Answer:
45,000
Explanation:
Hey! Let's help you with your question here!
So, let's think about this logically. The only limit we have here is that the leftmost digit cannot be zero. This makes sense because there would be no five-digit number if the leftmost is zero. In order to find the possible amount of even numbers, we need to take the possible numbers of each digit and have them multiplied to each other to get the total. (I will explain this soon).
First Digit:
Since, the rules state that the leftmost digit cannot be zero, this would be the digit that the rule affects. From here, we can have a possibility of the numbers 1 through 9 here. So, for the first digit, we have the possibility of 9 numbers that can be here.
Second, Third, Fourth Digit:
Now you're probably wondering as to why I've grouped up these 3 digits and not the last or the first one. We'll get to the last one in the next explanation, but we exclude the first digit because the rule that affects the first digit, does not affect these digits nor the last digit. With these 3 digits, we don't have that rule of it cannot be zero, so now our possibilities for what the numbers can be is 0 through 9. If we include 0 as a number too, then we have a possibility of 10 numbers that can be within these digits.
Fifth (Last) Digit:
For this last digit, there is an implicit rule being stated for the last digit. The question asks how many five-digit even numbers are possible if the leftmost digit cannot be zero. This rule affects the last digit only as that allows the whole five-digit number to be even and zero is included in this. So, the even numbers are 0, 2, 4, 6, and 8. In this case, we only have 5 possible numbers to choose from for the very last digit.
Answer Explanation:
Before I begin answering, back in the very first paragraph, I said we need to take the possible numbers of each digit and multiply them altogether to get the total amount of possible values. Why do we do this? This is the idea of possibility combination. We multiply because we are taking in account all of the possible values whereas if we just add, we're only taking in account the maximum possible value of each possibility. So, let's calculate the answer now! For the first digit, we have a possibility of 9 numbers being there (1-9). For the Second, Third, and Fourth digit, we have a possibility of 10 numbers being there (0-9). And finally for the last digit, we have a possibility of only 5 numbers (0, 2, 4, 6, and 8). So, the total possible combination is:
Therefore, we get 45,000 total possible five-digit even numbers where the leftmost digit cannot be zero.