Problem:
What is the largest 2-digit prime factor of the integer n=(100200​)?
Solution:
Let p be a prime less than 100. Then
n=(1⋅2⋅3⋯p⋯2p⋯⋅⋅⋅100)21⋅2⋅3⋯p⋯2p⋯3p⋯⋅⋅kp⋯200​.
Thus, if in addition p2>200, we have
n= (a p-free integer) ⋅p2jpk​
Now, such a prime divides n iff k>2j. Once we show that there is at least one prime meeting these conditions, our answer will be the largest such prime, because any p with p2≤200 will be smaller. Since kp≤200, the requirement that p be as large as possible leads to our choosing k as small as possible. Thus we take k=3 and j=1, for which the largest p is 61. Since 612>200, the second display above is correct for p=61. Thus 61​ divides n and is the largest 2-digit prime to do so.
The problems on this page are the property of the MAA's American Mathematics Competitions