Using only the digits 7, 8 and 9, how many positive seven-digit integers can be made that are palindromes?
To make a palindrome, the first and last digits must be the same, the second and second-to-last digits must be the same, and so on. Since there are only three choices for each digit, I can use the multiplication principle to count the number of palindromes. The first digit can be any of 7, 8 or 9, so there are 3 choices. The second digit can also be any of 7, 8 or 9, so there are 3 choices. The third digit can also be any of 7, 8 or 9, so there are 3 choices. The fourth digit is the middle digit, and it can be any of 7, 8 or 9, so there are 3 choices. The fifth digit must be the same as the third digit, so there is only 1 choice. The sixth digit must be the same as the second digit, so there is only 1 choice. The seventh digit must be the same as the first digit, so there is only 1 choice. By the multiplication principle, the total number of palindromes is 3 x 3 x 3 x 3 x 1 x 1 x 1 = 81. 