# Number of positive integral divisors

I understand in order to find number of divisors, you need to follow following method, But I don't seem to find why it works.

In order to find number of divisors a number has, you find the prime factorization, and add one to exponents and multiply them.

Eg:

The number 48 has how many positive integral divisors?

a. First find the prime factorization: $2^4$ x $3^1$.

b. Adding 1 to each exponent we get: 4+1 and 1+1 or 5 and 2.

c. Multiplying these numbers together we get 10.

Can anyone explain or give me resources behind the logic of this method.

• Do you understand the simplest case where $n$ has $m$ distinct prime factors then it has $2^m$ divisors? Eg, 30=2x3x5, so it has $2^3=8$ divisors. If not, consider the number of subsets of the set $\{2,3,5\}$. – PM 2Ring Feb 3 '16 at 7:59

The logic is simple:

• In each divisor, the factor $2$ can appear between $0$ and $4$ times, i.e., $5$ different combinations
• In each divisor, the factor $3$ can appear between $0$ and $1$ times, i.e., $2$ different combinations

Hence there are $5\cdot2=10$ divisors:

• $2^0\cdot3^0$
• $2^0\cdot3^1$
• $2^1\cdot3^0$
• $2^1\cdot3^1$
• $2^2\cdot3^0$
• $2^2\cdot3^1$
• $2^3\cdot3^0$
• $2^3\cdot3^1$
• $2^4\cdot3^0$
• $2^4\cdot3^1$