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I have got an exercise. 7 men are in a room, sitting at a table. In front of each people is a cup. In some cups are juice. The cups altogether have got 3 liters juice in them. Now the first man pour the juice in his cup equally in the other 6 cups. The second man will pour the juice, which is now in his cup, equally in the other 6 cups. And the third man will pour it in the other 6 and so on. When the last man (the $7^{th}$) did it already the man are surprised. The amount of juice they have all now in the cup is exactly as much as before they started this decant. Every men did the decant only once. Now i have to find out how much juice was in every cup before they started the decant.

Obviously the last man has got 0 liter juice because he is the last man who did the decant, so if the procedure ends he will have no juice in his cup, so when they began he had no juice.

My Problem is I don't know how I have to solve this problem. Hope you can help me. I have got after the procedure of decant.

$(1) = \frac{7^6 * (1) + 7^5 * 6 * (2) + 7^4 *6^2*(3) + 7^3 * 6^3*(4) + 7^2*6^4*(5) + 7*6^5(6)}{6^7}$

$(2) = \frac{9031*7*(1) + 9031*6*(2) + 7^4*6^2*(3) + 7^3*6^3*(4) + 7^2*6^4*(5) + 7*6^5*(6)}{6^7}$

$(3) = \frac{1105*7^2*(1) + 1105*7*6*(2) + 1105*6^2*(3) + 7^3*6^3*(4) + 7^2*6^4*(5) + 7*6^5*(6)}{6^7}$

$(4) = \frac{127*7^3*(1) + 127*7^2*6*(2) + 127*7*6^2*(3) + 127*6^3*(4) + 7^2*6^4(5) + 7*6^5*(6)}{6^7}$

$(5) = \frac{13*7^4*(1) + 13*7^3*6*(2) + 13*7^2*6^2*(3) + 13*7*6^3*(4) + 13*6^4*(5) + 7*6^5*(6)}{6^7}$

$(6) = \frac{7^5*(1) + 7^4*6*(2) + 7^3*6^2*(3) + 7^2*6^3*(4) + 7*6^4*(5) + 6^5*(6)}{6^7}$

$(7) = 0$

This is what I have got. My main problem is to show why $\frac{6}{7}, \frac{5}{7}, \frac{4}{7}, \frac{3}{7}, \frac{2}{7}, \frac{1}{7}$ and $0$ is the single solution for my problem. Thanks for your help.

Daifus
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  • Where did you got this problem from? – user5402 Aug 28 '14 at 15:17
  • My friend gave me this problem and he had it from his uncle and he got this problem in the school (if I remember right) but he hasn't got the solution anymore. – Daifus Aug 28 '14 at 15:21
  • The second man pours the juice in his cup equally in the other 6 cups or in the remaining 5 cups? – user5402 Aug 28 '14 at 15:24
  • @metacompactness It should be the other $6$ cups, if not the first $6$ men would end up with no juice in their cups. – Yiyuan Lee Aug 28 '14 at 15:26

2 Answers2

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If they start with $6,5,4,3,2,1,0$ units, then each round rotates the quantities around the table by one seat: $0,6,5,4,3,2,1$; $1,0,6,5,4,3,2$; etc.

TonyK
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You are correct, you need to work backwards. If you persist, it is not too hard to find that the starting amounts are 6/7, 5/7, 4/7, 3/7, 2/7, 1/7, 0.

@TonyK. Correct. It was problem 9 in the All-Soviet Union Math Competition for 1977 with elves and milk.

almagest
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    The problem is how I can show that this is the only solution for this problem? – Daifus Aug 28 '14 at 16:29
  • As I said. you need to work backwards. Label the elves E1-E7. Let the initial amounts be $a_1,\dots,a_7$. Obviously $a_7=0$. Immediately before the 7th distribution, E6 had nil. So in the 7th distribution E6 must have given $a_6$ to every other elf. So the amounts before the last distribution must have been $a_1-a_6,a_2-a_6,a_3-a_6,a_4-a_6,a_5-a_6,0,6a_6$. It is now slightly laborious but entirely straightforward to work all the way back to get $a_1=-6a_1+42a_6,a_2=7a_1-6a_2-7a_6,a_3=7a_2-6a_3-7a_6$, $a_4=7a_3-6a_4-7a_6,a_5=7a_4-6a_5-7a_6,a_6=7a_5-13a_6,a_7=0$. The answer follows. – almagest Aug 30 '14 at 14:35