Question

Find the digit at the unit’s place of ${\left( {107} \right)^{59}} \times {\left( {313} \right)^{87}} \times {\left( {438} \right)^{129}} \times {\left( {399} \right)^{99}}$
(A) 2
(B) 8
(C) 1
(D) 4

Answer 1

Hint: For this, we will consider the last digit of the given numbers. Compute their respective last digit using their cyclicity. Find the product of these last digits. The digit in the unit’s place of this product is the required answer.

Complete step by step solution:
We are given the expression${\left( {107} \right)^{59}} \times {\left( {313} \right)^{87}} \times {\left( {438} \right)^{129}} \times {\left( {399} \right)^{99}}$
We need to find the digit at the unit’s place (the last digit) of this expression.
So, we will use the concept of cyclicity of numbers.
Let us try to understand the cyclicity of a number through a simple example.
Following are the powers or the cyclicity chart of 2:
${2^1} = 2$ ${2^3} = 8$ ${2^5} = 32$
${2^2} = 4$ ${2^4} = 16$
The digit in the unit’s place of 2 is 2.
We can observe from the successive powers of 2 that the first such power whose unit’s place is 2 is the fifth power.
The digit in the unit’s place of ${2^5}$ is 2.
As the fifth power gives us the answer, the cyclicity of 2 will be $5 - 1 = 4$
Consider the last digits of 107, 313, 438, and 399.
So, the last digits are 7, 3, 8, and 9.
We know that the last digit of ${\left( {107} \right)^{59}} \times {\left( {313} \right)^{87}} \times {\left( {438} \right)^{129}} \times {\left( {399} \right)^{99}}$is same as the last digit of${\left( 7 \right)^{59}} \times {\left( 3 \right)^{87}} \times {\left( 8 \right)^{129}} \times {\left( 9 \right)^{99}}.........(I)$
Also, the cyclicity of 7, 3, 8, and 9 are 4, 4, 4, and 2 respectively.
Consider the cyclicity charts of 7, 3, 8, and 9
${7^1} = 7$, ${7^3} = 343$, ${7^2} = 49$. ${7^4} = 2401$
Thus, digit in the unit’s place of ${7^4} = 1$
Consider the power 59 of 7. We can write $59 = 4 \times 14 + 3$
\[ \Rightarrow {\left( 7 \right)^{59}} = {\left( 7 \right)^{4 \times 14 + 3}} = {({7^4})^{14}}{.7^3}\]
Therefore, last digit of \[{\left( 7 \right)^{59}}\]= last digit of \[{({7^4})^{14}}\]$ \times $ last digit of \[{7^3}\]
As the last digit of \[{7^4}\]is 1, the last digit of any power of\[{7^4}\] is 1.
Also, as seen from the cyclicity chart, last digit of \[{7^3}\]is 3\[ \ldots \ldots ..\left( 1 \right)\]
Similarly, with the help of cyclicity, we get the last digits of${3^4},{8^4},{9^2}$ as follows:
Last digit of ${3^4}$ is 1.
Last digit of ${8^4}$ is 6.
Last digit of${9^2}$ is 1.
Also,
\[{\left( 3 \right)^{87}} = {\left( 3 \right)^{4 \times 21 + 4}} = {({3^4})^{21}}{.3^3}\]
\[{\left( 8 \right)^{129}} = {\left( 8 \right)^{4 \times 32 + 1}} = {({8^4})^{14}}{.8^1}\]
\[{\left( 9 \right)^{99}} = {\left( 9 \right)^{2 \times 49 + 1}} = {({9^2})^{49}}{.9^1}\]
Thus, last digit of \[{\left( 3 \right)^{87}} = 7..........(2)\]
Last digit of\[{\left( 8 \right)^{129}} = 8..........(3)\]
Last digit of\[{\left( 9 \right)^{99}} = 9..........(4)\]
Thus, from (I), (1), (2), (3), and (4), we have${\left( {107} \right)^{59}} \times {\left( {313} \right)^{87}} \times {\left( {438} \right)^{129}} \times {\left( {399} \right)^{99}}$
= last digit of$3 \times 7 \times 8 \times 9$
= last digit of$(3 \times 7)$$ \times $last digit of$8 \times 9$
= last digit of 21$ \times $ last digit of 72
=$1 \times 2$
$ = 2$.
Hence the last digit of${\left( {107} \right)^{59}} \times {\left( {313} \right)^{87}} \times {\left( {438} \right)^{129}} \times {\left( {399} \right)^{99}}$ is 2.

Note: If the square of a number has the same digit in the unit place as the number, then the cyclicity of such numbers is 1. For example, consider 6. Here ${6^2} = 36$. Thus, the number 6 and its square 36 have the same digit 6 in their unit place. So, the cyclicity of 6 is 1.