The Last Digit Only Matters

Relevant wiki: Integer Number Bases

$1110_3=(3^3×1)+(3^2×1)+(3^1×1)+(3^0×0)=39$ which is divisible by 3.

We can see that the other options are not divisible by 3.

We see that $10_{10}$ is divisible by 10 and $10_{2}$ is divisible by 2. Hence, we can say that a number who ends with 0 is divisible by the base in which it is written. So, $1110_3$ is the only number to be divisible by 3

$1011_3 = 3^3 + 3^1 + 3^0\\ \dfrac{3^3 + 3^1 + 3^0}{3} = 3^2 + 1 + \dfrac{1}{3}\\ 1110_3 = 3^3 + 3^2 + 3^1\\ \dfrac{3^3 + 3^2 + 3^1}{3} = 3^2 + 3^1 + 1\\ 1101_3 = 3^3 + 3^2 + 3^0\\ \dfrac{3^3 + 3^2 + 3^0}{3} = 3^2 + 3 + \dfrac{1}{3}$

Therefore, it is obvious that $\boxed{1110_3}$ is divisible by $3$

Generally, for any positive number $x$ in base $n$

$x$ is divisible by $n$ if the last digit is $0$

(a+b)%base=(a%base+b%base)%base
By this, we know that except the units place all the other positions always yield a multiple of base...in case if a rightmost bit is set, then 1 will be added, which when base>1 will always yield a remainder of 1...To say that the number is divisible by some number, we need to make the remainder 0...Title is the hint to get solution...(1110) is the solution here...