Question 831380
{{{660=10*66=10*6*11=5*2*2*3*11=2^2*3*5*11}}}
so the prime factorization is {{{660=2^2*3*5*11}}}
Since the prime factorization for {{{6}}} and {{{12}}} are
{{{6=2*3}}} and {{{12=2^2*3}}} ,
the factors {{{5}}} and {{{11}}} must come from {{{n}}} .
There could be other factors of {{{n}}} included in {{{2^2*3*5*11}}} .
The prime factorization of {{{n}}} could have
{{{2}}} with exponents 0, 1, or 2,
and/or {{{3}}} with exponents 0 or 1,
in addition to {{{5}}} and {{{11}}}.
That gives us {{{3*2=6}}} possible values for {{{n}}} :
{{{2^0*3^0*5*11=highlight(55)}}}
{{{2^0*3^1*5*11=highlight(165)}}}
{{{2^1*3^0*5*11=highlight(110)}}}
{{{2^1*3^1*5*11=highlight(330)}}}
{{{2^2*3^0*5*11=highlight(220)}}}
{{{2^2*3^1*5*11=highlight(660)}}}