Difference between revisions of "2022 AMC 10B Problems/Problem 8"

Revision as of 14:44, 10 November 2024

The following problem is from both the 2022 AMC 10B #8 and 2022 AMC 12B #6, so both problems redirect to this page.

1 Problem
2 Solution 1 (Casework)
3 Solution 2 (Find A Pattern)
4 Solution 3 (Fastest)
5 Solution 4 (Simple Counting, Similar to Solution 1)
6 Solution 5 (Very Fast System of Equations)
7 Solution 6 (reasonable count)
8 Video Solution (🚀Under 3 min🚀)
9 Video Solution(1-16)
10 Video Solution by Interstigation
11 See Also

Problem

Consider the following $100$ sets of $10$ elements each: $\begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*}$ How many of these sets contain exactly two multiples of $7$ ?

$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$

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 $7$ are $1\pmod{10}$ and $8\pmod{10}.$ That is, the first and eighth elements of such sets are multiples of $7.$

The first element is $1+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=2,9,16,\ldots,93.$

The multiples of $7$ are $2\pmod{10}$ and $9\pmod{10}.$ That is, the second and ninth elements of such sets are multiples of $7.$

The second element is $2+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=4,11,18,\ldots,95.$

The multiples of $7$ are $3\pmod{10}$ and $0\pmod{10}.$ That is, the third and tenth elements of such sets are multiples of $7.$

The third element is $3+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=6,13,20,\ldots,97.$

Each case has $\left\lfloor\frac{100}{7}\right\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{\textbf{(B)}\ 42}.$

~MRENTHUSIASM

Solution 2 (Find A Pattern)

We find a pattern. $\begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*}$ We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The fourth set has $1$ multiple of $7$ . The fifth set has $2$ multiples of $7$ . The sixth set has $1$ multiple of $7$ . The seventh set has $1$ multiple of $7$ . The eighth set has $2$ multiples of $7$ . Disregarding the first set and then calculating this pattern further, we can see (reasonably) that it repeats for each $1$ sets. We see that the pattern for the number of multiples per $7$ sets (again, disregarding the first set) goes: $1,2,1,2,1,1,2.$ So, for every $7$ sets after the first, there are three sets with $2$ multiples of $7$ . We calculate $\left\lfloor\frac{100-1}{7}\right\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $1$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\cdot3= \boxed{\textbf{(B) }42}$ .

~(edited by) mihikamishra

~(edited by) MiniGlasses2009

Solution 3 (Fastest)

Each set contains exactly $1$ or $2$ multiples of $7$ .

There are $\dfrac{1000}{10}=100$ total sets and $\left\lfloor\dfrac{1000}{7}\right\rfloor = 142$ multiples of $7$ .Thus, there are $142-100=\boxed{\textbf{(B) }42}$ sets with $2$ multiples of $7$ .

~BrandonZhang202415

Solution 4 (Simple Counting, Similar to Solution 1)

Consecutive multiples of $7$ must differ by $7$ . So, if a set $\{\ldots1,\ldots2,\ldots3,(\ldots),\ldots8,\ldots9,\ldots0\}$ contains two multiples of $7$ , they must end with the digits $1$ and $8$ , $2$ and $9$ , or $3$ and $0$ . This reduces the problem to counting the amount of multiples of $7$ less than $1000$ that end with $1$ , $2$ , and $3$ .

The first multiple of $7$ that ends with $1$ is $21$ . The next multiple that ends with $1$ occurs $70$ later, since that is the smallest multiple of $7$ we can add to $21$ without affecting the last digit. The greatest number of $70$ 's we can add to $21$ while keeping it less than $1000$ is $13$ , because $21 + 70(13) = 931$ . Therefore, the set of multiples of $7$ less than $1000$ ending with $1$ is $\{21 + 70(0), 21+70(1),\ldots,21+70(13)\}$ , meaning there are $14$ of these particular multiples. We can use the same reasoning to count the multiples of $7$ that end with $2$ and $3$ .

The first multiple of $7$ that ends with $2$ is $42$ . The greatest number of $70$ 's we can add to $42$ here is also $13$ , since $42 + 70(13) = 952$ . The set of multiples of $7$ less than $1000$ ending with $2$ is $\{42 + 70(0), 42+70(1),\ldots,42+70(13)\}$ , giving $14$ multiples.

The first multiple of $7$ that ends with $3$ is $63$ . The greatest number of $70$ 's we can add to $63$ here is yet again $13$ , since $63 + 70(13) = 973$ . The set of multiples of $7$ less than $1000$ ending with $3$ is $\{63 + 70(0), 63+70(1),\ldots,63+70(13)\}$ , giving another $14$ multiples.

In total, there are $14 + 14 + 14 = 42$ of these multiples, and so $\boxed{\textbf{(B) }42}$ sets with two multiples of $7$ .

~marsus16112

Solution 5 (Very Fast System of Equations)

Let $a$ be the number of sets with $1$ multiple of $7$ and $b$ be the number of sets with $2$ multiples of $7$ . Note that it is impossible for a set to have more than two multiples of $7$ .

Since there are a total of $1000$ sets, $a+b=100$ . Also, since there are $\lfloor \frac{1000}{7}\rfloor = 142$ multiples of $7$ between $1$ and $1000$ , we must have $a+2b=142$ .

Solving the system of equations

$a+b=100$ $a+2b=142$

for $b$ gives us the answer of $\boxed{\textbf{(B) }42}$ .

~FIREDRAGONMATH16

Solution 6 (reasonable count)

(Similar to Solution 2, but a little more intuitive and less numbers.) Note that this system loops every cycle of length $70$ , or $7$ such sets. From $1$ to $70$ , the multiples of $7$ are $7$ , $14$ , $21$ , $28$ , $35$ , $42$ , $49$ , $56$ , $63$ , and $70$ ; note that $(21,28),(42,49),$ and $(63,70)$ are in the same sets. Thus, for every $7$ sets, we have $3$ sets with exactly two multiples of $7$ . We have $100$ sets, which is $98+2=7\cdot14+2$ ; the first $98$ sets contain $\dfrac37(7\cdot14)=42$ desired sets. The last two sets comprise the integers from $981$ to $999$ ; the multiples of $7$ here are $987$ and $994$ . Neither of the two last sets contains two multiples of $7$ , so our answer is simply $\boxed{\textbf{(D) }42}$ .

~Technodoggo

Video Solution (🚀Under 3 min🚀)

https://youtu.be/PdyKJ1p9Y2w

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

Video Solution by Interstigation

https://youtu.be/_KNR0JV5rdI?t=884

@@ Line 48: / Line 48: @@
 Each set contains exactly <math>1</math> or <math>2</math> multiples of <math>7</math>.
-There are <math>\dfrac{1000}{10}=100</math> total sets and <math>\left\lfloor\dfrac{1000}{7}\right\rfloor = 142</math> multiples of <math>7</math>.=Thus, there are <math>142-100=\boxed{\textbf{(B) }42}</math> sets with <math>2</math> multiples of <math>7</math>.
+There are <math>\dfrac{1000}{10}=100</math> total sets and <math>\left\lfloor\dfrac{1000}{7}\right\rfloor = 142</math> multiples of <math>7</math>.Thus, there are <math>142-100=\boxed{\textbf{(B) }42}</math> sets with <math>2</math> multiples of <math>7</math>.
 ~BrandonZhang202415

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

Page

Toolbox

Search