Problem:
Four positive integers a,b,c, and d have a product of 8 ! and satisfy
ab+a+b=524bc+b+c=146, and cd+c+d=104​
What is a−d ?
Answer Choices:
A. 4
B. 6
C. 8
D. 10
E. 12
Solution:
Note that
(a+1)(b+1)=ab+a+b+1=524+1=525=3â‹…52â‹…7
and
(b+1)(c+1)=bc+b+c+1=146+1=147=3â‹…72
Since (a+1)(b+1) is a multiple of 25 and (b+1)(c+1) is not a multiple of 5 , it follows that a+1 must be a multiple of 25 . Since a+1 divides 525,a is one of 24,74,174, or 524 . Among these only 24 is a divisor of 8 !, so a=24. This implies that b+1=21, and b=20. From this it follows that c+1=7 and c=6. Finally, (c+1)(d+1)=105=3⋅5⋅7, so d+1=15 and d=14. Therefore, a−d=24−14=10​.
The problems on this page are the property of the MAA's American Mathematics Competitions