128
Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.
The number is the square root of 128 times the square root of 128 = 128
999 is divisible by 9, but not by six; the next lower number divisible by 9 is 990, which is also divisible by 6, so that's the answer. Some shortcuts for divisibility: 0 is divisible by any number. If the last digit of a number is divisible by 2, the number itself is divisible by 2. If the sum of the digits of a number is divisible by 3, the number itself is divisible by 3. If the last TWO digits of a number are divisible by 4, the number itself is divisible by 4. If the last digit of a number is divisible by 5, the number itself is divisible by 5. If a number is divisible by both 2 and 3, it is divisible by 6. If the last THREE digits of a number are divisible by 8, the number itself is divisible by 8. If the sum of the digits of a number is divisible by 9, the number itself is divisible by 9. 990: 9+9+0=18, which is divisible by 9, so 990 is divisible by 9. 18 is also divisible by 3, so 990 is divisible by 3, and since 990 ends in 0 it's also divisible by 2, meaning that it's divisible by 6 as well.
There are 128 integers between 100 and 1000 that are divisible by seven.
1x128 2x64 4x32 8x16
Not to make a whole number- it works out at 42.7.
128
128/2 = 64
Yes and it is 128/2 = 64
Any number that 32 can go into is divisible (can be divided) by 32. That means 32, 64, 96, 128, etc, are all divisible by 32.
Yes
An incorrect statement.
All multiples of 64, which is an infinite number, including 64, 128, 192, 256, 320 . . .
Yes, the result is 32.
Yes. The quotient is 32.
128.