2009 AMC 8 Problem 16

Attempt Problem 16 of the 2009 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

How many 33-digit positive integers have digits whose product equals 24?24?
1212
1515
1818
2121
2424
Answer: D
Concepts:digitsmultiset permutationscasework
Small hint:
List the unordered triples of digits whose product is 24.24.
Big hint:
Triplets with three distinct digits have 66 permutations, while (2,2,6)(2,2,6) has 3.3.
Solution:
The only triples of integers less than 1010 that multiply to 2424 are (1,3,8),(1,4,6),(2,2,6), (1, 3, 8), (1, 4, 6), (2, 2, 6), (2,3,4). (2, 3, 4).
The triples with 33 distinct numbers can be rearranged to form 66 distinct 33-digit positive integers. The other triple can be arranged to form 33 distinct 33-digit positive integers.
This leaves a total of 3⋅6+3=213 \cdot 6 + 3 = 21 integers.
Thus, D is the correct answer.
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