Meadow Secondary Student Hello. I have a question that was assigned, but i really don't understand how to do it. I was wondering if you guys could help me. Here's the question. 5 sailors plan to divide a pile of coconuts among themselves in the morning. During the night, one of them decides to take his share. After throwing a coconut to a monkey to make the division come out even he takes 1/5 of the pile. The other four sailors repeat this procedure, each throwing a coconut to the monkey and taking 1/5 of the remaining coconuts. In the morning the 5 sailors throw a coconut to the monkey, and divide the remaining coconuts into 5 equal piles. What is the minimum number of coconuts that could have been in the pile originally? I am totally clueless of how start so please help! thanks! ~Meadow Meadow, Let n be the number of coconuts initially. The first sailor throws 1 to the monkey, leaving n-1 coconuts. This is divided into 5 equal piles, and the first sailor takes 1 pile, leaving 4 piles. That is, the first sailor takes 1/5 and leaves 4/5 of n-1. Thus, after the first sailor, the number of coconuts is 4/5(n-1) = (4/5)n - 4/5. The second sailor throws 1 to the monkey, leaving (4/5)n - 4/5 - 1 coconuts. This is divided into 5 equal piles, and the second sailor takes 1 pile, leaving 4 piles. That is, the first sailor takes 1/5 leaves 4/5 of (4/5)n - 4/5 - 1. Thus, after the second sailor, the number of coconuts is (4/5)[(4/5)n - 4/5 - 1] = (4/5)2 n - (4/5)2 - 4/5 Similarly, after the third sailor, the number of coconuts is (4/5)[(4/5)2 n - (4/5)2 - 4/5 - 1] = (4/5)3 n - (4/5)3 - (4/5)2 - 4/5 Similarly, after the fourth sailor, the number of coconuts is (4/5)[(4/5)3 n - (4/5)3 - (4/5)2 - 4/5 -1] = (4/5)4 n - (4/5)4 - (4/5)3 - (4/5)2 - 4/5 Similarly, after the fifth sailor, the number of coconuts is (4/5)[(4/5)4 n - (4/5)4 - (4/5)3 - (4/5)2 - 4/5 - 1] = (4/5)5 n - (4/5)5 - (4/5)4 - (4/5)3 - (4/5)2 - 4/5 = [(45)n]/(55) - 4^5/55 - (44) 5/55 - (43) (52)/55 - (4^2) (53)/55 - (4) (54)/55 = [1024n]/3125 - 1024/3125 - (256) 5/3125 - (64) (25)/3125 - (16) (125)/3125 - (4) (625)/3125 = 1024n/3125 - 1024/3125 - 1280/3125 - 1600/3125 - 2000/3125 - 2500/3125 = 1024n/3125 - 8404/3125 In the morning, the sailors give 1 to the monkey, leaving = 1024n/3125 - 8404/3125 - 1 = 1024n/3125 - 8404/3125 - 3125/3125 = 1024n/3125 - 11529/3125 This is divided into five equal piles, so each sailor gets: (1/5)(1024n/3125 - 11529/3125) = 1024n/15625 - 11529/15626 = (1024n - 11529)/15625 This number must be a whole number, so 1024n - 11529 is a multiple of 15625 In other words, there is a whole number k, such that 1024n - 11529 = 15625k Solving for n gives n = (15625k + 11529)/1024 n = 15625k/1024 + 11529/1024 n = 15k+ (265k/1024) + 11 + (265/1024) n = 15k + 11 + 265(k+1)/1024 n = 15k + 11 + 5 53(k+1)/(210) We must find the smallest k such that the equation above gives a whole number answer for n. In other words, we must find the smallest k such that 5 53(k+1)/(210) is a whole number. But 5 is not divisible by 2, and 53 is not divisible by 2. Therefore, (k+1) must be divisible by 210. So, the minimum value of k that works will come from: k+1 = 210 Hence, k = 1024 - 1 = 1023 Recall: n = (15625k + 11529)/1024 Therefore, n = (15625 1023 + 11529)/1024 = 15621 Therefore the minimum number of coconuts is 15621. Check that answer works: First sailor: 15620/5 = 3124 15620 - 3124 = 12496 coconuts left after the first sailor. Second sailor: 12495/5 = 2499 12495 - 2499 = 9996 coconuts left after the second sailor. Third sailor: 9995/5 = 1999 9995 - 1999 = 7996 coconuts left after the third sailor. Fourth sailor: 7995/5 = 1599 7995 - 1599 = 6396 coconuts left after the fourth sailor. Fifth sailor: 6395/5 = 1279 6395 - 1279 = 5116 coconuts left after the fifth sailor. In the morning, there are 5116 coconuts. Give 1 to the monkey, leaving 5115 coconuts. 5115/5 = 1023. We always get a whole number, so the answer works.