Sharing a sack of nuts.
A woman with four children bought a sack of peanuts. To the oldest child, a boy, she gave one peanut and 1/4 of what remained; and to each of the other children she did the same. (gave 1 peanut + 1/4 of the rest) The second child was a girl, the third a boy, and the last a girl. It was found that the boys had received 100 more peanuts than the girls.
What was the initial number of peanuts?
The answer is 1021.
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1 solution
Here is the answer. It was maddening keeping track of all the +1 and -1 terms, but I finally got it all straight.
Start with 1021 First boy gets 1 + 1020/4 = 1 + 255 = 256 Leaves 765 First girl gets 1 + 764/4 = 1 + 191 = 192 Leaves 573 Second boy gets 1 + 572/4 = 1 + 143 = 144 - Boys' total 400. Leaves 429 Second girl gets 1 + 428/4 = 1 + 107 = 108- Girls' total 300. Leaves 321 for the mother.
Here's how - after a very long time and many tries - I solved it:
Let a, b, c, d be the amounts each child got and e the number left. a+b+c+d+e = all the peanuts
a = (a + b + c + d + e - 1)/4 + 1
b = (b + c + d + e - 1)/4 + 1 a - b = a/4 b = 3/4 a
c = (c + d + e - 1)/4 + 1 b - c = b/4 c = 3/4 b = 9/16 a
d = (d + e - 1)/4 + 1 c - d = d/4 d = 3/4 c = 27/64 a
a + c - 100 = b + d a + 9/16 a - 100 = 3/4 a + 27/64 a 64a + 36a - 6400 = 48a + 27a 25a = 6400 a = 256 b = 192 = 3/4 a c = 144 = 3/4 b d = 108 = 3/4 c Total 700 d = (d + e - 1)/4 + 1 4d = d + e - 1 + 4 3d = e + 3 324 - 3 = 321 = e