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Suppose we have two numbers x and n. We have to check whether x is divisible by 2^n or not without using arithmetic operators.

So, if the input is like x = 32 n = 5, then the output will be True as 32 = 2^5.

To solve this, we will follow these steps −

- if x AND (2^n - 1) is 0, then
- return True

- return False

Let us see the following implementation to get better understanding −

def solve (x, n): if (x & ((1 << n) - 1)) == 0: return True return False x = 32 n = 5 print(solve(x, n))

32, 5

True

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