sum of divisors with abundancy ratio

[chelsea.j5021]

the abundancy ratio, a_n, of a positive integer n is the sum of all its positive divisors divided by n.
e.g. a₁₀ = 1+2+5+10/10 =18/10=9/5=1.8

show that a_n ≥ 2 if n is a multiple of 6

*my argument so far is that; if n is a multiple of 6 then it is divisible by 2 and 3 as well, which guarantees that n has at least 4 factors, which could increase the value of a_n
my argument isn't good at all, if someone could help with the explanation or possibly a mathematical rule, i would much appreciate it.

 

[soroban]

Hello, chelsea!

The abundancy ratio, a_n, of a positive integer n is the sum of all its positive divisors divided by n.
e.g. .a₁₀ .= .(1 + 2 + 5 + 10)/10 .= .18/10 .= .1.8

Show that a_n ≥ 2 if n is a multiple of 6.

If n is a multiple of 6, say, n = 6k for some positive integer k,
. . then the divisors of n are: .1, 2, 3, 6, and k. .(Note: k could be 1.)

Hence, the abundancy ratio is at least: .(1 + 2 + 3 + 6)/6 .= .2

 

[chelsea.j5021]

thanks for that. it seems simple now.