2023 AMC 10B Problems/Problem 5
Contents
Problem
https://www.youtube.com/watch?v=lYBUbBu4W08
How many positive integers not exceeding are multiples of
or
but not
?
Solutions
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 integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of .
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 positive integers. Determine that the numbers
and
exhibit the properties given in the question.
is a divisor of
, so there are
numbers satisfying the given conditions between
and
. Since
is a multiple of
, add
to
to get
.
~ mathmagical
Solution 5
By PIE, there are numbers that are multiples of
or
and less than or equal to
.
of them will not be divisible by
, and by far the closest number to
of
is
.
~ Fasolinka
Solution 5
Similar to some of the above solutions.
We can divide by
and
to find the number of integers divisible by
and
. Hence, we find that there are
numbers less than
that are divisible by
, and
numbers that are divisible by
. However, we will need to subtract the number of multiples of
from 667 and that of
from
, since they're also divisible by 5 which we don't want. There are
+
=
such numbers. Note that during this process, we've subtracted the multiples of
twice because they're divisible by both
and
, so we have to add
back to the tally (there are
multiples of
that does not exceed
). Lastly, we have to subtract multiples of both
AND
since we only want multiples of either
or
. This is tantamount to subtracting the number of multiples of
. And there are
such numbers. Let's now collect our numbers and compute the total:
+
-
-
+
-
=
.
~ PlainOldNumberTheory
Solution 6
Similar to @above:
Let the function return how many multiples of
are there not exceeding
. Then we have that the desired number is:
Evaluating each of these we get:
Thus, the answer is
-ConfidentKoala4
Video Solutions
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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