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If $\mathrm{n}$ is the smallest natural number such that $n+2 n+3 n+\ldots .+99 n$ is a perfect square, then the number of digits in $n^{2}$ is
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The correct answer is:
3
$$
\begin{array}{l}
\mathrm{n}+2 \mathrm{n}+3 \mathrm{n}+\ldots \ldots \ldots .+99 \mathrm{n} \\
\mathrm{n}(1+2+3+\ldots \ldots+99)
\end{array}
$$
$n \frac{(99)(100)}{2}=9 \times 25 \times 22 \times \mathrm{n}$ is a perfect square when $\mathrm{n}=22$
Number of digits in $\mathrm{n}=3$
\begin{array}{l}
\mathrm{n}+2 \mathrm{n}+3 \mathrm{n}+\ldots \ldots \ldots .+99 \mathrm{n} \\
\mathrm{n}(1+2+3+\ldots \ldots+99)
\end{array}
$$
$n \frac{(99)(100)}{2}=9 \times 25 \times 22 \times \mathrm{n}$ is a perfect square when $\mathrm{n}=22$
Number of digits in $\mathrm{n}=3$
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