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 Pretzel sticks (Posted on 2003-09-22)
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?

 See The Solution Submitted by Gamer Rating: 3.3333 (3 votes)

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 Explanation to Puzzle Answer Comment 5 of 5 |

Let the original number of sticks possessed by the individual be s.

Now, the individual has eaten 6 pretzels by the time he divides them evenly by 2
He has eaten 9 pretzels by the time he divides the rest by 3, etc.
Accordingly,  we must have:
(s - 3k) divisible by k, whenever k=2,3,4,5,6,7,8,9
=> s  must be divisible by k for k = 2 to 9 inclusively.
Therefore,  the least value of s must correspond to LCM( 2,3,4,5,6,7,8,9)=2520
By this time the individual has eaten 3*10=30 pretzels, so that:
The remaining 2520-30=2490 pretzels are divided into 10 piles of 2490/10= 249 each, so the # sticks possessed by each of the 10 individuals is 249.
The individual would have had all 2520 pretzels to himself if people did not come to join him.

Edited on July 7, 2022, 1:21 am
 Posted by K Sengupta on 2022-07-06 22:05:47

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