All of the digits of a three-digit integer are distinct and non-zero. Furthermore, the three-digit integer is divisible by 13. Find the largest three-digit integer that has these properties.
In order for a three-digit integer to be divisible by 13, the number formed by its last 2 digits must be divisible, while the first digit minus 1 times 10 must make a multiple of 13.
To find a 3-digit multiple of 13, we look for such numbers that go from 1 to 1000. To find the highest 3-digit number that is a multiple of 13, subtract 13 until you get 3-digit numbers. Since 1000 divided by 13 leaves a remainder of 3, the greatest 3-digit number divisible by 13 is $\boxed{988}$.
To find the largest three-digit integer with distinct, nonzero digits that is divisible by 13, we can work backwards from the largest possible three-digit number (999).
To determine if a number is divisible by 13, we can use the divisibility rule: "Take the difference between 10 times the last digit and the number formed by the remaining digits. If the difference is divisible by 13, then the original number is divisible by 13."
Starting with 999:
1. The last digit is 9. Let's check if 90 (10 times the last digit) minus the number formed by the remaining digits (99) is divisible by 13.
90 - 99 = -9, which is not divisible by 13. So 999 is not divisible by 13.
Next, we try the next largest three-digit number: 998.
2. The last digit is 8. Let's check if 90 minus the number formed by the remaining digits (99) is divisible by 13.
90 - 99 = -9, which is not divisible by 13. So 998 is not divisible by 13.
We continue this process until we find a number that satisfies the given conditions.
Going through the numbers in decreasing order, we find that the largest three-digit integer that meets these conditions is 987.