TITLE: Math Olympiad Divisor Problem
QUESTION [6 upvotes]: The sum of the two smallest positive divisors of an integer $N$ is $6$,
  while the sum of the two largest positive divisors of $N$ is $1122$.
  Find $N$.

I came across this question in a Math Olympiad Competition. I am able to find out that the two smallest positive divisors would be $1$ and $5$ but after that, I am not sure how to work on to find out the value of $N$. Can anyone help? Thanks.

REPLY [5 votes]: Hint $ $ The set of factors of $\,N\,$ enjoy a cofactor involution (reflection) $\ k\mapsto k' = N/k,\,$ so  $\,N = k\,k'.\,$  This gives a pairing of $\,k\,$ with its cofactor $\,k' = N/k.\,$ It's order reversing $\, j < k\,\Rightarrow\, N/j > N/k,\,$ so it pairs the least factor $\,k_1$ with the greatest $\,N/k_1;\,$  2nd least $\,k_2$ with the 2nd greatest $\,N/k_2,\,$ etc. 
Thus the least factor $\,k_1 = 1$ pairs with the greatest $\,N/1 = N.\ $ The 2nd least factor $\,k_2$ is the least prime factor $\,p\,$ (else some prime $\,q\mid k_2\mid n,\,$ and $\,q\le k_2< p,\,$ contra leastness of $\,p).$  Therefore this 2nd least factor $\,k_2 =p\,$ pairs with $\,N/p =\,$ 2nd greatest factor. The rest is straightforward.