3-Digit Even Numbers with Conditional Digits

If 5 appears in a 3-digit number, the digit immediately following it must be 7. Since the overall 3-digit number must be even, its unit digit must be an even digit (0, 2, 4, 6, 8).

• If 5 is in the hundreds place, the tens place must be 7: form $57\_$. The units digit can be any of $\{0, 2, 4, 6, 8\}$, giving $5$ numbers (570, 572, 574, 576, 578).
• If 5 is in the tens place, the units digit must be 7. But 7 is odd, so the number cannot be even! Hence 5 cannot be in the tens place.
• The units place cannot be 5 because the number is even. Therefore, there are only 5 such numbers.