Question:

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> Show that if $n$ is a positive integer such that $24$ divides into $n + 1$, then $24$ divides the sum of all divisors of $n$ (denoted in number theory by $\sigma(n)$ or $\sigma\_1(n)$). 
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For example if $n = 95$, then $n + 1 = 96 = 4 \times 24$ and the sum of the divisors of $n$ is $$1 + 5 + 19 + 95 = 120 = 5 \times 24.$$ (Note that the number $n$ is included among its divisors.)