A telephone dial has holes numbered from 0 to 9 inclusive, and eight of these holes are also lettered. How many different telephone numbers are possible with this dial, if each "number" consists of two letters followed by five numbers, where the first number is not zero?

To find the number of different telephone numbers possible with this dial, we need to count the number of choices for each digit of the telephone number.

For the first letter, there are 8 choices because there are eight holes that are lettered.

For the second letter, there are also 8 choices because there are eight lettered holes remaining.

For the first number, there are 9 choices because all the numbers from 1 to 9 are available options.

For the second number, third number, fourth number, and fifth number, all digits from 0 to 9 are available options, so each digit has 10 choices.

To find the total number of different telephone numbers, we multiply the number of choices for each digit:

Total number of telephone numbers = 8 (choices for the first letter) * 8 (choices for the second letter) * 9 (choices for the first number) * 10 * 10 * 10 * 10 (choices for the remaining four numbers)

= 8 * 8 * 9 * 10^4

= 57,600 telephone numbers

So, there are 57,600 different telephone numbers possible with this dial if each "number" consists of two letters followed by five numbers, where the first number is not zero.