Difference between revisions of "2021 AMC 10A Problems/Problem 22"

Revision as of 09:46, 29 October 2021

Problem

Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?

$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Solution 1

Suppose the roommate took sheets $a$ through $b$ , or equivalently, page numbers $2a-1$ through $2b$ . Because there are $(2b-2a+2)$ numbers taken, $\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50\cdot51}{2} \implies (2a+2b-39)(b-a+1)=\frac{50\cdot13}{2}=25\cdot13.$ The first possible solution that comes to mind is if $2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12$ , which indeed works, giving $b=22$ and $a=10$ . The answer is $22-10+1=\boxed{\textbf{(B)} ~13}$ .

~Lcz

Solution 2

Suppose the smallest page number borrowed is $k,$ and $n$ pages are borrowed. It follows that the largest page number borrowed is $k+n-1.$

We have the following preconditions:

$n$ pages are borrowed means that $\frac{n}{2}$ sheets are borrowed, from which $n$ must be even.

$k$ must be odd, as the smallest page number borrowed is on the right side (odd-numbered).

$1+2+3+\cdots+50=\frac{51(50)}{2}=1275.$

The sum of the page numbers borrowed is $\frac{(2k+n-1)n}{2}.$

Together, we have $\begin{align*} \frac{1275-\frac{(2k+n-1)n}{2}}{50-n}&=19 \\ 1275-\frac{(2k+n-1)n}{2}&=19(50-n) \\ 2550-(2k+n-1)n&=38(50-n) \\ 2550-(2k+n-1)n&=1900-38n \\ 650&=(2k+n-39)n. \end{align*}$ The factors of $650$ are $1,2,5,10,13,25,26,50,65,130,325,650.$ Since $n$ is even, we only have a few cases to consider: $\begin{array}{c|c|c} & & \\ [-2.25ex] \boldsymbol{n} & \boldsymbol{2k+n-39} & \boldsymbol{k} \\ [0.5ex] \hline & & \\ [-2ex] 2 & 325 & 181 \\ 10 & 65 & 47 \\ 26 & 25 & 19 \\ 50 & 13 & 1 \\ 130 & 5 & -43 \\ 650 & 1 & -305 \\ \end{array}$ Since $1\leq k \leq 49,$ only $k=47,19,1$ are possible:

If $k=47,$ then there will not be sufficient pages when we take $10$ pages out starting from page $47.$

If $k=1,$ then the average page number of all remaining sheets will be undefined, as there will be no sheets remaining after we take $50$ pages ( $25$ sheets) out starting from page $1.$

Therefore, the only possibility is $k=19.$ We conclude that $n=26$ pages, or $\frac n2=\boxed{\textbf{(B)} ~13}$ sheets, are borrowed.

~MRENTHUSIASM

Solution 3

Let $n$ be the number of sheets borrowed, with an average page number $k+25.5$ . The remaining $25-n$ sheets have an average page number of $19$ which is less than $25.5$ , the average page number of all $50$ pages, therefore $k>0$ . Since the borrowed sheets start with an odd page number and end with an even page number we have $k \in \mathbb N$ . We notice that $n < 25$ and $k \le (49+50)/2-25.5=24<25$ .

The weighted increase of average page number from $25.5$ to $k+25.5$ should be equal to the weighted decrease of average page number from $25.5$ to $19$ , where the weights are the page number in each group (borrowed vs. remained), therefore

$2nk=2(25-n)(25.5-19)=13(25-n) \implies 13 | n \text{ or } 13 | k$

Since $n, k < 25$ we have either $n=13$ or $k=13$ . If $n=13$ then $k=6$ . If $k=13$ then $2n=25-n$ which is impossible. Therefore the answer should be $n=\boxed{\textbf{(B)} ~13}$ .

~asops

Solution 4

Let $(2k-1)-2n$ be pages be borrowed, the sum of digits in those pages is $(2n+2k+1)(n-k)$ while the sum of the rest pages is $1275-(2n+2k+1)(n-k)$ and we know the average of the rest is $\frac{1275-(2n+2k+1)}{50-2n+2k}$ which equals to $19$ ; multiply this out we got $950-38(n-k)=1275-(2n+2k+1)(n-k)$ and we got $(2n+2k-37)(n-k)=325$ . As $325=25\cdot13$ , we can see $n-k=13$ and that is desired $\boxed{\textbf{(B)} ~13}$ .

~bluesoul

Solution 5

Since Hiram's notes have an average of $19$ , he could have all of his pages be consecutive like $1$ through $37$ . We can check this is true because $\frac {37 \cdot 38}{37 \cdot 2}$ = $19$ . Therefore, Hiram's friend took his notes from pages $38$ through $50$ . Therefore, his friend took $12+1$ pages of notes = $\boxed {(C)13}$

~Arcticturn

Video Solution by OmegaLearn (Arithmetic Sequences and System of Equations)

https://youtu.be/dWOLIdTxwa4

~ pi_is_3.14

Video Solution by MRENTHUSIASM (English & Chinese)

https://www.youtube.com/watch?v=28te8OUiVxE

~MRENTHUSIASM

@@ Line 67: / Line 67: @@
 ==Solution 5==
+Since Hiram's notes have an average of <math>19</math>, he could have all of his pages be consecutive like <math>1</math> through <math>37</math>. We can check this is true because <math>\frac {37 \cdot 38}{37 \cdot 2}</math> = <math>19</math>. Therefore, Hiram's friend took his notes from pages <math>38</math> through <math>50</math>. Therefore, his friend took <math>12+1</math> pages of notes = <math>\boxed {(C)13}</math>
+~Arcticturn
 == Video Solution by OmegaLearn (Arithmetic Sequences and System of Equations) ==

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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