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?
Answer Choices:
A. 10
B. 13
C. 15
D. 17
E. 20
Solution:
Let a and b be even positive integers with 1≤a≤b≤50 such that
1,2,…,a,b+1,b+2,…,50.
are the numbers of the pages that remain. The sum of these numbers is
(1+2+⋯+a)+((b+1)+⋯+50)
=(1+2+⋯+50)−((a+1)+⋯+b)
=250⋅51−(1+2+⋯+b)+(1+2+⋯+a)
=250⋅51−2b(b+1)+2a(a+1)
The number of pages that remain is 50−b+a. Therefore
19=50−b+a250⋅51−2b(b+1)+2a(a+1)=2(50−b+a)50⋅51−b(b+1)+a(a+1)
Simplifying yields
50⋅13=(b2−a2)−37(b−a)=(b−a)(b+a−37)
Observe that 50⋅13=2⋅52⋅13, and that b−a and b+a−37 are two positive integers with sum equal to 2b−37≤100−37=63. The only pairs of factors that sum to at most 63 are {13,50} and {26,25}. Testing these gives the four possible solutions for (a,b), namely (0,50),(18,44),(19,44), and (37,50). The only solution in which both a and b are positive even integers is (18,44). Thus Hiram's roommate borrowed the sheets containing pairs of page numbers {19,20},{21,22},…,{43,44}. There are 244−18=(B)13 sheets in all.
The problems on this page are the property of the MAA's American Mathematics Competitions