## Sum of digits untill reach single digit

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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:
1. Store the input in an array of individual numbers as strings.
2. Create the summation string using `"+".join(nums)` - for the output print.
3. Calculate the sum using `eval(op)` which works on strings (a built-in function) - store in `n`.
4. 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.

• 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:`