Can18consecutive 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!

**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,⋯,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.

**Attribution***Source : Link , Question Author : Hawk , Answer Author : Marc van Leeuwen*