Solution 1: Sum of Digits using a While Loop Code: importjava.util.Scanner;publicclassSumOfDigits{//Method to calculate thesumof digits using awhileloop public staticintcalculateSum(intnumber){intsum=0;//Loop u
A program to determine the sum of digits of a given non-negative integer number using a while loop is presented in Program. The program segment given below does the same thing using a do...while loop.
functiongetSumOfDigits(num){letinitialNumber=num;letsum=0;while(initialNumber){sum+=initialNumber%10;initialNumber=Math.floor(initialNumber/10);}returnsum;}console.log(getSumOfDigits(1));// 👉️ 1console.log(getSumOfDigits(123));// 👉️ 6console.log(getSumOfDigits(1234));// 👉...
using namespace std; unsigned getUnsigned(const string&); unsigned sumDigits(unsigned); int main(int argc, char *argv[]) { int n; while (true) { n = getUnsigned("Enter a positive integer: "); cout << "sum of digits = " << sumDigits(n) << endl << endl; ...
Haskell Program to Find the Sum of Natural Numbers using Recursion Java program to find the product of two numbers using recursion Java program to find sum of digits of a number using recursion Java Program to Find Sum of First N Odd numbers and Even numbers How to Find Sum of Natural Num...
(n, sum);// returning sum for print} }intmain() { n = 1, sum = 0;while(n > 0) { cout <<"Enter a non-negative integer (enter 0 to end): \n"; cin >> n; sumDigits (n, sum);// calling sumDigitscout <<"The sum of all digits ", n," is: ", sum,"\n"; }return...
Inside the "sumDigits()" method: It initializes an integer variable 'sum' to store the sum of the digits, starting with 0. It enters a while loop that continues as long as 'n' is not equal to 0. Inside the loop, it calculates the last digit of 'n' using the modulus operator (...
Given a number N, calculate the sum of all the digits in 1–NIt’s easy to misread/misinterpret this question. People quick off the blocks interpret the question as “sum 1–N”, which is a much simpler problem. An inefficient way of performing this is a simple loop, but most people...
If you divide any number by integer number, it’ll only return integer value and discard the digits after decimal point. So make sure to divide the number by floating point value. To convert integer to float, make use of typecasting syntax. Typecasting int N = 5; sum = 2 + 4 + 6...
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