Split the Candies

Here is a problem by J.A.H. Hunter.

Sam divided up the candies into equal shares, one share for each of them. 'There you are,' he said, taking his quota, 'and you can't say I'm not fair.' But Hilda had other ideas. She undid what remained of Sam's work, and divided them again into equal shares, but for the girls only. 'Boys shouldn't want candies, not real boys,' she very nearly said. But none of her cousins protested, so she took one of those little heaps of luscious chocolate creams and stood back from the table. 'And that leaves eight each for the rest of you,' she announced. Hilda had certainly taken more than her fair share of the 70 candies their uncle had given them for their party, but then she is like that. Maybe you can figure out how many girls and boys shared those candies.

Please give the answer or answers and the method to get the answer(s).

Peter Heichelheim

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I don't get it...

So there are 70 total, and they divide evenly. There are at least two children, so our possibilities for Sam's piles are:

cousins per pile left after Sam takes 2 35 70 - 35 = 35 5 14 70 - 14 = 56 7 10 70 - 10 = 60 10 7 70 - 7 = 63 14 5 70 - 5 = 65 35 2 70 - 2 = 68

But Sam's share is certainly less than eight

You seem to have missed the bit where Hilda also takes a share.

I reckon there are 7 cousins, 4 boys and 3 girls.

Sam makes 7 piles of 10 and takes one pile leaving 60 candies. Hilda makes 3 piles of 20 and takes one pile leaving 40 candies. There remain 5 cousins that haven't already grabbed a pile, and they get 8 each.

Nope...I missed the fact that after Hilda takes, the candies are redivided *again* (for boys and girls both). I supposed that the 'each of you' in Hilda's rejoinder indicated only the girls.

You got J.A.H. Hunter's answer. However you did not show there were no other answers. I think there is a least one other. Can you or anyone else show that you have all the answers.

Let b and g represent the number of girls and boys; s be the number of candies Sam takes, and h be the number Hilda takes. We know:

(1) s * (b + g) = 70 (2) h * g = 70 - s (3) 8 * (b + g - 2) = 70 - s - h (3' 8 * (b + g) = 86 - s - h

From (1) and (3':

(4) 70 / s = (86 - s - h) / 8

Now, s has to be one of {35, 14, 10, 7, 5, 2, 1}, since Sam took a factor of 70, and we know (b + g) is at least 1. If we plug those in to (4) and solve for h:

2 = (51 - h) / 8

I'm with you on this one. Why would she start off saying 'girls only', then suddenly switch to 'girls & boys'? It makes more sense your way.

A very good solution. Looking at my other possible solution again I realized I made a mistake and now realize there is only one solution.

Well, the boys later beat the heck out of her, but that's another story.

70 = 7 * 5 * 2, and we know the total number of people N divides 70. We also know that 8 > 70/N. We also know that 70 - 70/N is divisible by 8. The possibilities are

N = 10, 70 - 70/N = 63, not divisible by 8. N = 14, 70 - 70/N = 65, not divisible by 8. N = 35, 70 - 70/N = 68, not divisible by 8.

Therefore this problem is deeply flawed, and you owe me a chocolate