start with the 100's place
0 = 0+0
1 = 1+0 = 0+1
2 = 1+1=2+0=0+2
3 = 3+0 = 2+1 = 1+2 = 0+3
...
with the 100's digit, the number of such numbers is one greater than that number
0 : 1
1 : 2
...
therefore there are 1+2+3+4+5+6+7+8+9+10 = 5*11 = 55 such numbers
unless you don't count 000 as a three digit number, then there are 54 such numbers.
There are one thousand. Starting at 2000 and going to 2999
-48
-22
1 x 10 x 10 x 10 x 9 = 9000 1 = it must be 2 10's = any digits 9 = any digit BUT zero
Select all the numbers that $4221462$ is divisible by.
54
There are one thousand. Starting at 2000 and going to 2999
18 positive integers and 36 integers (negative and positive)
There are 320 such numbers.
90
44
1339
There are 10 one digit positive integers (0 - 9) and 9 one digit negative integers (-9 to -1) making 19 in all.
the range of three-digit integers is from 100 to 999. Therefore, there are 300 positive three-digit integers that are divisible by neither 2 nor 3.1 day ago
To determine the number of positive integers less than 1000 with distinct digits and are even, we need to consider the possible combinations of digits. Since the number must be even, the last digit must be even, giving us 5 options (0, 2, 4, 6, 8). For the hundreds digit, we have 9 options (1-9), and for the tens digit, we have 8 options (0-9 excluding the hundreds digit and the last digit). Therefore, the total number of such integers is 5 * 9 * 8 = 360.
the answer would be 900
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