2001 AMC 12 Problems/Problem 12
- The following problem is from both the 2001 AMC 12 #12 and 2001 AMC 10 #25, so both problems redirect to this page.
Contents
Problem
How many positive integers not exceeding are multiples of
or
but not
?
Solution 1
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.
This is correct.
Solution 2
We can solve this problem by finding the cases where the number is divisible by or
, then subtract from the cases where none of those cases divide
. To solve the ways the numbers divide
or
we find the cases where a number is divisible by
and
as separate cases. We apply the floor function to every case to get
,
, and
. The first two floor functions were for calculating the number of individual cases for
and
. The third case was to find any overlapping numbers. The numbers were
,
, and
, respectively. We add the first two terms and subtract the third to get
. The first case is finished.
The second case is more or less the same, except we are applying and
to
. We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions
,
, and
yields the numbers
,
, and
. The first two numbers counted all the numbers that were multiples of either four with five or three with five less than
. The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach
. Subtracting this number from the original
numbers procures
.
Solution 3
First find the number of such numbers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of 3.
There are numbers that are not multiples of
.
are not multiples of
or
, so
numbers are.
Solution 4
Take a good-sized sample of consecutive integers; for example, the first 25 positive integers. Determine that the numbers 3, 4, 6, 8, 9, 12, 16, 18, 21, and 24 exhibit the properties given in the question. 25 is a divisor of 2000, so there are numbers satisfying the given conditions between 1 and 2000. Since 2001 is a multiple of 3, add 1 to 800 to get
.
Edit by mathmagical~
Video Solution
https://www.youtube.com/watch?v=XHmKu-ZoRxI&feature=youtu.be
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 25 |
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 |
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