Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #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

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

How to solve

Strategy Convert to Algebra

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(linear in a, c) = 325 = 5² · 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.

1STEP 1

Set up the removed sum

Let c sheets from sheet a be removed; the 2c removed pages are consecutive, summing to c(4a + 2c - 3).

removed sum = c(4a + 2c - 3)

2STEP 2

Apply the mean condition

Setting the remaining mean to 19 over the total sum 1275 and clearing the fraction collapses to c(4a + 2c - 41) = 325.

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

3STEP 3

Factor 325 into divisors

Since 325 = 5² · 13, its divisors that stay under 25 sheets leave only c ∈ {1, 5, 13} to test.

c ∈ {1, 5, 13}

4STEP 4

Test c equals 1

Testing c = 1 gives 4a = 364, so a = 91 — impossible with only 25 sheets, so drop it.

c=1 → a = 91 (impossible, a ≤ 25)

5STEP 5

Test c equals 5

Testing c = 5 gives a = 24, but then the last removed sheet is 28 ＞ 25 — out of range, so drop it.

c=5 → a = 24, a+c-1 = 28 ＞ 25 (impossible)

6STEP 6

Test c equals 13

Testing c = 13 gives a = 10, removing sheets 10 to 22 — both ends interior, matching "from the middle", so it works.

c = 13, a = 10, block = [10, 22] ⊂ [2, 24]

7STEP 7

Verify the final mean

Removed pages 19 to 44 sum to 819, so remaining 456 over 24 pages gives mean $\frac{456}{24}$ = 19, confirming the count.

remaining mean = (1275 - 819)/(50 - 26) = 456/24 = 19

Answer

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.

💡Key takeaway

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 = 5² · 13 and test c ∈ {1, 5, 13} — only c = 13 gives a valid block (sheets 10 to 22, pages 19 to 44).

Set up the removed sum
Apply the mean condition
Factor 325 into divisors
Test c equals 1
Test c equals 5
Test c equals 13
Verify the final mean

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