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Question: If \[n\] is the smallest number such that \[n+2n+3n+...+99n\] is a perfect square, then the number o...

If nn is the smallest number such that n+2n+3n+...+99nn+2n+3n+...+99n is a perfect square, then the number of digits in n2{{n}^{2}} is
a.11
b.22
c.33
d.None of these

Explanation

Solution

Hint: To find the minimum value of nn such that the value of sum n+2n+3n+...+99nn+2n+3n+...+99n is a perfect square, use the formula for the sum of kk consecutive positive integers as i=1ki=k(k+1)2\sum\limits_{i=1}^{k}{i}=\dfrac{k\left( k+1 \right)}{2} to find the sum of n+2n+3n+...+99nn+2n+3n+...+99n. Find the terms needed to be multiplied to make the given value of sum a perfect square. Square the calculated value of nn and count the digits in the value of n2{{n}^{2}}.

Complete step-by-step answer:
We have to find the smallest value of integer nn such that the value of n+2n+3n+...+99nn+2n+3n+...+99n is a perfect square. Further, we have to calculate the digits in the number n2{{n}^{2}}.
We can rewrite n+2n+3n+...+99nn+2n+3n+...+99n as n(1+2+3+...+99)n\left( 1+2+3+...+99 \right).
We have to find the value of 1+2+3+...+991+2+3+...+99.
We know that the formula for sum of kk consecutive positive integers is i=1ki=k(k+1)2\sum\limits_{i=1}^{k}{i}=\dfrac{k\left( k+1 \right)}{2}.
Substituting k=99k=99, we have 1+2+3+...+99=99×1002=99×501+2+3+...+99=\dfrac{99\times 100}{2}=99\times 50.
Thus, we have n+2n+3n+...+99n=n(99×50)n+2n+3n+...+99n=n\left( 99\times 50 \right).
We observe that n(99×50)n\left( 99\times 50 \right) is not a perfect square. We have to make it a perfect square. Factorizing the term n(99×50)n\left( 99\times 50 \right), we have n(99×50)=n(9×11×2×25)n\left( 99\times 50 \right)=n\left( 9\times 11\times 2\times 25 \right).
We observe that 9×259\times 25 is already a perfect square. Thus, the minimum value of nn should be 11×211\times 2 to make n(9×11×2×25)n\left( 9\times 11\times 2\times 25 \right) a perfect square.
Thus, we have the value of nn as n=11×2=22n=11\times 2=22.
So, the value of n2{{n}^{2}} will be n2=484{{n}^{2}}=484.
Hence, the number of digits in n2{{n}^{2}} is 33, which is option (c).

Note: It’s necessary to use the formula for calculating the sum of kk consecutive positive integers. Also, it’s necessary to keep in mind that the value of nn has to be minimum to get a perfect square, otherwise, we will get an incorrect answer. A perfect square is a number obtained by multiplying a whole number by itself. The perfect square numbers must end with digits 1,4,5,6,91,4,5,6,9. Perfect squares never end with digits 2,3,7,82,3,7,8.

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