# Number of factors

There’s a neat trick for finding the number of factors of a number.  First find the prime factorization of the number, for example:

$756 = 2^2 \cdot 3^3 \cdot 7^1$

Add $+1$ to the power of each prime factor and then multiply those numbers together.  In this example,

$(2+1)\cdot(3+1)\cdot(1+1)= 3\cdot 4 \cdot 2 = 24$ factors.

This reflects the fact that each factor of $756$ contains $2^n$ where $n= 0, 1$, or $2$ and $3^m$, where $m = 0, 1, 2$, or $3$ and $7^p$ where $p = 0$ or $1$.

I used this property to solve a problem in a number theory class I’m taking.  The problem asks to compute the sum of all positive integers $k$ such that $1984k$ has  $21$ positive factors.

Since $21 = 3\cdot7$, working backward we are looking for a prime factorization with 2 primes raised to the powers of 2 and 6.  The prime factorization of $1984 = 2^6\cdot 31$ so we have one prime raised to the sixth power.  All we need is the other prime to be squared.  Setting $k=31$ gives us

$1984k = 2^6\cdot 31^2$ with $(6+1)(2+1)= 21$ divisors.