2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 6 Easy

There is a real n such that (n +1)!+(n+2)!=n!\cdot 440. What is the sum of the digts of n? (2019 AMC 10B Problem, Question#6)

  • A.

    3

  • B.

    8

  • C.

    10

  • D.

    11

  • E.

    12

Answer:C

(n+1)n!+(n+2)(n+1)n!=440\cdot n!

\Rightarrow n![n+1+(n+2)(n+1)]=440\cdot n!

\Rightarrow n+1+n^2+3n+2=440

\Rightarrow n^2+4n-437=0,

solving by the quadratic formula. n= \frac{-4 \pm \sqrt{16+437 \cdot 4}}{2}= \frac{-4 \pm 42}{2}= \frac{38}{2}=19 (since clearly n\geqslant 0)The answer is therefore 1+9=\text {(C)}10.

Dividing both sides by n! gives (n+1)+(n+2)(n+1) = 440 ⇒n^2+4n-437=0\Rightarrow (n-19)(n+23) =0. since n is non-negative, n=19. The answeris 1+9 =\text {(C)} 10.

Dividing both sides by n! as before gives (n +1)+(n+1)(n+2)=440, Now factor out (n+1), giving (n +1)(n +3) = 440. By considering the prime factorization of 440, a bit of experimentation gives us n +1=20 and n +3 =22, so n =19 so the answer is 1+9 =\text {(C)} 10.

Related Problems
Problem 4 2019 AMC 10 B
Problem 5 2019 AMC 10 B
Problem 7 2019 AMC 10 B
Problem 8 2019 AMC 10 B
Think Academy Powered by ChatGPT