Please explain how to solve factorial equations. Or tell me how to approach this kind of problem if there is no special way to solve factorial equations.

Q: There is a real number *n* such that \((n+1)!+(n+2)!=n!*440\), What is the sum of the digits of *n?*

(A) 3

(B) 8

(C) 10

(D) 11

(E) 12

CalculatorUser Feb 18, 2019

#1**+3 **

Not as tough as it seems, CU

(n + 1) ! + ( n + 2) ! = n! * 440 divide both sides by n !

(n + 1) ! (n + 2) !

______ + _______ = 440

n ! n !

Note that the first term simplifies to ( n + 1)

And the second term simplifies to ( n + 2) ( n + 1)

So....we really have

(n + 1) + ( n + 2)(n + 1) = 440 factor

(n + 1) ( 1 + n + 2) = 440

(n + 1) ( n + 3) = 440

We could solve this with the quad formula....but......a little logic might lead to the answer much more quickly

Note that

20 * 22 = 440

This would imply that n = 19

And the sum of its digits = 10

CPhill Feb 18, 2019