2021 AMC 10A Problems/Problem 22

Revision as of 20:30, 11 February 2021 by Lcz (talk | contribs) (Solution)

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

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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