Difference between revisions of "2022 AMC 10B Problems/Problem 8"
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We can figure out that the first set has <math>1</math> multiple of <math>7</math>. The second set also has <math>1</math> multiple of <math>7</math>. The third set has <math>2</math> multiples of <math>7</math>. The fourth set has <math>1</math> multiple of <math>7</math>. The fifth set has <math>2</math> multiples of <math>7</math>. The sixth set has <math>1</math> multiple of <math>7</math>. The seventh set has <math>1</math> multiple of <math>7</math>. The eighth set has <math>2</math> multiples of <math>7</math>. Disregarding the first set and then calculating this pattern further, we can see (reasonably) that it repeats for each <math>1</math> sets. | We can figure out that the first set has <math>1</math> multiple of <math>7</math>. The second set also has <math>1</math> multiple of <math>7</math>. The third set has <math>2</math> multiples of <math>7</math>. The fourth set has <math>1</math> multiple of <math>7</math>. The fifth set has <math>2</math> multiples of <math>7</math>. The sixth set has <math>1</math> multiple of <math>7</math>. The seventh set has <math>1</math> multiple of <math>7</math>. The eighth set has <math>2</math> multiples of <math>7</math>. Disregarding the first set and then calculating this pattern further, we can see (reasonably) that it repeats for each <math>1</math> sets. | ||
− | We see that the pattern for the number of multiples per <math>7</math> sets (again, disregarding the first set) goes: <math>1,2,1,2,1,1,2.</math> So, for every <math>7</math> sets after the first, there are three sets with <math>2</math> multiples of <math>7</math>. We calculate <math>\left\lfloor\frac{100}{7}\right\rfloor</math> and multiply that by <math>3</math>. (We also disregard the remainder of <math>2</math> since it doesn't add any extra sets with <math>2</math> multiples of <math>7</math>.). We get <math>14\cdot3= \boxed{\textbf{(B) }42}</math>. | + | We see that the pattern for the number of multiples per <math>7</math> sets (again, disregarding the first set) goes: <math>1,2,1,2,1,1,2.</math> So, for every <math>7</math> sets after the first, there are three sets with <math>2</math> multiples of <math>7</math>. We calculate <math>\left\lfloor\frac{100-1}{7}\right\rfloor</math> and multiply that by <math>3</math>. (We also disregard the remainder of <math>2</math> since it doesn't add any extra sets with <math>2</math> multiples of <math>7</math>.). We get <math>14\cdot3= \boxed{\textbf{(B) }42}</math>. |
~(edited by) mihikamishra | ~(edited by) mihikamishra |
Revision as of 14:05, 31 August 2024
- The following problem is from both the 2022 AMC 10B #8 and 2022 AMC 12B #6, so both problems redirect to this page.
Contents
Problem
Consider the following sets of
elements each:
How many of these sets contain exactly two multiples of
?
Solution 1 (Casework)
We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
- The multiples of
are
and
That is, the first and eighth elements of such sets are multiples of
- The multiples of
are
and
That is, the second and ninth elements of such sets are multiples of
- The multiples of
are
and
That is, the third and tenth elements of such sets are multiples of
The first element is for some integer
It is a multiple of
when
The second element is for some integer
It is a multiple of
when
The third element is for some integer
It is a multiple of
when
Each case has sets. Therefore, the answer is
~MRENTHUSIASM
Solution 2 (Find A Pattern)
We find a pattern.
We can figure out that the first set has
multiple of
. The second set also has
multiple of
. The third set has
multiples of
. The fourth set has
multiple of
. The fifth set has
multiples of
. The sixth set has
multiple of
. The seventh set has
multiple of
. The eighth set has
multiples of
. Disregarding the first set and then calculating this pattern further, we can see (reasonably) that it repeats for each
sets.
We see that the pattern for the number of multiples per
sets (again, disregarding the first set) goes:
So, for every
sets after the first, there are three sets with
multiples of
. We calculate
and multiply that by
. (We also disregard the remainder of
since it doesn't add any extra sets with
multiples of
.). We get
.
~(edited by) mihikamishra
~(edited by) MiniGlasses2009
Solution 3 (Fastest)
Each set contains exactly or
multiples of
.
There are total sets and
multiples of
.
Thus, there are sets with
multiples of
.
~BrandonZhang202415
Solution 4 (Simple Counting, Similar to Solution 1)
Consecutive multiples of must differ by
. So, if a set
contains two multiples of
, they must end with the digits
and
,
and
, or
and
. This reduces the problem to counting the amount of multiples of
less than
that end with
,
, and
.
The first multiple of that ends with
is
. The next multiple that ends with
occurs
later, since that is the smallest multiple of
we can add to
without affecting the last digit. The greatest number of
's we can add to
while keeping it less than
is
, because
. Therefore, the set of multiples of
less than
ending with
is
, meaning there are
of these particular multiples. We can use the same reasoning to count the multiples of
that end with
and
.
The first multiple of that ends with
is
. The greatest number of
's we can add to
here is also
, since
. The set of multiples of
less than
ending with
is
, giving
multiples.
The first multiple of that ends with
is
. The greatest number of
's we can add to
here is yet again
, since
. The set of multiples of
less than
ending with
is
, giving another
multiples.
In total, there are of these multiples, and so
sets with two multiples of
.
~marsus16112
Video Solution (🚀Under 3 min🚀)
~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=884
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.