Math Fundamentals18 sections · 814 units
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Counting Divisors

Using prime factorization

If n=p1a1×p2a2×...×pkakn = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}, then the number of divisors is (a1+1)(a2+1)...(ak+1)(a_1 + 1)(a_2 + 1)...(a_k + 1).

For 72=23×3272 = 2^3 \times 3^2, the divisor count is (3+1)(2+1)=12(3+1)(2+1) = 12. Each divisor is formed by choosing a power of 2 (0 to 3) and a power of 3 (0 to 2). Product rule in action.