
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
Answer
615k+ views
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 \[2,3,7,8\].
.
Recently Updated Pages
Which cell organelles are present in white blood C class 11 biology CBSE

What is the molecular geometry of BrF4 A square planar class 11 chemistry CBSE

How can you explain that CCl4 has no dipole moment class 11 chemistry CBSE

Which will undergo SN2 reaction fastest among the following class 11 chemistry CBSE

The values of mass m for which the 100 kg block does class 11 physics CBSE

Why are voluntary muscles called striated muscles class 11 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

State the laws of reflection of light

Difference Between Prokaryotic Cells and Eukaryotic Cells

Show that total energy of a freely falling body remains class 11 physics CBSE

