No.
To test if a number is divisible by 8:
* first all multiples of 8 are even, so the number must be even;
* then: add 4 times the hundreds digit to twice the tens digit to the ones digit - if this sum is divisible by 8, then so is the original number.
As the test can be applied to the sum, repeating this summing until a single digit remains, only if this single digit is 8 is the original number divisible by 8.
For 100: 4x1 + 2x0 + 0 = 4 which is not 8, so 100 is not divisible by 8.
100 ÷ 4 = 25 → 24 numbers between 0 and 100 exclusive are divisible by 4 100 ÷ 6 = 16 2/3 → 16 numbers between 0 and 100 are divisible by 6 lcm(4, 6) = 12 → 100 ÷ 12 = 8 1/3 → 8 numbers between 0 and 100 are divisible by 4 and 6.
8
112
There are 8 possibles.
101
There are (500-100)/2 = 200 numbers divisible by 2 between 100 and 500 counting 100 but not 500. Of these (500-100)/8 = 50 are divisible by 8. So there are 150 numbers between 100 and 500 divisible by two but not by 8. By relative primeness exactly 50 out of these 150 are divisible by 3 and therefore these 50 are exactly the ones divisible by 6 but not by 8.
100 ÷ 4 = 25 → 24 numbers between 0 and 100 exclusive are divisible by 4 100 ÷ 6 = 16 2/3 → 16 numbers between 0 and 100 are divisible by 6 lcm(4, 6) = 12 → 100 ÷ 12 = 8 1/3 → 8 numbers between 0 and 100 are divisible by 4 and 6.
A number that is divisible by 4 and 6 must be divisible by the LCM of 4 and 6 which is 12. 100/12 = 8.33 200/12 = 16.67 So, there are 16 numbers below 200 that are divisible by 4 and 6 and, of these, 8 are smaller than 100. That gives 16 - 8 = 8 between 100 and 200.
8 of them.
12. 13, if you count 200
8
100 / 8 = 12.5. Rounding this up gives you 13, therefore 13 x 8 is the first multiple of 8 (i.e., number divisible by 8) in the desired range. After that, simply add 8 at a time to get additional multiples.
112
There are 8 possibles.
101
Two of them.
If the last 3 digits are divisible by 8, the number is divisible by 8.