Math Olympiad Divisor Problem 
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.
 A: Thanks to the hints from lhf, I am now able to solve:
Smallest Positive Divisors: 1 and 5
Largest Positive Divisors: $N$ and $N/5$.
Basically: 
$N$ $+$ $N/5$ = 1122
$6N$ = 5610
$N$ = 935
A: 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.
A: Let 
$$
N=p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}, p_i>p_{i+1}
$$
be the canonical decomposition of the integer number $N.$ 
The first two smallest divisors are $1$ and $p_1$.
The largest  two divisors are $p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}$ and $p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}$. Thus we obtain the following system
$$
1+p_1=6\\
p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}+p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}=1122
$$
From the fist equation we obtain that $p_1=5.$  Rewrite the second one
$$
p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}+p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}=p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}(1+p_1)=6p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}=1122.
$$
At  last 
$$
N=p_1 \frac{1122}{6}=935.
$$
A: Hint: The two largest positive divisors of $N$ are $N$ and $N/5$.
