Difference between revisions of "2001 AMC 12 Problems/Problem 12"
(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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Revision as of 21:52, 16 March 2011
Problem
How many positive integers not exceeding 
are multiple of 
or 
but not 
?
Solution
Out of the numbers 
to 
four are divisible by and three by , counting twice.
Hence 
out of these numbers are multiples of or .
The same is obviously true for the numbers 
to 
for any positive integer 
.
Hence out of the numbers to 
there are 
numbers that are divisible by or .
Out of these 
, the numbers 
, 
, , 
, 
and 
are divisible by .
Therefore in the set 
there are precisely 
numbers that satisfy all criteria from the problem statement.
Again, the same is obviously true for the set 
for any positive integer .
We have 
, hence there are 
good numbers among the numbers to 
. At this point we already know that the only answer that is still possible is 
, as we only have numbers left.
By examining the remaining by hand we can easily find out that exactly 
of them match all the criteria, giving us 
good numbers.
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| 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 | |
| All AMC 12 Problems and Solutions | |
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Question | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
