no
No, it cannot. Just these: 1, 3, 43, 129.
21
6 times, remainder 15
6 ÷ 129 = 0.046511628
21.5
129
Twelve of them
129 times 10 to the negative 6 (129 x 10-6) in standard notation = 0.000129
1,3,14,88,89, and 129
No. All multiples of 6 are even and 129 is odd so it cannot be divisible by 6.
These are all the whole numbers that go into 129 evenly: 1, 3, 43, 129.