Question
Question: If \[n\] is the smallest number such that \[n+2n+3n+...+99n\] is a perfect square, then the number o...
If n is the smallest number such that n+2n+3n+...+99n is a perfect square, then the number of digits in n2 is
a.1
b.2
c.3
d.None of these
Solution
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 i=1∑ki=2k(k+1) 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 n2.
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 n2.
We can rewrite n+2n+3n+...+99n as n(1+2+3+...+99).
We have to find the value of 1+2+3+...+99.
We know that the formula for sum of k consecutive positive integers is i=1∑ki=2k(k+1).
Substituting k=99, we have 1+2+3+...+99=299×100=99×50.
Thus, we have n+2n+3n+...+99n=n(99×50).
We observe that n(99×50) is not a perfect square. We have to make it a perfect square. Factorizing the term n(99×50), we have n(99×50)=n(9×11×2×25).
We observe that 9×25 is already a perfect square. Thus, the minimum value of n should be 11×2 to make n(9×11×2×25) a perfect square.
Thus, we have the value of n as n=11×2=22.
So, the value of n2 will be n2=484.
Hence, the number of digits in n2 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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