2021 AMC 10A Problems/Problem 22

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

Suppose the roommate took pages $a$ through $b$ , or equivalently, page numbers $2a-1$ through $2b$ . Because there are $(2b-2a+2)$ numbers taken, $\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50*51}{2} \implies (2a+2b-39)(b-a+1)=\frac{50*13}{2}=25*13.$ 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 (Different Variable Choice, Similar Logic)

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

Remarks:

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

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

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

4. The sum of the page numbers removed 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:

$\begin{array}{ c c c } \boldsymbol{n} & \boldsymbol{2k+n-39} & \boldsymbol{k} \\ \hline 2 & 325 & 181 \\ 10 & 65 & 47 \\ 26 & 25 & 19 \\ 50 & 13 & 1 \\ 130 & 5 & \text{negative} \\ 650 & 1 & \text{negative} \\ \end{array}$

Since $1\leq k \leq 50,$ only $k=47,19,1$ are possible:

If $k=47,$ then the notebook will run out if we take $10$ pages starting from page $47.$

If $k=1,$ then the average page number of the remaining pages will be undefined, as there is no page remaining (after taking $50$ pages starting from page $1$ ).

So, the only possibility is $k=19,$ from which $n=26$ pages are taken out, which is $\frac n2=\boxed{\textbf{(B)} ~13}$ sheets.

~MRENTHUSIASM

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

https://youtu.be/dWOLIdTxwa4

~ pi_is_3.14

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

Page

Toolbox

Search

2021 AMC 10A Problems/Problem 22

Contents

Problem

Solution

Solution 2 (Different Variable Choice, Similar Logic)

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

See also