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# How to solve factorial equations

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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

Feb 18, 2019

#1
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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   Feb 18, 2019
#2
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Thanks CPhill!

Feb 18, 2019