant. So in the stationary case, that's straightforward. But when the box is in free fall, both the box and the mass m are accelerating downward with acceleration g. So, effectively, they're in a state of weightlessness, right? Because in free fall, everything inside the box would experience apparent weightlessness. Like astronauts in the International Space Station—they're in free fall around the Earth, so they feel weightless.So, if the box is in free fall, then from the perspective of someone inside the box, the mass m would not feel the force of gravity anymore. Because the box and the mass are both accelerating downward at g. So, the effective gravity inside the box is zero. So, how does that affect the spring?Wait, in the stationary case, the spring is stretched due to the weight of the mass m. The equilibrium position is where the spring force balances mg. So, when the box is in free fall, that equilibrium position would change, right? Because now, there's no effective gravity. So, the spring wouldn't be stretched at equilibrium? Or would it?Wait, if the box is in free fall, then the mass m is also in free fall. So, relative to the box, the mass m would experience zero net force. But the spring was initially stretched when the box was stationary. If the box is let go, as it starts to accelerate downward, the spring might start to contract because the mass is no longer needing to counteract gravity. Hmm. But once the entire system is in free fall, then the equilibrium length of the spring would be the natural length, since there's no effective gravity.So, in the free-falling scenario, the mass is attached to the spring, but there's no net external force acting on the mass except the spring force. Because in free fall, the gravitational force is exactly canceled by the inertial force, or something like that. So, if the mass was displaced from the equilibrium position (which is now the natural length of the spring), it would oscillate around that point. But since the angular frequency depends only on the spring constant and the mass, which are both unchanged, then the angular frequency ω should remain the same, right?Wait, but wait. When the box is stationary, the equilibrium position is determined by balancing the spring force and gravity. The effective "restoring force" in that case is still governed by the spring constant. Because when you displace the mass from equilibrium, the net force is -k times the displacement. So, even though gravity sets the equilibrium position, the oscillations around that position still have the same ω = sqrt(k/m).But if the box is in free fall, then the equilibrium position is the natural length of the spring. But when you displace the mass from that position, the restoring force is still -kx. So, the equation of motion is the same: m d²x/dt² = -kx. Therefore, the angular frequency should still be sqrt(k/m). So, in that case, the angular frequency doesn't change. So, answer A?But wait, the answer could be D) Oscillations are impossible. Wait, why would that be? If the entire system is in free fall, then if you pull the mass and let go, would it oscillate? If there's no damping, then in the absence of air resistance (since it's a massless spring and inside a box), maybe it would oscillate. But from the reference frame of the box, which is in free fall, the effective gravity is zero, so the only force is the spring. So, the oscillations should still occur with the same frequency. Hmm.Alternatively, maybe when the box is in free fall, the spring can't support the oscillations because both the box and the mass are accelerating at the same rate. Wait, but the spring is massless, so any force applied to one end is instantly transmitted to the other. If the box is accelerating downward at g, and the mass is attached to the spring, which is attached to the box, then the spring can still stretch and compress as the mass moves relative to the box.Let me think. Suppose the box is in free fall. Let's model this. Let's take the box's reference frame, which is a non-inertial frame because it's accelerating downward at g. In this frame, there's an effective gravitational field upward with acceleration g. But wait, in free fall, the equivalence principle says that the frame is locally equivalent to an inertial frame. So, in the box's frame, the effective gravity is zero. So, the mass would just oscillate as if there's no gravity. Therefore, the equation of motion would be the same as in the case where gravity is absent. So, the angular frequency is sqrt(k/m), same as before. Wait, but in the original problem, when the box is stationary, the angular frequency is sqrt(k/m) as well, because gravity only affects the equilibrium position, not the frequency.Wait, that's right. The frequency of a mass-spring system doesn't depend on gravity. Whether you have gravity or not, the frequency is determined by the mass and spring constant. Because when you add a constant force like gravity, it changes the equilibrium position, but the oscillations around that equilibrium are still governed by the same k and m. So, whether the box is in free fall or stationary, the frequency remains sqrt(k/m). So, the answer should be A.But wait, the problem says "angular frequency". So, in the stationary case, ω is sqrt(k/m). In free fall, the angular frequency is still sqrt(k/m). So, the answer is A. So, why is option D there? Maybe someone might think that since both the box and the mass are in free fall, the spring can't exert any force? But that's not true. The spring is massless, so if the mass moves relative to the box, the spring will exert a force to oppose that displacement. For example, if you have a spring in free fall and you compress it, it will push back and oscillate. So, oscillations are possible. So, D is not correct.Alternatively, maybe in free fall, the restoring force is different? But no, the restoring force is still -kx. So, the equation of motion is the same. So, the frequency is the same. Therefore, answer A.Wait, but let me check with an example. Suppose we have a spring in space, far from any gravity. If you attach a mass to it, it will oscillate with ω = sqrt(k/m). If we then take that same spring and mass to Earth's surface, hang it vertically, the frequency is still sqrt(k/m), even though there's gravity. Because the equilibrium position is shifted, but the frequency remains the same. So, yes, the frequency is independent of gravity. So, in free fall, even though the equilibrium position is shifted to the natural length, the frequency remains sqrt(k/m). Therefore, the answer is A.Wait, but the problem says when the box is dropped, so both the box and the mass are in free fall. So, from the perspective of the box, the mass is weightless, so the spring isn't stretched. If you displace the mass from the equilibrium (which is now the natural length), then it will oscillate with the same frequency. Therefore, the angular frequency is unchanged. So, answer A.Therefore, the correct answer should be A. So, options: A) ω will be unchanged. Therefore, the answer is A.Final Answerboxed{A}</think>When the box is held stationary, the angular frequency of the mass-spring system is given by (omega = sqrt{frac{k}{m}}), where (k) is the spring constant and (m) is the mass. This frequency depends only on the spring constant and the mass, not on gravity, as gravity only affects the equilibrium position.When the box is dropped and falls freely under gravity, both the box and the mass (m) are in free fall. In this state, the effective gravity inside the box is zero. From the perspective of the box, the mass (m) experiences weightlessness. However, the spring can still exert forces as it is massless. Any displacement of the mass from the new equilibrium position (the natural length of the spring) results in a restoring force (-kx), leading to oscillations. The equation of motion remains (m frac{d^2x}{dt^2} = -kx), which has the same angular frequency (omega = sqrt{frac{k}{m}}).Thus, the angular frequency (omega) remains unchanged. The correct answer is (boxed{A}).]

🔑:1. Understanding the problem: We have a mass ( m ) hanging from a massless spring connected to the roof of a box of mass ( M ). When the box is stationary, the mass-spring system oscillates with angular frequency ( omega ). We need to determine how the angular frequency changes when the box is dropped and falls freely under gravity.2. Angular frequency in a stationary frame: When the box is stationary, the angular frequency ( omega ) of the mass-spring system is given by: [ omega = sqrt{frac{k}{m}} ] where ( k ) is the spring constant.3. Free-fall condition: When the box is dropped, it falls freely under gravity. In this scenario, both the box and the mass ( m ) experience the same gravitational acceleration ( g ). This means that in the frame of the box, the mass ( m ) is effectively in a gravity-free environment.4. Oscillations in a freely falling frame: In the freely falling frame of the box, the only force acting on the mass ( m ) is the restoring force of the spring. The gravitational force does not affect the relative motion between the mass ( m ) and the box.5. Angular frequency in the freely falling frame: Since the gravitational force is not affecting the relative motion, the angular frequency of the mass-spring system in the freely falling frame remains the same as it was when the box was stationary. Therefore, the angular frequency ( omega ) in the freely falling frame is still: [ omega = sqrt{frac{k}{m}} ]6. Conclusion: The angular frequency ( omega ) of the mass-spring system does not change when the box is dropped and falls freely under gravity. Thus, the correct answer is: [ boxed{text{A}} ]

🧠:[<think>Alright, let's try to tackle this problem. So, the question is: For all positive integers from 1 to 1,000,000,000, we need to repeatedly sum the digits of each number until we get a single-digit number. Then, among all these resulting single-digit numbers, we have to determine whether there are more 1s or 2s. Hmm, okay.First, let me make sure I understand the process correctly. For each number from 1 to 1 billion, we sum its digits. If the result is more than one digit, we sum those digits, and repeat until we get a single digit. This is essentially finding the digital root of each number, right? Because the digital root is the single-digit number obtained by repeatedly summing the digits of a number until only one digit remains. So, digital root of n is given by the formula: dr(n) = 1 + ((n - 1) mod 9). Wait, is that correct?Let me recall the formula for digital roots. Yes, the digital root of a number n is equal to n mod 9, unless n is a multiple of 9, in which case the digital root is 9. So, dr(n) = n mod 9 if n mod 9 ≠ 0, else 9. So, that formula is equivalent to 1 + ((n - 1) mod 9). Let me check with an example. Take n=10: 10 mod 9 is 1, which is correct. If n=9: 9 mod 9 is 0, so dr(9)=9. So yes, the digital root can be calculated as dr(n) = 1 + (n - 1) mod 9. That works.Therefore, the problem reduces to finding how many numbers from 1 to 1,000,000,000 have a digital root of 1 or 2, and which of these counts is higher.But wait, the problem says "until obtaining 1 billion single-digit numbers". So, each number from 1 to 1 billion is converted to its digital root. Then we have 1 billion single-digit numbers (each being the digital root of the original number). Then, among these 1 billion digital roots, we need to count how many are 1s and how many are 2s, and compare those counts.So, the question is equivalent to: In the numbers from 1 to 1,000,000,000, how many have a digital root of 1, and how many have a digital root of 2? Which is more?Therefore, to solve the problem, we can compute the number of integers between 1 and 1,000,000,000 (inclusive) that are congruent to 1 mod 9 and those congruent to 2 mod 9, since digital root 1 corresponds to n ≡ 1 mod 9, and digital root 2 corresponds to n ≡ 2 mod 9.Wait, but 1,000,000,000 is 10^9. Let's check what 10^9 mod 9 is. Since 10 ≡ 1 mod 9, so 10^k ≡ 1^k = 1 mod 9. Therefore, 10^9 ≡ 1 mod 9, so 10^9 = 9k + 1 for some integer k. Therefore, the numbers from 1 to 10^9 inclusive can be partitioned into complete residue classes mod 9, except for the last one.Wait, more precisely, the numbers from 1 to N can be divided into floor(N/9) full cycles of residues 1 to 9, and then a remaining part. But since N = 10^9, which is congruent to 1 mod 9, as we saw, 10^9 = 9 * 111,111,111 + 1. So, there are 111,111,111 complete cycles of 9 numbers, and one extra number (which is 10^9 itself, which is 1 mod 9). So, each residue from 1 to 9 occurs exactly 111,111,111 times, except residue 1, which occurs one more time (because the last number, 10^9, is 1 mod 9). Therefore, the count of numbers congruent to 1 mod 9 is 111,111,111 + 1 = 111,111,112, and counts for residues 2 to 9 are 111,111,111 each.Therefore, in the range 1 to 1,000,000,000, the number of numbers with digital root 1 is 111,111,112, and the number with digital root 2 is 111,111,111. Therefore, there are more 1s than 2s.But wait, wait. Wait, hold on. Let's verify this step by step.First, digital root 1 corresponds to numbers ≡1 mod 9. Similarly for digital root 2. So, the count of numbers from 1 to N with digital root k is equal to the count of numbers in that range congruent to k mod 9, except when k=9, which corresponds to ≡0 mod 9. But in our case, k=1 and k=2, so they correspond directly to ≡1 mod 9 and ≡2 mod 9.Now, N = 1,000,000,000. Let's compute how many numbers from 1 to N are ≡1 mod 9 and ≡2 mod 9.The formula for the number of integers in [1, N] congruent to a mod m is floor((N - a)/m) + 1 if a <= N mod m, otherwise floor((N - a)/m) + 1. Wait, perhaps an easier way is:Number of numbers ≡r mod 9 in [1, N] is floor((N + (9 - r))/9). Wait, maybe not. Let's think.Let me consider N divided by 9. Let N = 9q + s, where 0 ≤ s < 9.Then, the number of numbers congruent to r mod 9 in [1, N] is q + 1 if r ≤ s, else q.But wait, if N = 9q + s, then for each residue 1 to 9 (since mod 9 residues are 0 to 8, but we are considering numbers from 1 to N, so residues 1 to 9? Wait, no, numbers from 1 to N have residues 1 to 9 only if N is a multiple of 9. Wait, perhaps confusion here.Wait, in mod 9, residues are 0,1,2,...,8. But in numbers from 1 to N, residues would be 1 to 9 only if N is a multiple of 9. Wait, no. Let's correct that.Actually, when we divide any integer by 9, the possible remainders are 0,1,2,...,8. So residues mod 9 are 0 through 8. However, numbers from 1 to N will have residues 1 through 8 and 0 (for multiples of 9). So, for example, the number 9 has residue 0 mod 9, 10 has residue 1, etc.Therefore, digital roots correspond to residues as follows: digital root 9 corresponds to residue 0 mod 9, and digital roots 1 through 8 correspond to residues 1 through 8 mod 9. So, actually, digital root k is equal to n mod 9 if n mod 9 ≠ 0, else 9.Therefore, the count of digital root k in [1, N] is equal to the count of numbers ≡k mod 9 if k ≠9, and equal to the count of numbers ≡0 mod 9 if k=9.So, coming back, for digital roots 1 and 2, their counts are equal to the number of numbers ≡1 mod 9 and ≡2 mod 9 in [1, N], respectively.Now, to compute how many numbers from 1 to N are ≡1 mod 9. Let N = 1,000,000,000.As we saw earlier, 10^9 divided by 9 is equal to 111,111,111.111..., since 9*111,111,111 = 999,999,999. Then 10^9 = 9*111,111,111 + 1. Therefore, N = 9q + 1, where q = 111,111,111.Therefore, the number of numbers ≡1 mod 9 in [1, N] is equal to q + 1 (since the remainder is 1, which is equal to the residue we are counting). Similarly, for residues 2 to 9, but since N = 9q +1, the residues in [1, N] are:For the first 9q numbers (1 to 999,999,999), each residue from 1 to 9 occurs exactly q times. Then, the remaining number (1,000,000,000) has residue 1 mod 9. Therefore, residues 1 occurs q +1 times, and residues 2 to 9 occur q times each.Therefore, the number of numbers with digital root 1 is q +1 = 111,111,111 +1 = 111,111,112.Similarly, the number of numbers with digital root 2 is q = 111,111,111.Therefore, there are more 1s than 2s in the resulting single-digit numbers.Wait, but let me double-check. Let's take a smaller example to verify.Suppose N = 10. Then numbers from 1 to 10. Let's compute digital roots:1:1, 2:2, 3:3, 4:4, 5:5, 6:6, 7:7, 8:8, 9:9, 10:1.So digital roots: two 1s, one each of 2-9. So counts: 1:2, 2:1, ..., 9:1. So, indeed, residue 1 occurs one more time. Since 10 = 9*1 +1. Therefore, q=1, so residue 1 occurs q+1=2 times, others q=1.Similarly, if N=9, which is 9*1, then numbers 1-9 have digital roots 1-9 each once. So counts equal.If N=9q + r, where 0 ≤ r <9, then residues 1 to r occur q+1 times, and residues r+1 to 9 occur q times. But wait, in our case, N=9q +1, so r=1. Therefore, residue 1 occurs q+1 times, and residues 2-9 occur q times each. So, yes.Therefore, applying to N=10^9, which is 9*111,111,111 +1, residue 1 occurs 111,111,112 times, and residues 2-9 occur 111,111,111 times each. Therefore, there are more 1s than 2s.Hence, the answer should be that there are more 1s.But wait, wait a second. Let me check if the original problem counts numbers from 1 to 1,000,000,000 inclusive. So yes, 1 to 10^9. So, including 10^9, which is 1 followed by 9 zeros, so its digit sum is 1, hence digital root 1. So, yes, that contributes to the count of 1s. So, indeed, the digital roots are periodic with period 9, and since 10^9 is 1 more than a multiple of 9, the residue 1 gets an extra count. Therefore, more 1s.Therefore, the answer is that there are more 1s.But wait, wait again. Wait, the problem says "for all positive integers from 1 to 1000000000". So, numbers 1 through 10^9, inclusive. So total of 10^9 numbers. When we compute digital roots, each number is converted to its digital root, so we have 10^9 digital roots. Now, digital roots 1 through 9. The counts for each digital root would be floor(10^9 /9) or ceil(10^9 /9). Since 10^9 divided by 9 is 111,111,111.111..., which is 111,111,111 + 1/9. Therefore, each digital root from 1 to 9 would occur 111,111,111 times, and then there is an extra 1 digital root because 10^9 = 9*111,111,111 +1, so the remaining 1 number (10^9) has digital root 1. Therefore, total counts: digital root 1 occurs 111,111,112 times, others 111,111,111 times. Therefore, more 1s than 2s.But let's make sure that the number 10^9 is included. The problem says "from 1 to 1000000000", so yes, inclusive. So, 10^9 is included. Therefore, the count of 1s is one more than the count of 2s. Therefore, the answer is that there are more 1s.Therefore, the answer is that there are more 1s than 2s.Final Answerboxed{1}</think>To determine whether there are more 1s or 2s in the resulting single-digit numbers after repeatedly summing the digits of each number from 1 to 1,000,000,000, we need to find the digital root of each number. The digital root of a number ( n ) is given by ( 1 + ((n - 1) mod 9) ).1. Digital Root Calculation: The digital root of a number ( n ) is equivalent to ( n mod 9 ), unless ( n mod 9 = 0 ), in which case the digital root is 9. This means the digital roots cycle through 1 to 9.2. Range Analysis: We need to analyze the numbers from 1 to 1,000,000,000. Since ( 1,000,000,000 = 9 times 111,111,111 + 1 ), this range includes 111,111,111 complete cycles of 9 numbers and one additional number (1,000,000,000 itself).3. Distribution of Digital Roots: Each digital root from 1 to 9 appears exactly 111,111,111 times in the complete cycles. The additional number, 1,000,000,000, has a digital root of 1. Therefore, the count of digital root 1 is ( 111,111,111 + 1 = 111,111,112 ), while the counts for digital roots 2 to 9 remain 111,111,111 each.Thus, there are more 1s than 2s in the resulting single-digit numbers.[boxed{1}],<think>Okay, let's try to tackle this problem. So, we need to consider all positive integers from 1 to 1,000,000,000. For each number, we have to repeatedly sum its digits until we get a single-digit number. Then, among all those resulting single-digit numbers, we have to determine whether there are more 1s or 2s. Hmm, interesting.First off, I remember that repeatedly summing the digits of a number until you get a single digit is called finding the digital root of the number. The digital root of a number is congruent to the number modulo 9. If the number is a multiple of 9, the digital root is 9; otherwise, it's the remainder when divided by 9. So, for example, the digital root of 1234 is 1 because 1+2+3+4=10, then 1+0=1. Similarly, 987 would be 9+8+7=24, then 2+4=6. So, the digital roots cycle from 1 to 9.Therefore, each number from 1 to 1,000,000,000 has a digital root between 1 and 9. So, the problem reduces to finding how many numbers in that range have a digital root of 1 versus a digital root of 2. Then, whichever is larger will determine whether there are more 1s or 2s.But wait, the problem statement says "for the resulting 1 billion numbers, find the sum of the digits of each number, ⋯, until obtaining 1 billion single-digit numbers." So, actually, maybe they are considering multiple iterations? Let me check again.Wait, the problem says: "For all positive integers from 1 to 1000000000, find the sum of the digits of each number, then for the resulting 1 billion numbers, find the sum of the digits of each number, ⋯, until obtaining 1 billion single-digit numbers." Hmm, so the process is: first sum the digits of each original number, resulting in 1 billion numbers (which are the digit sums), then sum the digits of each of those numbers, resulting in another 1 billion numbers, and repeat this until all numbers are single-digit. So, it's like iteratively taking digit sums until everything is single-digit. But actually, the first iteration would already reduce some numbers to single digits. For example, numbers from 1 to 9 would stay the same. Numbers from 10 to 99 would have their digits summed once, resulting in numbers from 1 to 18, but then in the next iteration, those would be summed again to get single digits. Wait, but the problem says "until obtaining 1 billion single-digit numbers". So, regardless of how many iterations it takes, each original number eventually becomes a single-digit number through repeated digit summing. However, the question is whether there are more 1s or 2s in the final single-digit numbers.But, hold on, but isn't this process equivalent to just computing the digital root? Because the digital root is the single-digit number obtained by repeatedly summing the digits. So, even if you do multiple iterations, the end result is the digital root. Therefore, the problem is equivalent to asking: Among the numbers from 1 to 1,000,000,000, how many have a digital root of 1 and how many have a digital root of 2? Then compare those counts.Therefore, the answer hinges on the distribution of digital roots in the numbers from 1 to 1,000,000,000. If the distribution is uniform, then each digital root from 1 to 9 would occur approximately the same number of times, but we need to check for exact counts.But first, let's confirm that the process described in the problem is indeed equivalent to digital roots. Let's take an example. Take the number 1234. Summing its digits once gives 1+2+3+4=10. Then summing the digits of 10 gives 1+0=1. So, after two iterations, we get 1. But the digital root is also 1. So, regardless of how many iterations we take, as long as we keep summing until we get a single digit, the result is the digital root. Therefore, yes, the final single-digit numbers are precisely the digital roots of the original numbers.Therefore, we can rephrase the problem as: Among the numbers from 1 to 1,000,000,000, how many have a digital root of 1, and how many have a digital root of 2? Which is larger?Now, digital roots are congruent to the number modulo 9. So, for a number n, digital root is dr(n) = 1 + ((n - 1) mod 9). Wait, actually, more precisely, dr(n) = n mod 9, unless n is a multiple of 9, in which case dr(n) = 9. So, digital roots cycle every 9 numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, etc.Therefore, if we consider numbers from 1 to N, where N is a multiple of 9, the distribution of digital roots would be exactly N/9 for each digit from 1 to 9. However, if N is not a multiple of 9, then the counts would be slightly different. In our case, N is 1,000,000,000.So, let's compute 1,000,000,000 divided by 9. Let's see, 9*111,111,111 = 999,999,999. Then 1,000,000,000 is 999,999,999 +1, so 1,000,000,000 = 9*111,111,111 +1.Therefore, the numbers from 1 to 1,000,000,000 consist of 111,111,111 complete cycles of 9 numbers, each cycle containing the digital roots 1 through 9, and then one extra number (which is 1,000,000,000).Therefore, for each digital root from 1 to 9, there are 111,111,111 numbers with that digital root. Then, the remaining number (1,000,000,000) has a digital root of 1 (since 1,000,000,000 is 1 followed by 9 zeros, so sum of digits is 1, so digital root is 1). Therefore, the total count for digital root 1 is 111,111,111 +1 = 111,111,112, and the counts for digital roots 2 through 9 are each 111,111,111. Therefore, there is one extra 1 compared to 2s. Therefore, there are more 1s than 2s.Wait, but wait a second. Wait, is the number 1,000,000,000 included in the range? The problem says "all positive integers from 1 to 1000000000". So, yes, 1,000,000,000 is included. So, that is the 1 followed by nine zeros. The sum of its digits is 1, so digital root is 1. Therefore, as I calculated, the digital root counts are 111,111,112 for 1 and 111,111,111 for 2 through 9. Therefore, there is exactly one more 1 than each of the others. Therefore, there are more 1s than 2s.But wait, let me double-check that division. Let's compute 1,000,000,000 divided by 9. 9*111,111,111 = 999,999,999. Then 1,000,000,000 - 999,999,999 =1. Therefore, 1,000,000,000 = 9*111,111,111 +1. Therefore, the quotient is 111,111,111 and the remainder is 1. So, in terms of digital roots, each cycle of 9 numbers contributes one of each digital root. Therefore, for the first 999,999,999 numbers, there are 111,111,111 numbers with digital roots 1 through 9. Then, the last number, 1,000,000,000, has a digital root of 1. Therefore, total digital roots: 111,111,112 ones, and 111,111,111 for the others. Therefore, yes, there is one more 1 than each of the other digits. Therefore, there are more 1s than 2s.But let me verify with a smaller example to see if this logic holds. Let's take N=10. So numbers 1 through 10. Digital roots: 1,2,3,4,5,6,7,8,9,1. So, two 1s and one of each from 2-9. Therefore, counts: 2 ones, 1 of each other. So, yes, when N=9*1 +1=10, there's one extra 1. Similarly, N=9*2 +1=19. Then, the first 18 numbers have two of each digital root 1-9, and the 19th number (19) has digital root 1 (1+9=10→1). So, three 1s, two of each other. So, that seems consistent.Therefore, the general formula is that for N = 9k + r, where 0 ≤ r <9, then the count for digital root d is k +1 if d ≤ r (but wait, no, actually, the remainder when divided by 9 is different). Wait, no. Wait, actually, when N = 9k + m, where m is from 1 to 9, then the digital roots from 1 to m will have k+1 counts, and digital roots from m+1 to 9 will have k counts. Wait, but in our case, N=1,000,000,000, which is 9*111,111,111 +1. So, m=1. Therefore, digital root 1 gets k +1 counts (111,111,111 +1), and digital roots 2-9 get k counts (111,111,111). Therefore, yes, this matches.Therefore, in the problem, since N=1,000,000,000=9*111,111,111 +1, the digital root 1 occurs one more time than the others, which all occur 111,111,111 times. Therefore, the number of 1s is 111,111,112 and the number of 2s is 111,111,111. Hence, there are more 1s than 2s.But wait, wait a second. Wait, digital roots are from 1 to 9. If N is congruent to m mod 9, where m is from 1 to 9, then digital roots 1 through m will have one extra count. Wait, but in our case, N=1,000,000,000. Let's compute 1,000,000,000 mod 9. Sum of digits of 1,000,000,000 is 1, so 1 mod 9 is 1. Therefore, the remainder is 1. Therefore, according to the formula, digital roots 1 through 1 (i.e., just 1) will have one extra count.Therefore, the count for digital root 1 is k +1 =111,111,111 +1=111,111,112, and others are 111,111,111. Therefore, yes, there is exactly one more 1 than each of the other digits.Therefore, the conclusion is that there are more 1s than 2s.But wait, let me think again. Wait, let's take another example where N is 10. As I did before, numbers 1-10. The digital roots are 1,2,3,4,5,6,7,8,9,1. So, two 1s and one of each other. So, when N=10, which is 9*1 +1, there is one extra 1. Similarly, if N=19, which is 9*2 +1=19, the digital roots would be 1-9,1-9,1. So, three 1s, two of each other. So, yeah, same pattern.Therefore, this seems to hold. So, generalizing, for N=9k +m, where m is the digital root of N, then digital roots 1 through m will have k +1 counts, and m+1 through 9 will have k counts. Wait, but in the example with N=10, m=1, so digital roots 1 through 1 (just 1) get k +1=1 +1=2 counts, others get k=1. Which matches. Similarly, for N=19, m=1 (since digital root of 19 is 1), so again digital root 1 gets 3 counts, others 2. Wait, but N=19 is 9*2 +1, so k=2, m=1. Then, digital roots 1 through m=1 get k +1=3, others get 2. Correct.Wait, but if N=18, which is 9*2, then digital roots 1-9 each occur exactly 2 times, right? 1-9,1-9. So, yes, two each. So, if N is a multiple of 9, then digital roots are equally distributed.Therefore, in our problem, N=1,000,000,000 which is 9*111,111,111 +1, so digital root 1 gets an extra count. Therefore, answer is there are more 1s than 2s.But wait, let me check the problem statement again to make sure I haven't missed anything. It says: "For all positive integers from 1 to 1000000000, find the sum of the digits of each number, then for the resulting 1 billion numbers, find the sum of the digits of each number, ⋯, until obtaining 1 billion single-digit numbers. Ask whether there are more 1s or 2s in the resulting numbers?"Wait a second, so the first step is to take each number from 1 to 1e9, sum its digits, resulting in 1e9 numbers. Then, take those numbers, sum their digits, resulting in another 1e9 numbers, and repeat until all numbers are single-digit. The question is about the counts of 1s and 2s in the final 1e9 single-digit numbers.But earlier, I assumed that this process is equivalent to computing the digital roots of the original numbers, which would result in 1e9 single-digit numbers. However, is this actually the case?Wait, for example, take the number 1234. First iteration: sum digits to get 10. Second iteration: sum digits of 10 to get 1. So, the final result is 1. But if we have a number like 19. First iteration: 1+9=10. Second iteration: 1+0=1. So, digital root is 1. So, yes, regardless of the number of iterations, the final result is the digital root. Therefore, even though we have to perform multiple iterations until all numbers are single-digit, the result is the same as the digital root. Therefore, the final counts are indeed the counts of digital roots from 1 to 9 in the original numbers.Therefore, my initial conclusion holds: there's one more 1 than the other digits, so more 1s than 2s.But wait, let me verify this with a different example. Let's take numbers from 1 to 10. If we apply the process:Original numbers: 1,2,3,4,5,6,7,8,9,10.First iteration (sum digits):1,2,3,4,5,6,7,8,9,1 (since 1+0=1).Now, these are the numbers after first iteration. Already, all except 10 are single-digit. Then, in the next iteration, we take the numbers: 1,2,3,4,5,6,7,8,9,1. All are single-digit, so we stop. So, the final counts are two 1s and one of each other. So, same as digital roots.Another example: numbers 1 to 19.First iteration:1,2,3,4,5,6,7,8,9,1 (10→1), 2 (11→2), 3 (12→3), 4 (13→4), 5 (14→5), 6 (15→6), 7 (16→7), 8 (17→8), 9 (18→9), 1 (19→10→1). Wait, no. Wait, 19: first iteration sum is 1+9=10. Then, next iteration sum is 1+0=1. So, yes, after two iterations, 19 becomes 1. Therefore, in the final counts, we have three 1s (from 1,10,19), two of each other digits (2-9 from 2-9 and 11-18). So, digital roots: 1 occurs three times, others two. Therefore, same as digital roots counts.Therefore, this confirms that the process described in the problem is equivalent to computing digital roots.Therefore, returning to the original problem, since 1,000,000,000 is congruent to 1 modulo 9, the digital roots will have one extra 1 compared to the other digits. Therefore, there are more 1s than 2s.Therefore, the answer should be that there are more 1s.But to ensure thoroughness, let's consider the process step by step. Suppose we start with numbers 1 to N. Each iteration replaces each number with the sum of its digits. We repeat until all numbers are single-digit. The question is about the counts in the final step.But in each iteration, numbers are replaced by their digit sums. For numbers that are already single-digit, they remain the same. For multi-digit numbers, they get replaced by smaller numbers. So, the process is equivalent to digital roots.For example, take 1234. First iteration: 1+2+3+4=10. Second iteration: 1+0=1. So, two iterations. But 1 is the digital root. So, regardless of how many iterations, the end result is the digital root.Therefore, the final counts are exactly the counts of digital roots in the original numbers. Hence, since N=1e9, which is 1 mod 9, digital root 1 occurs once more than the others. Therefore, more 1s.But wait, hold on. Wait, when we do the iterations, each number is replaced by its digit sum. So, for the number 1,000,000,000, first iteration gives 1, so it's done. For the number 999,999,999, first iteration gives 9*9=81, next iteration gives 8+1=9. So, its digital root is 9. So, indeed, the digital roots are preserved through iterations.Therefore, the conclusion is correct. Therefore, the answer is that there are more 1s than 2s.Final Answerboxed{1}</think>To determine whether there are more 1s or 2s in the resulting single-digit numbers after repeatedly summing the digits of each number from 1 to 1,000,000,000, we need to consider the concept of digital roots. The digital root of a number is the single-digit number obtained by repeatedly summing its digits until a single digit remains. This digital root is equivalent to the number modulo 9, with the exception that multiples of 9 have a digital root of 9.For the range from 1 to 1,000,000,000, we note that 1,000,000,000 divided by 9 gives a quotient of 111,111,111 and a remainder of 1. This means there are 111,111,111 complete cycles of the digits 1 through 9, and one additional number (1,000,000,000) which has a digital root of 1. Thus, the digital root counts are as follows:- The digital root 1 occurs 111,111,111 + 1 = 111,111,112 times.- Each of the digital roots 2 through 9 occurs 111,111,111 times.Therefore, there are more 1s than 2s in the resulting single-digit numbers.[boxed{1}]]

🔑:[Solution] Since any positive integer and the sum of its digits have the same remainder when divided by 9.Therefore, the numbers whose digit sum is 1 are all numbers of the form 9k+1, i.e.,1,10,19,28, cdots, 999999991,1000000000 .There are 111111112 in total.And the numbers whose digit sum is 2 are all numbers of the form 9k+2, i.e.,2,11,20,29, cdots, 999999992 text {. }There are 111111111 in total.Therefore, the numbers with a digit sum of 1 are 1 more than those with a digit sum of 2.

🧠:[<think>Okay, so I have this problem here: If the complex number z satisfies (1 + i)z = 1 - 2i, then what is |z|? The options are A through D. Hmm, let me think. Alright, complex numbers can be tricky, but let me recall what I know. The equation given is (1 + i) multiplied by z equals 1 - 2i. And we need to find the modulus of z, which is |z|. First, maybe I should solve for z here. If I can express z in terms of the other complex numbers, then I can compute its modulus. To solve for z, I need to divide both sides of the equation by (1 + i). So, z would be (1 - 2i) divided by (1 + i). But dividing complex numbers isn't straightforward like real numbers. I remember that to divide complex numbers, we multiply the numerator and denominator by the conjugate of the denominator. The conjugate of (1 + i) is (1 - i). So, if I do that, it should help me simplify the expression. Let me write that out: z = (1 - 2i)/(1 + i) multiplied by (1 - i)/(1 - i). That way, the denominator becomes a real number. Let me compute the numerator first: (1 - 2i)(1 - i). Let's expand that. Multiplying the first terms: 1 * 1 = 1.Then the outer terms: 1 * (-i) = -i.Then the inner terms: (-2i) * 1 = -2i.And the last terms: (-2i) * (-i) = 2i². Wait, so combining all those terms: 1 - i - 2i + 2i². Now, i² is -1, so 2i² is -2. So replacing that, we have 1 - i - 2i - 2. Combining like terms: (1 - 2) + (-i - 2i) = (-1) + (-3i) = -1 - 3i. Now the denominator is (1 + i)(1 - i). Let me compute that. Using the difference of squares formula: (1)^2 - (i)^2 = 1 - (-1) = 1 + 1 = 2. So the denominator is 2. Therefore, z = (-1 - 3i)/2. Which can be written as (-1/2) - (3/2)i. Now, to find |z|, the modulus of z. The modulus of a complex number a + bi is sqrt(a² + b²). So here, a is -1/2 and b is -3/2. Let's compute that. First, square a: (-1/2)² = 1/4. Then square b: (-3/2)² = 9/4. Adding those together: 1/4 + 9/4 = 10/4 = 5/2. Then the modulus is sqrt(5/2). Wait, sqrt(5/2) can be written as sqrt(10)/2 because sqrt(5/2) is equal to sqrt(10)/sqrt(4) which is sqrt(10)/2. Wait, let me check that again. sqrt(5/2) is the same as sqrt(10)/2? Hmm. Because sqrt(5/2) = sqrt(5)/sqrt(2) = (sqrt(5)*sqrt(2))/sqrt(2)*sqrt(2)) = sqrt(10)/2. Yes, that's correct. So sqrt(5/2) is sqrt(10)/2. Looking back at the options, option A is sqrt(10)/2. So the answer should be A. Let me double-check my steps to make sure I didn't make a mistake. Starting with (1 + i)z = 1 - 2i. Solving for z by dividing both sides by (1 + i). Then multiplying numerator and denominator by (1 - i), the conjugate. Expanding the numerator: (1 - 2i)(1 - i) gives 1 - i - 2i + 2i², which simplifies to -1 - 3i. Denominator is 2. So z = (-1 -3i)/2. Then modulus squared is (-1/2)^2 + (-3/2)^2 = 1/4 + 9/4 = 10/4 = 5/2. Square root of that is sqrt(10)/2. Yes, that seems right. Alternatively, another way to compute |z| is using the property of moduli. If (1 + i)z = 1 - 2i, then |1 + i||z| = |1 - 2i|. Therefore, |z| = |1 - 2i| / |1 + i|. Let me try that approach. Compute |1 + i|: that's sqrt(1² + 1²) = sqrt(2). Compute |1 - 2i|: sqrt(1² + (-2)^2) = sqrt(1 + 4) = sqrt(5). Therefore, |z| = sqrt(5)/sqrt(2) = sqrt(10)/2. Same answer as before. This confirms that the answer is indeed option A. So, both methods lead to the same result. Therefore, I can be confident that the correct answer is A: sqrt(10)/2.Final Answerboxed{A}</think>To solve for the complex number ( z ) given the equation ((1 + i)z = 1 - 2i), we first isolate ( z ) by dividing both sides by ( 1 + i ):[z = frac{1 - 2i}{1 + i}]To simplify this, we multiply the numerator and the denominator by the conjugate of the denominator, ( 1 - i ):[z = frac{(1 - 2i)(1 - i)}{(1 + i)(1 - i)}]Expanding the numerator:[(1 - 2i)(1 - i) = 1 cdot 1 + 1 cdot (-i) + (-2i) cdot 1 + (-2i) cdot (-i) = 1 - i - 2i + 2i^2]Since ( i^2 = -1 ):[1 - i - 2i + 2(-1) = 1 - 3i - 2 = -1 - 3i]Expanding the denominator:[(1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2]Thus, we have:[z = frac{-1 - 3i}{2} = -frac{1}{2} - frac{3}{2}i]To find the modulus ( |z| ):[|z| = sqrt{left(-frac{1}{2}right)^2 + left(-frac{3}{2}right)^2} = sqrt{frac{1}{4} + frac{9}{4}} = sqrt{frac{10}{4}} = sqrt{frac{5}{2}} = frac{sqrt{10}}{2}]Alternatively, using the property of moduli:[|(1 + i)z| = |1 - 2i| implies |1 + i||z| = |1 - 2i|]Calculating the moduli:[|1 + i| = sqrt{1^2 + 1^2} = sqrt{2}, quad |1 - 2i| = sqrt{1^2 + (-2)^2} = sqrt{5}]Thus:[|z| = frac{|1 - 2i|}{|1 + i|} = frac{sqrt{5}}{sqrt{2}} = frac{sqrt{10}}{2}]The answer is (boxed{A}).,<think>Okay, let's try to solve this problem. So, we have a complex number z that satisfies the equation (1 + i)z = 1 - 2i. We need to find the modulus of z, which is |z|, and then choose the correct answer from the given options. Alright, let me remember how to handle complex numbers here.First, I know that to solve for z, I can divide both sides of the equation by (1 + i). That should isolate z on one side. So, z = (1 - 2i)/(1 + i). But dividing complex numbers can be a bit tricky. I think the standard method is to multiply the numerator and the denominator by the complex conjugate of the denominator to rationalize it. The complex conjugate of (1 + i) is (1 - i). Let me try that.So, multiplying numerator and denominator by (1 - i):z = [(1 - 2i)(1 - i)] / [(1 + i)(1 - i)]Let me compute the denominator first. (1 + i)(1 - i) is a difference of squares, right? So, that's 1² - (i)² = 1 - (-1) = 1 + 1 = 2. So, the denominator simplifies to 2.Now the numerator: (1 - 2i)(1 - i). Let's expand that. Multiply 1 by (1 - i): 1*(1) - 1*(i) = 1 - i. Then multiply -2i by (1 - i): -2i*1 + (-2i)*(-i) = -2i + 2i². Remember that i² is -1, so 2i² is 2*(-1) = -2. So, combining those parts:1 - i - 2i - 2. Combine like terms. The constants: 1 - 2 = -1. The terms with i: -i -2i = -3i. So the numerator is -1 - 3i.Therefore, z = (-1 - 3i)/2. That simplifies to -1/2 - (3/2)i.Now, we need to find the modulus of z, which is |z|. The modulus of a complex number a + bi is sqrt(a² + b²). So, here, a is -1/2 and b is -3/2. Let's compute that.First, square a: (-1/2)² = 1/4. Square b: (-3/2)² = 9/4. Add them together: 1/4 + 9/4 = 10/4 = 5/2. Then take the square root of that: sqrt(5/2). Wait, but that's sqrt(10)/2, because sqrt(5/2) is sqrt(10)/2. Let me check that.Yes, sqrt(5/2) can be written as sqrt(10)/2 because sqrt(5)/sqrt(2) = (sqrt(5)*sqrt(2))/2 = sqrt(10)/2. So, |z| is sqrt(10)/2. Looking at the options, that's option A.Wait, let me make sure I didn't make any mistakes in the calculation. Let's go through it again step by step.Starting with (1 + i)z = 1 - 2i. Solving for z by dividing both sides by (1 + i):z = (1 - 2i)/(1 + i). Multiply numerator and denominator by (1 - i):Numerator: (1 - 2i)(1 - i). Let's compute this again. 1*1 = 1, 1*(-i) = -i, -2i*1 = -2i, -2i*(-i) = 2i² = 2*(-1) = -2. Adding all together: 1 - i -2i -2 = (1 - 2) + (-i -2i) = -1 -3i. Denominator: (1 + i)(1 - i) = 1 - i² = 1 - (-1) = 2. So z = (-1 -3i)/2. Correct.Then modulus is sqrt[(-1/2)^2 + (-3/2)^2] = sqrt[(1/4) + (9/4)] = sqrt[10/4] = sqrt(10)/2. Yes, that's correct. So the answer is A: sqrt(10)/2. That's one of the options. Let me check the options again:A: sqrt(10)/2B: 3/2C: sqrt(2)/2D: 1/2Yes, so A is the right answer. I don't see any mistakes in the steps. Alternatively, maybe there's another way to compute |z| without finding z explicitly? Let me think.We know that for complex numbers, |ab| = |a||b|. So, if (1 + i)z = 1 - 2i, then |(1 + i)z| = |1 - 2i|. Which means |1 + i| |z| = |1 - 2i|. Therefore, |z| = |1 - 2i| / |1 + i|. Maybe this is a faster way?Let's compute |1 + i|. The modulus of 1 + i is sqrt(1^2 + 1^2) = sqrt(2). The modulus of 1 - 2i is sqrt(1^2 + (-2)^2) = sqrt(1 + 4) = sqrt(5). Therefore, |z| = sqrt(5)/sqrt(2) = sqrt(10)/2. Same result. So this confirms that the answer is indeed A.So both methods give the same answer. That gives me more confidence that it's correct. So, yeah, the modulus of z is sqrt(10)/2. The answer is option A.Final Answerboxed{A}</think>To solve for the complex number ( z ) given the equation ((1 + i)z = 1 - 2i), we start by isolating ( z ). We divide both sides by ( 1 + i ):[z = frac{1 - 2i}{1 + i}]Next, we multiply the numerator and the denominator by the complex conjugate of the denominator, ( 1 - i ), to rationalize it:[z = frac{(1 - 2i)(1 - i)}{(1 + i)(1 - i)}]The denominator simplifies as follows:[(1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2]Expanding the numerator:[(1 - 2i)(1 - i) = 1 cdot 1 + 1 cdot (-i) - 2i cdot 1 - 2i cdot (-i) = 1 - i - 2i + 2i^2]Since ( i^2 = -1 ), this simplifies to:[1 - 3i + 2(-1) = 1 - 3i - 2 = -1 - 3i]Thus, the expression for ( z ) becomes:[z = frac{-1 - 3i}{2} = -frac{1}{2} - frac{3}{2}i]To find the modulus ( |z| ), we use the formula ( |a + bi| = sqrt{a^2 + b^2} ):[|z| = sqrt{left(-frac{1}{2}right)^2 + left(-frac{3}{2}right)^2} = sqrt{frac{1}{4} + frac{9}{4}} = sqrt{frac{10}{4}} = sqrt{frac{10}{2}} = frac{sqrt{10}}{2}]Alternatively, using the property of moduli ( |(1 + i)z| = |1 - 2i| ):[|1 + i| cdot |z| = |1 - 2i|]Calculating the moduli:[|1 + i| = sqrt{1^2 + 1^2} = sqrt{2}, quad |1 - 2i| = sqrt{1^2 + (-2)^2} = sqrt{5}]Thus,[|z| = frac{sqrt{5}}{sqrt{2}} = frac{sqrt{10}}{2}]The final answer is (boxed{A}).]

🔑:AnalysisThis question examines the arithmetic operations of complex numbers and the modulus of a complex number.Using the arithmetic operations of complex numbers, we get z=- dfrac{1}{2}- dfrac{3}{2}i, and then we use the modulus of a complex number to obtain the conclusion.AnswerSolve: From (1+i)z=1-2i, we get z= dfrac{1-2i}{1+i}=- dfrac{1}{2}- dfrac{3}{2}i,Therefore, |z|= sqrt{left(- dfrac{1}{2}right)^{2}+left(- dfrac{3}{2}right)^{2}}= dfrac{ sqrt{10}}{2}.Hence, the correct choice is boxed{A}.

🧠:[<think>Okay, so I need to solve this problem about two disks. The larger one is a clock face with a radius of 20 cm, and the smaller disk has a radius of 10 cm. They're externally tangent at the 12 o'clock position initially. The smaller disk has a pointer that starts pointing straight up. Then, the smaller disk rolls clockwise around the larger one, staying tangent the whole time. When the pointer on the smaller disk points up again, we need to find where the point of tangency is on the larger clock face. The options are 2, 3, 4, 6, or 8 o'clock.Hmm, let me visualize this. The big clock is fixed, right? And the small disk is rolling around it without slipping. Since they're externally tangent, the center of the small disk is outside the big clock. The pointer on the small disk starts pointing upward, which is the same direction as the 12 o'clock on the big clock. After rolling around, when the pointer points up again, we need to figure out where the point of contact is.First, maybe I should recall how rolling without slipping works. When a circle rolls around another, the distance it travels (the circumference) relates to the rotation of the moving circle. But in this case, it's rolling externally, so the path the center of the small disk takes is a circle around the center of the big clock.Wait, the radius of the big clock is 20 cm, and the small disk is 10 cm. So the distance between their centers when they're tangent is 20 + 10 = 30 cm. So the center of the small disk moves along a circle of radius 30 cm around the center of the big clock.To make the pointer point upward again, the small disk must have rotated an integer number of full rotations. But since it's rolling around the big clock, the total rotation is a combination of the rotation due to moving around the big clock and its own spinning.Wait, maybe I need to compute the total angle the small disk rotates as it goes around the big clock. The total rotation angle of the small disk is equal to the angle due to its orbit around the big clock plus the angle due to its own spinning. But in the case of rolling without slipping, the rotation of the small disk is related to the distance traveled along the circumference of the big clock.Wait, when a circle rolls externally around another circle without slipping, the total rotation of the rolling circle is given by (circumference of the path / circumference of the rolling circle). But the circumference of the path here is the circumference of the circle that the center of the small disk follows. Since the center is moving along a circle of radius 30 cm, the circumference is 2π*30 = 60π cm. The circumference of the small disk is 2π*10 = 20π cm. So the number of rotations the small disk makes as it goes around once is 60π / 20π = 3. But since it's rolling externally, I think there's an additional factor. Wait, actually, when rolling externally, the direction of rotation is such that the small disk rotates in the opposite direction relative to its orbit. Wait, maybe I need to think about it more carefully.Let me recall that formula for the number of rotations a circle makes when rolling around another. If a circle of radius r rolls around a circle of radius R without slipping, the number of rotations the rolling circle makes is (R ± r)/r, where the sign is positive if it's rolling internally and negative if externally. Wait, is that right?Wait, let's see. If it's rolling externally, the center moves along a circle of radius R + r. The distance the center travels is 2π(R + r). The circumference of the rolling circle is 2πr. So the number of rotations would be (R + r)/r = (R/r) + 1. So for our case, R is 20 cm, r is 10 cm. So number of rotations would be (20 + 10)/10 = 3. So three full rotations. But also, since it's rolling around, the rotation is in the opposite direction. Wait, no. If the small disk is rolling clockwise around the big clock, then the rotation of the small disk itself is also clockwise, right? Because when you roll something, the direction of rotation is the same as the direction of movement. Wait, but when you go around a circle, the center is moving along the circumference, so the rotation of the small disk is both due to its revolution around the big clock and its own spinning.Wait, maybe this is where the +/- comes in. For external rolling, the rolling circle rotates in the same direction as its orbital motion. Wait, actually, no. Let me think. If you have a gear outside another gear, and they are meshing, then the direction of rotation is opposite. But in this case, the small disk is rolling on the outside, like a coin rolling around another coin. If you roll a coin around another coin of the same size, it makes two rotations by the time it gets back. Wait, but in this case, the big clock is twice the radius of the small disk. Wait, radius 20 vs 10. So the circumference ratio is 2:1.Wait, let's take the standard example. If a coin rolls around another coin of the same size without slipping, the moving coin makes two rotations: one due to its own circumference and one due to the path. So in that case, the number of rotations is 1 + 1 = 2. So perhaps in general, when rolling externally around a circle of radius R, a circle of radius r makes (R/r + 1) rotations. Therefore, in our case, R = 20, r = 10, so 20/10 + 1 = 3. So three rotations. Therefore, when the small disk rolls all the way around the big clock back to the starting position, it would have rotated three times. Therefore, if the pointer starts pointing up, after one full orbit, it would have rotated three times, so the pointer would be pointing up again. But the problem says "until the pointer on the smaller disk points vertically upward again". So does that mean when it completes enough of its path such that the rotation brings the pointer back to up?Wait, but maybe it doesn't have to go all the way around. If the small disk rolls such that its own rotation brings the pointer back to vertical, which might happen before completing a full orbit. Hmm, so maybe we need to find the point where the total rotation of the small disk is an integer multiple of 2π, so that the pointer is back to its original orientation.So, the total rotation angle of the small disk is the angle due to its orbital motion plus the angle due to its spinning. Wait, but in rolling without slipping, the rotation angle of the disk is directly related to the distance rolled. So if the small disk rolls along a path, the rotation in radians is equal to the arc length rolled divided by the radius. So if the small disk rolls along a circular path of radius 30 cm (distance between centers), then the total angle it rotates is (arc length along the path) divided by its own radius.But also, the small disk is moving around the big clock, so the angle it has orbited is θ, which would correspond to an arc length on the big clock's circumference. Wait, but the path of the center is a circle of radius 30 cm, so the arc length the center moves is 30θ, where θ is the angle in radians swept from the center of the big clock.But the rotation of the small disk is (arc length rolled)/r = (30θ)/10 = 3θ. So the small disk rotates 3θ radians clockwise due to its orbital motion.But also, as the small disk moves around the big clock, the contact point changes. The question is, when does the pointer on the small disk point upward again? That would happen when the total rotation of the small disk is a multiple of 2π. So 3θ ≡ 0 mod 2π. Therefore, θ = 2π/3 radians. So θ is 120 degrees. Since the big clock is a circle, 120 degrees corresponds to 12 o'clock plus 120 degrees clockwise, which is 4 o'clock. Because each hour on the clock is 30 degrees (360/12). 120 degrees is 120/30 = 4 hours. So starting at 12, moving 4 hours clockwise is 4 o'clock. So the point of tangency is at 4 o'clock. Therefore, answer C.Wait, but let me check that again. The total rotation of the small disk is 3θ, and we set that equal to 2π. Then θ = 2π/3. So the center of the small disk has moved 2π/3 radians around the big clock. Since the big clock's circumference is 2π*20 = 40π cm, but we are measuring the angle at the center, so θ in radians is the angle swept by the center. Therefore, 2π/3 radians is equivalent to 120 degrees. Converting that to the clock position, starting at 12, moving 120 degrees clockwise. Since 360 degrees is 12 hours, each hour is 30 degrees. 120 divided by 30 is 4. So 4 o'clock. That seems right.But wait, let me confirm with another approach. Suppose the small disk rolls around the big one. The circumference it's effectively rolling along is the circumference of the large circle plus the small circle? Wait, no. When rolling externally, the path is a circle with radius R + r, so the circumference is 2π(R + r). For our case, that's 2π(30) = 60π. The small disk's circumference is 2π(10) = 20π. Therefore, the number of rotations the small disk makes is 60π / 20π = 3. So if it goes all the way around, it makes 3 rotations. So when it gets back to the starting point, the pointer would have spun 3 times, so pointing up again. But the problem says "until the pointer points vertically upward again". So that could be after 1 full rotation of the pointer, which might happen before completing the full orbit.Wait, but in my previous reasoning, when the center moves θ radians, the disk rotates 3θ radians. So if we want 3θ = 2π, then θ = 2π/3. Therefore, the center moves 2π/3 radians, which is 120 degrees, so 4 o'clock. But if it goes all the way around, θ = 2π, then 3θ = 6π, which is 3 full rotations, so pointer points up again. So the first time it points up again is at θ = 2π/3, which is 4 o'clock. Therefore, answer C.Alternatively, maybe I need to account for the fact that the rotation of the small disk is a combination of its orbital motion and its spinning. Wait, but in rolling without slipping, the rotation is purely due to the contact with the surface. So the rotation of the disk is entirely determined by the movement along the path. So if the disk is moving along a circular path, the rotation is the arc length divided by radius. So yes, that should be correct.Wait, but maybe there's a different way to think about it. Imagine the small disk rolling around the big one. Each point on the small disk's circumference will trace out a curve called an epicycloid. The number of cusps (points where the point touches the big circle) is equal to the ratio of the radii if it's an integer. In our case, the ratio is 20/10 = 2, so the epicycloid would have 2 cusps. Wait, but epicycloid cusps occur when the rolling circle has made a full rotation relative to the center. Wait, maybe not exactly. Let me recall. For an epicycloid, the parametric equations are:x = (R + r) cos θ - r cos((R + r)/r θ)y = (R + r) sin θ - r sin((R + r)/r θ)The number of cusps is given by the number of times the rolling circle rotates as it goes around once, which is (R + r)/r. So for R = 20, r = 10, that's 30/10 = 3. So three cusps. So the epicycloid would have three cusps, meaning that after going around the big circle once, the small circle has rotated three times. Therefore, when it completes one full orbit, the pointer has rotated three times, thus pointing up again. But the problem is asking for when the pointer points up again, which could be before completing the full orbit. Wait, but the first time it points up again is after one full rotation of the pointer, which would correspond to the small disk rotating 2π radians. So according to the previous formula, total rotation is 3θ = 2π, so θ = 2π/3. Therefore, after moving 2π/3 radians around the big clock, which is 120 degrees, the pointer is back up. Therefore, the point of tangency is at 4 o'clock. So answer C.Alternatively, someone might think that since the small disk has to make a full revolution, but since it's rolling around, the total rotation is 3θ, so θ = 2π/3. So that seems correct.Wait, but another way to check: the length of the path that the small disk rolls over is the arc length on the big clock's circumference. Wait, no. Actually, when rolling externally, the contact point is moving along the circumference of the big clock. But the distance rolled on the big clock's circumference would be equal to the arc length on the big clock. However, the small disk's rotation is determined by the distance it rolls. So if the small disk rolls a distance s along the big clock's circumference, then the rotation of the small disk is s / r_small, where r_small is 10 cm. But when rolling externally, the center of the small disk is moving along a circle of radius R + r_small = 30 cm. Therefore, the relationship between the angle θ (in radians) that the center has moved around the big clock and the arc length rolled on the big clock's circumference is a bit different.Wait, actually, when the small disk rolls around the big clock, the contact point moves along the big clock's circumference. The distance rolled on the big clock is equal to the arc length along the big clock's circumference. However, the rotation of the small disk is determined by the arc length it rolls divided by its own circumference. So if the small disk rolls a distance s along the big clock, then the rotation in radians is s / r_small.But the arc length s on the big clock is equal to R * θ, where θ is the angle in radians that the contact point has moved around the big clock. So s = 20 * θ. Therefore, the rotation of the small disk is (20 * θ) / 10 = 2θ. Wait, but this contradicts my previous result. Hmm, now I'm confused.Wait, perhaps the problem is whether the distance rolled is the same as the arc length on the big clock. Let me think. If the small disk is rolling on the big clock, then the point of contact moves along the big clock's circumference. So the distance the small disk has rolled is equal to the arc length on the big clock. Therefore, the rotation of the small disk is s / (2πr_small) * 2π = s / r_small. So in radians, rotation angle is s / r_small.But s, the arc length on the big clock, is R * θ, where θ is the angle moved around the big clock. Therefore, rotation angle of small disk is (R * θ) / r_small = (20 / 10) * θ = 2θ.But earlier, when considering the center moving along a circle of radius R + r_small = 30 cm, the rotation angle was 3θ. Which one is correct?Wait, this is a conflict. So which approach is right? Let's analyze both.First approach: Center moves along a circle of radius 30 cm. The distance the center moves is 30θ, where θ is the angle in radians. The rotation of the small disk is (distance moved by center) / r_small = 30θ / 10 = 3θ.Second approach: The arc length rolled on the big clock is 20θ, so rotation of small disk is 20θ / 10 = 2θ.These two approaches give conflicting results. So which is correct?Wait, this is a classic problem in rolling without slipping. The key is to realize that when a circle rolls around another circle, the total rotation of the rolling circle is a combination of its orbital rotation and its spinning rotation.Wait, let me look up the formula for the number of rotations when rolling around another circle. According to some sources, when a circle of radius r rolls without slipping around a circle of radius R, the number of rotations the rolling circle makes is (R + r)/r for external rolling. So in our case, (20 + 10)/10 = 3. So three full rotations when going around once. Therefore, if the small disk rolls all the way around the big clock (which is 360 degrees), it makes three rotations. Therefore, the first time it points up again is after one third of the orbit, because 3 rotations would bring it back. Wait, but 1/3 of 360 degrees is 120 degrees, which is 4 o'clock. Therefore, answer C.But why the discrepancy between the two approaches?Ah, perhaps the first approach (center moving) is considering the rotation due to both the orbit and the spin, whereas the second approach (arc length rolled) is only considering the spin. Wait, no. When you roll without slipping, the rotation of the small disk is purely due to the movement along the path. So the rotation is equal to the arc length rolled divided by the circumference. But when moving around the circle, the center is moving along a circular path, so there is both the rotation of the center around the big clock (orbital motion) and the rotation of the small disk around its own center (spin). However, in the case of rolling without slipping, the spin rotation is directly related to the orbital motion.Wait, perhaps the key is that when rolling externally, the direction of rotation is such that the total number of rotations is (R + r)/r. Therefore, in this case, three rotations. So after going around 360 degrees, the small disk has rotated three times. Therefore, the first time the pointer points up again is after one full rotation of the small disk, which occurs after 1/3 of the full orbit, i.e., 120 degrees. Therefore, the point of tangency is at 4 o'clock. So answer C.Alternatively, let's parameterize the motion. Let's say the small disk starts at the top (12 o'clock). As it rolls clockwise, the point of contact moves clockwise around the big clock. The center of the small disk moves along a circle of radius 30 cm. Let θ be the angle that the center has moved from the top. So θ is the angle between the vertical (12 o'clock) and the line connecting the centers of the two disks.As the center moves θ radians around the big clock, the small disk must rotate in order to maintain contact without slipping. The rotation of the small disk is such that the arc length it has rolled over the big disk equals the arc length along its own circumference. The arc length on the big disk is R * θ = 20θ. The arc length on the small disk is r * φ = 10φ, where φ is the angle the small disk has rotated. Since these must be equal (no slipping), 20θ = 10φ ⇒ φ = 2θ. So the small disk rotates 2θ radians clockwise.But wait, in this case, the rotation of the small disk is 2θ. However, from the center's perspective, moving θ radians around the big clock, the orientation of the small disk is also affected by the orbital motion. Wait, that is, the small disk is both moving around the big clock and rotating. So the total rotation of the pointer is the sum of the orbital rotation and the spinning rotation?Wait, no. The pointer is fixed to the small disk, so its orientation is determined solely by the rotation of the small disk around its own center. However, as the small disk moves around the big clock, the direction of "up" for the pointer changes relative to the big clock. Wait, maybe not. Let me clarify.The pointer is fixed on the small disk. Initially, the pointer points upward (12 o'clock on the big clock). As the small disk rolls, the pointer rotates with the small disk. To have the pointer point upward again (in the same absolute direction as initially, i.e., 12 o'clock on the big clock), the small disk must have rotated an integer multiple of 2π radians. However, due to the orbital motion, the "up" direction for the pointer is also changing. Wait, no. The pointer is fixed relative to the small disk. If the small disk rotates, the pointer rotates with it. The problem says "points vertically upward again". So does "vertically upward" refer to the same absolute direction (12 o'clock on the big clock), or relative to the small disk's position?The problem states: "the pointer on the smaller disk points vertically upward". Initially, this is at 12 o'clock. After rolling, we need the pointer to point "vertically upward again". So I think this is in the absolute sense, meaning pointing towards 12 o'clock on the big clock. Therefore, the pointer must have rotated such that it's aligned with the upward direction again. But since the small disk is moving around the big clock, the "up" direction for the pointer (relative to the big clock) changes as the small disk orbits.Wait, this is critical. If "vertically upward" is relative to the big clock's 12 o'clock, then the pointer must point in that fixed direction. Therefore, the small disk must have rotated such that the pointer is pointing towards 12 o'clock on the big clock again. However, as the small disk moves around the big clock, the direction from the small disk's center to 12 o'clock is changing. Therefore, the pointer's orientation is a combination of the disk's rotation and its orbital position.Alternatively, maybe the pointer is fixed on the small disk, and "vertically upward" is relative to the small disk's local vertical. Wait, that is, the pointer is fixed relative to the small disk's own orientation. Initially, the small disk is at 12 o'clock, and the pointer points upward (toward 12 o'clock on the big clock). As the small disk rolls, it rotates, so the pointer's direction relative to the big clock changes. But the problem says "points vertically upward again". If "vertically upward" is relative to the small disk's local vertical (i.e., pointing away from the point of contact), then the pointer is always pointing vertically upward in its own frame. But the problem mentions "vertically upward" which is probably in the absolute sense, i.e., the same direction as the original 12 o'clock.Wait, the problem says "the pointer on the smaller disk points vertically upward". Since the smaller disk is externally tangent at 12 o'clock initially, the pointer points upward (same as 12 o'clock). After rolling, the pointer points upward again, so it must again align with the absolute upward direction. So to achieve this, the small disk must have rotated such that the pointer has completed full rotations, considering both its own spinning and its orbital motion.Wait, perhaps the key is that the total rotation of the pointer relative to the fixed stars (i.e., the absolute rotation) is the sum of the rotation due to its orbit around the big clock and its own spinning. But since the small disk is rolling without slipping, its own spinning is related to its orbital motion.Wait, let's think in terms of the rotation of the pointer. The pointer's absolute rotation is the sum of the rotation of the small disk around the big clock and the rotation of the small disk around its own center.When the small disk moves θ radians around the big clock, the center of the small disk has moved θ radians. The small disk itself has rotated φ radians clockwise (due to rolling without slipping). The total rotation of the pointer is the combination of these two rotations.But how do these rotations combine? If the small disk is orbiting the big clock by θ radians clockwise, then the pointer has rotated θ radians clockwise due to the orbital motion. Additionally, the small disk has rotated φ radians clockwise due to its own spin. So the total rotation of the pointer relative to the fixed background is θ + φ radians clockwise.But we need the total rotation to be an integer multiple of 2π radians for the pointer to point upward again. So θ + φ = 2π * k, where k is an integer.But from the rolling without slipping condition, we have φ = (arc length rolled) / r_small. The arc length rolled is equal to the arc length on the big clock, which is R * θ. So φ = (R * θ) / r_small = (20θ)/10 = 2θ.Therefore, θ + 2θ = 3θ = 2π * k.So 3θ = 2π * k ⇒ θ = (2π/3) * k.We need the smallest positive θ where this occurs, which is k = 1 ⇒ θ = 2π/3 radians, which is 120 degrees. Converting that to the clock position, starting from 12 o'clock, moving 120 degrees clockwise, each hour is 30 degrees (since 360/12 = 30). So 120/30 = 4 hours. So 4 o'clock. Therefore, the answer is C) 4.Yes, this makes sense. The key was realizing that the total rotation of the pointer is the sum of the orbital rotation (θ) and the spinning rotation (2θ), leading to 3θ. Setting that equal to 2π gives θ = 2π/3, which is 120 degrees or 4 o'clock.Therefore, the correct answer is C) 4.Final Answerboxed{C}</think>The problem involves a larger clock face with a radius of 20 cm and a smaller disk with a radius of 10 cm that is externally tangent to the clock face at the 12 o'clock position. The smaller disk has a fixed pointer initially pointing vertically upward. We need to determine the point of tangency on the larger clock face when the pointer on the smaller disk points vertically upward again after rolling clockwise without slipping.1. Key Concepts: - The smaller disk rolls without slipping around the larger clock face. - The total rotation of the smaller disk is a combination of its orbital motion around the larger clock and its own spinning. - The distance between the centers of the two disks is 30 cm (20 cm + 10 cm).2. Rotations Analysis: - The circumference of the path traced by the center of the smaller disk is (2pi times 30 = 60pi) cm. - The circumference of the smaller disk is (2pi times 10 = 20pi) cm. - The number of rotations the smaller disk makes while rolling around the larger clock is (frac{60pi}{20pi} = 3). However, this is when it completes a full orbit.3. Total Rotation: - The total rotation of the pointer is the sum of the orbital rotation and the spinning rotation. - Let (theta) be the angle (in radians) the center of the smaller disk moves around the larger clock. The spinning rotation of the smaller disk is (2theta) (due to rolling without slipping). - The total rotation of the pointer is (theta + 2theta = 3theta). - For the pointer to point vertically upward again, (3theta) must be an integer multiple of (2pi). The smallest positive (theta) is (frac{2pi}{3}).4. Conversion to Clock Position: - (frac{2pi}{3}) radians is equivalent to 120 degrees. - Each hour on the clock corresponds to 30 degrees (since (360^circ / 12 = 30^circ)). - Converting 120 degrees to hours: (120^circ / 30^circ = 4).Thus, the point of tangency is at the 4 o'clock position on the larger clock face. The final answer is (boxed{C}).,<think>Okay, let me try to figure out this problem step by step. So, there's a big clock with a radius of 20 cm, and a smaller disk with radius 10 cm that's externally tangent to the big clock at the 12 o'clock position. The smaller disk has a pointer that starts pointing straight up. Then, the smaller disk rolls clockwise around the big clock, staying tangent to it. We need to find where on the big clock the point of tangency is when the pointer on the smaller disk is pointing up again. The options are 2, 3, 4, 6, or 8 o'clock.First, I need to visualize this. The big clock is twice as big as the small disk. When the small disk rolls around the big one without slipping, the path it takes is called a circle around the big clock. But since it's rolling externally, the center of the small disk has to move along a circle that's outside the big clock. The radius of that path should be the sum of both radii, right? So the big clock is 20 cm, the small is 10 cm, so the center of the small disk moves along a circle of radius 20 + 10 = 30 cm.Now, the problem says the small disk rolls until the pointer points upward again. So, we need to figure out how far the small disk has rolled around the big clock such that the small disk itself has made a full rotation. Because when it makes a full rotation, the pointer will be pointing up again. But since it's rolling around the big clock, the rotation of the small disk is a combination of its movement around the big clock and its own spinning.Wait, this is similar to the idea of a coin rolling around another coin. If you roll a coin around another of the same size, it makes one full rotation by the time it gets back to the starting point. But in this case, the coins are different sizes. There's a formula for the number of rotations a coin makes when rolling around another. The formula is (circumference of the path / circumference of the rolling circle) minus 1 if it's external. Wait, I need to recall that.Actually, when a circle rolls around another without slipping, the number of rotations the rolling circle makes is equal to the ratio of the circumferences plus or minus 1 depending on internal or external. For external, it's (R + r)/r, so (20 + 10)/10 = 3. So, the small disk would make 3 rotations as it goes around the big clock once. But in this problem, we don't need to go all the way around; we just need to go until the pointer is pointing up again. Since the pointer is fixed on the small disk, pointing up again would mean the small disk has rotated an integer number of full rotations. So, the first time the pointer points up again is after 1 full rotation of the small disk. But as the small disk is rolling around, its rotation is due to both the translation around the big clock and its own spinning.Wait, this might be confusing. Let's think about the total rotation. When the small disk rolls without slipping, the distance it rolls is equal to the arc length covered. So, if the small disk rolls along a path, the rotation angle in radians is equal to the arc length divided by the radius of the small disk. So, if the center of the small disk moves along a circle of radius 30 cm (20 + 10), the circumference of that path is 2π*30 = 60π cm. If the small disk were to roll that entire distance, it would rotate 60π / (2π*10) = 60π / 20π = 3 times. So, 3 full rotations. But here, we need the small disk to rotate just once, so that the pointer points up again. So, how far does it need to roll for the small disk to make one full rotation?If 60π cm of rolling causes 3 rotations, then each rotation corresponds to 60π / 3 = 20π cm of rolling. But the circumference of the small disk is 20π cm (since circumference is 2πr = 2π*10 = 20π). Wait, that seems redundant. Wait, if the small disk rolls without slipping, then the rotation is directly related to the distance rolled. So, for each circumference length rolled, the disk makes one full rotation. So, if the disk rolls 20π cm, it makes one full rotation. But the path circumference is 60π cm. So, to make one full rotation, the disk needs to roll 20π cm along the path.But we need to find where on the big clock the point of tangency is when the small disk has rolled such that it has rotated once. So, the center of the small disk would have moved 20π cm along the path. The angle that corresponds to on the path is the arc length divided by the radius of the path. Wait, the path's radius is 30 cm, so the angle θ in radians is arc length / radius. So θ = 20π / 30 = (2/3)π radians. Converting that to degrees, (2/3)π * (180/π) = 120 degrees. Since the center is moving clockwise around the big clock, starting at 12 o'clock, moving 120 degrees clockwise would bring it to what position?Starting at 12 o'clock, moving 120 degrees clockwise. A full circle is 360 degrees, which corresponds to 12 hours, so each hour is 30 degrees (360/12). So 120 degrees is 120/30 = 4 hours. So starting at 12, moving 4 hours clockwise would be 4 o'clock. But wait, the point of tangency. Wait, the center of the small disk is moving along a circle of radius 30 cm. The point of tangency is where the two circles touch. Initially, at 12 o'clock. When the center moves to a new position, the point of tangency is the point on the big clock that is closest to the center of the small disk. Wait, the center of the small disk is 30 cm from the center of the big clock. The radius of the big clock is 20 cm, so the point of tangency is 20 cm from the center of the big clock towards the center of the small disk.Alternatively, the point of tangency can be thought of as a point on the circumference of the big clock, and when the small disk rolls, this point moves around the big clock. Wait, maybe another way: If the center of the small disk is at angle θ from the original position (12 o'clock), then the point of tangency is at angle θ on the big clock. Wait, let me think.Suppose the center of the big clock is at point O, and the center of the small disk is at point C. Initially, at 12 o'clock, point C is 30 cm above O. The point of tangency T is 20 cm from O towards C. So, T is at 12 o'clock. When the small disk rolls, point C moves along the circle of radius 30 cm. The point of tangency T is always the point on the big clock that is in the direction of point C from O. So, if point C has moved θ degrees around the big clock, then the point of tangency T is also at θ degrees on the big clock.Therefore, if the center moves 120 degrees around the path, the point of tangency is at 120 degrees from 12 o'clock, which is 4 o'clock. Therefore, the answer is C) 4.But wait, let me verify this again. Because when a circle rolls around another, the number of rotations is (R + r)/r for external. So, here R = 20, r = 10, so (20 + 10)/10 = 3. So, if the small disk goes all the way around the big one, it makes 3 rotations. Therefore, to make 1 full rotation, it needs to go 1/3 of the way around. 1/3 of 360 degrees is 120 degrees, which is 4 o'clock. That seems to check out.But let me think again. Suppose the small disk rolls without slipping. The rotation of the small disk is due to both its revolution around the big clock and its own spinning. The total rotation is the sum of these two. Wait, no. In rolling without slipping, the rotation of the small disk is purely due to the translation. The contact point is instantaneously at rest relative to the big clock. So, the rotation of the small disk is such that the point in contact has no relative motion. Therefore, the angular speed of the small disk is related to the tangential speed of its center.Let me try to model this. Let the center of the big clock be O, the center of the small disk be C, and the point of tangency be T. As the small disk rolls clockwise around the big clock, the center C moves along a circle of radius 30 cm. The angular speed of C around O is ω. Then, the tangential speed of C is v = ω * 30 cm.Since the small disk is rolling without slipping, the tangential speed at the point of contact T must be equal to the tangential speed of the small disk's edge. The small disk's angular speed around its own center C is ω_s. The tangential speed at the edge of the small disk is ω_s * 10 cm. But since the point T is instantaneously at rest relative to the big clock, the speed of point T on the small disk must be zero. Wait, no. Wait, in external rolling, the point of tangency is moving relative to both disks. Wait, maybe I need to clarify.When two circles are externally tangent and rolling without slipping, the direction of rotation of the small disk is opposite to the direction of its revolution. So, if the small disk is rolling clockwise around the big clock, it must be rotating counterclockwise. But wait, actually, no. If you have a gear outside another gear, the rotation directions are opposite. But in this case, the small disk is rolling on the outside, like a coin rolling around another. Wait, actually, when a coin rolls around another without slipping, the direction of rotation is such that if it rolls clockwise around the fixed coin, it rotates counterclockwise. But how does this affect the total rotation?Alternatively, the total rotation of the small disk is a combination of its orbital motion around the big clock and its own spin. The total number of rotations is given by the ratio of the path circumference to the disk circumference. But in this case, since it's external, it's (R + r)/r, so 3. So, when it completes the full circle, it has made 3 rotations. Therefore, when it gets back to the starting point, the pointer would have rotated 3 times. Therefore, to have the pointer point up again, it could be after 1 full rotation, 2, 3, etc. So, the first time it points up again is after 1 full rotation of the small disk, which corresponds to 1/3 of the full path around the big clock. So, 1/3 of 360 degrees is 120 degrees, which is 4 o'clock.Alternatively, if the small disk rolls without slipping, the rotation angle of the small disk is equal to the angle swept by the center moving around the big clock plus the angle due to its own spinning. Wait, maybe that's the wrong way to think. Let me try another approach.Imagine the small disk rolling around the big one. The center of the small disk moves along a circle of radius 30 cm. The circumference of this path is 2π*30 = 60π cm. The circumference of the small disk is 2π*10 = 20π cm. So, if the small disk were to roll that distance on flat ground, it would rotate 60π / 20π = 3 times. However, since it's rolling around the big disk, the direction of rotation is such that as it goes around, each point on the small disk's circumference touches the big disk once per orbit. Wait, but this might not be the same as rolling on flat ground. Wait, in the case of rolling around another circle externally, the number of rotations is indeed (R + r)/r. So here, 3. So, three rotations when going around once. Therefore, to get the pointer back to its original position, the small disk needs to make an integer number of rotations. The first time is after 1 full rotation, which would correspond to 1/3 of the full path. Therefore, moving 120 degrees around the big clock. Therefore, 120 degrees corresponds to 4 o'clock.So, the answer should be C) 4.Wait, but let me double-check. Let's imagine the small disk rolling 120 degrees around the big clock. The point of tangency moves from 12 o'clock to 4 o'clock. But also, the small disk has rotated once. So, the pointer, which was pointing up, after the small disk rotates once, points up again. So, yes, when the small disk has rolled 120 degrees around the big clock, the point of tangency is at 4 o'clock, and the pointer is back to pointing up. That seems to make sense.Alternatively, if we consider the rotation of the small disk. If the center moves 120 degrees around the big clock, that's an arc length on the path of (120/360)*60π = 20π cm. The small disk, with circumference 20π cm, would have rolled exactly once. Therefore, the rotation is 1 full rotation, so the pointer is back to pointing up. Therefore, the point of tangency is at 4 o'clock. Therefore, the answer is C) 4.I think that's correct. Let me see if any of the other answer choices make sense. If someone thought it was 3 o'clock, maybe they thought 90 degrees, but 90 degrees is 3 hours, but according to our calculation, it's 120 degrees, which is 4 hours. So 4 o'clock is the right answer. So, option C.Final Answerboxed{C}</think>The radius of the larger clock face is 20 cm, and the radius of the smaller disk is 10 cm. The smaller disk is externally tangent to the larger clock face at the 12 o'clock position initially. We need to determine the point of tangency when the smaller disk rolls clockwise around the larger clock face until its pointer points vertically upward again.1. Path of the Center of the Smaller Disk: The center of the smaller disk moves along a circle with a radius of 30 cm (sum of the radii of the larger clock and the smaller disk).2. Circumference Calculation: The circumference of the path of the center of the smaller disk is (2pi times 30 = 60pi) cm. The circumference of the smaller disk is (2pi times 10 = 20pi) cm.3. Number of Rotations: When the smaller disk rolls without slipping, the number of rotations it makes is given by the ratio of the path circumference to the smaller disk's circumference. For the pointer to point upward again, the smaller disk must complete one full rotation. This requires rolling a distance equal to its circumference, (20pi) cm.4. Angle Calculation: The angle corresponding to an arc length of (20pi) cm on the path of radius 30 cm is (theta = frac{20pi}{30} = frac{2pi}{3}) radians. Converting this to degrees, (theta = frac{2pi}{3} times frac{180}{pi} = 120^circ).5. Position on the Clock: Starting from the 12 o'clock position, moving 120 degrees clockwise corresponds to (frac{120}{30} = 4) hours, which is the 4 o'clock position.Thus, the point of tangency between the two disks is located at the (boxed{C}) o'clock position.]

🔑:14. C.As shown in Figure 3.Let the angular displacement of the large circle be theta. Then the angular displacement of the small circle is 360^{circ}-theta.Thus, 20 theta=10left(360^{circ}-thetaright) Rightarrow theta=120^{circ}.Therefore, the two disks are externally tangent at the 4 o'clock position on the large dial.

🧠:[<think>Okay, let's tackle part a) first. The problem is asking how many digits were used to paint the numbers on the first hundred lockers. Each locker is numbered with the square of its position, so the first locker is 1²=1, the second is 2²=4, up to the 100th locker, which is 100²=10,000. We need to count the total number of digits in all these squared numbers.First, I need to figure out how many digits each of these squares has. Let's break it down by the number of digits:1. Single-digit numbers (1² to 9²): These squares will be from 1 (1²) up to 81 (9²). Let's confirm: 1² = 1 (1 digit) 2² = 4 (1 digit) ... 9² = 81 (2 digits). Wait, hold on. 9² is 81, which is actually a two-digit number. So maybe my initial thought is wrong here. Let me check each number:Wait, actually, 1²=1 (1 digit)2²=4 (1 digit)3²=9 (1 digit)4²=16 (2 digits)5²=25 (2 digits)...9²=81 (2 digits). Hmm, so actually, the squares from 1² to 3² are single-digit numbers, and starting from 4² up, they become two-digit numbers. So that's an error in my initial breakdown. Let me correct that:- Numbers from 1 to 3 (1² to 3²): 1, 4, 9. Each of these is a single-digit number. So 3 numbers with 1 digit each. That's 3*1=3 digits.- Numbers from 4 to 9 (4² to 9²): 16, 25, 36, 49, 64, 81. Each of these is a two-digit number. There are 6 numbers here (4 to 9 inclusive is 6 numbers). So 6*2=12 digits.Wait, but 4 to 9 is 6 numbers, correct. So 3 single-digit squares and 6 two-digit squares in the first 9 lockers. Then locker 10 is 10²=100, which is a three-digit number. Wait, but we're going up to the first hundred lockers, which is 100²=10,000. So I need to consider all squares from 1² to 100².Wait, actually, the lockers are numbered with the squares of the first 2,019 numbers, but part a) is only about the first hundred lockers, so numbers 1² to 100².So let me approach this systematically by ranges where the number of digits in the square changes.First, determine for numbers n from 1 to 100, when does n² transition from 1-digit, 2-digits, 3-digits, 4-digits, and 5-digits.Wait, the maximum is 100²=10,000, which is a 5-digit number. So possible digits in squares here are 1, 2, 3, 4, or 5 digits.But let's find the ranges:1-digit squares: n² < 10. So n < sqrt(10) ≈ 3.16. So n=1,2,3. Their squares are 1,4,9 (1 digit each).2-digit squares: n² >=10 and n² <100. So n >=4 and n <10. Because 10²=100. So n=4 to 9.3-digit squares: n² >=100 and n² <1000. So n >=10 and n < sqrt(1000). sqrt(1000) is approximately 31.62, so n=10 to 31.4-digit squares: n² >=1000 and n² <10000. So n >=32 (since 32²=1024) up to n=99 (since 99²=9801).5-digit squares: n² >=10000, which is n=100, since 100²=10,000.So breaking down the first 100 lockers:1. n=1 to 3 (1,2,3): 1-digit squares. 3 numbers, 1 digit each. Total digits: 3*1=3.2. n=4 to 9 (4,5,6,7,8,9): 2-digit squares. 6 numbers, 2 digits each. Total digits: 6*2=12.3. n=10 to 31: 3-digit squares. From 10 to 31 inclusive is 22 numbers (31-10+1=22). Each square is 3 digits. Total digits: 22*3=66.4. n=32 to 99: 4-digit squares. From 32 to 99 inclusive is 68 numbers (99-32+1=68). Each square is 4 digits. Total digits: 68*4=272.5. n=100: 100²=10,000. 5 digits. Just 1 number. Total digits: 1*5=5.Now sum them up: 3 + 12 + 66 + 272 + 5.Let's compute:3 + 12 = 1515 + 66 = 8181 + 272 = 353353 + 5 = 358.Wait, so total digits used would be 358? Let me check my calculations again to make sure.First segment: n=1-3: 3 lockers, 1 digit each: 3*1=3. Correct.Second segment: n=4-9: 6 lockers, 2 digits each: 6*2=12. Correct.Third segment: n=10-31: 22 numbers (31-10+1=22). Each has 3 digits: 22*3=66. Correct.Fourth segment: n=32-99: 99-32+1=68 numbers. Each 4 digits: 68*4=272. Correct.Fifth segment: n=100: 1 locker, 5 digits: 5. Correct.Total digits: 3 +12=15; 15+66=81; 81+272=353; 353+5=358.Hmm. So the answer for part a) is 358 digits.But wait, let me verify with a specific example. Let's take n=10: 10²=100 (3 digits). n=31: 31²=961 (3 digits). Correct. Then n=32: 32²=1024 (4 digits). n=99: 99²=9801 (4 digits). Correct. Then n=100: 100²=10,000 (5 digits). Correct.So the calculation seems right. So 358 digits. Therefore, the answer to part a) is 358.Now part b): Adding all the numbers on the lockers, what is the units digit of this result?We need to find the units digit of the sum S = 1² + 2² + 3² + ... + 2019².But wait, the problem says "Adding all the numbers on the lockers", which are the squares of the first 2019 positive integers. So the sum is 1² + 2² + ... + 2019². The question is, what is the units digit of this sum.To find the units digit of the sum, we can consider that the units digit of the sum is equal to the sum of the units digits of each term, modulo 10. So, if we compute the sum of the units digits of each square from 1² to 2019², then take that sum modulo 10, we'll get the units digit of the total sum.So, first, we need to find the pattern of the units digits of squares. Let's recall that the units digit of a square number depends only on the units digit of the original number being squared.For example:0² = 0 --> units digit 01² = 1 --> 12² = 4 --> 43² = 9 --> 94² = 16 --> 65² = 25 --> 56² = 36 --> 67² = 49 --> 98² = 64 --> 49² = 81 --> 1Therefore, the units digits of squares cycle in a pattern based on the units digit of n. The possible units digits for squares are 0,1,4,5,6,9, and the cycle repeats every 10 numbers. So for each block of 10 consecutive numbers, their squares will have units digits following the pattern [0,1,4,9,6,5,6,9,4,1].So, for numbers 1 to 10:1²:1, 2²:4, 3²:9, 4²:6, 5²:5, 6²:6, 7²:9, 8²:4, 9²:1, 10²:0.Adding these units digits: 1+4+9+6+5+6+9+4+1+0 = let's compute step by step:1+4=5; 5+9=14; 14+6=20; 20+5=25; 25+6=31; 31+9=40; 40+4=44; 44+1=45; 45+0=45. So sum per 10 numbers is 45.But wait, each block of 10 numbers contributes 45 to the total sum of units digits. Then, since the units digits cycle every 10 numbers, we can compute how many complete blocks of 10 there are in 2019 numbers, and then the remainder.2019 divided by 10 is 201 with a remainder of 9. So 201 complete blocks and 9 extra numbers.Therefore, total sum of units digits is (201 * 45) + sum of units digits of the first 9 squares.First, compute 201 * 45:200 *45 = 90001*45 =45Total: 9000 +45=9045Then sum the first 9 squares' units digits:From 1² to 9²:1,4,9,6,5,6,9,4,1Sum: 1+4=5; 5+9=14; 14+6=20; 20+5=25; 25+6=31; 31+9=40; 40+4=44; 44+1=45.Wait, the first 9 squares' units digits sum to 45? But the first 10 sum to 45 as well. Wait, that's because the 10th is 0, so the first 9 sum to 45. Wait, but in the first 10, 1-10, the sum is 45. If we take the first 9, 1-9, the sum is 45 as well? Let's check:1 (1²) +4 (2²) +9 (3²) +6 (4²) +5 (5²) +6 (6²) +9 (7²) +4 (8²) +1 (9²) = 1+4=5; +9=14; +6=20; +5=25; +6=31; +9=40; +4=44; +1=45. Yes, the sum is 45. So the first 9 numbers also sum to 45. Wait, that's interesting. So each block of 10 numbers contributes 45, and the first 9 numbers (the remainder) also contribute 45. Therefore, the total sum of units digits is (201 blocks *45) + (remainder 9 numbers *45) = 201*45 + 45 = 202*45.But 2019 divided by 10 is 201.9, which is 201 full blocks and 9 extra numbers. But since the first 9 numbers contribute 45, same as a full block. Therefore, total sum of units digits is 202 *45.But let me confirm. Wait, if you have 201 blocks of 10 numbers (each contributing 45), that's 201*45. Then the remaining 9 numbers (from 2011 to 2019) will be 2011² to 2019². The units digits of these numbers depend on their own units digits. Wait, but 2011 has units digit 1, 2012 has 2, ..., 2019 has 9. Therefore, the units digits of these squares will be the same as 1², 2², ...,9². Therefore, the units digits of the remaining 9 numbers are the same as the first 9 numbers, so their sum is also 45. Therefore, total sum is 201*45 +45= (201+1)*45=202*45.Now compute 202 *45. Let's compute:200*45=90002*45=90Total:9000+90=9090.Therefore, the total sum of units digits is 9090. Then the units digit of the total sum is 9090 mod 10, which is 0. Wait, because 9090 ends with a 0. Therefore, the units digit is 0.Wait, but let me check again. Wait, the sum of the units digits is 9090, which is 9090. So when we add up all the numbers, the total sum's units digit is the same as the sum of all the units digits mod 10. Since 9090 mod10 is 0, the answer is 0.But let me verify this with a different approach. Alternatively, instead of summing all units digits, we can compute the sum of squares modulo 10.We need to compute S = sum_{k=1}^{2019} k² mod 10.There's a formula for the sum of squares: sum_{k=1}^n k² = n(n+1)(2n+1)/6.Therefore, S = 2019*2020*4039 /6. But since we need this modulo 10, let's compute each term modulo 10 and then multiply.First, compute each term modulo 10:2019 mod10=92020 mod10=04039 mod10=9So, S ≡ (9 * 0 *9)/6 mod10.But 9*0*9=0, so 0/6 mod10=0. However, division in modular arithmetic can be tricky. Wait, but since 0 divided by 6 is 0, which is 0 mod10.Therefore, the sum S mod10 is 0. So the units digit is 0. This corroborates the previous result.Therefore, the answer to part b) is 0.But let me check this formula approach again. The formula is n(n+1)(2n+1)/6. Let's compute each part modulo 10:n=2019.n mod10=9(n+1) mod10=2019+1=2020 mod10=0(2n+1) mod10=2*2019 +1=4038 +1=4039 mod10=9So, the product n(n+1)(2n+1) mod10=9*0*9=0.Therefore, 0 divided by 6. But 0 divided by any number is 0, so 0/6=0 mod10. Hence, yes, 0.Therefore, both methods give the same result. So the units digit is 0.Therefore, part b) answer is 0.Final Answera) boxed{358}b) boxed{0}</think>a) To determine the number of digits used to paint the numbers on the first hundred lockers, we analyze the number of digits in each square number from (1^2) to (100^2):1. 1-digit numbers (1² to 3²): 1, 4, 9. Total digits: (3 times 1 = 3).2. 2-digit numbers (4² to 9²): 16, 25, 36, 49, 64, 81. Total digits: (6 times 2 = 12).3. 3-digit numbers (10² to 31²): There are 22 numbers (10 to 31). Total digits: (22 times 3 = 66).4. 4-digit numbers (32² to 99²): There are 68 numbers (32 to 99). Total digits: (68 times 4 = 272).5. 5-digit number (100²): 10,000. Total digits: (1 times 5 = 5).Summing these digits: (3 + 12 + 66 + 272 + 5 = 358).Thus, the number of digits used to paint the first hundred lockers is (boxed{358}).b) To find the units digit of the sum of all numbers on the lockers (i.e., the sum of the squares of the first 2019 positive integers), we note that the units digit of a square number depends only on the units digit of the original number. The units digits of squares cycle every 10 numbers with the pattern [0, 1, 4, 9, 6, 5, 6, 9, 4, 1], summing to 45 per cycle.For 2019 numbers:- There are (201) complete cycles of 10 numbers, contributing (201 times 45 = 9045).- The remaining 9 numbers contribute the same units digits as the first 9 squares, summing to (45).Total sum of units digits: (9045 + 45 = 9090). The units digit of 9090 is 0.Thus, the units digit of the total sum is (boxed{0}).,<think>Okay, so I've got these two math problems here about lockers numbered with squares of numbers. Let's start with part a: figuring out how many digits were used to paint the numbers on the first hundred lockers. Hmm. Each locker is numbered with the square of its position, right? So locker number one is 1²=1, locker two is 2²=4, up to locker 100, which is 100²=10,000. Wait, so the numbers on the lockers are 1, 4, 9, 16, 25, ..., 10,000. I need to count how many digits are used in total for all these numbers.First, maybe I should break it down by how many digits each square has. For example, numbers from 1² to 9² (which is 81) are all one or two-digit numbers. Let's check:1² = 1 (1 digit)2² = 4 (1 digit)...9² = 81 (2 digits)Wait, actually, starting from 1² up to 9², the squares go from 1 to 81. But 1² to 3² (1, 4, 9) are single-digit numbers. Then starting from 4²=16 up to 9²=81, those are two-digit numbers. Wait, no: 4² is 16, which is two digits. So 1²=1 (1 digit), 2²=4 (1 digit), 3²=9 (1 digit), 4²=16 (2 digits), 5²=25 (2 digits), ..., 9²=81 (2 digits). So actually, the first three squares are single-digit, and the next six (4² to 9²) are two-digit numbers. Wait, 1 to 3 is three numbers, 4 to 9 is six numbers. So from locker 1 to 3, they have 1 digit each, lockers 4 to 9 have two digits each.Then locker 10 is 10²=100 (3 digits). So from locker 10 onward, the squares will be three-digit numbers? Wait, let's see: 10²=100 (3 digits), 11²=121 (3 digits), ..., 31²=961 (3 digits), 32²=1024 (4 digits). Wait, so actually, squares from 10²=100 up to 31²=961 are three-digit numbers, and 32²=1024 is the first four-digit square. But in our case, since we're only going up to locker 100, which is 100²=10,000 (5 digits). Wait, so we need to figure out how the number of digits increases as the square numbers get bigger.So let's categorize the lockers based on the number of digits their squares have:1. 1-digit squares: These are squares of numbers from 1 to 3, since 3²=9 (1 digit), and 4²=16 (2 digits). So lockers 1-3 have 1-digit numbers.2. 2-digit squares: From 4²=16 up to 9²=81. So lockers 4-9.Wait, 4²=16, which is 2 digits, and 9²=81, which is also 2 digits. So lockers 4 through 9 (6 lockers) have 2-digit numbers.3. 3-digit squares: From 10²=100 up to 31²=961. So lockers 10-31.Wait, 10² is 100 and 31² is 961. Let's check 31²: 31*31=961, which is 3 digits. Then 32²=1024, which is 4 digits. So lockers 10 to 31 (which is 22 lockers) have 3-digit numbers.4. 4-digit squares: From 32²=1024 up to 99²=9801. So lockers 32-99.Wait, 32²=1024 (4 digits), 99²=9801 (4 digits). Then locker 100 is 100²=10,000 (5 digits). So lockers 32 to 99 (68 lockers) have 4-digit numbers, and locker 100 has 5 digits.So to compute the total number of digits:- Lockers 1-3: 3 lockers × 1 digit = 3 digits- Lockers 4-9: 6 lockers × 2 digits = 12 digits- Lockers 10-31: 22 lockers × 3 digits = 66 digits- Lockers 32-99: 68 lockers × 4 digits = 272 digits- Locker 100: 1 locker × 5 digits = 5 digitsAdding them up: 3 + 12 + 66 + 272 + 5. Let me compute that step by step.3 + 12 = 1515 + 66 = 8181 + 272 = 353353 + 5 = 358Wait, so the total number of digits used is 358? Hmm, but let me verify the counts again because I might have miscalculated the number of lockers in each range.Starting with locker numbers:1-3: 3 lockers (1,2,3)4-9: 9 - 4 + 1 = 6 lockers (4 to 9 inclusive)10-31: 31 - 10 + 1 = 22 lockers (10 to 31 inclusive)32-99: 99 - 32 + 1 = 68 lockers (32 to 99 inclusive)100: 1 lockerYes, that's correct. So 3, 6, 22, 68, 1.So digits:3×1 = 36×2 = 1222×3 = 6668×4 = 2721×5 = 5Total: 3+12=15; 15+66=81; 81+272=353; 353+5=358.Therefore, part a) should be 358 digits.Wait, but let me double-check. Maybe I made a mistake in the number of lockers for each range.For example, from 1 to 3: lockers 1,2,3. That's 3 lockers. Correct.From 4 to 9: 4,5,6,7,8,9. That's 6 lockers. Correct.From 10 to 31: 31 - 10 +1 = 22 lockers. Correct.From 32 to 99: 99 -32 +1 = 68 lockers. Correct.And locker 100. Correct.So calculations are right. So 358 digits total. Therefore, the answer to part a is 358.Now moving on to part b: Adding all the numbers on the lockers, what is the units digit of this result?So we need to sum all the squares from 1² to 2019² and find the units digit of the total. That is, compute S = 1² + 2² + 3² + ... + 2019², find S mod 10.Hmm. Calculating the units digit of the sum of squares from 1 to 2019. The units digit of a sum is the sum of the units digits modulo 10. So if we can find the sum of the units digits of each square, modulo 10, that will give us the answer.But since the units digit of a square depends only on the units digit of the original number. For example, the units digit of n² is determined by the units digit of n. For example, numbers ending with 0: 0²=0, ending with 1: 1²=1, 2²=4, 3²=9, 4²=16 (units digit 6), 5²=25 (5), 6²=36 (6), 7²=49 (9), 8²=64 (4), 9²=81 (1). So each units digit of n leads to a specific units digit in n². So for numbers 0-9, their squares end with 0,1,4,9,6,5,6,9,4,1 respectively.Therefore, if we can figure out how many numbers in 1 to 2019 end with each digit from 0 to 9, then multiply each count by the corresponding units digit of their square, sum all those products, and take modulo 10, we can get the answer.Since the numbers from 1 to 2019 are 2019 numbers. Since the pattern of units digits repeats every 10 numbers, we can compute how many complete cycles of 10 there are in 2019, and the remainder.2019 divided by 10 is 201 with a remainder of 9. So there are 201 complete decades (groups of 10 numbers) and 9 extra numbers.In each complete decade (1-10, 11-20, ..., 2001-2010), the units digits are 0-9. Then the last 9 numbers are 2011-2019, whose units digits are 1-9.Wait, 2019 is the last number. Let's confirm:2010 is the last multiple of 10. Then 2011-2019 are 2011 (units digit 1), 2012 (2), ..., 2019 (9). So the last 9 numbers have units digits 1-9.In each complete group of 10 numbers, the units digits 0-9 each appear once. Except for the first group (1-10), the units digit 0 only appears in 10. But when we're considering numbers from 1 to 2019, the first number is 1, and the last is 2019. So in each group of 10 numbers, starting from 1-10, 11-20, etc., each units digit 0-9 appears exactly once per group, except that the first group 1-10 includes a 10 (units digit 0), and the last group 2011-2020 would be incomplete if there was a remainder, but here 2019 is 201*10 +9, so the last group is 2011-2019, units digits 1-9.Therefore, in total, the number of times each units digit appears:For digits 0-9:- Digit 0: appears in numbers 10, 20, ..., 2010. So total number of times is 201 (since 2010 is 201*10). So 201 times.- Digits 1-9: Each of these digits appears 201 times in the complete groups (each complete group has each digit once), plus an additional 1 time for the remaining 9 numbers (digits 1-9). Wait, no:Wait, 201 complete groups (each group contributing one of each digit 0-9). Wait, but the first group is 1-10, which includes 10 (digit 0). Then each subsequent group (11-20, etc.) includes a multiple of 10. So each complete group from 1-10, 11-20, ..., 2010 would have exactly 201 complete groups where each digit 0-9 appears once. Wait, but 1-10 is group 1, 11-20 group 2, ..., 2010 would be group 201. Then after that, we have 2011-2019, which is 9 numbers with digits 1-9.So for digit 0: appears once in each complete group (10, 20, ..., 2010). So 201 times.Digits 1-9: In each complete group, digits 1-9 appear once each. So 201 times each. Then in the last partial group (2011-2019), digits 1-9 appear once each. Wait, no. The last partial group has numbers 2011-2019, which is 9 numbers. The units digits are 1, 2, 3, 4, 5, 6, 7, 8, 9. So each digit from 1-9 appears once in the last partial group. Therefore, total for digits 1-9: 201 (from complete groups) +1 (from the partial group) = 202 times. Wait, that can't be. Wait, in each complete group, digits 1-9 appear once each. There are 201 complete groups. So digits 1-9 each appear 201 times. Then, in the last partial group (digits 1-9), they appear once more. So total 201 +1=202 times for digits 1-9. But digit 0 only appears in the complete groups, 201 times. Wait, that seems correct.Wait, total numbers: 201*10 (complete groups) +9 (partial group) = 2010 +9=2019. Correct. So digits:- 0: 201 times- 1-9: each 202 timesWait, wait: 201 complete groups (each contributing 0 once and 1-9 once each). Then the partial group (2011-2019) contributes 1-9 once each. Therefore:- Digit 0: 201 times- Digits 1-9: 201 (from complete groups) +1 (from partial group) = 202 times eachYes. So for each digit d from 1-9, count_d =202, and digit 0: count_0=201.Now, the units digit of each square depends on the units digit of the number. So for each digit d (0-9), when squared, gives a units digit of (as per before):0²: 01²:12²:43²:94²:65²:56²:67²:98²:49²:1So mapping each digit d to its square's units digit:0 → 01 →12→43→94→65→56→67→98→49→1Therefore, for each digit d in 0-9, the contribution to the total sum's units digit is count_d × (units digit of d²). Then sum all these contributions and take modulo 10.So let's compute this:First, list the units digits:- For digit 0: count=201, units digit contribution=0. So total contribution: 201×0=0- Digit 1: count=202, units digit=1. Contribution: 202×1=202- Digit 2: count=202, units digit=4. Contribution: 202×4=808- Digit 3: count=202, units digit=9. Contribution: 202×9=1,818- Digit 4: count=202, units digit=6. Contribution: 202×6=1,212- Digit 5: count=202, units digit=5. Contribution: 202×5=1,010- Digit 6: count=202, units digit=6. Contribution: 202×6=1,212- Digit 7: count=202, units digit=9. Contribution: 202×9=1,818- Digit 8: count=202, units digit=4. Contribution: 202×4=808- Digit 9: count=202, units digit=1. Contribution: 202×1=202Now, sum all these contributions:0 (from 0) +202 + 808 + 1,818 + 1,212 + 1,010 + 1,212 + 1,818 + 808 + 202Let me compute step by step:Start with 0.First, add the contributions from digits 1 to 9:202 (digit1) + 808 (digit2) = 1,0101,010 + 1,818 (digit3) = 2,8282,828 +1,212 (digit4)=4,0404,040 +1,010 (digit5)=5,0505,050 +1,212 (digit6)=6,2626,262 +1,818 (digit7)=8,0808,080 +808 (digit8)=8,8888,888 +202 (digit9)=9,090So total contributions from digits 1-9: 9,090Adding digit0's contribution (0): total sum is 9,090.Now, take 9,090 modulo 10. The units digit is 0. Wait, 9,090 divided by 10 is 909 with remainder 0. So the units digit is 0.But wait, this seems straightforward, but let me verify once more. Maybe I made a miscalculation when adding.Wait, let's recalculate the sum of contributions:Digits 1-9:202 (d1) +808 (d2) = 10101010 +1818 (d3) = 28282828 +1212 (d4) = 40404040 +1010 (d5) = 50505050 +1212 (d6) = 62626262 +1818 (d7) = 80808080 +808 (d8) = 88888888 +202 (d9) = 9090Yes, 9090. Then 9090 mod10=0.So the units digit is 0.But wait a second, that seems counterintuitive. Let me check with smaller numbers. For example, sum of squares from 1 to 10:1 +4 +9 +16 +25 +36 +49 +64 +81 +100 = ?Compute:1 +4=55 +9=1414 +16=3030 +25=5555 +36=9191 +49=140140 +64=204204 +81=285285 +100=385385 has units digit 5.But according to our method, for numbers 1-10, the count for digits:digit 0: 1 (from 10)digits1-9: each appears once (from 1-9), and in the last group (which is complete in this case, since 1-10 is a complete group). Wait, but in our previous logic, numbers 1-10 would have digit 0 once and digits1-9 once each. Then the total contributions would be:d0:1×0=0d1:1×1=1d2:1×4=4d3:1×9=9d4:1×6=6d5:1×5=5d6:1×6=6d7:1×9=9d8:1×4=4d9:1×1=1Sum these up:0 +1 +4 +9 +6 +5 +6 +9 +4 +1 =Let me compute:1+4=5; 5+9=14; 14+6=20; 20+5=25; 25+6=31; 31+9=40; 40+4=44; 44+1=45.But the actual sum of units digits is 385, units digit 5. 45 mod10=5. Which matches. So our method works here.But in that case, according to our method for 1-10, the sum of contributions is 45, which mod10=5, which is correct. So the method works.So for the total sum, 9090 mod10=0. Therefore, the units digit is 0.But let me check with another example. Let's take numbers 1-20. Then:Sum of squares from 1-20. Let's compute the sum:1²+2²+…+20² = (20×21×41)/6 = (420×41)/6=70×41=2870. Units digit 0.Using our method:Number of complete groups: 20/10=2 complete groups. Each digit0 appears twice, digits1-9 each appear twice. Then contributions:d0:2×0=0d1:2×1=2d2:2×4=8d3:2×9=18d4:2×6=12d5:2×5=10d6:2×6=12d7:2×9=18d8:2×4=8d9:2×1=2Sum contributions:0+2+8+18+12+10+12+18+8+2 =2+8=10; 10+18=28; 28+12=40; 40+10=50; 50+12=62; 62+18=80; 80+8=88; 88+2=90. 90 mod10=0. Which matches the actual sum's units digit (2870).So the method works. Therefore, for the original problem, the answer is 0.But let me think again. The key here is that the sum of all the units digits contributions is 9090, which ends with 0, so mod10 is 0. So the answer is 0.Therefore, part b answer is 0.Final Answera) boxed{358}b) boxed{0}</think>To solve the given problems, we break them down as follows: Part a: How many digits were used to paint the numbers on the first hundred lockers?1. Identify the ranges and number of digits for each range: - Lockers 1-3 (3 lockers): Each number is a 1-digit square. - Lockers 4-9 (6 lockers): Each number is a 2-digit square. - Lockers 10-31 (22 lockers): Each number is a 3-digit square. - Lockers 32-99 (68 lockers): Each number is a 4-digit square. - Locker 100 (1 locker): The number is a 5-digit square.2. Calculate the total digits for each range: - Lockers 1-3: (3 times 1 = 3) digits - Lockers 4-9: (6 times 2 = 12) digits - Lockers 10-31: (22 times 3 = 66) digits - Lockers 32-99: (68 times 4 = 272) digits - Locker 100: (1 times 5 = 5) digits3. Sum the digits: [ 3 + 12 + 66 + 272 + 5 = 358 ]Thus, the number of digits used to paint the numbers on the first hundred lockers is (boxed{358}). Part b: Units digit of the sum of all numbers on the lockers1. Determine the units digit contribution of each number: - The units digit of a square number depends only on the units digit of the original number. The mapping is: [ begin{aligned} 0 & rightarrow 0 1 & rightarrow 1 2 & rightarrow 4 3 & rightarrow 9 4 & rightarrow 6 5 & rightarrow 5 6 & rightarrow 6 7 & rightarrow 9 8 & rightarrow 4 9 & rightarrow 1 end{aligned} ]2. Count occurrences of each units digit in 1 to 2019: - Each digit 0-9 appears 201 times in complete groups of 10, except for digits 1-9 which appear once more in the partial group 2011-2019. - Total occurrences: - Digit 0: 201 times - Digits 1-9: 202 times each3. Calculate the contribution of each digit: - Sum the contributions modulo 10: [ begin{aligned} &0 times 201 + 1 times 202 + 4 times 202 + 9 times 202 + 6 times 202 + 5 times 202 + 6 times 202 + 9 times 202 + 4 times 202 + 1 times 202 &= 0 + 202 + 808 + 1818 + 1212 + 1010 + 1212 + 1818 + 808 + 202 &= 9090 end{aligned} ] - The units digit of 9090 is 0.Thus, the units digit of the sum of all numbers on the lockers is (boxed{0}).]

🔑:Solutiona) Let's divide into groups by the number of digits:I) 1 digit: 3 lockers left(1^{2}, 2^{2}, 3^{2}right);II) 2 digits: 9-3=6 lockers ( 4^{2} to 9^{2} );III) 3 digits: 31-9=22 lockers left(10^{2}right. to left.31^{2}right);IV) 4 digits: 99-31=68 lockers left(32^{2}right. to left.99^{2}right);V) 5 digits: only the locker 100^{2}.Therefore, 3+2 cdot 6+3 cdot 22+4 cdot 68+5 cdot 1=358 digits were painted on the first hundred lockers.b) The unit digit of the squares of the first 10 lockers are1,4,9,6,5,6,9,4,1,0This sequence of digits repeats every 10. Since the sum of one of these sequences ends in 5 and there are, up to 2020^{2} (let's consider 2020^{2} to have a whole number of sequences and as its unit digit is zero, it does not affect the result) 202 sums that end in 5, when added together, the result ends in 0.