2001 AMC 12 Problem 12

Attempt Problem 12 of the 2001 AMC 12 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 2001 AMC 12 solutions, or check the answer key.

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12.

How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5?$

$768$

$801$

$934$

$1067$

$1167$

Answer: B

Concepts:inclusion-exclusion counting integers in a range multiple

Difficulty rating: 1540

Solution:

Multiples of $3$ or $4$ up to $2001$ number $667 + 500 - 166 = 1001,$ using $\lfloor 2001/3 \rfloor = 667,$ $\lfloor 2001/4 \rfloor = 500,$ and $\lfloor 2001/12 \rfloor = 166.$

Among these, the ones divisible by $5$ are multiples of $15$ or $20$ : $133 + 100 - 33 = 200,$ using $\lfloor 2001/15 \rfloor = 133,$ $\lfloor 2001/20 \rfloor = 100,$ and $\lfloor 2001/60 \rfloor = 33.$

The count is $1001 - 200 = 801.$

Thus, the correct answer is B.

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