Question

A perfect square number is of n digits, then find the number of digits in its square root.
If (i) n is even (ii) n is odd.

Answer 1

A number with n -digits is said to have 2n-1 or 2n digits in its perfect square. Here 2n-1 is odd.

Given the perfect square has n digits.

If n is odd then number of digits in its square root = n+1 / 2 .

If n is even then the number of digits in its sauayee root = n/2

Answer 2

Heya User,

--> Let 'k' be the square root so that k² has 'n' digits ...

No. of digits in k² = $[ log_{10} k^{2} ] + 1 [ rounded \ to \ the \ integer ]= n$

=> 2 log k = ( n - 1 )
=> log k = ( n - 1 ) / 2

=> No. of digits in 'k' = [ log k + 1 ] rounded to integer = [ n + 1 ]/2

However, if 'n' is even --> No. of digits = n / 2 [ '.' the '-1' isn't required at 1st step ]

Hope it helps ^_^