Problem:
75=2!a2+3!a3+4!a4+5!a5+6!a6+7!a7,
where 0≤ai<i for i=2,3,…,7. Find a2+a3+a4+a5+a6+a7.
Answer Choices:
A. 8
B. 9
C. 10
D. 11
E. 12
Solution:
Multiply both sides of the equation by 7! to obtain
3600=2520a2+840a3+210a4+42a5+7a6+a7.
It follows that 3600−a7 is a multiple of 7, which implies that a7=2. Thus,
73598=514=360a2+120a3+30a4+6a5+a6.
Reason as above to show that 514−a6 is a multiple of 6, which implies that a6=4. Thus, 510/6=85=60a2+20a3+5a4+a5. Then it follows that 85−a5 is a multiple of 5, whence a5=0. Continue in this fashion to obtain a4=1,a3=1, and a2=1. Thus the desired sum is 1+1+1+0+4+2=9.