1991 AMC 8 Problem 9

Attempt Problem 9 of the 1991 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 1991 AMC 8 solutions, or check the answer key.

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

9.

How many whole numbers from 11 through 4646 are divisible by either 33 or 55 or both?
1818
2121
2424
2525
2727
Answer: B
Concepts:inclusion-exclusionmultiple
Small hint:
Count the multiples of 33 and the multiples of 55 separately
Big hint:
Numbers divisible by both 33 and 55 are multiples of 1515 and get counted twice, so subtract them once
Solution:
There are 1515 multiples of 33 and 99 multiples of 55 up to 46.46. The 33 multiples of 1515 (namely 15,15, 30,30, 4545) were counted twice.
By inclusion-exclusion, the count is 15+9−3=21.15 + 9 - 3 = 21.
Thus, the correct answer is B .
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