AMC 10 · 2017 · #16

Grade 7 probability
prime-factorizationdivisor-countprobability-basic legendre-formulacomplementary-counting ↑ Prerequisites: prime-factorization
📏 Medium solution 💡 2 insights
Problem
The number twenty-one factorial is the product of one through twenty-one. It has more than sixty thousand positive divisors. If one of those divisors is picked at random, each equally likely, find the probability that the divisor is odd.

Pick an answer.

(A)
$\frac{1}{21}$
(B)
$\frac{1}{19}$
(C)
$\frac{1}{18}$
(D)
$\frac{1}{2}$
(E)
$\frac{11}{21}$
How to solve
Strategy Change Focus / Count the Complement

Tool #16 (Change Focus): a divisor's oddness depends only on how many 2's it contains, so I ignore every other prime and watch only the power of 2. Tool #7 (Identify Subproblems): the one number I really need is how many 2's hide inside 21!, a self-contained counting job. Tool #4 (Introduce a Variable): I let the exponent of 2 in a divisor be a variable that ranges over a clear list of values, turning 'odd or not' into a clean count.

1STEP 1

Oddness depends only on 2's

Odd or even depends only on the prime two.

divisor = 2^a · m, m odd
2STEP 2

Count the 2's in 21!

Counting the twos inside gives 18.

10+5+2+1 = 18
3STEP 3

List the choices for a

The exponent of two has 19 choices, zero included.

a ∈ {0,1,2,…,18} → 19 choices
4STEP 4

Turn the count into a probability

Only one of them is odd, so one nineteenth.

P(odd) = 1/19 → (B)
Answer
1/19
Check with the full factorization 21! = 2¹⁸ · 3⁹ · 5⁴ · 7³ · 11 · 13 · 17 · 19. The total number of divisors is (18+1)(9+1)(4+1)(3+1)(1+1)⁴ = 19 · 10 · 5 · 4 · 16 = 60,800, matching 'over 60,000.' The odd divisors drop the 2's, giving 10 · 5 · 4 · 16 = 3,200. Then 3,200/60,800 = 1/19, confirming the answer. A probability near 0.05 is sensible since most divisors carry at least one factor of 2.
💡Key takeaway

A divisor is odd only when it grabs none of the 2's; 21! has 18 twos, so the power of 2 has 19 choices (0 through 18) and just one keeps it odd — probability 1/19, choice (B).

  • Oddness depends only on 2's
  • Count the 2's in 21!
  • List the choices for a
  • Turn the count into a probability