To add all the digits of a number till you get a single digit.
#include<bits/stdc++.h>
using namespace std;
int digSum(int n)
{
if (n == 0)
return 0;
return (n % 9 == 0) ? 9 : (n % 9);
}
int main()
{
int n = 9999;
cout<<digSum(n);
return 0;
}Write a function digitsum that calculates the digit sum of an integer. The digit sum of an integer is the sum of all its digits.
function digSum(n) {
let sum = 0;
let str = n.toString();
console.log(parseInt(str.substring(0, 1)));
for (let i = 0; i < str.length; i++) {
sum += parseInt(str.substring(i,i+1));
}
return sum;
}Write a function digitsum that calculates the digit sum of an integer. The digit sum of an integer is the sum of all its digits. Example: digitsum(192) should return 12.
function digitsum(n){
let sum = 0;
n = n.toString();
for(let i=0; i<n.length; i++) {
sum += parseInt(n[i]);
}
return sum;
}Given a long number, return all the possible sum of two digits of it. For example, 12345: all possible sum of two digits from that number are:
function digits(num){
let numArray = num.toString().split('');
let sumArray = [];
for (let i = 0; i < numArray.length; i++) {
for (let j = i+1; j < numArray.length; j++) {
let sum;
sum = Number(numArray[i]) + Number(numArray[j]);
sumArray.push(sum);
}
}
return sumArray;
}Copyright © 2021 Codeinu
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