2021 AMC 10A Problems/Problem 22

Revision as of 01:13, 16 April 2021 by MRENTHUSIASM (talk | contribs) (Solution 2 (Different Variable Choice, Similar Logic))

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 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 $k+n-1.$

We have the following preconditions:

  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:

\[\begin{array}{c|c|c} & & \\ [-2.25ex]  \boldsymbol{n} & \boldsymbol{2k+n-39} & \boldsymbol{k} \\ [0.5ex] \hline  & & \\ [-2ex]  2 & 325 & 181 \\    10 & 65 & 47 \\  26 & 25 & 19 \\  50 & 13 & 1 \\ 130 & 5 & -43 \\ 650 & 1 & -305 \\ \end{array}\]

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

If $k=47,$ then the note pages will run out when 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 $1$).

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

~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

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

https://youtu.be/dWOLIdTxwa4

~ pi_is_3.14

See also

2021 AMC 10A (ProblemsAnswer KeyResources)
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

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