# N has all divisors up to 31 except two

Let $$N$$ be a positive integer. It is true that $$1|N, 2|N, 3|N, \dots, 31|N$$, except two. Which are false?

It seems like the problem is impossible. For example, any two consecutive integers must have an even integer E, and this even integer is divisible by 2 and an integer less than it, and since 2 and that integer divide N, E must divide $$N$$, which is a contradiction. For example, if we chose 31 and 30, then since 2 and 15 divide N, it must be the case that 30 divides $$N$$, which is a contradiction. So I'm not sure where to go with this problem.

• I understand the problem as follows: Of the 31 statements $i|N, i=1,2,\ldots,31$ 29 are true and 2 are false. Which are the 2 that are false? As your example shows, an even divisor is likely not among them. – Ingix Mar 19 at 21:42
• It seems like there is more than one solution. We could have $$N= {31!\over 17\cdot19}$$ but we could choose any two primes $17\leq p,q\leq31$ in the denominator. – saulspatz Mar 19 at 22:10