venmic wrote:A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 2, 4
and 5, which can be repeated. If the product of the digits in the area code must be even, how many different
codes can be created?

can you please help 26

For the product of the digits to be even, at least one of the digits needs to be even: either a 2, or a 4, or both. That leaves too many wanted scenarios, so we go for the opposite:

Total number of options (no limitations) - number of forbidden scenarios. The forbidden scenario is the scenario where the product is not even - an odd product.


Total number of options for 3 digits:
Break down into 1st, 2nd 3rd choice of digit
1st choice: 3 options for digits (2, 4 or 5)
2nd choice: still 3 options for digits (since digits can repeat)
3rd choice: still 3 options.

Thus, total number of options is 3*3*3 = 27.

Forbidden option: an odd product means means that all three digits are odd (since only odd*odd*odd will give an odd product). Basically, the only forbidden code is 5,5,5 - a single forbidden scenario.

Thus, total number of 3 digit, repeating, even-product area codes is 27-1=26.

The 1st digit can be selected in 3 ways
The second digit can be selected in 3 ways
The 3rd digit can be selected in 3 ways
Total number of ways a 3 digit number can be formed is 3*3*3=27

Since the product of all the digit must be even there is only 1 combination in which the product is odd. 5-5-5.

Hence the answer [spoiler]27-1=26.[/spoiler]