Problem:
What is the largest 2-digit prime factor of the integer n=(100200β)?
Solution:
Let p be a prime less than 100. Then
n=(1β
2β
3β―pβ―2pβ―β
β
β
100)21β
2β
3β―pβ―2pβ―3pβ―β
β
kpβ―200β.
Thus, if in addition p2>200, we have
n= (a p-free integer) β
p2jpkβ
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 p2β€200 will be smaller. Since kpβ€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 61β 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