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if n is the smallest natural number such that `n+2n+3n+* * * * * * *+99n` is a perfect squre , then the number of digits in `n^(2)` is -

A

1

B

2

C

3

D

more than 3

Text Solution

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The correct Answer is:
C
`n+2n+3n+** *+99n=k^(2)`
`implies n(99.100)/(2)=k^(2) `
`implies n.99.50=k^(2)`
`implies 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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