2021 AMC 10 A Problem 22

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In This Page:

To help you master 2021 AMC 10 A Problem 22, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 A Problem 22

Solution

Solution 1

Variable-and-Equation Method

By expressing the sum of the removed pages in two different ways—first using the total sum minus the remaining sum, and second using the formula for an arithmetic sequence—we can create a single algebraic equation to solve for the number of sheets.

The sum of an arithmetic sequence is $S_{n} = \frac{n ( a _{1} + a _{n} )}{2}$ , where $n$ is the number of terms, $a_{1}$ is the first term, and $a_{n}$ is the last term.

Step 1: Calculate the total sum of all pages

The notebook contains 50 pages, numbered 1 through 50. The sum of the first 50 positive integers can be found using the arithmetic sum formula.

S_{total} = \frac{50 ( 1 + 50 )}{2} = 1275

Step 2: Define variables for the borrowed pages

Let $k$ be the number of sheets borrowed. Since each sheet contains 2 pages, the number of borrowed pages is $2 k$ . Because the pages are printed on sheets with pages 1 and 2 on the first sheet, 3 and 4 on the second, and so on, the first page of any borrowed sheet must be an odd number. Let this starting odd page number be $x$ .

Step 3: Express the sum of the remaining pages

The number of pages left in the notebook is $50 - 2 k$ . We are given that the average of these remaining pages is exactly 19. The sum of a set of numbers is its average multiplied by the number of values.

S_{remaining} = 19 (50 - 2 k) = 950 - 38 k

Step 4: Express the sum of the borrowed pages in two ways

First, the sum of the borrowed pages must be the total sum minus the sum of the remaining pages. Second, the borrowed pages form a sequence of $2 k$ consecutive integers starting at $x$ and ending at $x + 2 k - 1$ . We can calculate their sum using the arithmetic sequence formula.

S_{borrowed} = 1275 - (950 - 38 k) = 325 + 38 k

S_{borrowed} = \frac{2 k ( x + ( x + 2 k - 1 ))}{2} = k (2 x + 2 k - 1)

Step 5: Equate and simplify to find a divisibility condition

Set the two expressions for the sum of the borrowed pages equal to each other. Rearranging the equation allows us to factor out $k$ . Since $x$ and $k$ are both integers, $k$ must be a factor of 325.

k (2 x + 2 k - 1) = 325 + 38 k

k (2 x + 2 k - 1 - 38) = 325

k (2 x + 2 k - 39) = 325

Step 6: Test the factors of 325

The prime factorization of 325 is $5^{2} \times 13$ , so its positive factors are 1, 5, 13, 25, 65, and 325. We know $k$ must be less than 25 because the sheets were borrowed from the middle (meaning some sheets remain). Let’s test the reasonable candidates for $k$ . If $k = 5$ : $5 (2 x + 10 - 39) = 325 ⟹ 2 x - 29 = 65 ⟹ x = 47$ . The 10 borrowed pages would begin at 47 and end at 56, which is impossible because there are only 50 pages. If $k = 13$ : $13 (2 x + 26 - 39) = 325 ⟹ 2 x - 13 = 25 ⟹ x = 19$ . This works perfectly! The first page is 19 (which is odd, correctly starting on the 10th sheet), and the 26 borrowed pages go up to 44, which fits safely inside the 50-page notebook.

325 = 5^{2} \times 13

Correct answer: B. $13$
The number of borrowed sheets is 13 (Answer choice B). Final check: 13 sheets means 26 pages, from 19 to 44. Their sum is

26 \times \frac{19 + 44}{2} = 819

. The total original sum is 1275. The remaining sum is

1275 - 819 = 456

. There are

50 - 26 = 24

pages left. The average of the remaining pages is

456/24 = 19

. Everything perfectly matches.

Solution 2

Divisibility Analysis Method

Once we establish the algebraic equation $k (2 x + 2 k - 39) = 325$ , we can look directly at the answer choices. Since $k$ must be an integer divisor of 325, we can quickly identify the correct choice without solving for the starting page $x$ .

In a Diophantine equation of the form $A \times B = C$ where $A$ and $B$ are integers, both $A$ and $B$ must be exact divisors of $C$ .

Step 1: Set up the algebraic relationship

Following the same reasoning as the standard method, we set the sum of the $2 k$ removed pages equal to the total sum of the book minus the sum of the remaining pages.

k (2 x + 2 k - 1) = 1275 - 19 (50 - 2 k)

k (2 x + 2 k - 39) = 325

Step 2: Analyze divisibility requirements

Because the first borrowed page $x$ and the number of borrowed sheets $k$ are both integers, the expression $(2 x + 2 k - 39)$ must also evaluate to an integer. Therefore, $k$ must perfectly divide 325.

325 = 5^{2} \times 13

Step 3: Compare with answer choices

We can simply check the provided multiple-choice options to find which one is a factor of 325. The choices are 10, 13, 15, 17, and 20. The only number in this list that divides 325 is 13.

k \in {10, 13, 15, 17, 20}

Correct answer: B. $13$
The answer must be 13 (Answer choice B). This shortcut avoids testing multiple cases and directly leverages the constraints of integer divisibility and the multiple-choice format.

Solution 3

Global Excess (Deviation) Method

Instead of calculating absolute sums, we can evaluate how much the pages ‘deviate’ from the target average of 19. If the remaining pages average 19, their net deviation from 19 is zero. Therefore, the entire ‘excess’ sum of the 1 to 50 notebook over a baseline of 19 must be concentrated entirely within the removed pages.

The sum of a set of numbers equals a chosen ‘baseline’ value multiplied by the number of elements, plus the sum of the deviations of each element from that baseline.

Step 1: Calculate the global excess relative to 19

Suppose every page in the 50-page notebook had a value of exactly 19. The total sum would be 950. The actual sum of the integers 1 through 50 is 1275. The difference is the global ‘excess’.

S_{baseline} = 50 \times 19 = 950

S_{actual} = \frac{50 \times 51}{2} = 1275

Excess = 1275 - 950 = 325

Step 2: Determine the excess of the removed pages

After removing $2 k$ pages, the remaining $50 - 2 k$ pages have an average of exactly 19. This means their net excess over 19 is 0. Consequently, all 325 points of excess from the original notebook must reside exclusively in the $2 k$ removed pages.

Step 3: Write the sum of the removed pages

Because the removed pages carry all the excess, their sum is equal to their baseline value (which is 19 for each of the $2 k$ pages) plus the 325 excess.

S_{borrowed} = 2 k (19) + 325 = 38 k + 325

Step 4: Equate and solve

As derived previously, the removed pages are consecutive integers starting at an odd number $x$ . We set their sequence sum equal to the deviation-based sum and factor to solve.

k (2 x + 2 k - 1) = 38 k + 325

k (2 x + 2 k - 39) = 325

Step 5: Conclude the value of k

Testing the factors of 325 reveals that $k = 13$ uniquely yields a valid starting page ( $x = 19$ ).

Correct answer: B. $13$
The number of borrowed sheets is 13 (Answer choice B). This conceptual method skips directly to the core equation,

k (2 x + 2 k - 39) = 325

, with less algebraic substitution.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Sum of an Arithmetic Sequence:

S_{n} = \frac{n ( a _{1} + a _{n} )}{2}

$n$ is the number of terms, $a_{1}$ is the first term, and $a_{n}$ is the last term. Useful for summing consecutive integers. The removed pages form an arithmetic sequence of consecutive integers with $n = 2 k$ terms, first term $x$ , and last term $x + 2 k - 1$ .

Relationship Between Sum and Average:

Sum = Average \times Number of elements

Applies to any finite dataset. The problem gives the exact average (19) of the remaining pages, allowing us to immediately express their total sum as $19 (50 - 2 k)$ .

Diophantine Divisibility:

If $A \times B = C$ and $A, B$ are unknown integers while $C$ is a known integer, then $A$ and $B$ must belong to the factor pairs of $C$ .

Requires the variables to be integers. We reach the equation $k (2 x + 2 k - 39) = 325$ . Because $k$ and $x$ represent counts and page numbers, they are integers, forcing $k$ to divide 325 perfectly.

Deviation from a Baseline Average:

T o t a lE x cess = S u m o f El e m e n t s - (B a se l in e A v er a g e \times T o t a lN u mb ero f El e m e n t s)

Useful when dealing with subsets that have a specified average. By setting 19 as a baseline, the remaining pages contribute an excess of 0, meaning the difference between the total sum of the book and the baseline sum must be entirely absorbed by the removed pages.

Key Takeaway: Remember This Strategy for Next Time

A problem specifies the 'average' of an unknown number of items.

Try: Immediately translate the average into a sum using $Sum = Average \times Count$ . Equations involving sums are usually much easier to manipulate algebraically than equations involving fractions.

Why: It clears denominators and allows you to add or subtract groups of items (like 'total pages' minus 'remaining pages') using basic arithmetic.

An equation contains two integer variables multiplied together to equal a constant, such as $k (2 x + 2 k - 39) = 325$ .

Try: Find the prime factorization of the constant and test its positive divisors.

Why: Integers are rigidly constrained. A continuous equation has infinite solutions, but a Diophantine (integer) equation restricts the variables exclusively to the divisors of the constant.

A problem splits a group into two parts, and gives you the exact average of one part.

Try: Try using the given average as a 'baseline' for the whole group. Calculate the 'global excess' (Total Actual Sum - Baseline Sum).

Why: Because one part perfectly matches the baseline, its deviation is zero. This forces 100% of the calculated global excess into the other part, cleanly isolating it and drastically shortening the algebra.

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