Digital Root (repeated digital sum) of the given large integer
The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.
Given a number, the task is to find its digital root. The input number may be large and it may not be possible to store even if we use long long int.
Asked in ACM-ICPC
Input : num = "1234" Output : 1 Explanation : The sum of 1+2+3+4 = 10, digSum(x) == 10 Hence ans will be 1+0 = 1 Input : num = "5674" Output : 4
We have discussed a solution for numbers that can fit in long long it in below post.
Finding sum of digits of a number until sum becomes single digit
In this post, similar approaches are discussed for large numbers.
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Find Digital root of 65785412
- Find out all the digits of a number
- Add all the number one by one
- If the final sum is double digit, add again to make it single digit
- The result obtained in single digit is the Digital root of number
Find Digital root: (6 + 5 + 7 + 8 + 5 + 4 + 1 + 2) = 38 => 11 => (1 + 1) = 2
The idea is based on the fact that for a non-zero number num, digital root is 9 if number is divisible by 9, else digital root is num % 9. (Please see http://www.sjsu.edu/faculty/watkins/Digitsum0.htm for details)
Find the digital root of 65785412
- Sum of digits = 6 + 5 + 7 + 8 + 5 + 4 + 1 + 2 = 38
- Since 38 is not multiple of 9, digital root is 38 % 9 = 2.
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