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(C) The given expression is $S = n + 2n + 3n + \ldots + 99n$.This can be written as $S = n(1 + 2 + 3 + \ldots + 99)$.Using the sum formula for the fir

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$3$

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(C) The given expression is $S = n + 2n + 3n + \ldots + 99n$.This can be written as $S = n(1 + 2 + 3 + \ldots + 99)$.Using the sum formula for the first $k$ natural numbers,$\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$,we have:$S = n 	imes \frac{99 	imes 100}{2} = n 	imes 99 	imes 50 = n 	imes 4950$.Prime factorization of $4950$ is $4950 = 495 	imes 10 = (5 	imes 99) 	imes 10 = 5 	imes 9 	imes 11 	imes 2 	imes 5 = 2 	imes 3^2 	imes 5^2 	imes 11$.For $S$ to be a perfect square,$n 	imes 2 	imes 3^2 	imes 5^2 	imes 11$ must be a perfect square.This requires $n$ to be of the form $2 	imes 11 	imes k^2 = 22k^2$ for some integer $k$.Since we want the smallest natural number $n$,we set $k=1$,which gives $n = 22$.Then $n^2 = 22^2 = 484$.The number $484$ has $3$ digits.

If $n$ is the smallest natural number such that $n+2n+3n+\ldots+99n$ is a perfect square,then the number of digits of $n^2$ is

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