Question:
if n is the smallest natural no. such that n+2n+3n+....+99n is a perfect square,then the no. of digits in square of n is
Answer:
Given that
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
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