you need to do the same thing as 2 digit integers is really easy
18 positive integers and 36 integers (negative and positive)
Add the magnitudes of the integers (-4 has a magnitude of 4), then take the sign to the answer.
There are 4500 such numbers.
22 of them.
400/4= 100 and 800/4=200
Four random, positive, one-digit integers.
There are 10 one digit positive integers (0 - 9) and 9 one digit negative integers (-9 to -1) making 19 in all.
-4, -2, 0, and 2 are the four consecutive even integers. When you add them up they equal -4.
from 1000 to 9999, ie 9999 - 999 = 9000
There is a clever but tricky way to solve this. If we have to find the number of 4-digit integers that contains at least one 5, we can also find the number of all the 4-digit integers and the number of integers that do not contain any 5's and subtract it from the first number. This is called complementary counting.So, first of all, we must find the number of 4-digit numbers there are. There are 9000 of them.Now, we find the number of 4-digit integers without 5's. By thinking a little bit, we see the first digit must be from 0~9 excluding 5. That is a total of 9 numbers. This is the same for the next three digits. Therefore, there are 94 = 6561 4-digit numbers without a 5.Finally, we can subtract. 9000 - 6561 = 2439