AMC 10 · 2019 · #4

Grade 6 algebra
factorialpolynomial-factoringfactorsdigit-sum pattern-recognitionguess-and-check ↑ Prerequisites: factorialpolynomial-factoring
📏 Medium solution 💡 3 insights
Problem
Find a positive integer n for which the factorial of n plus one, added to the factorial of n plus two, equals n factorial times 440. Then add the digits of that n.

Pick an answer.

(A)
3
(B)
8
(C)
10
(D)
11
(E)
12
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

Factor out the common piece

Both terms share the same factorial.

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

Tidy the leftover

It tidies into a product of two numbers.

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

Split 440 into two factors

The two factors differ by two.

20 × 22 = 440 ✓
4STEP 4

Solve for n

That gives n immediately.

n + 1 = 20 → n = 19
5STEP 5

Add the digits

Adding the digits gives 10.

1 + 9 = 10 → (C)
Answer
10
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 12 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.