2022 AMC 12A Problem 10

Attempt Problem 10 of the 2022 AMC 12A 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 2022 AMC 12A solutions, or check the answer key.

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

10.

What is the number of ways the numbers from 11 to 1414 can be split into 77 pairs such that for each pair, the greater number is at least 22 times the smaller number?

108108

120120

126126

132132

144144

Answer: E
Concepts:arrangements with restrictionsmultiplication principle
Difficulty rating: 1570
Solution:

Any number 88 or larger cannot be a smaller element (its double exceeds 1414), so 881414 are all larger elements and 1177 are all smaller elements.

Match each smaller ss to a larger g2s.g\ge2s. Processing from the most restrictive: s=7s=7 forces g=14g=14 (11 way); then s=6s=6 has {12,13}\{12,13\} left (22); s=5s=5 has 3;3; s=4s=4 has 4;4; s=3s=3 has 3;3; s=2s=2 has 2;2; s=1s=1 has 1.1.

The number of matchings is 1234321=144.1\cdot2\cdot3\cdot4\cdot3\cdot2\cdot1=144.

Thus, the correct answer is E.

← Problem 9#9
Full Exam

Problem 10 in Other Years