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*Can* $18$ *consecutive positive integers be separated into two groups, such that their product is equal? We cannot leave out any number and neither we can take any number more than once.*

My work:

When the smallest number is not $17$ or its multiple, there cannot exist any such arrangement as $17$ is a prime.

When the smallest number is a multiple of $17$ but not of $13$ or $11$, then no such arrangement exists.

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But what happens, when the smallest number is a multiple of $ 17 $ and $13$ or $11$ or both?

Please help!

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This is impossible.

At most one of the integers can be divisible by $19$. If there is such an integer, then one group will contain it and the other one will not. The first product is then divisible by $19$ whereas the second is not (since $19$ is prime) — a contradiction.

So if this possible, the remainders of the numbers after division by $19$ must be precisely $1,2,3,\cdots,18$.

Now let $x$ be the product of the numbers in one of the groups. Then

$x^2 \equiv 18! \equiv -1 \pmod{19}$

by Wilson’s Theorem. However $-1$ is not a quadratic residue mod $19$, because the only possible squares mod $19$ are $1,4,9,16,6,17,11,7,5$.

If $18$ consecutive positive integers could be separated into two groups with equal products, then the product of all $18$ integers would be a perfect square. However, the product of two or more consecutive positive integers can never be a perfect square, according to a famous theorem of P. Erdős, Note on products of consecutive integers, J. London Math. Soc. 14 (1939), 194-198.

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