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

a.\[1\]

b.\[2\]

c.\[3\]

d.None of these

Last updated date: 25th Mar 2023

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Answer

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

Complete step-by-step answer:

We have to find the smallest value of integer \[n\] such that the value of \[n+2n+3n+...+99n\] is a perfect square. Further, we have to calculate the digits in the number \[{{n}^{2}}\].

We can rewrite \[n+2n+3n+...+99n\] as \[n\left( 1+2+3+...+99 \right)\].

We have to find the value of \[1+2+3+...+99\].

We know that the formula for sum of \[k\] consecutive positive integers is \[\sum\limits_{i=1}^{k}{i}=\dfrac{k\left( k+1 \right)}{2}\].

Substituting \[k=99\], we have \[1+2+3+...+99=\dfrac{99\times 100}{2}=99\times 50\].

Thus, we have \[n+2n+3n+...+99n=n\left( 99\times 50 \right)\].

We observe that \[n\left( 99\times 50 \right)\] is not a perfect square. We have to make it a perfect square. Factorizing the term \[n\left( 99\times 50 \right)\], we have \[n\left( 99\times 50 \right)=n\left( 9\times 11\times 2\times 25 \right)\].

We observe that \[9\times 25\] is already a perfect square. Thus, the minimum value of \[n\] should be \[11\times 2\] to make \[n\left( 9\times 11\times 2\times 25 \right)\] a perfect square.

Thus, we have the value of \[n\] as \[n=11\times 2=22\].

So, the value of \[{{n}^{2}}\] will be \[{{n}^{2}}=484\].

Hence, the number of digits in \[{{n}^{2}}\] is \[3\], which is option (c).

Note: It’s necessary to use the formula for calculating the sum of \[k\] consecutive positive integers. Also, it’s necessary to keep in mind that the value of \[n\] 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 \[1,4,5,6,9\]. Perfect squares never end with digits \[2,3,7,8\].

.

Complete step-by-step answer:

We have to find the smallest value of integer \[n\] such that the value of \[n+2n+3n+...+99n\] is a perfect square. Further, we have to calculate the digits in the number \[{{n}^{2}}\].

We can rewrite \[n+2n+3n+...+99n\] as \[n\left( 1+2+3+...+99 \right)\].

We have to find the value of \[1+2+3+...+99\].

We know that the formula for sum of \[k\] consecutive positive integers is \[\sum\limits_{i=1}^{k}{i}=\dfrac{k\left( k+1 \right)}{2}\].

Substituting \[k=99\], we have \[1+2+3+...+99=\dfrac{99\times 100}{2}=99\times 50\].

Thus, we have \[n+2n+3n+...+99n=n\left( 99\times 50 \right)\].

We observe that \[n\left( 99\times 50 \right)\] is not a perfect square. We have to make it a perfect square. Factorizing the term \[n\left( 99\times 50 \right)\], we have \[n\left( 99\times 50 \right)=n\left( 9\times 11\times 2\times 25 \right)\].

We observe that \[9\times 25\] is already a perfect square. Thus, the minimum value of \[n\] should be \[11\times 2\] to make \[n\left( 9\times 11\times 2\times 25 \right)\] a perfect square.

Thus, we have the value of \[n\] as \[n=11\times 2=22\].

So, the value of \[{{n}^{2}}\] will be \[{{n}^{2}}=484\].

Hence, the number of digits in \[{{n}^{2}}\] is \[3\], which is option (c).

Note: It’s necessary to use the formula for calculating the sum of \[k\] consecutive positive integers. Also, it’s necessary to keep in mind that the value of \[n\] 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 \[1,4,5,6,9\]. Perfect squares never end with digits \[2,3,7,8\].

.

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