Problem: The number 695 is to be written with a factorial base of numeration, that is, 695=a1​+a2​⋅2!+a3​⋅3!+⋯+an​⋅n! where a1​,a2​,⋯,an​ are integers such that 0≦ak​≦k, and n! means n(n−1)(n−2)⋯2⋅1. Find a4​:
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
695=a1​+a2​(2⋅1)+a3​(3⋅2⋅1)+a4​(4⋅3⋅2⋅1)+a5​(5⋅4⋅3⋅2⋅1)
695=a1​+2a2​+6a3​+24a4​+120a5​ with 0≦ak​≦k.
Since a5​ must equal 5 (in order to obtain 695), a4â€‹î€ =4 because 5−120+4−24>695. Also a4​ cannot be less than 3 , since, for a4​=2, we have 2â‹…24+3â‹…6+2â‹…2+1<95.∴a4​=3
Check: 5â‹…120+3â‹…24+3â‹…6+2â‹…2+1=695