2021 AMC 10A Problem 22

Problem 22 of 25HarderAlgebra Number Theory

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?

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Solution

Suppose the borrowed sheets are sheets

a

through

b,

and let

s=b-a+1.

The borrowed pages run from

2a-1

through

2b,

so there are

2s

borrowed pages and their sum is

s(2a+2b-1).

The total sum of all page numbers is

\frac{50\cdot51}{2}=1275.

If the remaining pages have mean

19,

then

\begin{aligned} 1275&-s(2a+2b-1)\\ &=19(50-2s),\\ s(2a+2b-39)&=325. \end{aligned}

Because

b-a+1

is a positive divisor of

325

and is at most

25,

its only possibilities are

1,5,13,25.

The first two would force

b>25,

and

25

would remove every sheet. Thus the only valid possibility is

\begin{aligned} 2a+2b-39 &=25, \\ b-a+1 &=13. \end{aligned}

Thus

a+b=32

and

b-a=12,

a=10

and

b=22.

Therefore

13

sheets were borrowed. Thus, B is the correct answer.

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