No, it is not.
yes 1428/3 = 476 if you add numbers up and you can divide sum by 3 then it is divisible by 3 1+4+2+8 = 15 which is divisible by 3
6
No, not without remainder.
The factors of 476 are 1, 2, 4, 7, 14, 17, 28, 34, 68, 119, 238, and 476.
1904 is divisible by 4. It goes in 476 times.
Heres a few: 4760, 47600, 476000000000000.
158.6667
119 is divisible without remainder by 1, 7, 17, and 119.
No, it is divisible by 3.No, it is divisible by 3.No, it is divisible by 3.No, it is divisible by 3.
The multiples of 119 (which are infinite) are all divisible by 119 , including these: 119, 238, 357, 476, 595, 714, 833, 952, 1071 . . .
It is divisible by 3, for example.It is divisible by 3, for example.It is divisible by 3, for example.It is divisible by 3, for example.
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.