There are many shell programs that will find the sum of the square of individual digits of a number. A small example is: SD=3n=2, sum=2, and SD=2.
1
12.529964112.5, to the justified number of significant digits.
Rounded to ten digits, the number is 10.39230485
49
64
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.
The square root of pi is an irrational number, meaning it has an infinite number of digits after the decimal point. It is typically rounded to a limited number of digits for practical use, such as 3.1416. However, if you are looking for the actual number of digits in the square root of pi, it would be infinite.
I/you/we am 49
1225 = 352.
If the number with the digits reversed can have a leading 0 so that it is a 1-digit number, then 16. Otherwise 13.
There are none.
81
81
81
102
81...........