Given a number num and a value n, we have to set the value of nth bit from right in the binary representation of that number.

**Introduction to Left shift and Right shift operator**

- The left shift and right shift operators are used to shift the bits of a number either left or right as specified.
- RIGHT SHIFT(>>): it accepts to numbers, and shifts the first number to the right, number of times as specified by the second number. For example,
- Shifting a number to left n times, is equivalent to multiplying the number by 2n.

**# Approaching the problem**

Setting nth bit of a number means assigning the value 1 to its nth bit if it was 0 and if 1 leaving it as it is.

To set the nth bit of a number we have to operate it with a number such that it edits only the nth bit of that number.

Since we have to turn it to 1 if its zero and leave it as it is if its 1, its clear that we have to perform an **OR** operation with a number that has 1 at the nth bit. Rest of the bits should be zero so that it doesn’t change any other bit. The number will be obtained by shifting 1 to the right by n-1 times.

For example, we have to set 2nd bit 13

13 | 1 | 1 | 0 | 1 | |

1>>(n-1) | 0 | 0 | 1 | 0 | |

OR | 1 | 1 | 1 | 1 | =15 |

n-1

=2-1

=1

**# Algorithm**

- Input num and n from the user
- Right shift 1, n-1 times
- Take its OR with num

**Code**

```
#include <iostream>
using namespace std;
int main()
{
int num, n;
cout << "Enter number\n";
cin >> num;
cout << "Enter bit number you wish to set\n";
cin >> n;
cout << "Bit set Successfully\n";
cout << "Answer:" << (num | (1 << (n - 1)));
}
```

**Output**

```
Enter Number
13
Enter bit number you wish to set
2
Bit set Successfully
Answer: 15
```

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