No - 896/3 = 298.6 recurring (that is, 298.6666...)
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NO!!! Because the last number is '6'. NB Any number ending in '5' or '0' is divisible by '5'. So 896 is not divisible by '5'. However, if you divide '896' by '5' you will have an answer with a remainder. Hence 896/5 = 179 remainder '1'.
Well 800 is divisible by 8 (100), and so is 96 (12), so we can say 896 = 800 + 96 = 8 x 100 + 8 x 12 = 8 x 112. You can divide 112 by 4: 112 / 4 = 28. And 28 = 4 x 7. So we have 896 = 8 x 4 x 4 x 7 = (2^3) x (2^2) x (2^2) x 7 = (2^7) x 7.
3.7333
Yes, if x is an integer divisible by 3, then x^2 is also divisible by 3. This is because for any integer x, x^2 will also be divisible by 3 if x is divisible by 3. This can be proven using the property that the square of any integer divisible by 3 will also be divisible by 3.
All numbers divisible by 3 are NOT divisible by 9. As an example, 6, which is divisible by 3, is not divisible by 9. However, all numbers divisible by 9 are also divisible by 3 because 9 is divisible by 3.