I don"t get it, what about the number after 9000.9000-9999 are still positive 4-digit integers right?

You are watching: How many 4-digit positive integers are there in which all 4 digits are even? The first 999 numbers are only three or fewer digits long; 9999 is the last four digit number. Hence 9000 is the number of up-to-4 digit numbers minus the three-or-fewer digit numbers. You can do this by determining valid values for each position or your numbers:

Call the first number the left-most digit. The second, third, and fourth digit appearing to the left of the preceding digit.

The first digit of a four-digit number must be greater than zero. (We have to rule out numbers with fewer than $4$ digits, e.g., we don"t want to count $(0)123, (00)29, (0)999$, etc.)

So that leaves $9$ values $(1, 2, 3, 4, 5, 6, 7, 8, 9)$ that can take the first position. The second, third, and fourth digit can include zero, so there are $10$ possible values, ranging from $0$ to $9$ for each of those positions.

That gives us the total number of 4-digit numbers less than 1000: $$9 \times 10 \times 10\times 10$$

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answered Nov 14 "13 at 18:39 amWhyamWhy
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