Any of its multiples! Simply calculate any of the following: 57 x 0 57 x 1 57 x 2 57 x 3 57 x (-1) 57 x (-2) etc., and you get a number that is a multiple of 57, and therefore, divisible by 57.
57 = 3 x 19 so 57 is divisible by 1, 3, 19 and 57.
The numbers that are divisible by 57 are infinite. The first four are: 57, 114, 171, 228 . . .
No.
No.
Any of its multiples! Simply calculate any of the following: 57 x 0 57 x 1 57 x 2 57 x 3 57 x (-1) 57 x (-2) etc., and you get a number that is a multiple of 57, and therefore, divisible by 57.
57 = 3 x 19 so 57 is divisible by 1, 3, 19 and 57.
114 is divisible by: 1, 2, 3, 6, 19, 38, 57, 114
The numbers that are divisible by 57 are infinite. The first four are: 57, 114, 171, 228 . . .
No. 741 is divisible by: 1, 3, 13, 19, 39, 57, 247, 741.
No. 57 is divisible by 3.
1, 3, 19, 57.
57 divided by 3 is 19. (Any number in which all the digits add up to equal a number divisible by 3 is divisible by 3. 5+7=12 which is divisible by 3.) And since 57 is divisible by more numbers than just 1 and itself it is a composite number.
No.
No.
No. it is composite. 57 is divisible by 3 and 19 :)
57 is divisible by 1, 3, 19 and 57.