Final answer:
The largest three-digit palindrome divisible by 18 is 882. It satisfies the conditions of being a palindrome, being even, and having a digit sum divisible by 9, which makes it divisible by 18.
Step-by-step explanation:
The largest three-digit palindrome divisible by 18 is 882. A palindrome is a number that reads the same forwards and backwards. To find the largest three-digit palindrome divisible by 18, we need to consider both the properties of a palindrome and the divisibility rule for 18. As we know, for a number to be divisible by 18, it must be divisible by both 2 and 9. This means the number must be even, and the sum of its digits must be divisible by 9.
Starting from the largest three-digit palindrome, 999, and moving downwards, we check each palindrome to see if it meets the criteria. Since 999 is not even, it's not divisible by 2, and so not divisible by 18. The largest even three-digit palindrome less than 999 is 988, but its digits don't sum up to a multiple of 9.
We continue checking, and when we reach 882, we find that it is both even and the sum of its digits (8+8+2) equals 18, which is divisible by 9. Thus, 882 is divisible by 18 and is the largest three-digit palindrome with this property.