Question

If n = 67 then the units digit of $\left[ {{3^n} + {2^n}} \right]$ is
A) 1
B) 10
C) 5
D) None

Answer 1

Hint:
To find the unit digit of a number ${x^y}$ at first we need to find the cyclicity of the number and divide the power by the cyclicity and let the remainder be I. The units place is given by the units place of the number x raised to the power I. Using this we can find the units place for both the numbers and add them to get the unit digit of the required number.

Complete step by step solution:
As we are given that n = 67
We need to find the unit digit of $\left[ {{3^{67}} + {2^{67}}} \right]$
First let's find the unit digit of ${3^{67}}$
Now we need to find the cyclicity of three
Cyclicity of any number is about the last digit and how they appear in a certain defined manner
We know that ,
$
   \Rightarrow {3^1} = 3 \\
   \Rightarrow {3^2} = 9 \\
   \Rightarrow {3^3} = 27 \\
   \Rightarrow {3^4} = 81 \\
   \Rightarrow {3^5} = 273 \\
 $
Therefore the cyclicity of three is 4
So now we need to divide 67 by 4
$ \Rightarrow \frac{{67}}{4} = 16\frac{3}{4}$
From this we get that the remainder is three
Hence the last digit will be 7 as ${3^3} = 27$
Now let's find the unit digit of ${2^{67}}$
Now we need to find the cyclicity of three
Cyclicity of any number is about the last digit and how they appear in a certain defined manner
We know that ,
$
   \Rightarrow {2^1} = 2 \\
   \Rightarrow {2^2} = 4 \\
   \Rightarrow {2^3} = 8 \\
   \Rightarrow {2^4} = 16 \\
   \Rightarrow {2^5} = 32 \\
 $
Therefore the cyclicity of two is 4
So now we need to divide 67 by 4
$ \Rightarrow \frac{{67}}{4} = 16\frac{3}{4}$
From this we get that the remainder is three
Hence the last digit will be 8 as ${2^3} = 8$
As the unit digit of ${3^{67}}$is 7 and unit digit of ${2^{67}}$is 8
When we add them we get 15 and the units place of it is 5
Therefore the units place of $\left[ {{3^{67}} + {2^{67}}} \right]$is 5.

The correct option is c.

Note:
Here the operation between the numbers was added.
If it was subtraction then we need to subtract the unit place of ${3^{67}}$ and ${2^{67}}$
If it was multiplication then we need to multiply the unit place of ${3^{67}}$ and ${2^{67}}$
The cyclicity of number 4 and 5 is 2.
The cyclicity of number 6 , 7, and 8 is 5.
The cyclicity of number 10 is 1.
The cyclicity of number 9 is 10.