Problem: The number 695 is to be written with a factorial base of numeration, that is, 695=a1β+a2ββ
2!+a3ββ
3!+β―+anββ
n! where a1β,a2β,β―,anβ are integers such that 0β¦akββ¦k, and n! means n(nβ1)(nβ2)β―2β
1. Find a4β:
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
695=a1β+a2β(2β
1)+a3β(3β
2β
1)+a4β(4β
3β
2β
1)+a5β(5β
4β
3β
2β
1)
695=a1β+2a2β+6a3β+24a4β+120a5β with 0β¦akββ¦k.
Since a5β must equal 5 (in order to obtain 695), a4βξ =4 because 5β120+4β24>695. Also a4β cannot be less than 3 , since, for a4β=2, we have 2β
24+3β
6+2β
2+1<95.β΄a4β=3
Check: 5β
120+3β
24+3β
6+2β
2+1=695