Question-1

Python

What is the output of the following snippet of code?

def foo(n):
    if n % 2 == 0:
        return n // 2
    return 3 * n + 1

count = 0
x = 10
while x != 1:
    x = foo(x)
    count += 1
print(count)

Answer

\(6\)

Solution

For a given integer n, foo(n) returns n // 2 if n is even and 3 * n + 1 if n is odd. Starting with x initialized to 10, the while loop repeatedly applies foo(x) until x equals 1, incrementing count with each iteration.

First Iteration:

x = 10
Since 10 % 2 == 0, x becomes 10 // 2 = 5
Increment count to 1

Second Iteration:

x = 5
Since 5 % 2 != 0, x becomes 3 * 5 + 1 = 16
Increment count to 2

iteration	previous value of x	x % 2 == 0	current value of x	count
first	10	True	10 // 2 = 5	1
second	5	False	3 * 5 + 1 = 16	2
third	16	True	16 // 2 = 8	3
fourth	8	True	8 // 2 = 4	4
fifth	4	True	4 // 2 = 2	5
sixth	2	True	2 // 2 = 1	6

As the value of x is 1, the loop terminates. So, the loop executes six times, making the final value of count 6.

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