Problem 11 · 1997 AJHSME
Hard
Number Theory
divisor-counting
Let [N] mean the number of whole number divisors of N. For example, [3] = 2 because 3 has two divisors, 1 and 3. Find the value of
[ [11] × [20] ].Show answer
Answer: A — 6.
Show hints
Hint 1 of 2
This is a function-inside-a-function: resolve the innermost brackets to plain numbers first, then move outward — never try to do it all at once.
Still stuck? Show hint 2 →
Hint 2 of 2
To count a number's divisors, prime-factorize it, then multiply (each exponent + 1). A divisor is built by choosing how many of each prime to include, and there are (exponent + 1) choices per prime.
Show solution
Approach: count divisors via prime factorization, inside-out
- Innermost: [11] = 2 since 11 is prime (only divisors 1 and 11). And [20] = [2² · 5] = (2+1)(1+1) = 6.
- Replace the inner brackets: [ [11] × [20] ] = [ 2 × 6 ] = [12].
- Finally [12] = [2² · 3] = (2+1)(1+1) = 6.
- Why the formula works: for 12 = 2²·3, a divisor picks 0, 1, or 2 twos (3 ways) and 0 or 1 three (2 ways), giving 3 × 2 = 6 divisors — and indeed 1, 2, 3, 4, 6, 12.
- Curiosity: the answer also equals [20], since both 12 and 20 are 'p²q' shaped — same exponent pattern means same divisor count.
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