Question

Find the digit at the unit place of the number \[{{12345}^{6789}}+{{6789}^{12345}}\] and \[{{44}^{44}}\times {{99}^{99}}\times {{66}^{66}}\].

Answer 1

Hint: In this problem, we have to find the unit place of the given expressions. Here we have larger power term, so we have to use the cyclicity concept to find the unit place of each terms and add them, then again we can find the unit place of the added value which will be the answer for the first expression and we can multiply the unit place for each term in the second expression and we can again find the unit place of the multiplied value which is the answer for the second expression.

Complete step by step solution:
Here we have to find the unit places for,
1. \[{{12345}^{6789}}+{{6789}^{12345}}\]
2. \[{{44}^{44}}\times {{99}^{99}}\times {{66}^{66}}\]
We should know that, here we can use the cyclicity concept to solve this problem.
We should know that the concept of cyclicity of any number is about the last digit and how they appear in a certain defined manner. For example, the number four has
\[\begin{align}
  & \Rightarrow {{4}^{1}}=\underline{4},{{4}^{3}}=6\underline{4} \\
 & \Rightarrow {{4}^{2}}=1\underline{6},{{4}^{4}}=25\underline{6} \\
\end{align}\]
Here we can see that the unit place is 4 for every odd power of 4 and the unit place will be 6 for every even power of 4.
We can now solve this problem,
1.\[{{12345}^{6789}}+{{6789}^{12345}}\]
Here we can see that the first term has 5 as the unit place with powers, where the unit place of 5 will be 5 itself.
We can now see that the second term has 9 as unit place, where for even power of 9, will give 1 as the unit place and for odd power of 9, will give 9 as unit place.
We can see that 9 has odd power, then the unit place will be 9.
We can now add the unit place of both terms,
\[\Rightarrow 5+9=14\]
Here, 4 is the unit number of 14.
Therefore, the unit place of \[{{12345}^{6789}}+{{6789}^{12345}}\] is 4.
2.\[{{44}^{44}}\times {{99}^{99}}\times {{66}^{66}}\]
Here we can see that, 4 has even power, then it will have 6 as a unit place.
For the second term, 9 has odd power, so it will give 9 as a unit place.
For the third term, 6 has the term 6 as its unit place irrespective of any positive integer power.
We can now multiply the unit place of three terms, we get
\[\Rightarrow 6\times 9\times 6=324\]
Here the unit place is 4.

Therefore, the unit place of \[{{44}^{44}}\times {{99}^{99}}\times {{66}^{66}}\] is 4.

Note: We should always remember the concept of cyclicity of any number is about the last digit and how they appear in a certain defined manner. For example, the number four has\[\begin{align}
  & \Rightarrow {{4}^{1}}=\underline{4},{{4}^{3}}=6\underline{4} \\
 & \Rightarrow {{4}^{2}}=1\underline{6},{{4}^{4}}=25\underline{6} \\
\end{align}\]
Similarly, we have to use the cyclicity concept to solve these types of problems.