2001 AMC 10 Problem 25

Attempt Problem 25 of the 2001 AMC 10 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 10 solutions, or check the answer key.

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

25.

How many positive integers not exceeding 20012001 are multiples of 33 or 44 but not 5?5?
768768
801801
934934
10671067
11671167
Answer: B
Concepts:inclusion-exclusioncounting integers in a rangemultiple
Small hint:
Count the multiples of 33 or 4,4, then remove those that are also multiples of 55
Big hint:
Use inclusion-exclusion with multiples of 3,3, 4,4, 1212 and then 15,15, 20,20, 6060
Solution:
Multiples of 33 or 44 up to 2001:2001: ⌊20013⌋\left\lfloor\tfrac{2001}{3}\right\rfloor +⌊20014⌋+\left\lfloor\tfrac{2001}{4}\right\rfloor −⌊200112⌋-\left\lfloor\tfrac{2001}{12}\right\rfloor =667+500−166=667+500-166 =1001.=1001.
Among these, remove the multiples of 5:5: there are 133133 multiples of 1515 and 100100 multiples of 20,20, re-adding the 3333 multiples of 60:60: 133+100−33=200.133+100-33=200.
So the count is 1001−200=801.1001-200=801. Thus, the correct answer is B.
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