Problem:
How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?
Answer Choices:
A. 768
B. 801
C. 934
D. 1067
E. 1167
Solution:
For integers not exceeding 2001, there are ⌊2001/3⌋=667 multiples of 3 and ⌊2001/4⌋=500 multiples of 4. The total, 1167, counts the ⌊2001/12⌋=166 multiples of 12 twice, so there are 1167−166=1001 multiples of 3 or 4. From these we exclude the ⌊2001/15⌋=133 multiples of 15 and the ⌊2001/20⌋=100 multiples of 20, since these are multiples of 5. However, this excludes the ⌊2001/60⌋=33 multiples of 60 twice, so we must re-include these. The number of integers satisfying the conditions is 1001−133−100+33=801.
Answer: B​.
The problems on this page are the property of the MAA's American Mathematics Competitions