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2001 AMC 12 Problems/Problem 12

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Problem

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

\text{(A) }768\qquad\text{(B) }801\qquad\text{(C) }934\qquad\text{(D) }1067\qquad\text{(E) }1167

Solution

Out of the numbers 1 to 12 four are divisible by 3 and three by 4, counting 12 twice. Hence 6 out of these 12 numbers are multiples of 3 or 4.

The same is obviously true for the numbers 12k+1 to 12k+12 for any positive integer k.

Hence out of the numbers 1 to 60=5\cdot 12 there are 5\cdot 6=30 numbers that are divisible by 3 or 4. Out of these 30, the numbers 15, 20, 30, 40, 45 and 60 are divisible by 5. Therefore in the set \{1,\dots,60\} there are precisely 30-6=24 numbers that satisfy all criteria from the problem statement.

Again, the same is obviously true for the set \{60k+1,\dots,60k+60\} for any positive integer k.

We have 1980/60 = 33, hence there are 24\cdot 33 = 792 good numbers among the numbers 1 to 1980. At this point we already know that the only answer that is still possible is \boxed{\text{(B)}}, as we only have 20 numbers left.

By examining the remaining 20 by hand we can easily find out that exactly 9 of them match all the criteria, giving us 792+9=\boxed{801} good numbers.

See Also

2001 AMC 12 (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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