Question

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

Answer 1

n+2n+3n+....+99n

= n*(1+2+3+........+99)

= (n*99*100)/2

= n*99*50

= n*9*11*2*25

To make it perfect square we need 2*11

So n = 2*11 = 22

Now n2 = 22*22 = 484

So number of digit in n2 = 3