AMC 10 · 2017 · #20
Grade 7 probabilityPick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Oddness depends only on 2's
Write any divisor as 2^a· m with m odd; it is odd exactly when a=0, because no other prime (3,5,7,…) can flip parity.
A number is odd only when it has zero factors of 2, so just track the 2's.
4.OA.B.4Count The ComplementCount the 2's in 21!
In 21!, multiples of 2, 4, 8, 16 give 10+5+2+1 = 18 factors of 2, so 21!=2¹⁸·(odd part).
Each multiple of 2 donates a 2, each multiple of 4 donates an extra one, and so on.
6.NS.B.4Identify SubproblemsList the choices for a
The exponent a can be any value from 0 to 18, giving 19 choices — one free power of 2 for each odd part m.
Counting 0 through 18 inclusive gives 18+1=19 possible powers of 2.
4.OA.A.3Use Matrix LogicTurn the count into a probability
Group divisors by odd part m: each group has 19 divisors and exactly one (a=0) is odd, so P = 1/19, choice (B).
Inside every family that shares an odd part, only the a=0 member is odd, so the odds are 1 in 19.
7.SP.C.7Count The ComplementA 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