Yes, but not evenly, quotient is 162 and remainder is 1
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No - 487/9 = 54.1 recurring (that is, 54.1111...)
No but it is divisible by: 1, 2, 487, 974, 10267, 20534, 5000029, and 10000058.
To determine what numbers are divisible by 28751, we need to find the factors of 28751. The factors of 28751 are 1, 59, 487, and 28751 itself. Therefore, any multiple of these factors (including 28751) will be divisible by 28751. For example, 59 x 487 = 28733, so 28733 is also divisible by 28751.
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.