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$n \Rightarrow 20$ Positive factors

$12n \Rightarrow 42$ Positive factors

$45n \Rightarrow 80$ Positive factors

$\Rightarrow 12n = 2^2 \cdot 3 \cdot n$

$\Rightarrow 45n = 3^2 \cdot 5 \cdot n$

Many positive factors of ${ a }_{ 1 }^{ { b }_{ 1 } }\cdot { a }_{ 2 }^{ { b }_{ 2 } }\cdot { a }_{ 3 }^{ { b }_{ 3 } }\cdot ...\cdot { a }_{ n }^{ { b }_{ n } }$ with ${ \left\{ { a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },....,{ a }_{ n } \right\} }$ $\in$ prime and ${ \left\{ { b }_{ 1 },{ b }_{ 2 },{ b }_{ 3 },....,{ b }_{ n } \right\} }$ $\in$ Original number, can be determined with ${ (b }_{ 1 }+1)({ b }_{ 2 }+1)({ b }_{ 3 }+1)...({ b }_{ n }+1)$ . Suppose the factorization $n$ $=$ $m$ $=$ ${ c }_{ 1 }^{ { d }_{ 1 } }\cdot { c }_{ 2 }^{ { d }_{ 2 } }\cdot { c }_{ 3 }^{ { d }_{ 3 } }\cdot ...\cdot { c }_{ n }^{ { d }_{ n } }$ with ${ \left\{ { c }_{ 1 },{ c }_{ 2 },{ c }_{ 3 },....,{ c }_{ n } \right\} }$ $\in$ prime and ${ \left\{ { d }_{ 1 },d_{ 2 },{ d }_{ 3 },....,{ d }_{ n } \right\} }$ $\in$ original number.

$\Rightarrow 12n = 2^2 \cdot 3 \cdot { c }_{ 1 }^{ { d }_{ 1 } }\cdot { c }_{ 2 }^{ { d }_{ 2 } }\cdot { c }_{ 3 }^{ { d }_{ 3 } }\cdot ...\cdot { c }_{ n }^{ { d }_{ n } }$

$\Rightarrow 42 = (2+1)(1 + 1)({ d }_{ 1 }+1)({ d }_{ 2 }+1)...({ d }_{ n }+1)$ $\Rightarrow 42 = 3 \cdot 2 \cdot ({ d }_{ 1 }+1)({ d }_{ 2 }+1)...({ d }_{ n }+1)$

$\Rightarrow 7= ({ d }_{ 1 }+1)({ d }_{ 2 }+1)...({ d }_{ n }+1)$

Because 7 is prime, so there is only 1. So we get $\Rightarrow 7 = d + 1$ $\Rightarrow 6 = d$

So, for a while we get $m^6$ .

$\Rightarrow 45n = 3^2 \cdot 5 \cdot { c }_{ 1 }^{ { d }_{ 1 } }\cdot { c }_{ 2 }^{ { d }_{ 2 } }\cdot { c }_{ 3 }^{ { d }_{ 3 } }\cdot ...\cdot { c }_{ n }^{ { d }_{ n } }$

$\Rightarrow 80 = (2 + 1)(1 + 1)({ d }_{ 1 }+1)({ d }_{ 2 }+1)...({ d }_{ n }+1)$ $\Rightarrow 80 = 3 \cdot 2 ({ d }_{ 1 }+1)({ d }_{ 2 }+1)...({ d }_{ n }+1)$

$\Rightarrow 15 = ({ d }_{ 1 }+1)({ d }_{ 2 }+1)...({ d }_{ n } + 1)$

So, the form we get there are two factors. $\Rightarrow 15 = ({ d }_{ 1 }+1)({ d }_{ 2 }+1)$ .

Value $n$ is possible ${ 2 }^{ x+2 }\cdot { 3 }^{ y+1 }\cdot { 5 }^{ z+1 }$ . An example is $2^4 \cdot 3 \cdot 5$ . $2^4 \cdot 3 \cdot 5$ and satisfying $12n$ , but not satisfying $45n$ .

Another possible form is $2^4 \cdot 3 \cdot 7$ .

$2^4 \cdot 3 \cdot 7$ satisfying all. So, $n = 2^4 \cdot 3 \cdot 7$ .

$21n^3 = 21 \cdot (2^4 \cdot 3 \cdot 7)^3 = 3 \cdot 7 \cdot 2^{12} \cdot 3^3 \cdot 7^3 = 2^{12} \cdot 3^4 \cdot 7^4$ .. So, many positive factors of $21n^3$ are $(12 + 1)(4 + 1)(4 + 1) = 13 \cdot 5 \cdot 5 = 325$ . .