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

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Suppose the roommate took pages <math>a</math> through <math>b</math>, or equivalently, page numbers <math>2a-1</math> through <math>2b</math>. Because there are <math>(2b-2a+2)</math> numbers taken, <cmath>\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.</cmath> The first possible solution that comes to mind is if <math>2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12</math>, which indeed works, giving <math>b=22</math> and <math>a=10</math>. The answer is $22-10+1=\boxed{(\textbf{B})13}
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2021|ab=A|num-b=21|num-a=23}}
 
{{AMC10 box|year=2021|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:29, 11 February 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

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}

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

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