Problem: Given that
12!17!+13!16!+14!15!+15!14!+16!13!+17!12!+18!11!+19!10!=N1!18!\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!} 2!17!1+3!16!1+4!15!1+5!14!1+6!13!1+7!12!1+8!11!1+9!10!1=1!18!N
find the greatest integer that is less than N100\frac N{100}100N.
Solution:
The problems on this page are the property of the MAA's American Mathematics Competitions