2021 AMC 10A Problems/Problem 22
Contents
Problem
Hiram's algebra notes are pages long and are printed on
sheets of paper; the first sheet contains pages
and
, the second sheet contains pages
and
, 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
. How many sheets were borrowed?
Solution 1
Let be the number of consecutive sheets Hiram’s roommate borrows, and let
be the number of sheets preceding the
borrowed sheets (I.e. if the friend borrows sheets 3, 4 and 5, then
and
).
The sum of the page numbers up till sheets is $1+2+3+\cdots + 2b=\frac{2b\cdot(2b+1)}{2} = b(b+1)
~KingRavi
==Solution 2==
Suppose the roommate took sheets$ (Error compiling LaTeX. Unknown error_msg)ab
2a-1
2b
(2b-2a+2)
2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12
b=22
a=10
22-10+1=\boxed{\textbf{(B)} ~13}$.
~Lcz
==Solution 3==
Suppose the smallest page number borrowed is$ (Error compiling LaTeX. Unknown error_msg)k,n
k+n-1.
n
\frac{n}{2}
n
k
1+2+3+\cdots+50=\frac{51(50)}{2}=1275.
\frac{(2k+n-1)n}{2}.$</li><p>
</ol>
Together, we have <cmath>\begin{align*}
\frac{1275-\frac{(2k+n-1)n}{2}}{50-n}&=19 \\
1275-\frac{(2k+n-1)n}{2}&=19(50-n) \\
2550-(2k+n-1)n&=38(50-n) \\
2550-(2k+n-1)n&=1900-38n \\
650&=(2k+n-39)n.
\end{align*}</cmath>
The factors of$ (Error compiling LaTeX. Unknown error_msg)650
n
1\leq k \leq 49,
k=47,19,1$are possible:
- If$ (Error compiling LaTeX. Unknown error_msg)k=47,
10
47.
k=1,
50
25
1.
k=19.
n=26
\frac n2=\boxed{\textbf{(B)} ~13}$sheets, are borrowed.
~MRENTHUSIASM
==Solution 4==
Let$ (Error compiling LaTeX. Unknown error_msg)nk+25.5
25-n
19
25.5
50
k>0
k \in \mathbb N
n < 25
k \le (49+50)/2-25.5=24<25$.
The weighted increase of average page number from$ (Error compiling LaTeX. Unknown error_msg)25.5k+25.5
25.5
19$, where the weights are the page number in each group (borrowed vs. remained), therefore
<cmath>2nk=2(25-n)(25.5-19)=13(25-n) \implies 13 | n \text{ or } 13 | k</cmath>
Since$ (Error compiling LaTeX. Unknown error_msg)n, k < 25n=13
k=13
n=13
k=6
k=13
2n=25-n
n=\boxed{\textbf{(B)} ~13}$.
~asops
==Solution 5==
Let$ (Error compiling LaTeX. Unknown error_msg)(2k-1)-2n(2n+2k+1)(n-k)
1275-(2n+2k+1)(n-k)
\frac{1275-(2n+2k+1)}{50-2n+2k}
19
950-38(n-k)=1275-(2n+2k+1)(n-k)
(2n+2k-37)(n-k)=325
325=25\cdot13
n-k=13
\boxed{\textbf{(B)} ~13}$.
~bluesoul
Video Solution by OmegaLearn (Arithmetic Sequences and System of Equations)
~ pi_is_3.14
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=28te8OUiVxE
~MRENTHUSIASM
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.