The number of digits in the square root of a number depends on the number. If it is a square number, the square root will have a finite number of digits. If the number is not a square number then the square root will be an irrational number with an infinite, non-repeating decimal representation. In both cases, the number of digits before the decimal point, in the square root of x will be the rounded value of 1+0.5*log10(x)
I cannot see how odd or even will determine the number of digits in the square root. For example:100 is an even number, and its square root is 10 (2 digits)121 is an odd number, and its square root is 11 (2 digits)Maybe the question is not phrased properly for what the questioner is asking.
If the number with the digits reversed can have a leading 0 so that it is a 1-digit number, then 16. Otherwise 13.
I assume you mean three-digit number. The individual symbols used to write a number (0, 1, ... 9) are called digits.Here is how you can solve this: The highest number you can write with three digits (that doesn't necessarily fulfil your conditions) is 999. Calculate its square root; round the result downwards to the nearest integer; square it again; add 1.
The square root of 55 is an irrational number, meaning it never ends and doesn't have a repeating digit pattern, so it has an infinite number of digits.
I/you/we am 49
1225 = 352.
There are none.
square and cube caculator
There can't be a prime number that has a square root because the square root would be a factor of the number.
It is an irrational number, but rounded to 10 digits, the square root of 3.43 is 1.852025918.
Rounded to ten digits, the number is 10.39230485
Pi is an irrational number and has an infinite number of digits after the decimal point. So does its square root.