KVPY SA Exam 2015 Question Paper

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Question : 2 of 80

Marks: +1, -0

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

1
2
3
more than 3

Solution:

n + 2n + 3n + ... + 99n = k^2

⇒

n99.100/2 = k^2

⇒

n.99.50 = k^2

⇒

n.9.11.25.2 = k^2

So

n = 11.2 = 22

n^2 = 484

No. of digits in

n^2 = 3.

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