AMC 10 · 2019 · #6

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

factorialpolynomial-factoringfactorsdigit-sum pattern-recognitionguess-and-check ↑ Prerequisites: factorialpolynomial-factoring

📏 Medium solution 💡 3 insights

Problem

Find a positive integer n for which (n+1)! + (n+2)! = n! · 440. Then add the digits of that n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

Factorials are scary, but the same n! hides in every term. Divide it out (Tool #9 simplifies the equation), look at the shape that remains (Tool #5: it becomes a product of two consecutive-step integers), then try a few values (Tool #6) and confirm by checking the answer choices' digit sums (Tool #3).

1STEP 1

Every term hides the same n!, so factor that common n! out of the whole equation.

(n+1)!+(n+2)! = n![(n+1)+(n+2)(n+1)]

2STEP 2

Cancel n! from both sides; what remains factors to (n+1)(n+3) = 440.

(n+1) + (n+2)(n+1) = 440 → (n+1)[1+(n+2)] = 440 → (n+1)(n+3) = 440

3STEP 3

Two factors two apart multiply to 440; near √440 ≈ 21, the pair 20 × 22 works.

20 × 22 = 440 ✓

4STEP 4

Set n+1 = 20 (and n+3 = 22 agrees), giving n = 19.

n + 1 = 20 → n = 19

5STEP 5

The digits of n = 19 add to 1 + 9 = 10, which is choice (C).

1 + 9 = 10 → (C)

Answer

Quick check: (19+1)!+(19+2)! = 20! + 21! = 20!·(1+21) = 20!· 22, and 19!· 440 = 19!· 20· 22 = 20!· 22. Both sides equal. Answer n=19 is verified and digit sum 10 matches choice (C).

💡Key takeaway

This AMC 10 problem only needs Grade 6 factoring you already know: pull the shared n! out of both terms, get (n+1)(n+3)=440, spot 20× 22, so n=19 and the digit sum is 10.

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