Two digits are correct but in the wrong places.
One digit is correct and in the right place.
Two digits are correct with one of them being in the right place and the other one in the wrong place.

One digit is correct but in the wrong place.
None of the digits is correct.
The statements above give clues to the identity of a four-digit number.
What is the last digit of the four-digit number?
A 0
B 1
C 3
D 5
E 9
Solution
C

Let's call the four-digit number $N$. The last clue tells us that none of the digits 7, 6, 4 or 2 is a digit in $N$. Then the fourth clue shows that 1 is a digit in $N$, but it is not the fourth digit. The first clue now tells us that $N$ involves a 3 but not as its third digit. It also shows that 1 is not the second digit. The second clue now tells us that either 8 is the second digit of $N$ and 9 is not one of its digits or else 9 is the first digit of $N$ and 8 is not one of its digits. Suppose that 8 were the correct second digit. Then the third clue would tell us that both 0 and 5 were correct digits. But this would mean that all of $1,3,8,0$ and 5 were digits of the four-digit number $N$. Therefore 8 is incorrect and so 9 is correct and is the first digit. Knowing this, the third clue shows us that exactly one of 5 and 0 is correct and, moreover, it is in the right place. It can't be 5 because the first digit of $N$ is 9 . So 0 is the correct second digit. We already know 3 is correct, but is not the third digit; so the last digit of $N$ is 3 and $N$ is 9013 .