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

When 2155{{2}^{155}} is divided by 9, the remainder is:
A. 8
B. 7
C. 5
D. 6
E. 1

Explanation

Solution

We will look at the trend of 2x{{2}^{x}} if it shows any common sequence in their last digit. It is like 21=2,22=4,23=8,24=16,25=32,26=64,27=128{{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 2155{{2}^{155}} is divided by 9. To find this we will try to find if any sequence is present in 2x{{2}^{x}} where x is any natural number. Putting natural numbers one by one we get –

& {{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{aligned}$$ 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. $4\overset{38}{\overline{\left){\begin{aligned} & 155 \\\ & \underline{12} \\\ & 035 \\\ & \underline{032} \\\ & \underline{003} \\\ \end{aligned}}\right.}}$ 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.