To solve this problem, we can use the bitwise AND operator. The property of numbers that are powers of 2 is that they have exactly one bit set in their binary representation. If we subtract 1 from such a number, we get a number with all bits set after the position of the bit in the original number. So, if we do a bitwise AND of the number and the number minus 1, we should get 0. If we don't, the number is not a power of 2.

Here is a Python solution for the problem:

```python
def is_power_of_two(n):
if n == 0:
return 'NO'
else:
return 'YES' if (n & (n - 1)) == 0 else 'NO'

# Test the function
print(is_power_of_two(20)) # Output: NO
print(is_power_of_two(64)) # Output: YES
print(is_power_of_two(98)) # Output: NO
```

In this code, the function `is_power_of_two` checks if the number `n` is a power of 2. It first checks if `n` is 0, because 0 is not a power of 2. If `n` is not 0, it performs a bitwise AND operation between `n` and `n - 1`. If the result is 0, it means `n` is a power of 2, so it returns 'YES'. Otherwise, it returns 'NO'.

Question

To solve this problem, we can use the bitwise AND operator. The property of numbers that are powers of 2 is that they have exactly one bit set in their binary representation. If we subtract 1 from such a number, we get a number with all bits set after the position of the bit in the original number. So, if we do a bitwise AND of the number and the number minus 1, we should get 0. If we don't, the number is not a power of 2.

Here is a Python solution for the problem:

```python
def is_power_of_two(n):
    if n == 0:
        return 'NO'
    else:
        return 'YES' if (n & (n - 1)) == 0 else 'NO'

# Test the function
print(is_power_of_two(20))  # Output: NO
print(is_power_of_two(64))  # Output: YES
print(is_power_of_two(98))  # Output: NO
```

In this code, the function `is_power_of_two` checks if the number `n` is a power of 2. It first checks if `n` is 0, because 0 is not a power of 2. If `n` is not 0, it performs a bitwise AND operation between `n` and `n - 1`. If the result is 0, it means `n` is a power of 2, so it returns 'YES'. Otherwise, it returns 'NO'.

Knowee AI · Accepted Answer

To solve this problem, we can use the bitwise AND operator. The property of numbers that are powers of 2 is that they have exactly one bit set in their binary representation. If we subtract 1 from such a number, we get a number with all bits set after the position of the bit in the original number. So, if we do a bitwise AND of the number and the number minus 1, we should get 0. If we don't, the number is not a power of 2.

Here is a Python solution for the problem:

```python
def is_power_of_two(n):
    if n == 0:
        return 'NO'
    else:
        return 'YES' if (n & (n - 1)) == 0 else 'NO'

# Test the function
print(is_power_of_two(20))  # Output: NO
print(is_power_of_two(64))  # Output: YES
print(is_power_of_two(98))  # Output: NO
```

In this code, the function `is_power_of_two` checks if the number `n` is a power of 2. It first checks if `n` is 0, because 0 is not a power of 2. If `n` is not 0, it performs a bitwise AND operation between `n` and `n - 1`. If the result is 0, it means `n` is a power of 2, so it returns 'YES'. Otherwise, it returns 'NO'.

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