answer：Okay, so I have this problem where four characters—Smeshariki Kros, Yozhik, Nyusha, and Barash—ate a total of 86 candies. Each of them ate at least 5 candies. There are two specific pieces of information: Nyusha ate more candies than any of the others, and Kros and Yozhik together ate 53 candies. I need to figure out how many candies Nyusha ate.First, let me break down the information:1. Total candies eaten by all four: 86.2. Each person ate at least 5 candies.3. Nyusha ate more than any other individual.4. Kros and Yozhik together ate 53 candies.So, let's denote the number of candies each person ate as follows:- Kros: K- Yozhik: Y- Nyusha: N- Barash: BFrom the problem, we know that:1. K + Y + N + B = 862. K + Y = 533. N > K, N > Y, N > B4. K, Y, N, B ≥ 5So, substituting equation 2 into equation 1, we can find the total eaten by Nyusha and Barash:N + B = 86 - (K + Y) = 86 - 53 = 33So, N + B = 33.We also know that N > K, N > Y, and N > B. Since K and Y together ate 53, and each ate at least 5, let's think about what K and Y could be.Since K + Y = 53, and both are at least 5, the minimum each could have eaten is 5, but since they sum to 53, which is a relatively large number, they must have eaten more than that.But wait, Nyusha has to eat more than both K and Y. So, N must be greater than both K and Y. Since K + Y = 53, the maximum that either K or Y could have eaten is 53 - 5 = 48, but that's if one ate 5 and the other ate 48. However, Nyusha has to eat more than both, so N must be greater than 48.But wait, let's think again. If K and Y together ate 53, the maximum either could have eaten is 53 - 5 = 48, because the other has to eat at least 5. So, if K is 48, Y is 5, or vice versa. But Nyusha has to eat more than both, so N must be greater than 48. But N + B = 33, so N can't be more than 33 because B has to be at least 5.Wait, that's a contradiction. If N has to be greater than 48, but N + B = 33, which is less than 48, that can't be. So, I must have made a mistake in my reasoning.Let me check. If K + Y = 53, and each of them ate at least 5, then the maximum either could have eaten is 53 - 5 = 48, but Nyusha has to eat more than both. So, N must be greater than 48. But N + B = 33, which is only 33. So, N can't be more than 33 because B has to be at least 5. So, N must be less than or equal to 33 - 5 = 28.But wait, that's a problem because N has to be greater than both K and Y, which could be up to 48. So, how can N be greater than 48 if N is only 28 at maximum? That doesn't make sense.Hmm, maybe I misapplied the logic. Let's think differently.Since K + Y = 53, and each of them ate at least 5, the minimum they could have eaten is 5 each, but that would only sum to 10, which is way less than 53. So, actually, K and Y must have eaten more than 5 each.Wait, but the problem says each of them ate at least 5, not exactly 5. So, K and Y could have eaten more than 5, but their total is 53.Now, Nyusha has to eat more than both K and Y. So, N must be greater than both K and Y. Therefore, N must be greater than the maximum of K and Y.But since K + Y = 53, the maximum either K or Y can be is 53 - 5 = 48, as I thought earlier. So, N must be greater than 48.But N + B = 33, so N can't be more than 33 - 5 = 28. That's a contradiction because 28 is less than 48.Wait, so maybe my initial assumption is wrong. Perhaps K and Y didn't eat 5 each, but more. Let me think again.If K + Y = 53, and each ate at least 5, but Nyusha has to eat more than both K and Y, which are part of the 53.So, let's denote that K and Y are both less than N. So, N > K and N > Y.Since K + Y = 53, and both K and Y are less than N, then N must be greater than the maximum of K and Y.But since K + Y = 53, the maximum of K and Y is at least 53/2 = 26.5, so at least 27.So, N must be greater than 27.But N + B = 33, so N must be less than 33 - 5 = 28.Wait, so N must be greater than 27 but less than 28. But N has to be an integer, so N must be 28.But wait, if N is 28, then B is 33 - 28 = 5.So, B = 5, which is acceptable because each must eat at least 5.But does N = 28 satisfy N > K and N > Y?Since K + Y = 53, and both K and Y are less than 28, is that possible?Wait, if both K and Y are less than 28, then their maximum possible sum would be 27 + 27 = 54, which is more than 53. So, it's possible that K and Y are both 26.5, but since they have to be integers, let's see.Wait, 26 + 27 = 53. So, if K = 26 and Y = 27, then both are less than N = 28. That works.Alternatively, K = 25 and Y = 28, but Y would be equal to N, which is not allowed because N has to be more than Y.Wait, so Y can't be 28 because N is 28. So, Y has to be less than 28.Similarly, K has to be less than 28.So, if K and Y are both less than 28, their maximum sum would be 27 + 26 = 53.Wait, 27 + 26 is 53. So, that works. So, K = 26, Y = 27, N = 28, B = 5.Let me check:26 + 27 + 28 + 5 = 86. Yes, that adds up.Also, each person ate at least 5 candies: 26, 27, 28, 5. Yes.Nyusha ate 28, which is more than K (26), Y (27), and B (5). So, that works.Wait, but Y is 27, which is less than N = 28, so that's okay.Alternatively, could K and Y be 25 and 28? But Y can't be 28 because N is 28, so Y has to be less than 28.Similarly, K can't be 28 because N is 28.So, the only way is K = 26, Y = 27, N = 28, B = 5.Is there another possibility? Let's see.Suppose K = 25, Y = 28. But Y can't be 28 because N is 28. So, Y has to be less than 28.Similarly, K can't be 28.So, the only way is K = 26, Y = 27, N = 28, B = 5.Wait, but could K and Y be 27 and 26? That's the same as above.Alternatively, could K and Y be 24 and 29? But Y would be 29, which is more than N = 28, which contradicts N being the maximum.So, no, that's not possible.Similarly, K can't be higher than 27 because Y would have to be lower, but Y can't be lower than 5.Wait, let's see:If K is 27, then Y is 53 - 27 = 26.So, K = 27, Y = 26, N = 28, B = 5.Same as before, just swapping K and Y.So, in either case, N is 28.Is there a way for N to be higher than 28?If N were 29, then B would be 33 - 29 = 4, which is less than 5. That's not allowed because each person must eat at least 5.So, N can't be 29.Similarly, N can't be 30 because B would be 3, which is too low.So, N must be 28.Therefore, Nyusha ate 28 candies.

answer：Alright, so I have this system of equations to solve:begin{aligned}x^{3} + y^{3} + 3xy & = 1 x^{2} - y^{2} & = 1end{aligned}And I need to find the solutions where both ( x ) and ( y ) are non-negative. Also, I have to check if considering ( x + y > 0 ) instead of non-negative gives any different solutions. Hmm, okay, let me try to figure this out step by step.First, looking at the second equation: ( x^{2} - y^{2} = 1 ). I remember that ( x^{2} - y^{2} ) can be factored as ( (x - y)(x + y) ). So, maybe that could be helpful.So, rewriting the second equation:(x - y)(x + y) = 1That's interesting. Maybe I can express one variable in terms of the other. Let me see.Let me denote ( S = x + y ) and ( D = x - y ). Then, the second equation becomes:D cdot S = 1So, ( D = frac{1}{S} ). That might be useful later.Now, looking back at the first equation: ( x^{3} + y^{3} + 3xy = 1 ). I recall that ( x^{3} + y^{3} ) can be factored as ( (x + y)(x^{2} - xy + y^{2}) ). Let me try that.So, rewriting the first equation:(x + y)(x^{2} - xy + y^{2}) + 3xy = 1Hmm, maybe I can express this in terms of ( S ) and ( D ) as well. Let me see.First, ( x + y = S ), so that's already there. Now, ( x^{2} - xy + y^{2} ). Let me compute that.I know that ( x^{2} + y^{2} = (x + y)^2 - 2xy = S^2 - 2xy ). So, substituting that into ( x^{2} - xy + y^{2} ):x^{2} - xy + y^{2} = (x^{2} + y^{2}) - xy = (S^2 - 2xy) - xy = S^2 - 3xySo, substituting back into the first equation:S(S^2 - 3xy) + 3xy = 1Let me expand that:S^3 - 3S cdot xy + 3xy = 1Factor out the ( xy ):S^3 - 3xy(S - 1) = 1Hmm, okay. So, now I have an expression involving ( S ) and ( xy ). I need another equation to relate ( S ) and ( xy ). Maybe from the second equation.Earlier, I had ( D = frac{1}{S} ), where ( D = x - y ). Also, since ( S = x + y ), and ( D = x - y ), I can express ( x ) and ( y ) in terms of ( S ) and ( D ):x = frac{S + D}{2}, quad y = frac{S - D}{2}So, plugging ( D = frac{1}{S} ), we get:x = frac{S + frac{1}{S}}{2}, quad y = frac{S - frac{1}{S}}{2}Okay, so now I can express ( x ) and ( y ) in terms of ( S ). Maybe I can find ( xy ) in terms of ( S ) as well.Let me compute ( xy ):xy = left( frac{S + frac{1}{S}}{2} right) left( frac{S - frac{1}{S}}{2} right ) = frac{(S + frac{1}{S})(S - frac{1}{S})}{4}Multiplying out the numerator:(S + frac{1}{S})(S - frac{1}{S}) = S^2 - left( frac{1}{S} right)^2 = S^2 - frac{1}{S^2}So, ( xy = frac{S^2 - frac{1}{S^2}}{4} )Therefore, ( xy = frac{S^2 - frac{1}{S^2}}{4} )Now, going back to the first equation rewritten in terms of ( S ) and ( xy ):S^3 - 3xy(S - 1) = 1Substitute ( xy ):S^3 - 3 cdot left( frac{S^2 - frac{1}{S^2}}{4} right) (S - 1) = 1Let me simplify this step by step.First, compute the term ( 3 cdot frac{S^2 - frac{1}{S^2}}{4} ):frac{3}{4} left( S^2 - frac{1}{S^2} right )So, the equation becomes:S^3 - frac{3}{4} left( S^2 - frac{1}{S^2} right )(S - 1) = 1Now, let me expand ( left( S^2 - frac{1}{S^2} right )(S - 1) ):Multiply term by term:First, ( S^2 cdot S = S^3 )Second, ( S^2 cdot (-1) = -S^2 )Third, ( -frac{1}{S^2} cdot S = -frac{1}{S} )Fourth, ( -frac{1}{S^2} cdot (-1) = frac{1}{S^2} )So, altogether:S^3 - S^2 - frac{1}{S} + frac{1}{S^2}Therefore, the equation becomes:S^3 - frac{3}{4} left( S^3 - S^2 - frac{1}{S} + frac{1}{S^2} right ) = 1Let me distribute the ( frac{3}{4} ):S^3 - frac{3}{4}S^3 + frac{3}{4}S^2 + frac{3}{4S} - frac{3}{4S^2} = 1Combine like terms:- ( S^3 - frac{3}{4}S^3 = frac{1}{4}S^3 )- The other terms remain as they are.So, the equation is:frac{1}{4}S^3 + frac{3}{4}S^2 + frac{3}{4S} - frac{3}{4S^2} = 1Hmm, this looks a bit complicated. Maybe I can multiply both sides by ( 4S^2 ) to eliminate denominators. Let's try that.Multiplying each term by ( 4S^2 ):- ( frac{1}{4}S^3 cdot 4S^2 = S^5 )- ( frac{3}{4}S^2 cdot 4S^2 = 3S^4 )- ( frac{3}{4S} cdot 4S^2 = 3S )- ( -frac{3}{4S^2} cdot 4S^2 = -3 )- Right-hand side: ( 1 cdot 4S^2 = 4S^2 )So, the equation becomes:S^5 + 3S^4 + 3S - 3 = 4S^2Bring all terms to the left side:S^5 + 3S^4 + 3S - 3 - 4S^2 = 0Simplify:S^5 + 3S^4 - 4S^2 + 3S - 3 = 0Hmm, so now I have a fifth-degree polynomial in ( S ). Solving quintic equations is generally difficult because there's no formula for them, but maybe this factors nicely or has rational roots.Let me try to see if there are any rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term over factors of the leading coefficient. So, possible roots are ( pm1, pm3 ).Let me test ( S = 1 ):( 1 + 3 - 4 + 3 - 3 = 0 ). So, ( 1 + 3 = 4; 4 - 4 = 0; 0 + 3 = 3; 3 - 3 = 0 ). So, yes, ( S = 1 ) is a root.Great! So, ( (S - 1) ) is a factor. Let's perform polynomial division or factor it out.Let me divide the polynomial ( S^5 + 3S^4 - 4S^2 + 3S - 3 ) by ( S - 1 ).Using synthetic division:Coefficients: 1 (S^5), 3 (S^4), 0 (S^3), -4 (S^2), 3 (S), -3 (constant)Set up synthetic division with root 1:1 | 1   3   0   -4   3   -3   |     1   4   4   0   3   ----------------------------     1   4   4   0   3   0So, the result is ( S^4 + 4S^3 + 4S^2 + 0S + 3 ), which is ( S^4 + 4S^3 + 4S^2 + 3 ). So, the polynomial factors as:( (S - 1)(S^4 + 4S^3 + 4S^2 + 3) = 0 )Now, set each factor equal to zero:1. ( S - 1 = 0 ) => ( S = 1 )2. ( S^4 + 4S^3 + 4S^2 + 3 = 0 )Looking at the quartic equation ( S^4 + 4S^3 + 4S^2 + 3 = 0 ). Since ( S ) is non-negative (because ( x ) and ( y ) are non-negative, so ( S = x + y geq 0 )), all terms ( S^4, 4S^3, 4S^2, 3 ) are non-negative. Therefore, their sum is always positive. So, no real non-negative roots from the quartic.Therefore, the only real non-negative solution is ( S = 1 ).So, ( S = 1 ). Then, from earlier, ( D = frac{1}{S} = 1 ).So, ( D = x - y = 1 ), and ( S = x + y = 1 ).So, now we have:1. ( x + y = 1 )2. ( x - y = 1 )Let me solve these two equations.Adding both equations:( (x + y) + (x - y) = 1 + 1 )( 2x = 2 )( x = 1 )Subtracting the second equation from the first:( (x + y) - (x - y) = 1 - 1 )( 2y = 0 )( y = 0 )So, the solution is ( x = 1 ), ( y = 0 ).Wait, but we need to check if this satisfies the original equations.First equation: ( x^3 + y^3 + 3xy = 1^3 + 0^3 + 3*1*0 = 1 + 0 + 0 = 1 ). That works.Second equation: ( x^2 - y^2 = 1^2 - 0^2 = 1 - 0 = 1 ). That also works.So, ( (1, 0) ) is a solution.But wait, the problem says "in the set of non-negative numbers". So, ( x ) and ( y ) are both non-negative. So, ( y = 0 ) is acceptable.But the question also asks: Do we obtain other solutions if we only assume that ( x + y > 0 )?Hmm, so in the original problem, we assumed non-negative numbers, so ( x, y geq 0 ). If we relax that to ( x + y > 0 ), which would allow ( x ) or ( y ) to be negative, as long as their sum is positive.But in the second equation, ( x^2 - y^2 = 1 ), which is positive. So, ( x^2 > y^2 ). Therefore, ( |x| > |y| ). So, if ( x ) is positive, ( y ) could be negative, but ( x ) must be greater in magnitude.But let's see if there are other solutions.Wait, in our earlier steps, we found that ( S = 1 ) is the only real non-negative solution. But if we allow ( S ) to be negative, maybe?Wait, no, because ( S = x + y ). If ( x + y > 0 ), ( S ) is positive. So, even if ( x ) or ( y ) is negative, as long as their sum is positive, ( S ) is positive.But in our polynomial, ( S^5 + 3S^4 - 4S^2 + 3S - 3 = 0 ), we had only one real root at ( S = 1 ). The quartic didn't have any real roots, so even if we consider ( S ) positive, only ( S = 1 ) is a solution.Wait, but perhaps I made a mistake in assuming that the quartic doesn't have real roots. Let me check.The quartic is ( S^4 + 4S^3 + 4S^2 + 3 ). Let me see if it can be factored or if it has real roots.Let me try to factor it:Looking at ( S^4 + 4S^3 + 4S^2 + 3 ). Maybe group terms:( (S^4 + 4S^3) + (4S^2 + 3) )Factor ( S^3 ) from the first group:( S^3(S + 4) + (4S^2 + 3) ). Doesn't seem helpful.Alternatively, maybe write it as ( (S^4 + 4S^3 + 4S^2) + 3 ). Then, factor ( S^2 ):( S^2(S^2 + 4S + 4) + 3 = S^2(S + 2)^2 + 3 ). Hmm, that's interesting.So, ( S^2(S + 2)^2 + 3 ). Since ( S ) is real, ( S^2 ) and ( (S + 2)^2 ) are both non-negative. So, ( S^2(S + 2)^2 geq 0 ), so the entire expression is ( geq 3 ). Therefore, ( S^4 + 4S^3 + 4S^2 + 3 geq 3 ), so it can never be zero. Therefore, the quartic has no real roots.Therefore, regardless of whether ( x + y > 0 ) or ( x, y geq 0 ), the only solution is ( S = 1 ), leading to ( x = 1 ), ( y = 0 ).Wait, but hold on. If ( x + y > 0 ), could there be another solution where ( x ) and ( y ) are such that ( x + y ) is positive but not necessarily non-negative individually?Wait, but in the second equation, ( x^2 - y^2 = 1 ). So, ( x^2 = y^2 + 1 ). Therefore, ( x geq sqrt{1} = 1 ), since ( x^2 geq 1 ). So, ( x geq 1 ). Therefore, ( x ) is at least 1, so ( x ) is positive. Therefore, ( x ) must be positive, but ( y ) could be negative or positive.But in our earlier solution, ( y = 0 ). So, perhaps if ( y ) is negative, could there be another solution?Wait, let me think. If ( y ) is negative, then ( x + y ) is still positive because ( x geq 1 ) and ( y ) is negative, but maybe their sum is still positive.But in our earlier steps, we had ( S = x + y ), and ( S = 1 ). So, even if ( y ) is negative, ( x + y = 1 ). So, ( x = 1 - y ). Since ( x geq 1 ), ( 1 - y geq 1 ) => ( -y geq 0 ) => ( y leq 0 ).So, ( y leq 0 ). So, ( y ) is non-positive.But in the original problem, we considered non-negative numbers, so ( y = 0 ) is the only possibility. If we allow ( y ) to be negative, could there be another solution?Wait, let's see.If ( y ) is negative, then ( x = 1 - y ), which is greater than 1, since ( y ) is negative. So, ( x > 1 ).But then, let's plug into the first equation:( x^3 + y^3 + 3xy = 1 )But ( x > 1 ), ( y ) negative.Let me compute ( x^3 + y^3 + 3xy ). Since ( x > 1 ), ( x^3 ) is greater than 1. ( y^3 ) is negative, and ( 3xy ) is negative because ( y ) is negative.So, the total is ( x^3 + y^3 + 3xy ). Let me see if this can be equal to 1.But ( x^3 ) is greater than 1, and ( y^3 + 3xy ) is negative. So, the total could be 1 if the negative terms exactly cancel the excess of ( x^3 ) over 1.But let's see if that's possible.Given that ( x = 1 - y ), and ( y ) is negative, let me denote ( y = -k ), where ( k > 0 ).Then, ( x = 1 - (-k) = 1 + k ).So, substituting into the first equation:( (1 + k)^3 + (-k)^3 + 3(1 + k)(-k) = 1 )Compute each term:1. ( (1 + k)^3 = 1 + 3k + 3k^2 + k^3 )2. ( (-k)^3 = -k^3 )3. ( 3(1 + k)(-k) = -3k - 3k^2 )So, adding them together:1. ( 1 + 3k + 3k^2 + k^3 )2. ( -k^3 )3. ( -3k - 3k^2 )Combine like terms:- Constants: 1- ( k ) terms: ( 3k - 3k = 0 )- ( k^2 ) terms: ( 3k^2 - 3k^2 = 0 )- ( k^3 ) terms: ( k^3 - k^3 = 0 )So, the total is 1. So, ( 1 = 1 ). So, this holds for any ( k ).Wait, that's interesting. So, if ( y = -k ), ( x = 1 + k ), then the first equation is satisfied for any ( k ). But we also have the second equation: ( x^2 - y^2 = 1 ).Let me check if this holds.( x^2 - y^2 = (1 + k)^2 - (-k)^2 = (1 + 2k + k^2) - k^2 = 1 + 2k )But according to the second equation, this should equal 1. So,( 1 + 2k = 1 )Which implies ( 2k = 0 ) => ( k = 0 )Therefore, ( y = -k = 0 ), so ( y = 0 ), and ( x = 1 + 0 = 1 ).So, even if we allow ( y ) to be negative, the only solution is still ( x = 1 ), ( y = 0 ).Therefore, whether we assume non-negative numbers or just ( x + y > 0 ), the only solution is ( x = 1 ), ( y = 0 ).Wait, but in the case where ( x + y > 0 ), could there be another solution where ( x + y ) is positive, but ( x ) or ( y ) is negative?But as we saw, substituting ( y = -k ) led us back to ( k = 0 ), so no other solutions.Therefore, the only solution is ( x = 1 ), ( y = 0 ).So, to sum up, the system has only one solution in non-negative numbers, which is ( (1, 0) ). Even if we relax the condition to ( x + y > 0 ), we still only get the same solution because any other potential solutions would require ( y ) to be negative, but that leads back to ( y = 0 ) due to the second equation.**Final Answer**The only solution is boxed{(1, 0)}.

answer：Okay, so I need to figure out how many natural divisors of 11! are multiples of three. Hmm, let's break this down step by step. First, I remember that 11! means the product of all natural numbers from 1 to 11. So, 11! = 1 × 2 × 3 × ... × 10 × 11. To find the number of divisors that are multiples of three, I think I need to look at the prime factorization of 11! because divisors are related to the exponents in the prime factors.Alright, let me recall how to find the prime factorization of a factorial. For 11!, I need to find how many times each prime number up to 11 divides into 11!. The primes up to 11 are 2, 3, 5, 7, 11. Starting with 2: I need to count how many multiples of 2, 4, 8, etc., are in 1 to 11. - Number of multiples of 2: 11 ÷ 2 = 5.5, so 5.- Multiples of 4: 11 ÷ 4 = 2.75, so 2.- Multiples of 8: 11 ÷ 8 = 1.375, so 1.- Multiples of 16: That's beyond 11, so stop here.Adding these up: 5 + 2 + 1 = 8. So, exponent of 2 in 11! is 8.Next, prime 3:- Multiples of 3: 11 ÷ 3 ≈ 3.666, so 3.- Multiples of 9: 11 ÷ 9 ≈ 1.222, so 1.- Multiples of 27: Beyond 11, so stop.Total exponent for 3: 3 + 1 = 4.Prime 5:- Multiples of 5: 11 ÷ 5 = 2.2, so 2.- Multiples of 25: Beyond 11, so stop.Exponent for 5: 2.Prime 7:- Multiples of 7: 11 ÷ 7 ≈ 1.571, so 1.- Multiples of 49: Beyond 11, stop.Exponent for 7: 1.Prime 11:- Multiples of 11: 11 ÷ 11 = 1, so 1.- Multiples of 121: Beyond 11, stop.Exponent for 11: 1.So, putting it all together, the prime factorization of 11! is:11! = 2^8 × 3^4 × 5^2 × 7^1 × 11^1.Now, to find the number of divisors that are multiples of three. I remember that the total number of divisors of a number is found by adding one to each exponent in the prime factorization and then multiplying them together. So, for 11!, the total number of divisors is (8+1)(4+1)(2+1)(1+1)(1+1) = 9×5×3×2×2. Let me calculate that: 9×5=45, 45×3=135, 135×2=270, 270×2=540. So, 540 total divisors.But we need only those divisors that are multiples of three. Hmm, how do I find that? I think it's equivalent to finding the number of divisors of 11! that include at least one factor of 3. Alternatively, maybe it's easier to subtract the number of divisors that are not multiples of three from the total number of divisors. That is, total divisors minus the number of divisors that are coprime to 3. Yes, that makes sense. So, if I can find the number of divisors of 11! that are not divisible by 3, then subtracting that from the total will give me the number of divisors that are multiples of three.To find the number of divisors not divisible by 3, I need to consider the prime factorization without the factor of 3. So, if I remove all the 3s from 11!, the remaining number is 2^8 × 5^2 × 7^1 × 11^1. The number of divisors of this number is (8+1)(2+1)(1+1)(1+1) = 9×3×2×2. Let me compute that: 9×3=27, 27×2=54, 54×2=108. So, 108 divisors are not divisible by 3.Therefore, the number of divisors that are multiples of three is total divisors minus non-multiples: 540 - 108 = 432.Wait, is that right? Let me double-check. Another way to think about it is that for a divisor to be a multiple of three, it must include at least one 3 in its prime factors. So, in the prime factorization of the divisor, the exponent of 3 can be 1, 2, 3, or 4 (since in 11! it's 4). So, the number of choices for the exponent of 3 is 4 (from 1 to 4). For the other primes, they can be anything from 0 up to their respective exponents in 11!.So, the number of such divisors is (number of choices for 2) × (number of choices for 3) × (number of choices for 5) × (number of choices for 7) × (number of choices for 11).Which is (8+1) × (4) × (2+1) × (1+1) × (1+1). Wait, hold on, no. Because for the exponent of 3, we have to have at least 1, so it's 4 choices (1,2,3,4). For the others, they can be 0 up to their exponents.So, it's (8+1) × (4) × (2+1) × (1+1) × (1+1). Let me compute that:(9) × (4) × (3) × (2) × (2). Calculating step by step:9 × 4 = 3636 × 3 = 108108 × 2 = 216216 × 2 = 432.So, same result, 432. That seems consistent.Alternatively, another way to think about it is that the number of divisors divisible by 3 is equal to the total number of divisors of 11! divided by 3. But wait, that might not be accurate because 11! is not a multiple of 3^1 necessarily for all its divisors. Wait, no, 11! is divisible by 3, so dividing by 3 would give us the number of divisors of 11!/3, which is 2^8 × 3^3 × 5^2 × 7^1 × 11^1. Then, the number of divisors of 11!/3 is (8+1)(3+1)(2+1)(1+1)(1+1) = 9×4×3×2×2. Let me compute that: 9×4=36, 36×3=108, 108×2=216, 216×2=432. So, same answer again. Therefore, I think 432 is correct. But just to make sure, let me recap:1. Prime factorization of 11! is 2^8 × 3^4 × 5^2 × 7^1 × 11^1.2. Total number of divisors: (8+1)(4+1)(2+1)(1+1)(1+1) = 9×5×3×2×2 = 540.3. Number of divisors not divisible by 3: remove all 3s, so factorization becomes 2^8 × 5^2 × 7^1 × 11^1. Number of divisors: (8+1)(2+1)(1+1)(1+1) = 9×3×2×2 = 108.4. Therefore, number of divisors divisible by 3: 540 - 108 = 432.Alternatively, directly compute the number of divisors with at least one 3: (8+1)(4)(2+1)(1+1)(1+1) = 9×4×3×2×2 = 432.Both methods give the same result, so I feel confident that 432 is the correct number of natural divisors of 11! that are multiples of three.**Final Answer**The number of natural divisors of (11!) that are multiples of three is boxed{432}.

answer：Okay, so I have this problem here about a parabola. Let me try to understand it step by step. The parabola is given by the equation ( C: y^2 = 2px ) where ( p > 0 ). There's a point ( A ) on this parabola, and we know two things: the distance from ( A ) to the focus of the parabola is 12, and the distance from ( A ) to the y-axis is 9. We need to find the value of ( p ).First, let me recall some properties of parabolas. The standard form of a parabola that opens to the right is ( y^2 = 4ax ), where the focus is at ( (a, 0) ). But in our case, the equation is ( y^2 = 2px ). Comparing this to the standard form, it seems like ( 4a = 2p ), so ( a = frac{p}{2} ). That means the focus of this parabola is at ( left( frac{p}{2}, 0 right) ).Now, let's denote point ( A ) as ( (x, y) ). Since ( A ) lies on the parabola ( C ), it must satisfy the equation ( y^2 = 2px ). So, that's our first equation.Next, the distance from point ( A ) to the y-axis is given as 9. The distance from a point ( (x, y) ) to the y-axis is just the absolute value of its x-coordinate, which is ( |x| ). Since the parabola ( y^2 = 2px ) opens to the right and ( p > 0 ), the x-coordinate of any point on the parabola must be non-negative. Therefore, ( x = 9 ).So, substituting ( x = 9 ) into the equation of the parabola, we get ( y^2 = 2p times 9 = 18p ). Therefore, ( y = pm sqrt{18p} ). So, point ( A ) is ( (9, sqrt{18p}) ) or ( (9, -sqrt{18p}) ). Since distance is a positive quantity, the sign of ( y ) doesn't matter for the distance calculations, so I can just consider ( y = sqrt{18p} ).Now, the distance from point ( A ) to the focus is given as 12. The focus is at ( left( frac{p}{2}, 0 right) ). So, the distance between ( A(9, sqrt{18p}) ) and the focus ( left( frac{p}{2}, 0 right) ) is 12.Let me write the distance formula. The distance ( d ) between two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is ( sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). So, applying this to our points:( sqrt{left( frac{p}{2} - 9 right)^2 + left( 0 - sqrt{18p} right)^2} = 12 )Let me square both sides to eliminate the square root:( left( frac{p}{2} - 9 right)^2 + left( -sqrt{18p} right)^2 = 144 )Simplify each term:First term: ( left( frac{p}{2} - 9 right)^2 ). Let me expand this:( left( frac{p}{2} - 9 right)^2 = left( frac{p}{2} right)^2 - 2 times frac{p}{2} times 9 + 9^2 = frac{p^2}{4} - 9p + 81 )Second term: ( left( -sqrt{18p} right)^2 = (sqrt{18p})^2 = 18p )So, putting it all together:( frac{p^2}{4} - 9p + 81 + 18p = 144 )Combine like terms:The ( p ) terms: ( -9p + 18p = 9p )So, the equation becomes:( frac{p^2}{4} + 9p + 81 = 144 )Now, subtract 144 from both sides to set the equation to zero:( frac{p^2}{4} + 9p + 81 - 144 = 0 )Simplify the constants:( 81 - 144 = -63 )So, the equation is:( frac{p^2}{4} + 9p - 63 = 0 )To make this easier, let me multiply every term by 4 to eliminate the fraction:( p^2 + 36p - 252 = 0 )Now, we have a quadratic equation in terms of ( p ):( p^2 + 36p - 252 = 0 )Let me try to solve this quadratic equation. I can use the quadratic formula:( p = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Here, ( a = 1 ), ( b = 36 ), and ( c = -252 ).Plugging in the values:Discriminant ( D = b^2 - 4ac = 36^2 - 4 times 1 times (-252) )Calculate ( 36^2 ): 36*36 = 1296Calculate ( 4ac ): 4*1*(-252) = -1008So, ( D = 1296 - (-1008) = 1296 + 1008 = 2304 )Square root of discriminant: ( sqrt{2304} = 48 )So, the solutions are:( p = frac{-36 pm 48}{2 times 1} )Calculating both possibilities:First solution: ( p = frac{-36 + 48}{2} = frac{12}{2} = 6 )Second solution: ( p = frac{-36 - 48}{2} = frac{-84}{2} = -42 )But wait, the problem states that ( p > 0 ), so we discard the negative solution. Therefore, ( p = 6 ).Let me verify this solution to make sure I didn't make any mistakes.Given ( p = 6 ), the equation of the parabola is ( y^2 = 12x ). The focus is at ( left( frac{6}{2}, 0 right) = (3, 0) ).Point ( A ) is at ( x = 9 ), so plugging into the parabola equation, ( y^2 = 12*9 = 108 ), so ( y = pm sqrt{108} = pm 6sqrt{3} ).So, point ( A ) is ( (9, 6sqrt{3}) ) or ( (9, -6sqrt{3}) ).Now, let's compute the distance from ( A ) to the focus ( (3, 0) ):Distance = ( sqrt{(9 - 3)^2 + (6sqrt{3} - 0)^2} )Calculate each part:( (9 - 3)^2 = 6^2 = 36 )( (6sqrt{3})^2 = 36 * 3 = 108 )So, distance squared is ( 36 + 108 = 144 ), so distance is ( sqrt{144} = 12 ), which matches the given condition.Also, the distance from ( A ) to the y-axis is ( |9| = 9 ), which is also given.Therefore, the solution ( p = 6 ) is correct.I think that's thorough. I considered the properties of the parabola, set up the equations correctly, solved the quadratic, checked the solutions, and verified the answer. So, I'm confident that ( p = 6 ) is the correct answer.**Final Answer**The value of ( p ) is boxed{6}.

answer：Okay, so I have this problem here: If the graph of the power function ( f(x) ) passes through the point ( (2, frac{1}{4}) ), then I need to find ( fleft( frac{1}{2} right) ). Hmm, power function... I remember that power functions are of the form ( f(x) = kx^n ), where ( k ) is a constant and ( n ) is the exponent, which can be any real number. Alright, so the function passes through ( (2, frac{1}{4}) ). That means when ( x = 2 ), ( f(x) = frac{1}{4} ). So I can plug these values into the equation to find a relationship between ( k ) and ( n ). Let me write that out:( frac{1}{4} = k cdot 2^n )Hmm, okay. So I have one equation here with two variables, ( k ) and ( n ). That means I can't solve for both directly without another equation or some additional information. Wait, but maybe I can express ( k ) in terms of ( n ) or vice versa. Let me try that.From the equation ( frac{1}{4} = k cdot 2^n ), I can solve for ( k ):( k = frac{1}{4} cdot frac{1}{2^n} = frac{1}{4 cdot 2^n} )So ( k = frac{1}{4 cdot 2^n} ). Hmm, that's an expression for ( k ) in terms of ( n ). But without another point or some other condition, I can't find the exact values of ( k ) and ( n ). Wait, but maybe the problem assumes that ( n ) is an integer or a simple fraction? Or perhaps there's a standard power function that passes through that point?Alternatively, maybe I can think about common power functions. For example, if ( f(x) = x^{-2} ), then ( f(2) = frac{1}{4} ), which matches the given point. Let me check that:( f(2) = 2^{-2} = frac{1}{4} ). Yes, that works. So if ( n = -2 ), then ( k ) would be 1. So in that case, ( f(x) = x^{-2} ) or ( f(x) = frac{1}{x^2} ). Wait, but is that the only possibility? What if ( n ) is something else? For example, if ( n = -1 ), then ( f(2) = k cdot 2^{-1} = frac{k}{2} ). If that equals ( frac{1}{4} ), then ( k = frac{1}{2} ). So ( f(x) = frac{1}{2}x^{-1} ). Then ( fleft( frac{1}{2} right) = frac{1}{2} cdot left( frac{1}{2} right)^{-1} = frac{1}{2} cdot 2 = 1 ). Hmm, so that would give a different result.Alternatively, if ( n = 2 ), then ( f(2) = k cdot 4 = frac{1}{4} ), so ( k = frac{1}{16} ). Then ( fleft( frac{1}{2} right) = frac{1}{16} cdot left( frac{1}{2} right)^2 = frac{1}{16} cdot frac{1}{4} = frac{1}{64} ). So that's another possible answer.Wait, so depending on the value of ( n ), the result can be different. But the problem just says it's a power function. So is there a standard assumption about power functions? Maybe it's a monomial function, which typically has integer exponents, but they can be any real number. Hmm.Wait, but in the absence of additional information, how can we determine ( n )? Maybe the problem expects a specific form? Or perhaps I'm overcomplicating it. Let me think again.Power functions are of the form ( f(x) = kx^n ). If it passes through ( (2, 1/4) ), then ( 1/4 = k cdot 2^n ). So we can write ( k = frac{1}{4 cdot 2^n} ). So the function is ( f(x) = frac{1}{4 cdot 2^n} cdot x^n ). Hmm, but without another condition, we can't find a unique solution. So maybe the problem expects a general expression? But the question is asking for ( f(1/2) ), so maybe we can express it in terms of ( n )?Wait, but the problem is presented as a fill-in-the-blank, so it's expecting a numerical answer. That suggests that perhaps ( n ) is uniquely determined. Maybe I need to make an assumption here.Wait, perhaps the power function is of the form ( f(x) = x^n ), meaning ( k = 1 ). Then, ( f(2) = 2^n = 1/4 ). So ( 2^n = 1/4 ). Since ( 1/4 = 2^{-2} ), then ( n = -2 ). So ( f(x) = x^{-2} ). Then, ( f(1/2) = (1/2)^{-2} = (2)^2 = 4 ). So that would be the answer.But wait, earlier I thought ( k ) might not be 1. So is ( k ) necessarily 1? The problem says it's a power function, which is generally ( kx^n ). So unless specified otherwise, ( k ) can be any constant. So without knowing ( k ), we can't determine ( n ), and vice versa.But the problem is asking for ( f(1/2) ). Maybe we can express ( f(1/2) ) in terms of ( f(2) ) without knowing ( k ) and ( n ). Let me try that.Given ( f(x) = kx^n ), so ( f(2) = k cdot 2^n = 1/4 ), and we need ( f(1/2) = k cdot (1/2)^n ). Let me express ( f(1/2) ) in terms of ( f(2) ).Notice that ( f(1/2) = k cdot (1/2)^n = k cdot 2^{-n} ). But from ( f(2) = k cdot 2^n = 1/4 ), we can write ( k = frac{1}{4} cdot 2^{-n} ). Plugging that into ( f(1/2) ):( f(1/2) = left( frac{1}{4} cdot 2^{-n} right) cdot 2^{-n} = frac{1}{4} cdot 2^{-2n} )Hmm, but that still leaves us with ( n ). Unless we can find a relationship between ( f(2) ) and ( f(1/2) ) without ( n ). Let me think.Alternatively, maybe we can express ( f(1/2) ) as ( (f(2))^{m} ) for some exponent ( m ). Let's see:If ( f(2) = 1/4 ), then ( f(1/2) = k cdot (1/2)^n ). But ( k = frac{1}{4} cdot 2^{-n} ), so:( f(1/2) = frac{1}{4} cdot 2^{-n} cdot 2^{-n} = frac{1}{4} cdot 2^{-2n} )But ( 2^{-2n} = (2^n)^{-2} ). From ( f(2) = k cdot 2^n = 1/4 ), we have ( 2^n = frac{1}{4k} ). Wait, but that might not help directly.Alternatively, let me take the ratio of ( f(1/2) ) to ( f(2) ):( frac{f(1/2)}{f(2)} = frac{k cdot (1/2)^n}{k cdot 2^n} = frac{(1/2)^n}{2^n} = frac{2^{-n}}{2^n} = 2^{-2n} )So ( f(1/2) = f(2) cdot 2^{-2n} ). But we don't know ( n ), so this might not help.Wait, but if I can express ( 2^{-2n} ) in terms of ( f(2) ). From ( f(2) = k cdot 2^n = 1/4 ), so ( 2^n = frac{1}{4k} ). Then ( 2^{-2n} = left( 2^n right)^{-2} = left( frac{1}{4k} right)^{-2} = (4k)^2 = 16k^2 ). So:( f(1/2) = f(2) cdot 16k^2 ). But ( f(2) = 1/4 ), so:( f(1/2) = frac{1}{4} cdot 16k^2 = 4k^2 )Hmm, but we still have ( k ) in there. From ( f(2) = k cdot 2^n = 1/4 ), we can write ( k = frac{1}{4 cdot 2^n} ). So ( k^2 = frac{1}{16 cdot 4^n} ). Therefore:( f(1/2) = 4 cdot frac{1}{16 cdot 4^n} = frac{1}{4 cdot 4^n} = frac{1}{4^{n+1}} )Hmm, but this seems to be going in circles. Maybe I need a different approach.Wait, let's think about the properties of power functions. If ( f(x) = kx^n ), then ( f(a) = k a^n ) and ( f(b) = k b^n ). So if we have two points, we can solve for ( k ) and ( n ). But here, we only have one point. So unless there's another condition, like the function is even or odd, or something else, we can't uniquely determine ( k ) and ( n ).Wait, but the problem is asking for ( f(1/2) ). Maybe there's a relationship between ( f(2) ) and ( f(1/2) ) that can be expressed without knowing ( k ) and ( n ). Let me try to express ( f(1/2) ) in terms of ( f(2) ).Let me denote ( f(2) = frac{1}{4} ). So ( f(2) = k cdot 2^n = frac{1}{4} ). Then, ( f(1/2) = k cdot (1/2)^n = k cdot 2^{-n} ). Let me express ( f(1/2) ) in terms of ( f(2) ):From ( f(2) = k cdot 2^n = frac{1}{4} ), we can write ( k = frac{1}{4 cdot 2^n} ). Then, substituting into ( f(1/2) ):( f(1/2) = frac{1}{4 cdot 2^n} cdot 2^{-n} = frac{1}{4} cdot 2^{-2n} )Hmm, so ( f(1/2) = frac{1}{4} cdot (2^{-2n}) ). But ( 2^{-2n} = (2^n)^{-2} ). From ( f(2) = k cdot 2^n = frac{1}{4} ), we have ( 2^n = frac{1}{4k} ). So:( 2^{-2n} = left( frac{1}{4k} right)^{-2} = (4k)^2 = 16k^2 ). Therefore, ( f(1/2) = frac{1}{4} cdot 16k^2 = 4k^2 ).But we still don't know ( k ). Wait, but from ( f(2) = k cdot 2^n = frac{1}{4} ), we can express ( k = frac{1}{4 cdot 2^n} ). So ( k^2 = frac{1}{16 cdot 4^n} ). Therefore, ( f(1/2) = 4 cdot frac{1}{16 cdot 4^n} = frac{1}{4 cdot 4^n} = frac{1}{4^{n+1}} ).Hmm, this is getting complicated. Maybe I need to make an assumption here. If the problem is expecting a numerical answer, perhaps it's assuming that ( k = 1 ). So if ( f(x) = x^n ), then ( f(2) = 2^n = 1/4 ), so ( n = -2 ). Then ( f(1/2) = (1/2)^{-2} = 4 ). So the answer would be 4.Alternatively, if ( k ) isn't 1, but we don't have another condition, maybe the problem expects the simplest case where ( k = 1 ). That seems plausible because often in problems like this, unless specified otherwise, they assume the simplest form. So ( f(x) = x^{-2} ), so ( f(1/2) = 4 ).Wait, but let me verify. If ( f(x) = x^{-2} ), then ( f(2) = 1/4 ), which matches. And ( f(1/2) = (1/2)^{-2} = 4 ). So that works.Alternatively, if ( k ) isn't 1, say ( k = 1/2 ), then ( f(2) = (1/2) cdot 2^n = 1/4 ). So ( 2^n = 1/2 ), so ( n = -1 ). Then ( f(1/2) = (1/2) cdot (1/2)^{-1} = (1/2) cdot 2 = 1 ). So that's another possible answer.But since the problem is asking for a numerical answer, and it's presented as a fill-in-the-blank, it's expecting a specific number. So maybe the answer is 4, assuming ( k = 1 ). Alternatively, maybe the answer is 1, assuming ( k = 1/2 ). But without more information, it's ambiguous.Wait, but in the definition of a power function, is ( k ) allowed to be any constant, or is it typically assumed to be 1? I think ( k ) can be any constant, so without knowing ( k ), we can't determine ( n ), and thus can't find ( f(1/2) ). But the problem is asking for a numerical answer, so maybe I'm missing something.Wait, perhaps the function is a monomial function, which is a power function with ( k = 1 ). So if that's the case, then ( f(x) = x^n ), and ( f(2) = 2^n = 1/4 ), so ( n = -2 ). Therefore, ( f(1/2) = (1/2)^{-2} = 4 ). So that's the answer.Alternatively, maybe the function is ( f(x) = x^{-2} ), which is a power function, and that passes through ( (2, 1/4) ). So ( f(1/2) = 4 ).Wait, but let me check another approach. If I take the ratio of ( f(1/2) ) to ( f(2) ):( frac{f(1/2)}{f(2)} = frac{k cdot (1/2)^n}{k cdot 2^n} = frac{(1/2)^n}{2^n} = frac{2^{-n}}{2^n} = 2^{-2n} )So ( f(1/2) = f(2) cdot 2^{-2n} ). But ( f(2) = 1/4 ), so ( f(1/2) = frac{1}{4} cdot 2^{-2n} ). Hmm, but without knowing ( n ), I can't compute this. Unless I can express ( 2^{-2n} ) in terms of ( f(2) ).Wait, from ( f(2) = k cdot 2^n = 1/4 ), we can write ( 2^n = frac{1}{4k} ). Therefore, ( 2^{-2n} = left( 2^n right)^{-2} = left( frac{1}{4k} right)^{-2} = (4k)^2 = 16k^2 ). So, ( f(1/2) = frac{1}{4} cdot 16k^2 = 4k^2 ).But we still have ( k ) in the equation. From ( f(2) = k cdot 2^n = 1/4 ), we can write ( k = frac{1}{4 cdot 2^n} ). So ( k^2 = frac{1}{16 cdot 4^n} ). Therefore, ( f(1/2) = 4 cdot frac{1}{16 cdot 4^n} = frac{1}{4 cdot 4^n} = frac{1}{4^{n+1}} ).Hmm, this seems to be going in circles. Maybe I need to accept that without additional information, the problem can't be uniquely solved. But since it's a problem given to me, it's expecting an answer, so perhaps I need to make an assumption.Wait, maybe the function is its own inverse or something? If ( f(f(x)) = x ), but that might not necessarily be the case.Alternatively, maybe the function is symmetric in some way. For example, if ( f(2) = 1/4 ), then ( f(1/4) = 2 ). But that's not directly helpful here.Wait, another thought: if ( f(x) = kx^n ), then ( f(2) = k cdot 2^n = 1/4 ), and ( f(1/2) = k cdot (1/2)^n ). Let me denote ( y = 2^n ). Then, ( f(2) = k cdot y = 1/4 ), so ( k = frac{1}{4y} ). Then, ( f(1/2) = k cdot (1/2)^n = frac{1}{4y} cdot y^{-1} = frac{1}{4y^2} ). But ( y = 2^n ), so ( y^2 = 4^n ). Therefore, ( f(1/2) = frac{1}{4 cdot 4^n} = frac{1}{4^{n+1}} ).Hmm, but without knowing ( n ), I can't compute this. So maybe the problem is expecting a general expression, but the question is asking for a numerical value. So perhaps I'm missing something.Wait, maybe the function is such that ( f(2) = 1/4 ) and ( f(1/2) ) is the reciprocal or something. Let me test that.If ( f(2) = 1/4 ), then maybe ( f(1/2) = 4 ). That's a possibility. Alternatively, maybe ( f(1/2) = 1/16 ), if the function is squared or something.Wait, but without knowing the exponent, it's hard to say. Maybe the function is linear, but no, a power function with exponent 1 would be linear, but ( f(2) = 1/4 ) would imply ( k = 1/8 ), so ( f(1/2) = 1/16 ). But that's just one possibility.Alternatively, if the function is quadratic, ( f(x) = kx^2 ), then ( f(2) = 4k = 1/4 ), so ( k = 1/16 ), and ( f(1/2) = (1/16)(1/4) = 1/64 ). But again, that's another possibility.Wait, but the problem is asking for ( f(1/2) ), and it's expecting a numerical answer. So perhaps the answer is 4, assuming ( n = -2 ) and ( k = 1 ). That seems to be the most straightforward case.Alternatively, maybe the function is ( f(x) = x^{-2} ), which is a standard power function, and that would give ( f(1/2) = 4 ). So I think that's the answer they're expecting.So, to summarize, assuming the power function is ( f(x) = x^{-2} ), which passes through ( (2, 1/4) ), then ( f(1/2) = 4 ).**Final Answer**boxed{4}

answer：Okay, so I need to figure out Petya's apartment number. It's a three-digit number, and if you rearrange its digits, you get five other three-digit numbers. The sum of these five numbers is exactly 2017. Hmm, let me break this down step by step.First, let's denote Petya's apartment number as a three-digit number, which we can write as ABC, where A, B, and C are its digits. Since it's a three-digit number, A can't be zero. So, A is between 1 and 9, and B and C are between 0 and 9.Now, when we rearrange the digits, we get five other three-digit numbers. That means the original number ABC is one of the permutations, and the other five are the different arrangements of A, B, and C. So, in total, there are six permutations of the digits, and the sum of all six permutations would be the sum of all possible rearrangements.Wait, but the problem says the sum of the five other numbers is 2017. So, that means if we take the sum of all six permutations (including ABC) and subtract ABC, we get 2017. So, the sum of all six permutations is 2017 + ABC.Let me write that down:Sum of all six permutations = 2017 + ABCNow, I need to find the sum of all six permutations. How can I calculate that?I remember that for a three-digit number with digits A, B, and C, each digit appears in each place (hundreds, tens, ones) exactly two times across all permutations. So, for each digit, it appears twice in the hundreds place, twice in the tens place, and twice in the ones place.Therefore, the sum of all six permutations can be calculated as:Sum = 2*(A + B + C)*100 + 2*(A + B + C)*10 + 2*(A + B + C)*1 = 2*(A + B + C)*(100 + 10 + 1) = 2*(A + B + C)*111 = 222*(A + B + C)So, the sum of all six permutations is 222*(A + B + C). Therefore, according to the earlier equation:222*(A + B + C) = 2017 + ABCSo, we have:222*(A + B + C) - ABC = 2017Now, let me denote S = A + B + C. Then, the equation becomes:222*S - ABC = 2017So, ABC = 222*S - 2017But ABC is a three-digit number, so ABC must be between 100 and 999.Therefore, 100 ≤ 222*S - 2017 ≤ 999Let me solve for S:100 ≤ 222*S - 2017 ≤ 999Adding 2017 to all parts:2117 ≤ 222*S ≤ 3016Dividing all parts by 222:2117 / 222 ≤ S ≤ 3016 / 222Calculating 2117 / 222:222*9 = 19982117 - 1998 = 119So, 2117 / 222 = 9 + 119/222 ≈ 9.536Similarly, 3016 / 222:222*13 = 28863016 - 2886 = 130So, 3016 / 222 = 13 + 130/222 ≈ 13.585Since S is the sum of three digits, each between 0 and 9, so S can be at most 9 + 9 + 9 = 27, and at least 1 + 0 + 0 = 1. But from the above, S must be between approximately 9.536 and 13.585. Since S must be an integer, S can be 10, 11, 12, or 13.So, possible values for S are 10, 11, 12, 13.Let me compute ABC for each possible S:For S = 10:ABC = 222*10 - 2017 = 2220 - 2017 = 203For S = 11:ABC = 222*11 - 2017 = 2442 - 2017 = 425For S = 12:ABC = 222*12 - 2017 = 2664 - 2017 = 647For S = 13:ABC = 222*13 - 2017 = 2886 - 2017 = 869So, the possible apartment numbers are 203, 425, 647, and 869.Now, I need to check which of these numbers satisfy the condition that the sum of the five other permutations is 2017.Let me check each candidate:1. ABC = 203, S = 2 + 0 + 3 = 5. Wait, but earlier S was supposed to be 10, 11, 12, or 13. So, this is a contradiction. Therefore, 203 is invalid because S should be 10, but the actual sum is 5. So, 203 is not a valid number.Wait, hold on, that can't be right. Because if ABC = 203, then S = 2 + 0 + 3 = 5, but according to our earlier calculation, S should be 10. So, this is a problem. That means our initial assumption that S is 10, 11, 12, or 13 is conflicting with the actual digit sums.Wait, perhaps I made a mistake in my reasoning. Let me double-check.We had:222*S - ABC = 2017But S is the sum of the digits of ABC. So, if ABC is 203, then S should be 2 + 0 + 3 = 5, but according to the equation, 222*5 - 203 = 1110 - 203 = 907, which is not equal to 2017. So, 203 is invalid.Similarly, for ABC = 425, S = 4 + 2 + 5 = 11, which is in our range.So, let's check if 425 is valid.Compute the sum of all six permutations:Sum = 222*S = 222*11 = 2442Sum of five other numbers = 2442 - 425 = 2017Which is exactly what the problem states. So, 425 is a valid candidate.Wait, but let me check the other candidates as well.ABC = 647, S = 6 + 4 + 7 = 17. Wait, 17 is outside our earlier range of 10 to 13. So, that's a problem. Because according to our earlier calculation, S should be 10, 11, 12, or 13.But 647 has S = 17, which is outside that range. So, that's a contradiction.Similarly, ABC = 869, S = 8 + 6 + 9 = 23, which is way outside the range. So, that's also a problem.Wait, so only ABC = 425 has S = 11, which is within our initial range. The others have S outside the range, which contradicts our earlier calculation.So, that suggests that ABC must be 425.But let me double-check.If ABC = 425, then the sum of all six permutations is 222*11 = 2442.Therefore, the sum of the other five numbers is 2442 - 425 = 2017, which matches the problem statement.So, that seems correct.But just to be thorough, let me list all permutations of 425 and sum them up to ensure that the sum of the five others is indeed 2017.The digits are 4, 2, 5.All permutations:425, 452, 245, 254, 524, 542Now, let's compute the sum of all six:425 + 452 + 245 + 254 + 524 + 542Let me add them step by step:425 + 452 = 877877 + 245 = 11221122 + 254 = 13761376 + 524 = 19001900 + 542 = 2442So, the total sum is 2442, which is correct.Therefore, the sum of the five other numbers is 2442 - 425 = 2017, which matches.So, that confirms that 425 is the correct apartment number.But just to ensure there are no other possibilities, let me check why the other candidates didn't work.For ABC = 203, S = 5, which is too low, and the sum of all permutations would be 222*5 = 1110, which is way below 2017. So, that's invalid.For ABC = 647, S = 17, which would give a total sum of 222*17 = 3774. Then, the sum of the other five numbers would be 3774 - 647 = 3127, which is way higher than 2017.Similarly, ABC = 869, S = 23, total sum = 222*23 = 5106, sum of others = 5106 - 869 = 4237, which is way higher.So, only ABC = 425 works.Therefore, Petya's apartment number is 425.