Problem 13 · Exactly ten divisors
$10 = 2 \cdot 5$: comparing $p^{9}$ with $p^{4}q$.
Integer answer, at most 4 digitsWhat is the smallest positive integer with exactly 10 positive divisors (including 1 and itself)?
Copa Cangur · SCM
Easy
Closed answer
Reasoned solution
Key idea: if $n = p_1^{a_1}\cdots p_k^{a_k}$, the number of divisors is $(a_1+1)\cdots(a_k+1)$. We need this product to be $10 = 10$ or $2 \cdot 5$.
Case $10$: $n = p^{9}$, smallest is $2^{9} = 512$. Case $2 \cdot 5$: $n = p^{4}q$ with distinct primes $p, q$; the smallest comes from $p = 2$, $q = 3$:
$$n = 2^{4} \cdot 3 = 48.$$
Check: divisors of the form $2^{i}3^{j}$ with $0 \le i \le 4$, $0 \le j \le 1$: $5 \cdot 2 = 10$ divisors ✓. And $48 < 512$.
Answer: 48