Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Find the unit digit of $\left( {{7^{95}} - {3^{58}}} \right)$
A) 6
B) 4
C) 3
D) None of these

seo-qna
Last updated date: 25th Jul 2024
Total views: 348.6k
Views today: 3.48k
Answer
VerifiedVerified
348.6k+ views
Hint: To find the unit digit of $\left( {{7^{95}} - {3^{58}}} \right)$, first of all let it be equal to N. Now, break the power of 7 and 3 separately in simple terms so we can find their unit digits. After breaking their powers, we will get their unit digits and now we just need to subtract them to get our answer.

Complete step by step solution:
In this question, we are given an expression and we need to find the unit digit of that expression.
Given expression: $\left( {{7^{95}} - {3^{58}}} \right)$
Now, let this expression be equal to N. Therefore, we get
$ \Rightarrow N = \left( {{7^{95}} - {3^{58}}} \right)$
Now, as the power is very large, we cannot directly find the unit digit of these numbers. So, we need to break the power of these numbers in simple powers.
Here, we know that ${7^4} = 2401$. Therefore, the unit digit of ${7^4}$ will be equal to 1.
$
   \Rightarrow {7^4} = 2401 \equiv 1 \\
   \Rightarrow {\left( {{7^4}} \right)^{23}} \equiv {1^{23}} \\
   \Rightarrow {7^{92}} \equiv 1 \\
 $
Hence, the unit digit of ${7^{92}}$ will be equal to 1.
Now, we need to find the unit digit of ${7^{95}}$. Therefore, we multiply ${7^{92}}$ with ${7^3}$.
${7^3} = 343 \equiv 3$
$
   \Rightarrow {7^{92}} \cdot {7^3} \equiv 1 \cdot 3 \\
   \Rightarrow {7^{95}} \equiv 3 \\
 $
Hence, the unit digit of ${7^{95}}$ will be equal to 3.
Now, for ${3^{58}}$, we know that ${3^4} = 81$ and so the unit digit for ${3^4}$ will be equal to 1.
$
   \Rightarrow {3^4} \equiv 1 \\
   \Rightarrow {\left( {{3^4}} \right)^{14}} \equiv {\left( 1 \right)^{14}} \\
   \Rightarrow {3^{56}} \equiv 1 \\
 $
But we need to find the unit digit of ${3^{58}}$, so we multiply ${3^{56}}$ with ${3^2}$.
${3^2} = 9 \equiv 9$
$
   \Rightarrow {3^{56}} \cdot {3^2} \equiv 1 \cdot 9 \\
   \Rightarrow {3^{58}} \equiv 9 \\
 $
Hence, the unit digit of ${3^{58}}$ will be equal to 9. Therefore,
$ \Rightarrow N = 3 - 9$
Now, we cannot subtract 9 from 3, so we take 1 carry from the number at the tens place. So, the equation becomes
$ \Rightarrow N = 13 - 9 = 4$
Therefore, the unit digit for $\left( {{7^{95}} - {3^{58}}} \right)$ will be equal to 4.
Hence, option (B) is the correct answer.


Note:
Note that finding the unit digit with such a high power is not possible directly. We need to break the powers into simple powers.
Here, note that we have to break the power in such a way that we get the unit digit as 1 only because anything raised to 1 will be 1 only. For example: We took $ {3^4} = 81 \equiv 1$.