Fun little math puzzle?

Question

Here's an interesting puzzle I'm struggling to find an answer to, though I don't believe it's really very hard:

A group of seven boys - Jon, Tim, Tom, Bob, Bill, Ben and Will - were playing a game in which the counters were beans. Whenever a boy lost a game, from his pile of beans he had to give each of the other boys as many beans as they already had. They had been playing for some time and they all had different numbers of beans. They then had a run of seven games in which each boy lost a game in turn, in the order given above. At the end of this sequence of games, amazingly, they all had the same number of beans - 128. How many did each of them have at the start of this sequence of seven games?

Does anyone have any answers, and any explanations of how they got them? Once again, have fun...

Answer 1

N(n, i) stands for the number of beans of boy i after run n.
We want to know N(0, i), based on the information N(7, i) = 128.

If i<>n, boy i did not lose run n, so in that case his amount was doubled, and
[1] ... N(n, i) = 2 * N(n-1, i) ... (i =/= n)

If i = n, boy i did lose run n, so he had to double the amounts of the others. The total number of beans in the game is 7*128 = 896. If the losing boy had X beans before he lost, the others had 896 - X beans together. Their amounts are doubled, so afterward they have 1792 - 2 X beans, leaving 896 - (1792 - 2 X) = 2 X - 896 beans for the losing boy. Therefore,
[2] ... N(n, n) = 2 * N(n-1, n) - 896

Formulas 1 and 2 combined show that every boy doubled his amount of beans, except that in run n, 896 beans were subtracted. In subsequent runs, the term -896 is also doubled. For boy i that doubling takes place 7-i times. Therefore,
[3] N(7, i) = 2^7 * N(0,i) - 2^(7-i) * 896

This means that
[4] 128 = 128 * N(0,i) - 2^(7-i) * 896
[5] 128 = 128 * (N(0,i) - 896 / 2^i)
[6] 1 = N(0,i) - 896 / 2^i
[7] N(0,i) = 896 / 2^i + 1

We find
Jon: N(0,1) = 896 / 2 + 1 = 449
Tim: N(0,2) = 896 / 4 + 1 = 225
Tom: N(0,3) = 896 / 8 + 1 = 113
Bob: N(0,4) = 896 / 16 + 1 = 57
Bill: N(0,5) = 896 / 32 + 1 = 29
Ben: N(0,6) = 896 / 64 + 1 = 15
Will: N(0,7) = 896 / 128 + 1 = 7

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In general, if there are N boys and they all end up with 2^N beans, their original numbers would be

N(0, i) = N * 2^(N-i) + 1

That is, the last boy would start with N+1 beans, the second last boy with one more than double that amount, etc.

Answer 2

Work backwards - start with a column of 128s, then find the next column etc.

i.e.

128 64 32 16 8 4 2 449
128 64 32 16 8 4 550 225
128 64 32 16 8 552 226 113
128 64 32 16 456 228 114 57
128 64 32 464 232 116 58 29
128 64 480 240 120 60 30 15
128 512 256 128 64 32 16 8

The answer in order of boys is:

449,225,113,57,29,15 and 8

and add these up to get 128x7

Answer 3

Note that they give the beans to the other people FROM THEIR OWN PILE and therefore the total number of beans in the whole game remains a constant. We know that at the end all seven boys had 128 beans and assuming the game is fair, they all started with the same number of beans as each other and therefore at the start of the game they all had 128 beans each.

Answer 4

So it not really hard in that case why you post on this then? how stupid of me forgot you want us to do it for you! silly me lol you could have fooled me lol

Answer 5

Hard one

Answer 6

o

Answer 7

wow.ill get back to ya on this

Answer 8

this is wat i called complicate