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.