Question

Consider the following method.

public int mystery(int num)
{
int x = num;
while (x > 0)
{
if (x / 10 % 2 == 0)
return x;
x = x / 10;
}
return x;
}
What value is returned as a result of the call mystery(1034) ?
4
A
10
B
34
C
103
D
1034

Answer 1

The method mystery takes an integer as input and returns the last even digit of the number, or the original number if none of the digits are even. In the given call mystery(1034), the method returns 10 as the result.

Step-by-step explanation:

The method mystery takes an integer as input and returns the last digit of the number if it is even. It does this by continuously dividing the number by 10 and checking if the resulting number is even. If it is, the method returns that number. If none of the digits are even, the original number is returned.

In the given call mystery(1034), the method starts with x = 1034. It divides 1034 by 10 and checks if the resulting number, 103, is even. Since 103 is not even, the method divides 103 again by 10 to get 10. This time, 10 is even, so the method returns 10 as the result.

Answer 2

The mystery method checks if the largest place-value digit is even and the value returned by mystery(1034) is 103. So the correct option is D.

Step-by-step explanation:

The method in question takes an integer num and processes it in a loop. During each iteration of the loop, it checks if the largest place-value digit of x (which initially equals num) is even. If it is, the current value of x is returned. x is then divided by 10 (to remove the rightmost digit), and the loop continues until x becomes 0 or the condition is met.

For the call mystery(1034), we follow these steps:

1. Initially, x is 1034. x / 10 is 103, and 103 % 2 is not equal to 0; the loop continues.
2. Next, x becomes 103. x / 10 is 10, and 10 % 2 is equal to 0; the condition is met, and 103 is returned.

The value returned by mystery(1034) is thus 103, which corresponds to option D.