I have lots of pretzel sticks. I eat 3 of the pretzel sticks before somebody comes and joins me. When he arrives I eat three more sticks and then divide the rest equally between the two of us. The piles come out equally, but just as I am done dividing up the piles between the other person and me, another person joins us.

I combine the piles and divide them up for 3 people, again eating 3 pretzel sticks before doing so. The piles come out equally again.

But before anyone has a chance to eat any of them, another person comes. So I combine all the piles, eat 3 pretzel sticks, and divide them up among the 4 of us.

People keep coming in this fashion until there are 10 piles (including mine), and then nobody comes once there are 10 piles, so we all eat our pretzel sticks. I am happy that the piles came out evenly each time, and note that I have the least amount of pretzel sticks that this would happen with.

How many pretzel sticks did the people who joined me have (individually), and how many pretzel sticks would I have had if I didn't share them with anybody?

As you have eaten 6 pretzels by the time you divide them evenly by 2, and you have eaten 9 by the time you divide the rest by 3, etc., letting your original number of sticks be n,
n-6 must be divisible by 2
n-9 must be divisible by 3
n-12 must be divisible by 4
n-15 must be divisible by 5
n-18 must be divisible by 6
n-21 must be divisible by 7
n-24 must be divisible by 8
n-27 must be divisible by 9
n-30 must be divisible by 10

As each of the subtracted amounts is itself divisible by the number given on that line, the number n itself must be divisible by each of these values, so n is the least common multiple of 2,3,4,...,10. This is 2520. So at the end, after you have eaten 30 there are 2490 to be divided evenly into 10 piles of 249 each. You would have had the full 2520 if people hadn't come to join you. You ate 30 already, and each of you now have 249.

Posted by Charlie on 2003-09-22 13:27:43