2021 AMC 10A Problem 22

P21-25~8 minPrint

Diophantine Equations Sequences & Series Statistics & Data Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

Write the total of the pages left two ways, then set them equal. The equation becomes (sheets taken) times (a whole number) = 325, so try factors of 325.

Solution

1. All $50$ pages add up to $1 + 2 + \cdots + 50 = \frac{50 \times 51}{2} = 1275$ .
2. Say the roommate took $c$ sheets, starting just after sheet $b$ . Sheet number $k$ holds pages $2k - 1$ and $2k$ , so the missing pages run from $2b + 1$ to $2b + 2c$ . For example, if $b = 2$ and $c = 3$ , sheets $3, 4, 5$ are gone, which is pages $5$ to $10$ .
3. There are $50 - 2c$ pages left, with mean $19$ , so they add up to
$19(50 - 2c) = 950 - 38c.$
4. So the missing pages add up to $1275 - (950 - 38c) = 325 + 38c$ .
5. Add the missing pages another way. They are $2c$ numbers in a row from $2b + 1$ to $2b + 2c$ . Pair the first with the last, the second with the second-to-last, and so on: each of the $c$ pairs adds to $(2b + 1) + (2b + 2c) = 4b + 2c + 1$ . So the missing pages add up to $c(4b + 2c + 1)$ .
6. Set the two totals equal and move the $38c$ over:
$c(4b + 2c + 1) = 325 + 38c \quad\Longrightarrow\quad c(4b + 2c - 37) = 325.$
7. Both $c$ and $4b + 2c - 37$ are whole numbers, so $c$ must divide $325 = 5 \times 5 \times 13$ . Fewer than $25$ sheets were taken, so $c$ is $1$ , $5$ or $13$ .
- $c = 1$ : $4b + 2 - 37 = 325$ , so $4b = 360$ and $b = 90$ . There are only $25$ sheets, so this fails.
- $c = 5$ : $4b + 10 - 37 = 65$ , so $4b = 92$ and $b = 23$ . Then sheets $24$ to $28$ were taken, but sheets $26$ to $28$ don't exist. Fails.
- $c = 13$ : $4b + 26 - 37 = 25$ , so $4b = 36$ and $b = 9$ . Sheets $10$ to $22$ were taken, which is pages $19$ to $44$ . This works.
8. Check: the pages left are $1$ to $18$ (total $171$ ) and $45$ to $50$ (total $285$ ). That is $24$ pages adding to $456$ , and $456 \div 24 = 19$ .

The answer is $\boxed{\textbf{(B)}\ 13}$ .

Why this works

A mean is really a statement about a total, and totals of numbers in a row are easy to find by pairing. Because sheet numbers are whole numbers, the equation has to be solved in whole numbers. Arranging it as (unknown) times (something) $= 325$ means only the factors of $325$ need testing, and the fact that there are only $25$ sheets throws out the extra ones.

Alternative approach

Test the choices using step 4: the missing pages add to $325 + 38c$ , and by step 5 this must be $c$ times a whole number.
- $c = 10$ : $705 \div 10$ is not whole.
- $c = 13$ : $819 \div 13 = 63$ , and $4b + 26 + 1 = 63$ gives $b = 9$ , which fits.
- $c = 15$ : $895 \div 15$ , $c = 17$ : $971 \div 17$ , and $c = 20$ : $1085 \div 20$ are not whole.

Only $13$ works.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Suppose the roommate took $c$ consecutive sheets starting right after sheet $b$ , so the missing pages are $2b+1, 2b+2, \dots, 2b+2c$ : that is $2c$ pages with average $\frac{(2b+1) + (2b+2c)}{2}$ , hence total $c(4b + 2c + 1)$ .

All $50$ pages sum to $\frac{50 \cdot 51}{2} = 1275$ . The $50 - 2c$ remaining pages have mean $19$ , so their total is $19(50 - 2c) = 950 - 38c$ . Therefore
$1275 - c(4b + 2c + 1) = 950 - 38c \quad\Longrightarrow\quad c(4b + 2c - 37) = 325 = 5^2 \cdot 13.$

Now $c$ must be a divisor of $325$ with $1 \le c < 25$ : $c \in \{1, 5, 13\}$ .
- $c = 1$ : $4b - 35 = 325$ , $b = 90$ , impossible.
- $c = 5$ : $4b - 27 = 65$ , $b = 23$ , but then sheets $24$ to $28$ would be taken; only $25$ exist.
- $c = 13$ : $4b - 11 = 25$ , $b = 9$ : sheets $10$ through $22$ , pages $19$ to $44$ . Valid.

Check: the remaining $24$ pages ( $1$ to $18$ and $45$ to $50$ ) total $171 + 285 = 456 = 19 \cdot 24$ .

The answer is $\boxed{\textbf{(B)}\ 13}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Prealgebra, Chapter 13: Data and Statistics
Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.5)