Integers for Which the Sum of the Divisors is 24
\[n\]
is the sum of the divisors of \[n\]
equal to 24?The sum of the divisors of
\[n\]
is the divisor function \[\sigma (n)= \sum_{d | n} d\]
. \[\sigma (n)\]
is multiplicative so that if \[a, \; b\]
are relatively prime then \[\sigma (ab)= \sigma (a) \sigma (b)\]
.\[24=2 \times 12=3 \times 8=4 \times 6\]
\[\sigma (n)=24 \]
and 1 is always a divisor of \[n\]
so we can take 23 as the other divisor and \[n=23\]
.\[\sigma (n)=2 \times 12 =\sigma (a) \sigma (b)\]
. \[\sigma (a)=2\]
has no solution since \[\sigma (1)=1, \sigma (2)=1+2=3\]
so there is no such \[n\]
.\[\sigma (n)=3 \times 8 =\sigma (a) \sigma (b)\]
. \[\sigma (a)=3 \rightarrow a=2\]
. \[\sigma (b)=8 \rightarrow b=7\]
then \[n=ab=2 \times 7=14\]
. The divisors of 14 are 1, 2, 7, 14 and 1+2+7+14=24.\[\sigma (n)=4 \times 6 =\sigma (a) \sigma (b)\]
. \[\sigma (a)=4 \rightarrow a=3\]
. \[\sigma (b)=6 \rightarrow b=5\]
then \[n=ab=3 \times 5=15\]
. The divisors of 15 are 1, 3, 5, 15 and 1+3+5+15=24.
