+0  
 
+1
57
2
avatar+244 

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
avatar+98126 
+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

 

 

cool cool cool

 Feb 18, 2019
 #2
avatar+244 
0

Thanks CPhill!

 Feb 18, 2019

15 Online Users

avatar
avatar
avatar