Skip to main content

2021 AMC 10A Problem 22

Problem 22 of 25HarderAlgebraNumber TheoryProbability & Statistics

Hiram’s algebra notes are 5050 pages long and are printed on 2525 sheets of paper; the first sheet contains pages 11 and 2,2, the second sheet contains pages 33 and 4,4, 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 19.19. How many sheets were borrowed?

Answer choices

Show solution

Solution

Suppose the borrowed sheets are sheets aa through b,b, and let s=b−a+1.s=b-a+1. The borrowed pages run from 2a−12a-1 through 2b,2b, so there are 2s2s borrowed pages and their sum is s(2a+2b−1).s(2a+2b-1). The total sum of all page numbers is 50⋅512=1275.\frac{50\cdot51}{2}=1275. If the remaining pages have mean 19,19, then 1275−s(2a+2b−1)=19(50−2s),s(2a+2b−39)=325. \begin{aligned} 1275&-s(2a+2b-1)\\ &=19(50-2s),\\ s(2a+2b-39)&=325. \end{aligned} Because b−a+1b-a+1 is a positive divisor of 325325 and is at most 25,25, its only possibilities are 1,5,13,25.1,5,13,25. The first two would force b>25,b>25, and 2525 would remove every sheet. Thus the only valid possibility is 2a+2b−39=25,b−a+1=13. \begin{aligned} 2a+2b-39 &=25, \\ b-a+1 &=13. \end{aligned} Thus a+b=32a+b=32 and b−a=12,b-a=12, so a=10a=10 and b=22.b=22. Therefore 1313 sheets were borrowed. Thus, B is the correct answer.
AoPS wiki

Tagged: mean · Diophantine Equation · factoring

More practice