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.
But what happens, when the smallest number is a multiple of $ 17 $ and $13$ or $11$ or both?
Please help!
Best Answer
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 $1,4,9,16,6,17,11,7,5$.