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One day 3 men (Ed, Eddie and Eric) decided to go fishing. They agreed that everyone takes one third of the total number N of fish caught. After fishing they went to sleep. At 3:00 AM Ed woke up. He wanted to take his 1/3 and leave but he realized that the number N of all fishes caught is not divisible by 3. So he threw one fish into the river and took his 1/3. Then at 4:00 AM Eddie woke up thinking that Ed and Eric were still asleep. He did the same thing as Ed and left. Then at 5:00 Eric was awakened by his alarm clock. Eric did the same thing as Eddie and left.

Objective: Find the smallest number N.

My problem with solution: I solved a system of equations and I realized that it has more than 1 solution (it means $N_1,N_2,...$). But I do not know how to find the smallest N.

Thanks for help

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Let $a$ be the number of fish. Then you know that $a=3b+1$ for some non-negative integer $b$, since Ed saw that the remainder was one when he tried to divide the fish into thirds. He threw one back and took $b$ or himself, leaving $2b$. Similarly, $2b=3c+1$ and $2c=3d+1$, where $d$ is the number of fish Eric ended up taking at the end.

We know that $d$ must be odd, so equal to $2k+1$ for some non-negative $k$. Plugging that in we get $$2c=6k+4$$ or $$c=3k+2.$$ Therefore $$2b = 9k+7$$ and $k$ must also be odd, $k=2l+1$. Then $$2b = 18l+16$$ or $$b = 9l+8$$ and $$a = 27l +25.$$ The smallest value of $a$ occurs when $l=0$: Ed sees 25 fish, tosses one back, and takes 8 (leaving 16). Eddie tosses another back, taking five and leaving 10. Lastly Eric tosses one and takes three.

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Started off with $-2$. If you dump one, you get $-2-1=-3$. If you divide it in to $3$ equal parts, each part becomes $-1$. Take one away and you're left with $-2$. Then you can repeat the process all over again.

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    $\begingroup$ If you catch negative fish, did you bring fish from another place and throw them in the lake? $\endgroup$ – templatetypedef Feb 25 '16 at 6:41

protected by Zev Chonoles Feb 25 '16 at 6:27

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