[Math] In how many possible ways can we write $3240$ as a product of $3$ positive integers $a$, $b$ and $c$

combinatorics

In how many possible ways can we write $3240$ as a product of $3$ positive integers $a$, $b$ and $c$?

This is the question where I've been stuck. The answer is $450$, but I don't know why. I've tried taking out the number of factors, then applying the combination formula in different ways.

Best Answer

$$3240=2^3\cdot3^4\cdot5^1$$

The factor $2$ can be split among $3$ divisors in $10$ different ways:

  • $0,0,3$
  • $0,1,2$
  • $0,2,1$
  • $0,3,0$
  • $1,0,2$
  • $1,1,1$
  • $1,2,0$
  • $2,0,1$
  • $2,1,0$
  • $3,0,0$

The factor $3$ can be split among $3$ divisors in $15$ different ways:

  • $0,0,4$
  • $0,1,3$
  • $0,2,2$
  • $0,3,1$
  • $0,4,0$
  • $1,0,3$
  • $1,1,2$
  • $1,2,1$
  • $1,3,0$
  • $2,0,2$
  • $2,1,1$
  • $2,2,0$
  • $3,0,1$
  • $3,1,0$
  • $4,0,0$

The factor $5$ can be split among $3$ divisors in $3$ different ways:

  • $0,0,1$
  • $0,1,0$
  • $1,0,0$

Hence the total number of ways to write it as a product of $3$ divisors is $10\cdot15\cdot3=450$.

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