A number theory problem by Vishnu Kadiri
Number Theory
Level
3
What is the maximum value of so that the following is a perfect square:
The answer is 3.
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1 solution
Let us add 1!, 2!, 3!, 4!, 5!. After 5!, the unit digit will always be equal to zero because anything multiplied by zero is 0. 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120 .So, 1+2+6+24+120 =…..3
So if you add even after 5! then also the unit digit will be 3. And a perfect square can’t end in 3. So n should be smaller than or equal to 3. If n is 3, then it will be a perfect square. But it won't for 2. It will for 1. So n can be 1 and 3. Hence maximum value of n=3