Which of the following statement(s) is/are TRUE?
I. $$1^{99}+2^{99}+3^{99}+4^{99}+5^{99}$$ is exactly divisible by 5

II. $$31^{11} > 17^{14}$$

$$\frac{1^{99}}{5}\ =reminder\ of\ \left(1\right).$$

$$\frac{2^{99}}{5}\ =\frac{4^{49}\times2}{5}=reminder\ of\ \left(\left(-1\right)^{49}\times2\right)=reminder\ of\ 3.$$

$$\frac{3^{99}}{5}=\frac{9^{49}\times3}{5}=reminder\ of\ \left(-3\right)=reminder\ of\ 2.$$

$$\frac{4^{99}}{5}=\frac{4^{98}\times4}{5}=reminder\ of\ 4.$$

And, $$\frac{5^{99}}{5}=reminder\ of\ 0.$$

So, $$re\min der\ of\ \left(\frac{1+3+2+4+0}{5}\right)=0.$$

So, (I) is true.

Now, we can say that :

$$34^{11}>31^{11}.$$

or, $$\left(2\times17\right)^{11}>31^{11}.$$

or, $$17^{11}\times2^4\times2^4\times2^3>31^{11}.$$

Now, $$17^{11}\times\ 17\times17\times17>17^{11}\times2^4\times2^4\times2^3.$$

So, $$17^{11}\times\ 17\times17\times17>\ 31^{11}.$$

or, $$17^{14}>\ 31^{11}.$$

So, (II) is not correct.

A is correct choice.