I don"t acquire 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 number are just three or under digits long; 9999 is the last four digit number. Hence 9000 is the variety of up-to-4 digit numbers minus the three-or-fewer digit numbers.

You have the right to do this by determining valid worths for every

**or her numbers:**

*position*Call the first number the left-most digit. The second, third, and also fourth digit appearing to the left that the preceding digit.

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

So that pipeline $9$ values $(1, 2, 3, 4, 5, 6, 7, 8, 9)$ that deserve to take the first position. The second, third, and fourth digit can incorporate zero, therefore there room $10$ feasible values, varying from $0$ to $9$ for each the those positions.

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

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reply Nov 14 "13 in ~ 18:39

amWhyamWhy

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