Question

The value of \[{{\left( 12 \right)}^{{{3}^{x}}}}+{{\left( 18 \right)}^{{{3}^{x}}}},x\in N,\] ends with the digit
(a) 2
(b) 8
(c) 0
(d) Cannot be determined

Answer 1

Hint: To solve the given question, we will first assume that the value of \[{{3}^{x}}\] is t. As t is equal to some power of 3, t will always be odd. Then we will write 12 as (15 – 3) and 18 as (15 + 3). After doing this, we will apply the binomial expansion of these terms. On doing this, the terms which are coming at even places in their expansion will get canceled out and the terms which are at odd places will get added up. Then we will determine the last digit of them by taking out the appropriate terms common.

Complete step-by-step answer:
To start with, we will assume that the value of \[{{3}^{x}}\] is t. Thus, the expression changes to \[{{\left( 12 \right)}^{t}}+{{\left( 18 \right)}^{t}}.\] One important thing to note here is that as t equals some power of 3, it will always be odd. Now, the next thing we are going to do is to assume that the value of this expression is y. Thus, we will get,
\[y={{\left( 12 \right)}^{t}}+{{\left( 18 \right)}^{t}}\]
Now, we will write 12 as (15 – 3) and 18 as (15 + 3). Thus we will get,
\[y={{\left( 15-3 \right)}^{t}}+{{\left( 15+3 \right)}^{t}}\]
Now, we are going to expand the terms on the right-hand side by using the formula of the binomial expansion. According to this, we have the following
\[{{\left( a+b \right)}^{n}}={{\text{ }}^{n}}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{{\text{ }}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{{\text{ }}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+.........+{{\text{ }}^{n}}{{C}_{n}}{{a}^{0}}{{b}^{n}}\]
\[\begin{align}
  & \Rightarrow y=\left[ ^{t}{{C}_{0}}{{\left( 15 \right)}^{t}}{{\left( -3 \right)}^{0}}+{{\text{ }}^{t}}{{C}_{1}}{{\left( 15 \right)}^{t-1}}{{\left( -3 \right)}^{1}}+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-2}}{{\left( -3 \right)}^{2}}+.......+{{\text{ }}^{t}}{{C}_{t}}{{\left( 15 \right)}^{t-t}}{{\left( -3 \right)}^{t}} \right]+ \\
 & \left[ ^{t}{{C}_{0}}{{\left( 15 \right)}^{t}}{{\left( 3 \right)}^{0}}+{{\text{ }}^{t}}{{C}_{1}}{{\left( 15 \right)}^{t-1}}{{\left( 3 \right)}^{1}}+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-2}}{{\left( 3 \right)}^{2}}+.......+{{\text{ }}^{t}}{{C}_{t}}{{\left( 15 \right)}^{t-t}}{{\left( 3 \right)}^{t}} \right] \\
\end{align}\]
Now, we know that t is odd, so we will get,
\[\begin{align}
  & \Rightarrow y=\left[ ^{t}{{C}_{0}}{{\left( 15 \right)}^{t}}-{{\text{ }}^{t}}{{C}_{1}}{{\left( 15 \right)}^{t-1}}\left( 3 \right)+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-2}}{{\left( 3 \right)}^{2}}+.......-{{\text{ }}^{t}}{{C}_{t}}{{\left( 3 \right)}^{t}} \right]+ \\
 & \left[ ^{t}{{C}_{0}}{{\left( 15 \right)}^{t}}+{{\text{ }}^{t}}{{C}_{1}}{{\left( 15 \right)}^{t-1}}\left( 3 \right)+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-2}}{{\left( 3 \right)}^{2}}+.......+{{\text{ }}^{t}}{{C}_{t}}{{\left( 15 \right)}^{t-t}}{{\left( 3 \right)}^{t}} \right] \\
\end{align}\]
Now, we can see that the terms which are present in the even position of the expansion will cancel out and the terms which are present in the odd positions will add up. Thus, we will get,
\[\begin{align}
  & \Rightarrow y={{\text{ }}^{t}}{{C}_{0}}{{\left( 15 \right)}^{t}}-{{\text{ }}^{t}}{{C}_{1}}{{\left( 15 \right)}^{t-1}}{{\left( 3 \right)}^{1}}+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-2}}{{\left( 3 \right)}^{2}}+.......-{{\text{ }}^{t}}{{C}_{t}}{{\left( 3 \right)}^{t}}+ \\
 & ^{t}{{C}_{0}}{{\left( 15 \right)}^{t}}+{{\text{ }}^{t}}{{C}_{1}}{{\left( 15 \right)}^{t-1}}\left( 3 \right)+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-2}}{{\left( 3 \right)}^{2}}+.......+{{\text{ }}^{t}}{{C}_{t}}{{\left( 15 \right)}^{t-t}}{{\left( 3 \right)}^{t}} \\
\end{align}\]
\[\Rightarrow y=2\left[ ^{t}{{C}_{0}}{{\left( 15 \right)}^{t}}+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-2}}{{\left( 3 \right)}^{2}}+.......+{{\text{ }}^{t}}{{C}_{t-1}}\left( 15 \right){{\left( 3 \right)}^{t-1}} \right]\]
Now, we will take 15 common from all the terms on the RHS. Thus, we will get,
\[\Rightarrow y=2\times 15\left[ ^{t}{{C}_{0}}{{\left( 15 \right)}^{t-1}}+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-3}}{{\left( 3 \right)}^{2}}+.......+{{\text{ }}^{t}}{{C}_{t-1}}{{\left( 3 \right)}^{t-1}} \right]\]
\[\Rightarrow y=3\times 10\left[ ^{t}{{C}_{0}}{{\left( 15 \right)}^{t-1}}+{{\text{ }}^{t}}{{C}_{2}}{{\left( 15 \right)}^{t-3}}{{\left( 3 \right)}^{2}}+.......+{{\text{ }}^{t}}{{C}_{t-1}}{{\left( 3 \right)}^{t-1}} \right]\]
Now, we know that when 10 is multiplied by any number, the last digit of the product will be zero. Thus, the last digit of y will be zero. Therefore, the last digit of \[{{\left( 12 \right)}^{{{3}^{x}}}}+{{\left( 18 \right)}^{{{3}^{x}}}}\] will be zero.
Hence, the option (c) is the right answer.

Note: The alternate method of solving the above question is shown below. We have to find the last digit of \[{{\left( 12 \right)}^{t}}+{{\left( 18 \right)}^{t}}.\] The last digit of \[{{\left( 12 \right)}^{t}}\] will be equal to the last digit of \[{{2}^{t}}\] and the last digit of \[{{\left( 18 \right)}^{t}}\] will be equal to the last digit of \[{{8}^{t}}.\] Thus,
\[\text{Last digit}={{2}^{t}}+{{8}^{t}}\]
Now, we put t = 1, we will get,
Last digit = 2 + 8 = 10
Therefore, the last digit will be 0.
Now, we put t = 2, we will get,
Last digit = 4 + 16 = 20
Therefore, the last digit will be 0.
Now, for any value of \[t\in N,\] the last digit will be 0.