AMC 10 · 2021 · #22

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

sequences-arithmeticmean-median-mode-rangedivisibility-ruleslinear-equations-two-var convert-to-algebracasework ↑ Prerequisites: mean-median-mode-range

📏 Long solution 💡 3 insights

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?

Pick an answer.

(A)

~10

(B)

~13

(C)

~15

(D)

~17

(E)

~20

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A notebook has $50$ pages on $25$ sheets, where sheet $k$ holds pages $2k-1$ and $2k$ (so sheet $1$ has pages $1,2$; sheet $2$ has pages $3,4$; etc.). A consecutive block of sheets is removed from the middle. The remaining sheets' page numbers have mean exactly $19$. How many sheets were removed?

Givens: $50$ pages total, on $25$ sheets, with sheet $k$ holding pages $2k-1$ and $2k$; Sum of all page numbers $= 1 + 2 + \cdots + 50 = 1275$; A consecutive block of $c$ sheets (with sheet numbers $a, a+1, \ldots, a+c-1$) is removed; The mean of the page numbers on the remaining $25 - c$ sheets ($= 50 - 2c$ pages) equals $19$; Answer choices: (A) $10$, (B) $13$, (C) $15$, (D) $17$, (E) $20$

Unknowns: The number of sheets removed, $c$

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems, #6 Guess and Check, #3 Eliminate Possibilities

Tool #7 (Subproblems) — split the count and the sum: removed sheets contribute removed-pages count $= 2c$ and removed-page sum $=$ a known arithmetic-series formula. Tool #13 (Algebra) — set up one equation in $a$ (starting sheet) and $c$ (count) from "mean of remaining $= 19$", and use $325 = $ small factorization to enumerate cases. Tool #6 (Guess & Check) — once the equation reduces to $c(\text{linear in }a, c) = 325 = 5^2 \cdot 13$, test each divisor of $325$ as $c$. Tool #3 (Eliminate) — answer choices $10, 13, 15, 17, 20$ are themselves a tight set; only the divisors of $325$ among them survive.

Execute — Answer: B

#7 Identify Subproblems 6.EE.A.2 Step 1

Set up the algebra.
Let $c$ = number of sheets removed and $a$ = first removed sheet number, so sheets $a, a+1, \ldots, a+c-1$ are gone, and the removed pages run from $2a-1$ through $2(a+c-1) = 2a+2c-2$.
That's $2c$ consecutive page numbers.
Their sum is $\sum_{p=2a-1}^{2a+2c-2} p = \tfrac{2c \cdot [(2a-1) + (2a+2c-2)]}{2} = c(4a + 2c - 3)$.

$$\text{removed sum} = c(4a + 2c - 3)$$

💡 Sum of $2c$ consecutive integers starting at $2a-1$ — arithmetic-series formula gives $c \cdot (\text{first}+\text{last})$.

#13 Convert to Algebra 6.EE.B.7 Step 2

Apply the mean condition.
The total page-number sum is $\tfrac{50 \cdot 51}{2} = 1275$, and after removal the remaining sum is $1275 - c(4a+2c-3)$ over $50 - 2c$ remaining pages.
Mean equals $19$ gives $\tfrac{1275 - c(4a+2c-3)}{50 - 2c} = 19$.
Multiply: $1275 - c(4a + 2c - 3) = 19(50 - 2c) = 950 - 38c$, so $c(4a + 2c - 3) = 325 + 38c$, i.e., $4ac + 2c^2 - 3c - 38c = 325$, giving $c(4a + 2c - 41) = 325$.

$$c(4a + 2c - 41) = 325$$

💡 Translate "mean = 19" into a single equation, then collect the $c$ on one side as a factor.

#3 Eliminate Possibilities 6.NS.B.4 Step 3

Factor $325 = 5^2 \cdot 13$, so divisors of $325$ are $1, 5, 13, 25, 65, 325$.
Since $c$ is the number of removed sheets and the problem says sheets came "from the middle" (so $c < 25$), the candidates are $c \in \{1, 5, 13\}$ (rule out $25, 65, 325$ since $c \le 24$).

$$c \in \{1, 5, 13\}$$

💡 $325$ is small enough to factor by hand — list its divisors and keep the ones $\le 24$.

#6 Guess and Check 6.EE.B.7 Step 4

Test $c = 1$.
The equation becomes $4a + 2 - 41 = 325$, so $4a = 364$, $a = 91$.
Impossible since the notebook has only $25$ sheets.
Eliminate.

$c=1 \Rightarrow a = 91$ (impossible, $a \le 25$)

💡 Plug $c=1$ into the linear equation and check whether $a$ falls in $[2, 24]$.

#6 Guess and Check 6.EE.B.7 Step 5

Test $c = 5$.
$5(4a + 10 - 41) = 325 \Rightarrow 4a - 31 = 65 \Rightarrow 4a = 96 \Rightarrow a = 24$.
But then the last removed sheet would be $a + c - 1 = 28$, exceeding $25$.
Eliminate.

$c=5 \Rightarrow a = 24, a+c-1 = 28 > 25$ (impossible)

💡 $c=5$ forces the block to spill past sheet $25$ — out of range.

#6 Guess and Check 6.EE.B.7 Step 6

Test $c = 13$.
$13(4a + 26 - 41) = 325 \Rightarrow 4a - 15 = 25 \Rightarrow 4a = 40 \Rightarrow a = 10$.
Range check: removed sheets $10, 11, \ldots, 22$, so first $= 10 \ge 2$ and last $= 22 \le 24$ — both interior, matching "from the middle".
This works.

$$c = 13, \;a = 10, \;\text{block} = [10, 22] \subset [2, 24]$$

💡 $c = 13$ gives integer $a = 10$ inside the valid range — the unique surviving case.

#3 Eliminate Possibilities 6.SP.A.3 Step 7

Verify the mean.
Removed pages: $2 \cdot 10 - 1 = 19$ through $2 \cdot 22 = 44$.
That's $44 - 19 + 1 = 26 = 2 \cdot 13$ pages.
Sum of removed pages $= \tfrac{26 \cdot (19 + 44)}{2} = 13 \cdot 63 = 819$.
Remaining sum $= 1275 - 819 = 456$.
Remaining pages $= 50 - 26 = 24$.
Mean $= 456 / 24 = 19$.
\checkmark

$$\text{remaining mean} = \tfrac{1275 - 819}{50 - 26} = \tfrac{456}{24} = 19$$

💡 Always plug the chosen $c$ back into the original mean condition before declaring victory.

[1] #7 6.EE.A.2 Set up the algebra. Let $c$ = number of sheets removed and $a$ = first removed s

[2] #13 6.EE.B.7 Apply the mean condition. The total page-number sum is $\tfrac{50 \cdot 51}{2} =

[3] #3 6.NS.B.4 Factor $325 = 5^2 \cdot 13$, so divisors of $325$ are $1, 5, 13, 25, 65, 325$. S

[4] #6 6.EE.B.7 Test $c = 1$. The equation becomes $4a + 2 - 41 = 325$, so $4a = 364$, $a = 91$.

[5] #6 6.EE.B.7 Test $c = 5$. $5(4a + 10 - 41) = 325 \Rightarrow 4a - 31 = 65 \Rightarrow 4a = 9

[6] #6 6.EE.B.7 Test $c = 13$. $13(4a + 26 - 41) = 325 \Rightarrow 4a - 15 = 25 \Rightarrow 4a =

[7] #3 6.SP.A.3 Verify the mean. Removed pages: $2 \cdot 10 - 1 = 19$ through $2 \cdot 22 = 44$.

Review

Reasonableness: Quick sanity. The overall mean of all $50$ pages is $25.5$. Removing pages $19$ through $44$ (mean $31.5$, above $25.5$) should push the remaining mean DOWN — indeed $19 < 25.5$. Magnitude check: $25.5 - 19 = 6.5$ drop from the overall mean, achievable by removing a heavy upper block. Among the answer choices $\{10, 13, 15, 17, 20\}$, only $c = 13$ is a divisor of $325$ that fits the range $[2, 24]$ for the starting sheet $a$, ruling out neighbors immediately.

Alternative: Tool #3 (Eliminate Possibilities): test each answer choice $c \in \{10, 13, 15, 17, 20\}$ directly in $c(4a + 2c - 41) = 325$ — only $c = 13$ yields an integer $a$ in range. This is even faster than the factor-of-$325$ enumeration when the answer choices are visible.

CCSS standards used (min grade 6)

6.NS.B.4 Find the greatest common factor and least common multiple (Factoring $325 = 5^2 \cdot 13$ to enumerate divisors as candidate values of $c$.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Expressing the sum of removed pages as $c(4a + 2c - 3)$ in terms of $a$ and $c$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form $px + q = r$ (Reducing the mean condition to $c(4a + 2c - 41) = 325$ and solving for $a$ given each candidate $c$.)
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Verifying the remaining-page-number mean equals $19$ for $c=13, a=10$.)

⭐ This AMC 10 problem only needs Grade 6 mean and the equation tools you already know — write the total page sum ($1275$) and the removed-sum formula ($c(4a + 2c - 3)$), set the remaining mean to $19$ to get $c(4a + 2c - 41) = 325$, then factor $325 = 5^2 \cdot 13$ and test $c \in \{1, 5, 13\}$ — only $c = 13$ gives a valid block (sheets $10$ to $22$, pages $19$ to $44$).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

More like this