If n is the smallest natural number such that n+2n+3n+.....+99n is a perfect square then the number of digits in n² is?
​a.1  b.2  c.3  d. more than 3

Let P(n)=n+2n+...+99n
Answer is 3. We can get this answer through a good logical maths.
1. Take n common and you get n(1+2+3+...+99), which is perfect square., Apply Sum of AP formula to 1+.+99 and you get it as 99x100/2. or 99 x 50. In 99x50, we have 9x11x25x2xn for ​ n(1+2+3+...+99) 

2. Now we want P(n) as perfect square, we should have n as 11x2 to mate it 9x25x11x11x2x2, which is perfect square. So, n=22. and n²=484. Thus 3 digits in n square.