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 β2001/3β=667 multiples of 3 and β2001/4β=500 multiples of 4. The total, 1167, counts the β2001/12β=166 multiples of 12 twice, so there are 1167β166=1001 multiples of 3 or 4. From these we exclude the β2001/15β=133 multiples of 15 and the β2001/20β=100 multiples of 20, since these are multiples of 5. However, this excludes the β2001/60β=33 multiples of 60 twice, so we must re-include these. The number of integers satisfying the conditions is 1001β133β100+33=801.
Answer: Bβ.
The problems on this page are the property of the MAA's American Mathematics Competitions