2001 AMC 12 Problems/Problem 12

Revision as of 20:11, 15 February 2009 by Misof (talk | contribs) (New page: == Problem == How many positive integers not exceeding <math>2001</math> are multiple of <math>3</math> or <math>4</math> but not <math>5</math>? <math> \text{(A) }768 \qquad \text{(B) }...)
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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 12 Problems and Solutions
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