+2862 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
+128407 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
