If n is smallest natural number such that n+2 n+3 n+ldots+99 n is a perfect square, then number of d

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8. Sequences and Series

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

A

$1$

B

$2$

C

$3$

D

more than $3$

(KVPY-2015)

Solution

(c)

We have,

$n+2 n+3 n+\ldots+99 n$ is a perfect square

$n(1+2+\ldots+99), \frac{n \times 99 \times 100}{2}$

$n \times 11 \times 9 \times 2 \times 25$

$=(3)^2 \times(5)^2 \times 2 \times 11 \times n$ is a perfect square

$\therefore n$ must be 22 .

$\therefore \quad n^2=(22)^2=484$

Number of digit of $n^2$ is $3 .$

Standard 11

Mathematics

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