answer：First, I need to determine the total number of coins in the treasure, which I'll denote as ( m ).The first pirate received one-third of the total coins plus 1 coin, so that's ( frac{m}{3} + 1 ).The second pirate received one-fourth of the total coins plus 5 coins, which is ( frac{m}{4} + 5 ).The third pirate received one-fifth of the total coins plus 20 coins, so that's ( frac{m}{5} + 20 ).Since all the coins were distributed among the three pirates, I can set up the equation:[left( frac{m}{3} + 1 right) + left( frac{m}{4} + 5 right) + left( frac{m}{5} + 20 right) = m]Combining the fractions and constants separately:[frac{m}{3} + frac{m}{4} + frac{m}{5} + 1 + 5 + 20 = m]Simplifying the constants:[frac{m}{3} + frac{m}{4} + frac{m}{5} + 26 = m]To combine the fractions, I'll find a common denominator, which is 60:[frac{20m}{60} + frac{15m}{60} + frac{12m}{60} + 26 = m]Adding the fractions:[frac{47m}{60} + 26 = m]Isolating ( m ):[26 = m - frac{47m}{60}][26 = frac{13m}{60}]Multiplying both sides by 60:[1560 = 13m]Dividing both sides by 13:[m = 120]So, the total number of coins in the treasure is 120.

answer：First, I need to determine the number of slips drawn from the hat such that the probability of all slips having odd numbers is 0.023809523809523808.There are 10 slips in total, with 5 odd numbers (1, 3, 5, 7, 9) and 5 even numbers (0, 2, 4, 6, 8).The probability of drawing ( n ) slips all with odd numbers is given by the ratio of the number of ways to choose ( n ) odd slips to the number of ways to choose ( n ) slips from all 10 slips.This can be expressed using combinations:[P = frac{binom{5}{n}}{binom{10}{n}}]I need to find the value of ( n ) that satisfies:[frac{binom{5}{n}}{binom{10}{n}} = 0.023809523809523808]I'll test different values of ( n ):For ( n = 1 ):[P = frac{5}{10} = 0.5]This does not match the given probability.For ( n = 2 ):[P = frac{10}{45} approx 0.2222222222222222]This does not match the given probability.For ( n = 3 ):[P = frac{10}{120} approx 0.08333333333333333]This does not match the given probability.For ( n = 4 ):[P = frac{5}{210} approx 0.023809523809523808]This matches the given probability.Therefore, the number of slips drawn is 4.

answer：Alright, so I have this problem with complex numbers z and w. The given conditions are |3z - w| = 15, |z + 3w| = 9, and |z + w| = 5. I need to find |z|. Hmm, okay, complex numbers can be tricky, but I remember that the modulus squared of a complex number can be expanded using the property that |k|^2 = k multiplied by its conjugate. Maybe I can use that here.Let me write down the given equations:1. |3z - w| = 152. |z + 3w| = 93. |z + w| = 5Since these are moduli, I can square both sides to make them easier to work with. So, squaring each equation:1. |3z - w|² = 2252. |z + 3w|² = 813. |z + w|² = 25Now, expanding each of these using the property |a + b|² = |a|² + |b|² + 2Re(aoverline{b}), where Re denotes the real part. But since dealing with real parts might complicate things, maybe I can represent these in terms of zoverline{z}, woverline{w}, and the cross terms.Let me denote:- Let a = zoverline{z} = |z|²- Let b = woverline{w} = |w|²- Let c = zoverline{w} + woverline{z} = 2Re(zoverline{w})Wait, actually, c is twice the real part of zoverline{w}, but maybe I can just keep it as c for simplicity.So, expanding each equation:1. |3z - w|² = (3z - w)(3overline{z} - overline{w}) = 9zoverline{z} - 3zoverline{w} - 3woverline{z} + woverline{w} = 9a - 3c + b = 2252. |z + 3w|² = (z + 3w)(overline{z} + 3overline{w}) = zoverline{z} + 3zoverline{w} + 3woverline{z} + 9woverline{w} = a + 3c + 9b = 813. |z + w|² = (z + w)(overline{z} + overline{w}) = zoverline{z} + zoverline{w} + woverline{z} + woverline{w} = a + c + b = 25So now I have three equations:1. 9a - 3c + b = 2252. a + 3c + 9b = 813. a + b + c = 25Hmm, okay, so I have three equations with three variables: a, b, c. Maybe I can solve this system.Let me write them again:1. 9a - 3c + b = 225 ...(1)2. a + 3c + 9b = 81 ...(2)3. a + b + c = 25 ...(3)I need to solve for a, which is |z|², so eventually, I can take the square root to find |z|.Let me see. Maybe I can express c from equation (3) in terms of a and b, and substitute into the other equations.From equation (3):c = 25 - a - bSo, c = 25 - a - bNow, substitute this into equations (1) and (2):Equation (1):9a - 3(25 - a - b) + b = 225Let me expand that:9a - 75 + 3a + 3b + b = 225Combine like terms:(9a + 3a) + (3b + b) - 75 = 22512a + 4b - 75 = 225Bring constants to the right:12a + 4b = 225 + 75 = 300Divide both sides by 4:3a + b = 75 ...(4)Equation (2):a + 3(25 - a - b) + 9b = 81Expand:a + 75 - 3a - 3b + 9b = 81Combine like terms:(a - 3a) + (-3b + 9b) + 75 = 81-2a + 6b + 75 = 81Bring constants to the right:-2a + 6b = 81 - 75 = 6Divide both sides by 2:- a + 3b = 3 ...(5)Now, I have equations (4) and (5):(4): 3a + b = 75(5): -a + 3b = 3I can solve this system. Let me use substitution or elimination. Maybe elimination is easier here.Let me multiply equation (5) by 3 to make the coefficients of a opposites:-3a + 9b = 9 ...(5a)Now, add equation (4) and (5a):(3a + b) + (-3a + 9b) = 75 + 9Simplify:0a + 10b = 84So, 10b = 84Therefore, b = 84 / 10 = 8.4Hmm, 8.4 is 42/5. Maybe I can keep it as a fraction for accuracy.So, b = 42/5Now, substitute back into equation (5):- a + 3*(42/5) = 3Compute 3*(42/5) = 126/5So, -a + 126/5 = 3Convert 3 to fifths: 3 = 15/5So, -a + 126/5 = 15/5Subtract 126/5 from both sides:-a = 15/5 - 126/5 = (15 - 126)/5 = (-111)/5Multiply both sides by -1:a = 111/5Wait, 111 divided by 5 is 22.2. Hmm, but let me check my calculations because 111/5 seems a bit messy.Wait, let me go back.From equation (5):- a + 3b = 3We found b = 42/5So, -a + 3*(42/5) = 3Compute 3*(42/5) = 126/5So, -a + 126/5 = 3Convert 3 to 15/5:- a + 126/5 = 15/5Subtract 126/5:- a = 15/5 - 126/5 = (-111)/5So, a = 111/5Hmm, 111/5 is 22.2, which is 22 and 1/5. That seems a bit unusual, but maybe it's correct.Wait, let me check equation (4):3a + b = 75If a = 111/5 and b = 42/5, then:3*(111/5) + 42/5 = (333 + 42)/5 = 375/5 = 75Yes, that checks out.So, a = 111/5, b = 42/5Now, from equation (3):a + b + c = 25So, c = 25 - a - b = 25 - 111/5 - 42/5Convert 25 to 125/5:c = 125/5 - 111/5 - 42/5 = (125 - 111 - 42)/5 = (-28)/5So, c = -28/5Okay, so now I have a = 111/5, b = 42/5, c = -28/5But wait, a is |z|², so |z| = sqrt(a) = sqrt(111/5)But sqrt(111/5) is sqrt(22.2), which is approximately 4.71, but the problem might expect an exact value or maybe I made a mistake somewhere.Wait, let me double-check my equations.From equation (1):9a - 3c + b = 225Substitute a = 111/5, b = 42/5, c = -28/5:9*(111/5) - 3*(-28/5) + 42/5Compute each term:9*(111/5) = 999/5-3*(-28/5) = 84/542/5 is 42/5So, total: 999/5 + 84/5 + 42/5 = (999 + 84 + 42)/5 = 1125/5 = 225Yes, that's correct.Equation (2):a + 3c + 9b = 81Substitute a = 111/5, b = 42/5, c = -28/5:111/5 + 3*(-28/5) + 9*(42/5)Compute each term:111/53*(-28/5) = -84/59*(42/5) = 378/5Total: 111/5 - 84/5 + 378/5 = (111 - 84 + 378)/5 = (27 + 378)/5 = 405/5 = 81Correct.Equation (3):a + b + c = 25111/5 + 42/5 -28/5 = (111 + 42 -28)/5 = 125/5 = 25Correct.So, all equations are satisfied. Therefore, a = 111/5, which is |z|².Thus, |z| = sqrt(111/5). Let me simplify that.sqrt(111/5) can be written as sqrt(111)/sqrt(5). Rationalizing the denominator:sqrt(111)/sqrt(5) = (sqrt(111)*sqrt(5))/5 = sqrt(555)/5But 555 factors into 5*111, which is 5*3*37. So, sqrt(555) doesn't simplify further.Alternatively, sqrt(111/5) is approximately 4.71, but since the problem might expect an exact value, I think sqrt(111/5) is acceptable, but maybe it can be written differently.Wait, 111/5 is 22.2, but perhaps there's a mistake in my earlier steps because 111/5 seems a bit messy. Let me go back and see if I made any calculation errors.Looking back at equation (5):- a + 3b = 3We had b = 42/5, so:- a + 3*(42/5) = 3Which is -a + 126/5 = 3Convert 3 to 15/5:- a + 126/5 = 15/5Subtract 126/5:- a = 15/5 - 126/5 = (-111)/5So, a = 111/5Yes, that seems correct.Alternatively, maybe I can express a as 22.2, but I think it's better to leave it as a fraction.So, |z| = sqrt(111/5). But let me see if this can be simplified or if I made a miscalculation earlier.Wait, another approach: Maybe instead of introducing c as zoverline{w} + woverline{z}, which is 2Re(zoverline{w}), I could have used vectors or another method, but I think the current approach is solid.Alternatively, perhaps I can use linear algebra to solve the system more carefully.We have:Equation (4): 3a + b = 75Equation (5): -a + 3b = 3Let me solve equation (4) for b:b = 75 - 3aSubstitute into equation (5):- a + 3*(75 - 3a) = 3Compute:- a + 225 - 9a = 3Combine like terms:-10a + 225 = 3Subtract 225:-10a = 3 - 225 = -222Divide by -10:a = (-222)/(-10) = 222/10 = 111/5Yes, same result.So, a = 111/5, which is 22.2.Therefore, |z| = sqrt(111/5). Let me rationalize it:sqrt(111/5) = sqrt(111)/sqrt(5) = (sqrt(111)*sqrt(5))/5 = sqrt(555)/5But 555 is 5*111, and 111 is 3*37, so no further simplification.Alternatively, sqrt(111/5) is approximately 4.71, but since the problem likely expects an exact value, I think sqrt(111/5) is the answer.Wait, but looking back, the problem gives all moduli as integers: 15, 9, 5. It's possible that |z| is an integer as well. Did I make a mistake somewhere?Let me check the equations again.From equation (3):a + b + c = 25From equation (1):9a - 3c + b = 225From equation (2):a + 3c + 9b = 81I solved for c = 25 - a - bSubstituted into (1) and (2), leading to:12a + 4b = 300 => 3a + b = 75-2a + 6b = 6 => -a + 3b = 3Solving these gave a = 111/5, which is 22.2, and b = 42/5 = 8.4But 22.2 and 8.4 are not integers, which seems odd given the problem's context.Wait, maybe I made a mistake in expanding the equations.Let me re-examine the expansion of |3z - w|²:(3z - w)(3overline{z} - overline{w}) = 9zoverline{z} - 3zoverline{w} - 3woverline{z} + woverline{w} = 9a - 3c + bYes, that's correct.Similarly, |z + 3w|² = zoverline{z} + 3zoverline{w} + 3woverline{z} + 9woverline{w} = a + 3c + 9bCorrect.And |z + w|² = a + c + bCorrect.So, the expansions are correct.Then, substituting c = 25 - a - b into equations (1) and (2) seems correct.So, perhaps the answer is indeed sqrt(111/5). Alternatively, maybe I can write it as (sqrt(555))/5, but that's the same thing.Wait, sqrt(111/5) can also be written as sqrt(22.2), but that's not helpful.Alternatively, maybe I can rationalize it differently, but I don't think so.Alternatively, perhaps I made a mistake in the initial setup.Wait, another approach: Let me consider z and w as vectors in the complex plane. Maybe I can set up a system of equations using their real and imaginary parts.Let me denote z = x + yi and w = u + vi, where x, y, u, v are real numbers.Then, the given moduli become:1. |3z - w| = |3(x + yi) - (u + vi)| = |(3x - u) + (3y - v)i| = sqrt[(3x - u)² + (3y - v)²] = 152. |z + 3w| = |(x + 3u) + (y + 3v)i| = sqrt[(x + 3u)² + (y + 3v)²] = 93. |z + w| = |(x + u) + (y + v)i| = sqrt[(x + u)² + (y + v)²] = 5So, squaring each:1. (3x - u)² + (3y - v)² = 225 ...(1)2. (x + 3u)² + (y + 3v)² = 81 ...(2)3. (x + u)² + (y + v)² = 25 ...(3)Now, I have three equations with four variables: x, y, u, v. But I need to find |z| = sqrt(x² + y²). So, maybe I can find x² + y² from these equations.Let me expand each equation:Equation (1):(3x - u)² + (3y - v)² = 9x² - 6xu + u² + 9y² - 6yv + v² = 9x² + 9y² + u² + v² - 6xu - 6yv = 225Equation (2):(x + 3u)² + (y + 3v)² = x² + 6xu + 9u² + y² + 6yv + 9v² = x² + y² + 9u² + 9v² + 6xu + 6yv = 81Equation (3):(x + u)² + (y + v)² = x² + 2xu + u² + y² + 2yv + v² = x² + y² + u² + v² + 2xu + 2yv = 25Now, let me denote:Let A = x² + y²Let B = u² + v²Let C = xu + yvSo, now, rewrite the equations:Equation (1):9A + B - 6C = 225 ...(1)Equation (2):A + 9B + 6C = 81 ...(2)Equation (3):A + B + 2C = 25 ...(3)Now, I have three equations with three variables: A, B, C.This looks similar to the previous system, but now A is |z|², which is what we need.So, let's write them:1. 9A + B - 6C = 2252. A + 9B + 6C = 813. A + B + 2C = 25Now, let's solve this system.First, from equation (3):A + B + 2C = 25Let me solve for one variable, say B:B = 25 - A - 2CNow, substitute B into equations (1) and (2):Equation (1):9A + (25 - A - 2C) - 6C = 225Simplify:9A + 25 - A - 2C - 6C = 225Combine like terms:(9A - A) + (-2C - 6C) + 25 = 2258A - 8C + 25 = 225Subtract 25:8A - 8C = 200Divide both sides by 8:A - C = 25 ...(4)Equation (2):A + 9*(25 - A - 2C) + 6C = 81Expand:A + 225 - 9A - 18C + 6C = 81Combine like terms:(A - 9A) + (-18C + 6C) + 225 = 81-8A - 12C + 225 = 81Subtract 225:-8A - 12C = -144Divide both sides by -4:2A + 3C = 36 ...(5)Now, we have equations (4) and (5):(4): A - C = 25(5): 2A + 3C = 36Let me solve equation (4) for A:A = C + 25Substitute into equation (5):2*(C + 25) + 3C = 36Expand:2C + 50 + 3C = 36Combine like terms:5C + 50 = 36Subtract 50:5C = -14So, C = -14/5 = -2.8Now, substitute back into equation (4):A - (-14/5) = 25So, A + 14/5 = 25Convert 25 to fifths: 25 = 125/5So, A = 125/5 - 14/5 = 111/5So, A = 111/5, which is |z|² = 111/5Therefore, |z| = sqrt(111/5)Hmm, same result as before. So, it seems correct.But wait, 111/5 is 22.2, and sqrt(22.2) is approximately 4.71, which is not an integer. But the problem gives all moduli as integers, so maybe I made a mistake in interpreting the problem or in the calculations.Wait, let me check the substitution again.From equation (3):A + B + 2C = 25So, B = 25 - A - 2CThen, equation (1):9A + B - 6C = 225Substitute B:9A + (25 - A - 2C) - 6C = 225Simplify:9A +25 - A -2C -6C = 2258A -8C +25 = 2258A -8C = 200A - C =25Equation (4): A = C +25Equation (2):A +9B +6C =81Substitute B:A +9*(25 - A -2C) +6C =81Expand:A +225 -9A -18C +6C =81Combine:-8A -12C +225 =81-8A -12C = -144Divide by -4:2A +3C =36Now, substitute A = C +25 into 2A +3C =36:2*(C +25) +3C =362C +50 +3C =365C +50 =365C = -14C= -14/5Then, A= C +25= -14/5 +125/5=111/5Yes, same result.So, it seems that |z|²=111/5, so |z|=sqrt(111/5). But let me see if this can be simplified or if I made a mistake in the initial setup.Wait, another approach: Maybe I can use the fact that |z + w| =5, |3z -w|=15, and |z +3w|=9 to set up a system of equations using inner products or something similar.Alternatively, perhaps I can consider these as vectors and use the properties of vector addition and subtraction.But I think the approach I took is correct, and the result is |z|=sqrt(111/5). Since the problem didn't specify that |z| has to be an integer, I think this is the correct answer.But just to be thorough, let me check if sqrt(111/5) is indeed the correct value.Compute sqrt(111/5):111 divided by 5 is 22.2sqrt(22.2) ≈ 4.714Let me see if this makes sense with the given moduli.Given that |z + w|=5, which is relatively small, and |3z -w|=15, which is larger, it suggests that z and w are such that when scaled and combined, they produce these moduli.But perhaps there's a different approach to solve this problem without getting into fractions.Wait, another idea: Let me consider the given equations as a system and try to eliminate variables.We have:1. 9A + B -6C =2252. A +9B +6C =813. A + B +2C =25Let me add equations (1) and (2):(9A + B -6C) + (A +9B +6C) =225 +8110A +10B =306Divide by 10:A + B =30.6But from equation (3):A + B +2C =25So, 30.6 +2C =25Wait, that can't be right because 30.6 +2C =25 would imply 2C= -5.6, so C= -2.8, which is consistent with earlier results.But 30.6 is 153/5, so A + B =153/5From equation (3):A + B +2C =25So, 153/5 +2C =25Convert 25 to 125/5:153/5 +2C =125/5Subtract 153/5:2C =125/5 -153/5 = (-28)/5So, C= (-28)/10= -14/5= -2.8Which is consistent.So, A + B =153/5=30.6And from equation (4):A - C=25C= -14/5, so A=25 + (-14/5)=125/5 -14/5=111/5=22.2So, A=111/5, B=153/5 -111/5=42/5=8.4So, same results.Therefore, I think the answer is indeed |z|=sqrt(111/5). But let me see if this can be expressed differently.sqrt(111/5)=sqrt(22.2)=approximately 4.714But maybe the problem expects an exact value, so I'll leave it as sqrt(111/5).Alternatively, rationalizing the denominator:sqrt(111/5)=sqrt(555)/5But 555=5×111=5×3×37, so it doesn't simplify further.Therefore, the exact value is sqrt(111/5), which is approximately 4.714.But wait, let me check if I can write it as a mixed number or something, but I don't think that's necessary.Alternatively, maybe I made a mistake in the initial setup by not considering that z and w could be colinear or something, but I don't think that's the case.Alternatively, perhaps I can use the method of solving for z and w directly.Let me assume that z and w are complex numbers, so they can be represented as vectors in 2D space.Let me denote z = a + bi and w = c + di.Then, the given moduli become:1. |3z - w| =152. |z +3w|=93. |z +w|=5So, writing these out:1. sqrt[(3a -c)^2 + (3b -d)^2]=152. sqrt[(a +3c)^2 + (b +3d)^2]=93. sqrt[(a +c)^2 + (b +d)^2]=5Squaring each:1. (3a -c)^2 + (3b -d)^2=2252. (a +3c)^2 + (b +3d)^2=813. (a +c)^2 + (b +d)^2=25Now, expanding each:Equation (1):9a² -6ac +c² +9b² -6bd +d²=225Equation (2):a² +6ac +9c² +b² +6bd +9d²=81Equation (3):a² +2ac +c² +b² +2bd +d²=25Now, let me denote:Let A = a² + b²Let B = c² + d²Let C = ac + bdSo, now, rewrite the equations:Equation (1):9A + B -6C=225Equation (2):A +9B +6C=81Equation (3):A + B +2C=25Which is the same system as before.So, solving this system gives A=111/5, B=42/5, C=-14/5Therefore, |z|=sqrt(A)=sqrt(111/5)So, I think this confirms that the answer is indeed sqrt(111/5).But just to be thorough, let me compute sqrt(111/5):111 divided by 5 is 22.2sqrt(22.2) is approximately 4.714But since the problem didn't specify that |z| has to be an integer, I think this is acceptable.Alternatively, maybe I can write it as a fraction with a square root:sqrt(111)/sqrt(5)= (sqrt(111)*sqrt(5))/5= sqrt(555)/5But 555=5×111=5×3×37, so it doesn't simplify further.Therefore, the exact value is sqrt(111/5), which is approximately 4.714.So, I think that's the answer.

answer：Okay, so I'm trying to figure out how much money the farmer will make from the pig's bacon this month. Let me start by understanding the problem step by step.First, the farmer usually gets 20 pounds of bacon from a pig. That's the average amount. He sells each pound for 6 at the farmer's market. So, normally, he would make 20 pounds multiplied by 6 per pound. Let me calculate that: 20 times 6 is 120. So, normally, he makes 120 from a pig.But this month, the pig is a runt, which means it's smaller than average. Specifically, it's only half the size of the average pig. If the average pig gives 20 pounds of bacon, then half the size would mean half the amount of bacon. So, I need to find half of 20 pounds. Half of 20 is 10. So, this month's pig will give him 10 pounds of bacon.Now, he sells each pound for 6, just like usual. So, to find out how much money he makes this month, I need to multiply the amount of bacon he gets by the price per pound. That would be 10 pounds times 6 per pound. Let me do that multiplication: 10 times 6 is 60.So, the farmer will make 60 from the bacon of this month's pig.Wait, let me double-check to make sure I didn't make a mistake. The average is 20 pounds, and the runt is half that size, so 10 pounds. Selling each pound for 6, so 10 times 6 is indeed 60. Yeah, that seems right.I think I've got it. The farmer will make 60 from the pig's bacon this month.

answer：To determine how many zeros follow the number 1 when (10000^{50}) is expanded, I start by expressing 10000 as a power of 10. Since (10000 = 10^4), I can rewrite the expression as ((10^4)^{50}).Next, I apply the power of a power rule, which states that ((a^m)^n = a^{m times n}). Applying this rule, ((10^4)^{50}) becomes (10^{4 times 50}), which simplifies to (10^{200}).The number (10^{200}) is represented as a 1 followed by 200 zeros. Therefore, when (10000^{50}) is expanded, it results in a 1 followed by 200 zeros.

answer：Okay, so I have this math problem here: Are there strictly positive integers ( n ) and ( m ) such that ( 5^n ) and ( 6^m ) end with the same four digits? Hmm, interesting. I need to figure out if there are any positive integers ( n ) and ( m ) where the last four digits of ( 5^n ) and ( 6^m ) are identical. First, I should probably understand what it means for two numbers to end with the same four digits. That means when you look at the number in base 10, the last four digits are exactly the same. For example, if ( 5^n ) ends with 1234, then ( 6^m ) should also end with 1234. So, I need to explore the patterns of the last four digits of powers of 5 and powers of 6. Maybe there's a repeating cycle or some property that can help me determine if they can ever coincide.Let me start by looking at the last four digits of ( 5^n ). I know that multiplying by 5 repeatedly affects the last digits. Let's compute a few powers of 5 modulo 10000 to see if there's a pattern.( 5^1 = 5 ) ( 5^2 = 25 ) ( 5^3 = 125 ) ( 5^4 = 625 ) ( 5^5 = 3125 ) ( 5^6 = 15625 ) ( 5^7 = 78125 ) ( 5^8 = 390625 ) ( 5^9 = 1953125 ) ( 5^{10} = 9765625 ) Wait, but I'm only interested in the last four digits. So, let's write them out modulo 10000:( 5^1 mod 10000 = 5 ) ( 5^2 mod 10000 = 25 ) ( 5^3 mod 10000 = 125 ) ( 5^4 mod 10000 = 625 ) ( 5^5 mod 10000 = 3125 ) ( 5^6 mod 10000 = 15625 mod 10000 = 5625 ) ( 5^7 mod 10000 = 5625 times 5 = 28125 mod 10000 = 8125 ) ( 5^8 mod 10000 = 8125 times 5 = 40625 mod 10000 = 625 ) ( 5^9 mod 10000 = 625 times 5 = 3125 ) ( 5^{10} mod 10000 = 3125 times 5 = 15625 mod 10000 = 5625 ) Hmm, I notice a pattern here. After ( 5^4 ), the last four digits cycle through 625, 3125, 5625, 8125, and then back to 625. So, the cycle length is 4. That means every 4 exponents, the last four digits repeat. So, the cycle is 625, 3125, 5625, 8125, and then repeats.Now, let's look at the last four digits of ( 6^m ). I'll compute a few powers of 6 modulo 10000.( 6^1 = 6 ) ( 6^2 = 36 ) ( 6^3 = 216 ) ( 6^4 = 1296 ) ( 6^5 = 7776 ) ( 6^6 = 46656 mod 10000 = 6656 ) ( 6^7 = 6656 times 6 = 39936 mod 10000 = 9936 ) ( 6^8 = 9936 times 6 = 59616 mod 10000 = 9616 ) ( 6^9 = 9616 times 6 = 57696 mod 10000 = 7696 ) ( 6^{10} = 7696 times 6 = 46176 mod 10000 = 6176 ) ( 6^{11} = 6176 times 6 = 37056 mod 10000 = 7056 ) ( 6^{12} = 7056 times 6 = 42336 mod 10000 = 2336 ) ( 6^{13} = 2336 times 6 = 14016 mod 10000 = 4016 ) ( 6^{14} = 4016 times 6 = 24096 mod 10000 = 4096 ) ( 6^{15} = 4096 times 6 = 24576 mod 10000 = 4576 ) ( 6^{16} = 4576 times 6 = 27456 mod 10000 = 7456 ) ( 6^{17} = 7456 times 6 = 44736 mod 10000 = 4736 ) ( 6^{18} = 4736 times 6 = 28416 mod 10000 = 8416 ) ( 6^{19} = 8416 times 6 = 50496 mod 10000 = 496 ) ( 6^{20} = 496 times 6 = 2976 mod 10000 = 2976 ) Hmm, this seems more complicated. The last four digits of ( 6^m ) don't seem to follow an obvious cycle quickly. Maybe I need to find the cycle length or see if there's a repeating pattern.Alternatively, perhaps I can use Euler's theorem to find the cycle length. Euler's theorem states that if ( a ) and ( n ) are coprime, then ( a^{phi(n)} equiv 1 mod n ), where ( phi(n) ) is Euler's totient function.In this case, ( n = 10000 ). Let's compute ( phi(10000) ). Since 10000 = ( 2^4 times 5^4 ), we have:( phi(10000) = 10000 times (1 - 1/2) times (1 - 1/5) = 10000 times 1/2 times 4/5 = 10000 times 2/5 = 4000 ).So, ( phi(10000) = 4000 ). That means that for any number ( a ) coprime to 10000, ( a^{4000} equiv 1 mod 10000 ).Now, 5 and 10000 are not coprime because 5 divides 10000. Similarly, 6 and 10000 are coprime because 6 and 10000 share no common factors other than 1. So, for 6, we can say that ( 6^{4000} equiv 1 mod 10000 ). That means the cycle length of ( 6^m ) modulo 10000 divides 4000.But for 5, since it's not coprime to 10000, Euler's theorem doesn't apply directly. Instead, we can look at the powers of 5 modulo 10000. Earlier, we saw that the last four digits cycle every 4 exponents after ( 5^4 ). So, the cycle length is 4.Now, if I want ( 5^n equiv 6^m mod 10000 ), I need to find ( n ) and ( m ) such that this congruence holds. Given that the cycle lengths are different (4 for 5 and 4000 for 6), it's not immediately obvious if there's a common solution.Alternatively, maybe I can look at the last four digits more carefully. For ( 5^n ), as we saw, the last four digits cycle through 625, 3125, 5625, 8125. So, the possible last four digits are limited to these four numbers.For ( 6^m ), the last four digits are more varied, but perhaps none of them match 625, 3125, 5625, or 8125.Looking back at the computations for ( 6^m ) modulo 10000, I didn't see any of these numbers. Let me check a few more exponents to be sure.Continuing from where I left off:( 6^{20} mod 10000 = 2976 ) ( 6^{21} = 2976 times 6 = 17856 mod 10000 = 7856 ) ( 6^{22} = 7856 times 6 = 47136 mod 10000 = 7136 ) ( 6^{23} = 7136 times 6 = 42816 mod 10000 = 2816 ) ( 6^{24} = 2816 times 6 = 16896 mod 10000 = 6896 ) ( 6^{25} = 6896 times 6 = 41376 mod 10000 = 1376 ) ( 6^{26} = 1376 times 6 = 8256 mod 10000 = 8256 ) ( 6^{27} = 8256 times 6 = 49536 mod 10000 = 9536 ) ( 6^{28} = 9536 times 6 = 57216 mod 10000 = 7216 ) ( 6^{29} = 7216 times 6 = 43296 mod 10000 = 3296 ) ( 6^{30} = 3296 times 6 = 19776 mod 10000 = 9776 ) ( 6^{31} = 9776 times 6 = 58656 mod 10000 = 8656 ) ( 6^{32} = 8656 times 6 = 51936 mod 10000 = 1936 ) ( 6^{33} = 1936 times 6 = 11616 mod 10000 = 1616 ) ( 6^{34} = 1616 times 6 = 9696 mod 10000 = 9696 ) ( 6^{35} = 9696 times 6 = 58176 mod 10000 = 8176 ) ( 6^{36} = 8176 times 6 = 49056 mod 10000 = 9056 ) ( 6^{37} = 9056 times 6 = 54336 mod 10000 = 4336 ) ( 6^{38} = 4336 times 6 = 26016 mod 10000 = 6016 ) ( 6^{39} = 6016 times 6 = 36096 mod 10000 = 6096 ) ( 6^{40} = 6096 times 6 = 36576 mod 10000 = 6576 ) Still, I don't see any of the last four digits matching 625, 3125, 5625, or 8125. Maybe I need to check more exponents, but this could take a while. Alternatively, perhaps there's a mathematical reason why they can't match.Looking back at the properties of ( 5^n ) and ( 6^m ), I know that ( 5^n ) is always divisible by 5, while ( 6^m ) is not divisible by 5 because 6 and 5 are coprime. Therefore, ( 6^m ) modulo 5 is always 1, since 6 ≡ 1 mod 5, and 1^m ≡ 1 mod 5.On the other hand, ( 5^n ) modulo 5 is 0, because it's a multiple of 5. So, if ( 5^n equiv 6^m mod 10000 ), then ( 5^n equiv 6^m mod 5 ). But ( 5^n equiv 0 mod 5 ) and ( 6^m equiv 1 mod 5 ). Therefore, we have 0 ≡ 1 mod 5, which is impossible.Wait, that seems like a contradiction. If ( 5^n equiv 6^m mod 10000 ), then they must also be congruent modulo 5. But ( 5^n ) is 0 mod 5, and ( 6^m ) is 1 mod 5. Therefore, 0 ≡ 1 mod 5, which is not possible. So, there can be no such positive integers ( n ) and ( m ) where ( 5^n ) and ( 6^m ) end with the same four digits.Therefore, the answer is no, such integers ( n ) and ( m ) do not exist.