**Convert Binary to Octal program**

**Given:** Binary number as input and we have to convert it to octal number.

This can be done by multiplying each digit of binary number starting from LSB with powers of 2 respectively, converting it to decimal and then diving it with 8 until it can be divided and print the reverse of remainder to get the octal value.

**Example:**

Binary number: 100101

(1*2^5) + (0*2^4)+ (0*2^3)+ (1*2^2)+ (0*2^1)+ (1*2^0) = 37

Decimal number =37

Divide 37 successively by 8 until the remainder is 0

37/8 = 4, remainder is 5

4/8 = 0, remainder is 4

Read from the bottom (MSB) to top (LSB) as 45

Octal number =45

**# Algorithm**

- Binary number is taken as input.
- Multiply each digit of the binary number (starting from the last) with the powers of 2 respectively.
- Add all the multiplied digits.
- We get the decimal form, now divide it with 8 and store the remainder.
- Repeat step 4 until number can be divided.
- Print the reverse of the remainder
- We get the octal value

**Code:**

```
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
//as binary numbers can be long
long binary, binaryinput;
int remainder, decimal_output, quotient, i, j, octal_output[100];
cout << " Enter a binary number : ";
cin >> binaryinput;
binary = binaryinput;
i = 1;
decimal_output = 0;
//converting binary input to decimal
while (binaryinput > 0)
{
remainder = binaryinput % 10;
decimal_output = decimal_output + remainder * i;
i = i + i;
binaryinput = binaryinput / 10;
}
i = 1;
//converting decimal to octal
quotient = decimal_output;
while (quotient > 0)
{
octal_output[i++] = quotient % 8;
quotient = quotient / 8;
}
//printing the output
cout << " The equivalent octal value of binary number " << binary << " is : ";
for (j = i - 1; j > 0; j--)
{
cout << octal_output[j];
}
return 0;
}
```

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