When ${{2}^{155}}$ is divided by 9, the remainder is:
A. 8
B. 7
C. 5
D. 6
E. 1

Hint: We will look at the trend of ${{2}^{x}}$ if it shows any common sequence in their last digit. It is like \[{{2}^{1}}=2,{{2}^{2}}=4,{{2}^{3}}=8,{{2}^{4}}=16,{{2}^{5}}=32,{{2}^{6}}=64,{{2}^{7}}=128\] and so on. Now if we observe the last digit of every equation/ power of 2, we get a sequence of 2,4,8, which are repeating after every third term. With the help of this concept we will find the remainder.

Complete step-by-step answer:
It is given in the question that we have to find the remainder when ${{2}^{155}}$ is divided by 9. To find this we will try to find if any sequence is present in ${{2}^{x}}$ where x is any natural number. Putting natural numbers one by one we get –
 \[\begin{align}
  & {{2}^{1}}=2, \\
 & {{2}^{2}}=4, \\
 & {{2}^{3}}=8, \\
 & {{2}^{4}}=16, \\
 & {{2}^{5}}=32, \\
 & {{2}^{6}}=64, \\
 & {{2}^{7}}=128, \\
 & {{2}^{8}}=256, \\
 & {{2}^{9}}=512, \\
 & {{2}^{10}}=1024 \\
\end{align}\]
And so on. Now, if we look at the last digit of every number we get a sequence of 2,4,8,6 ; 2,4,8,6… repeated. It means that after every four numbers this pattern of 2,4,8,6 is repeated in the expression of ${{2}^{x}}$ last digit. So, to find the last digit of ${{2}^{155}}$, we will divide it by 4.
$4\overset{38}{\overline{\left){\begin{align}
  & 155 \\
 & \underline{12} \\
 & 035 \\
 & \underline{032} \\
 & \underline{003} \\
\end{align}}\right.}}$
We get $\dfrac{155}{4}=38\dfrac{3}{4}$, because $\dfrac{155}{4}=\dfrac{38\times 4+3}{4}$, this means that the third number in the pattern of 4, which is 8. As we know that $8<9$, therefore when ${{2}^{155}}$ is divided by 9, we get 8 as a remainder and thus, option a) is the correct answer.

So, the correct answer is “Option A”.

Note: This type of question is a bit tricky and we have to apply some tricks and understand to solve such problems. Student may get stuck initially because expanding ${{2}^{155}}$ is not possible manually but if we apply our own common sense and try to find only the last digit of ${{2}^{155}}$ instead of finding whole number, we get our result soon.