Suppose that n is an integer divisible by 24. Show that the sum of all the positive divisors of n−1 (including 1 and n−1) is also divisible by 24. Suppose that n is an integer divisible by 24. Show that the sum of all the positive divisors of n−1 (including 1 and n−1) is also divisible by 24.
 A: Note that $n-1\equiv -1\pmod{3}$ and $n-1\equiv -1\pmod{8}$.
It is clear that $n-1$ is not a perfect square. If $ab=n-1$, where $a$ and $b$ are positive, call $a$ and $b$ a couple, or, in the more business-oriented language of today, partners.  
If $\{a,b\}$ are a couple, then since $ab\equiv -1\pmod{3}$, one of $a$ and $b$ is congruent to $1$ modulo $3$, and the other is congruent to $-1$. So their sum is divisible by $3$.
Now we work modulo $8$. Either one of $a$ and $b$ is congruent to $1$, and the other to $-1$, or one is congruent to $3$ and the other to $5$. In either case their sum is divisible by $8$.
Thus the sum of the two members of any couple is divisible by $24$, and therefore the sum of the divisors of $n-1$ is divisible by $24$.
A: There are multiple approaches to this problem. Here is a hint for the approach I developed.
Since n is divisible by 24,
n = 24a, for some integer a.
Now, consider the sum 1 + 2 + ... + n = n(n+1)/2. Thus The sum from 1 to (n-1) is:
n(n+1)/2 - n = n(n-1)/2 = 24a(24a-1)/2
