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2020 AMC 12B Problem 6

Below is the professionally curated solution for Problem 6 of the 2020 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:factorialfactoringperfect square

Difficulty rating: 1270

6.

For all integers n≥9,n \ge 9, the value of

(n+2)!−(n+1)!n!\frac{(n + 2)! - (n + 1)!}{n!}

is always which of the following?

a multiple of 44

a multiple of 1010

a prime number

a perfect square

a perfect cube

Solution:

Factor (n+1)!(n + 1)! from the numerator: (n+2)!−(n+1)!=(n+1)! [(n+2)−1]=(n+1)! (n+1).(n + 2)! - (n + 1)! = (n + 1)!\,\big[(n + 2) - 1\big] = (n + 1)!\,(n + 1).

Dividing by n!n! leaves (n+1)! (n+1)n!=(n+1)(n+1)=(n+1)2,\dfrac{(n + 1)!\,(n + 1)}{n!} = (n + 1)(n + 1) = (n + 1)^2, which is always a perfect square.

Thus, the correct answer is D.

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