Any element of the set of numbers of the form 195738*k where k is an integer.
Any multiple of 114.
GCF(95, 38, 57) = 19.
The greatest common factor of 38 and 57 is 19. The prime factorization of 38 is 2*19 The prime factorization of 57 is 3*19 So the GCF is 19.
The GCF is 19.
19, 38, 57, 76, 95, 114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 304, 323, 342, 361, 380.
19, 38, 57, 76, 95.
114 is divisible by: 1, 2, 3, 6, 19, 38, 57, 114
57 = 3 x 19 so 57 is divisible by 1, 3, 19 and 57.
GCF(95, 38, 57) = 19.
38 is divisible by 1, 2, 19, 38.
38 + 19 = 57.
1, 3, 19, 57.
There are an infinite series of numbers that are divisible by 228. Two of them are 456 and 684.On the other hand, if the questioner had meant "what is 228 divisible by?", then the answer is 2, 3, and 19, of which 2 applies twice.1, 2, 3, 4, 6, 12, 19, 38, 57, 76, 114, 228
The greatest common factor of 38 and 57 is 19. The prime factorization of 38 is 2*19 The prime factorization of 57 is 3*19 So the GCF is 19.
No. it is composite. 57 is divisible by 3 and 19 :)
57 is divisible by 1, 3, 19 and 57.
Yes, 57/3 = 19
1, 3, 19, 57.