Divisor Count of 35

Example of Use of Divisor Counting Function

 * $\map \tau {35} = 4$

where $\tau$ denotes the divisor counting (tau) function.

Proof
From Divisor Counting Function from Prime Decomposition:
 * $\ds \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$

where:
 * $r$ denotes the number of distinct prime factors in the prime decomposition of $n$
 * $k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.

We have that:
 * $35 = 5 \times 7$

Thus:
 * $\map \tau {35} = \map \tau {5^1 \times 7^1} = \paren {1 + 1} \paren {1 + 1} = 4$

Thus:

The divisors of $35$ can be enumerated as:
 * $1, 5, 7, 35$