Divisor Count of 35

Example of Use of $\tau$ Function

 * $\tau \left({35}\right) = 4$

where $\tau$ denotes the $\tau$ Function.

Proof
From Tau Function from Prime Decomposition:
 * $\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$

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:
 * $\tau \left({35}\right) = \tau \left({5^1 \times 7^1}\right) = \left({1 + 1}\right) \left({1 + 1}\right) = 4$

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