Exercise 925Number theory › Divisibility and gcdL1Added 3 Nov 2026

Number of divisors

#divisors#prime factorization#counting

Statement

Let $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ . Show that the number of positive divisors of $n$ is $(\alpha_1+1)\cdots(\alpha_k+1)$ .

Hint

Each divisor has the form $p_1^{\beta_1}\cdots p_k^{\beta_k}$ with $0\le\beta_i\le\alpha_i$ .

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Divisors. By the fundamental theorem, $d\mid n$ iff $d=p_1^{\beta_1}\cdots p_k^{\beta_k}$ with $0\le\beta_i\le\alpha_i$ .

Counting. There are $\alpha_i+1$ choices for each $\beta_i$ , so the total is $\prod_i(\alpha_i+1)$ .

Conclusion. $\tau(n)=\prod_i(\alpha_i+1)$ .

Example. $60=2^2\cdot3\cdot5$ , $\tau(60)=3\cdot2\cdot2=12$ .