If $n$ is the smallest natural number such that $n + 2 n + 3 n + . . . . . + 99 n$ is a perfect square, then the number of digits in $n^{2}$ is

1

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2

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3

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More than

3

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Solution

The correct option is B $3$
$n + 2 n + 3 n + . . . . . . + 99 n$
$n (1 + 2 + 3 + . . . . . . . . . . + 99)$
$n \frac{(99) (100)}{2} = 9 \times 25 \times 22 \times n$ is a perfect square when $n = 22$
Number of digits in $n = 3$

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