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

Revision as of 23:51, 27 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.

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 $2$ multiples of $7$ . Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets. We see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$ . We calculate $\left\lfloor\frac{100}{7}\right\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\cdot3= \boxed{\textbf{(B) }42}$ .

~(edited by) mihikamishra

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

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 52: / Line 52: @@
 ~BrandonZhang202415
-== Solution 4 (Simple Counting) ==
+== Solution 4 (Simple Counting, Similar to Solution 1) ==
 Consecutive multiples of <math>7</math> must differ by <math>7</math>. So, if a set <math>\{\ldots1,\ldots2,\ldots3,(\ldots),\ldots8,\ldots9,\ldots0\}</math> contains two multiples of <math>7</math>, they must end with the digits <math>1</math> and <math>8</math>, <math>2</math> and <math>9</math>, or <math>3</math> and <math>0</math>. This reduces the problem to counting the amount of multiples of <math>7</math> less than <math>1000</math> that end with <math>1</math>, <math>2</math>, and <math>3</math>.

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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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

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