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You could use something like this :
Lets say you have a number in num and want to add its digits and store in sum . for .e.g.
1. while num is greater than 0 do following else goto step 5
2. calculate sum as :
sum=sum+num mod 10;
3. modify num
num=num div 10;
4. goto step 1
5. print sum as result
I think the asker was interested in finding the answer when there is only an expression for the original number like 4^10 rather than 23897
rivermaker, perhaps - however it would be easy enough to write some code to take in the string "4^10" and then turn that into a real number, wouldn't it? Then the above algorithm would apply, yes?
Also worth noting that the above algorithm will only work for whole numbers. If you have decimal points involved, I believe things get hairy.
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i agree with farmdawgnation . i thought asker meant to find sum of digits of any number in general. sorry for misunderstanding that . but offcourse one anconvert that into a real number. And secondly, yes this algo will not work correctly for floating numbers.