Question

What is the unit digit of ${{\left( 2 \right)}^{5}}{{\left( 3 \right)}^{3}}{{\left( 4 \right)}^{2}}$?
(a) 6
(b) 4
(c) 8
(d) 2

Answer 1

We solve this problem by converting the given numbers into some other numbers such that any power of that number gives the same unit digit like 5, 6, 0. Then, we take the nit digit of those numbers and multiply them to get the unit digit of the whole number.
We use the standard result of exponents that is ${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$

Complete step by step answer:
We are asked to find the unit digit of ${{\left( 2 \right)}^{5}}{{\left( 3 \right)}^{3}}{{\left( 4 \right)}^{2}}$
Let us assume that the value of given number as,
$\Rightarrow A={{\left( 2 \right)}^{5}}{{\left( 3 \right)}^{3}}{{\left( 4 \right)}^{2}}$
Let us convert the above numbers into some other numbers such that they give the same unit digit when raised to any power.
We know that there are digits which give the same unit digit when raised to any power they are 0, 5, 6.
Here, we can see that there is no 5 in the given number so that getting 0, 5 is not possible.
So, let us try to convert as the power of numbers having unit digit 6.
We know that the number 6 is obtained by multiplying 2 and 3.
So, let us divide the given number in such a way that the power of 2 and 3 will be equal then we get,
$\Rightarrow A={{\left( 2 \right)}^{3}}{{\left( 3 \right)}^{3}}{{\left( 2 \right)}^{2}}{{\left( 4 \right)}^{2}}$
We know that the standard result of exponents that is ${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$
By using this standard result in the above equation then we get,
$\begin{aligned} & \Rightarrow A={{\left( 2\times 3 \right)}^{3}}{{\left( 2 \right)}^{2}}{{\left( 4 \right)}^{2}} \\\ & \Rightarrow A={{\left( 6 \right)}^{3}}{{\left( 2 \right)}^{2}}{{\left( 4 \right)}^{2}} \\\ \end{aligned}$
Now, we can see that the other two numbers have less power and the value can be written directly.
We know that the value of ${{\left( 2 \right)}^{2}}$ is 4 and the value of ${{\left( 4 \right)}^{2}}$ is 16.
By replacing the required values in above number then we get,
$\Rightarrow A={{\left( 6 \right)}^{3}}\times 4\times 16$
Here, we can see that we know that unit digits of all numbers that are in multiplication.
Now, we know that the unit digit of the whole number will be the unit digit of product of individual unit digits.
Here, we can see that there are 3 numbers in multiplication which are having 6, 4 and 6 as unit digits.
Now, let us take the product of individual unit digits then we get,
$\Rightarrow 6\times 4\times 6=144$
Here, we can see that the unit digit is 4.

$\therefore$ we can conclude that the unit digit of given number ${{\left( 2 \right)}^{5}}{{\left( 3 \right)}^{3}}{{\left( 4 \right)}^{2}}$ is 4.

Note: Here, we can see that the powers of each digit of the given number are very less such that we can directly find the given number.
We are asked to find the unit digit of ${{\left( 2 \right)}^{5}}{{\left( 3 \right)}^{3}}{{\left( 4 \right)}^{2}}$
Here, we know that values of each number that is
$\begin{aligned} & \Rightarrow {{2}^{5}}=32 \\\ & \Rightarrow {{3}^{3}}=27 \\\ & \Rightarrow {{4}^{2}}=16 \\\ \end{aligned}$
Now, let us multiply the unit digit of each number then we get,
$\Rightarrow 2\times 7\times 6=84$
Here, we can see that the unit digit is 4.
$\therefore$ Therefore, we can conclude that the unit digit of given number ${{\left( 2 \right)}^{5}}{{\left( 3 \right)}^{3}}{{\left( 4 \right)}^{2}}$ is 4.
But this method is possible in this question only because of less power. But sometimes the power of digits will be in the order of 100’s then we need to use the solution mentioned above to get the required answer easily.