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2019 AMC 12B Problem 4

Below is the professionally curated solution for Problem 4 of the 2019 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 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:factorialquadratic

Difficulty rating: 1200

4.

A positive integer nn satisfies the equation (n+1)!+(n+2)!=440⋅n!.(n+1)!+(n+2)!=440\cdot n!. What is the sum of the digits of n?n?

22

55

1010

1212

1515

Solution:

Factor the left side: (n+1)!+(n+2)!=(n+1)! [1+(n+2)]=(n+1)! (n+3). (n+1)!+(n+2)!=(n+1)!\,[1+(n+2)]=(n+1)!\,(n+3).

Dividing both sides by n!n! and using (n+1)!=(n+1) n!(n+1)!=(n+1)\,n! gives (n+1)(n+3)=440. (n+1)(n+3)=440.

So n2+4n−437=0,n^2+4n-437=0, which factors as (n−19)(n+23)=0,(n-19)(n+23)=0, giving n=19.n=19. Its digit sum is 1+9=10.1+9=10.

Thus, C is the correct answer.

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