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2019 AMC 10B Problem 6

Problem 6 of 25EasierAlgebraNumber TheoryArithmetic

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

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Solution

We can rewrite the left side as (n+1)n!+(n+2)(n+1)n!(n+1)n!+(n+2)(n+1)n!=((n+2)2−1)n!,=((n+2)^2-1)n!, so ((n+2)2−1)n!=440n!((n+2)^2-1) n! = 440 n! Therefore, (n+2)2=441,(n+2)^2=441, so n=19.n=19. The sum of its digits is 10.10. Thus, the answer is C .
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Tagged: factorial · algebraic manipulation · digits

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