Daniel is giving out all of his elephants to people. Overall, $28$ men and $37$ women showed up. Each man received the same number of elephants; ditto with women. Turns out, Daniel could only do it in a unique way. What is the largest number of elephants which Daniel could have? Prove your answer.
So from this, we know that the number of elephants is some number $k$ where $k=28a+37b$, where $a,b$ is the number of elephants that each man and woman receive respectively. Since the ways to to do are unique, then the exponents of the prime factorization of $k$ must be one. Is my logic correct here, and where do I go from here?
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Gerard L.
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The theorem is, if $u$ and $v$ are coprime positive integers, $N = uv-u-v$ is the largest integer such that $ux + vy = N$ has no non-negative integral solution $(x,y)$. Applying this, the answer to your question will be $2\times 28\times37 - 28 - 37$. – Hw Chu Dec 03 '17 at 23:20
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1You haven't stated the constraints $a \geq 0$, $b \geq 0$, since something like these are needed to get uniqueness. – Eric Towers Dec 03 '17 at 23:22
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@HwChu : Since OP demands at least one solution, "... $+1$"? – Eric Towers Dec 03 '17 at 23:23
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You need to add $28\times 37$ instead of $+1$. – Hw Chu Dec 03 '17 at 23:25
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If there are two ways to do it, he has to trade $37$ per man for $28$ per woman or the other way around. The most he could have and not be able to do this is $36$ per man and $27$ per woman, for a total of $36 \cdot 28+27\cdot 37=2007$. Is that the year the problem was asked?
Ross Millikan
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