## Sum of digits untill reach single digit

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write ac program to find sum of digits of a number until a single digit is occurred using while loop

sum of digits of a number in c++

write ac program to find the sum of digits of a given number until the sum becomes a single digit

I set an algorithm which sum a number's digits but I couldn't make it till single digit. It only work for one step.

For example:

a=2, b=8 a^b=256 = 6+5+2 = 13

But I want to reach single digit, like:

`a^b=256 = 6+5+2 = 13 = 3+1 = 4`

Below you can see my codes.

a = int(input("Enter a value")) b = int("Enter second value") number = pow(a, b) sum= 0 while float(number) / 10 >= .1: m = number % 10 sum += m number = number // 10 if float(number) / 10 > .1: print(m, end=" + ") else: print(m, "=", sum)

you can try this.

a = int(input("Enter a value")) b = int(input("Enter second value")) number = pow(a, b) result = str(a)+'^'+str(b) + ' = ' + str(number) while number > 9: digit_sum = sum(map(int, str(number))) result += ' = ' + '+'.join(str(number)) + ' = ' + str(digit_sum) number = digit_sum print ( result )

for a=2, b=8 result:

2^8 = 256 = 2+5+6 = 13 = 1+3 = 4

**Sum of Digits in a^n till a single digit,** Given two numbers a and n, the task is to find the single sum of digits of a^n (pow (a, n)). In single digit sum, we keep doing sum of digit until a� I set an algorithm which sum a number's digits but I couldn't make it till single digit. It only work for one step. For example: a=2, b=8 a^b=256 = 6+5+2 = 13 But I want to reach single digit, like: a^b=256 = 6+5+2 = 13 = 3+1 = 4. Below you can see my codes.

Here you go:

n = 256 while n > 9: n = sum(int(i) for i in str(n)) print(n)

So whats going on? `str(n)`

converts `n`

to a string, strings in python can be iterated over so we can access digit by digit. We do this in a generator, converting each digit back to a integer, `int(i) for i in str(n)`

, we use `sum`

to sum the elements in the generator. We repeat this process until n is a single digit.

Added a solution that gives the calculation explicitly:

def sum_dig(n): _sum = sum(int(i) for i in str(n)) explained = "+".join(list(str(n))) return _sum, explained n = 256 s = "" while n > 10: n, e = sum_dig(n) s+= f'{e}=' s += str(n) print(s)

yields:

2+5+6=1+3=4

**Sum of digits untill reach single digit,** str(n) converts n to a string, strings in python can be iterated over so we can access digit by digit. We do this in a generator, converting each digit back to a integer, int(i) for i in str(n) , we use sum to sum the elements in the generator. We repeat this process until n is a single digit. This Program is used to Find the sum of the digits of a number repeatedly until we get a single digit output. For example if we take a number : 1234, it will be calculated as 1+2+3+4 = 10 then This program will be useful to calculate the sum of digits using a simple C Program.

This produces the output in the format OP asked for:

a = int(input("Enter a value: ")) b = int(input("Enter second value: ")) n = pow(a, b) while n >= 10: nums = [i for i in str(n)] op = "+".join(nums) n = eval(op) print("{}={}".format(op, n))

##### Logic:

- Store the input in an array of individual numbers as strings.
- Create the summation string using
`"+".join(nums)`

- for the output print. - Calculate the sum using
`eval(op)`

which works on strings (a built-in function) - store in`n`

. - Print the summation string and what it equals.

##### Output:

Enter a value: 2 Enter second value: 8 2+5+6=13 1+3=4

Enter a value: 2 Enter second value: 6 6+4=10 1+0=1

Enter a value: 2 Enter second value: 50 1+1+2+5+8+9+9+9+0+6+8+4+2+6+2+4=76 7+6=13 1+3=4

**Program to find the sum of digits of a given number until the sum ,** This sounds like homework, so no code! But this is really simple: 1) Create a " total" variable, and set a "working" variable to the value you want� This video explains one more example of nested loop. Nested loop is used to calculate sum of digits of a given number till it will reduces to single digit.

sol = 0 if (a^b)%9==0: sol = 9 else: sol = (a^b)%9

**C Program,** C Program to print the sum of digits till single digit. Sum of Digits Till Single Digit : SumOfDigits.c. #include<stdio.h>� In single digit sum, we keep doing sum of digit until a single digit is left. Examples: Input : a = 5, n = 4 Output : 4 5^4 = 625 = 6+2+5 = 13 Since 13 has two digits, we sum again 1 + 3 = 4.

**How to write a program to add digits of a number until it becomes a ,** What u can do is call a function recursively until u get a single digit. int sumdigit( int n). {. int sum = 0;. while(n){. C Program - Sum of digits of given number till single digit. Here we are printing sum of given digits till we get the single digit.

**Sum of Digits Until Single Digit in Java – KNOW PROGRAM,** Find the sum of digits until single digit in Java. Example:- number = 123456, The sum of digits of 123456 = 1+2+3+4+5+6 = 21 = 2+1 = 3. Program to find the sum of digits of a given number until the sum becomes a single digit.. Answer / ashutosh shashi int n = 123456789; //any numer of you want sum

**Python: Add the digits of a positive integer repeatedly until the result ,** Python Exercises, Practice and Solution: Write a Python program to add the digits of a positive integer repeatedly until the result has a single digit. In this program, we are going to implement logic to find sum of digits until the number is a single digits in C++ programming language. Submitted by Abhishek Pathak , on October 05, 2017 Given an integer number and we have to keep adding the digits until a single digit is not found.

##### Comments

- Does this answer your question? How to make loop repeat until the sum is a single digit?
- Thank you, sir. Can you explain me what is not working with my code? And there is a point I did not understand about your code, can you please clarify it for me? I don't get the int function here "sum(map(int, str(number)))"
- You're welcome! your code needs an iteration until remaining one digit. but it's working just once.
`sum(map(int, str(number)))`

converts all digits to an integer list and sum. - You may want to go
`while n > 9:`

- I think he wanted to print the formula itself. you're giving him only the result
- Very elegant! You should also try to explain your code so that OP understands why your method works and learns something from it.
- I see, then i misunderstood
- I provided the correct formatting for OP's question. Largely based on this answer.