
What is the unit digit in $\left( {{7^{95}} - {3^{58}}} \right)$ ?
Answer
509.1k+ views
Hint: After a period, the unit digits of powers of a number will be repeated as a series. Let the length of the series be ‘d’. Divide the power of the number with d you will be left with a remainder. Find the unit digit of the number when the power is retained.
Complete step-by-step Solution:
We are given $\left( {{7^{95}} - {3^{58}}} \right)$ and we have to find its unit digit.
Consider \[{7^{95}}\] .
We know that
$
{7^1} = 7 \\
{7^2} = 49 \\
{7^3} = 343 \\
{7^4} = 2401 \\
{7^5} = 16807 \\
{7^6} = 117649 \\
{7^7} = 823543 \\
{7^8} = 5764801 \\
$
The unit digit of 7 power 1 is 7, 7 power 2 is 9, 7 power 3 is 3, 7 power 4 is 1, 7 power 5 is 7, 7 power 6 is 9, 7 power 7 is 3, 7 power 8 is 1.
As we can observe, the series of the unit digits of powers of 7 is 7, 9, 3, 1.
After every 4 powers the unit digits get repeated.
So divide 95 by 4.
$\dfrac{{95}}{4}$ Gives remainder 3. This means that 95 is 3rd in the series of unit digits as (92+3). And the unit digit of the 3rd term is 3.
So, the unit digit of \[{7^{95}}\] is 3.
Consider \[{3^{58}}\] .
We know that
$
{3^1} = 3 \\
{3^2} = 9 \\
{3^3} = 27 \\
{3^4} = 81 \\
{3^5} = 243 \\
{3^6} = 729 \\
{3^7} = 2187 \\
{3^8} = 6561 \\
$
The unit digit of 3 power 1 is 3, 3 power 2 is 9, 3 power 3 is 7, 3 power 4 is 1, 3 power 5 is 3, 3 power 6 is 9, 3 power 7 is 7, 3 power 8 is 1.
As we can observe, the series of the unit digits of powers of 3 is 3, 9, 7, 1.
After every 4 powers the unit digits get repeated here also.
So divide 58 by 4.
$\dfrac{{58}}{4}$ Gives remainder 2. This means that 58 is 2nd in the series of unit digits as (56+2). And the unit digit of the 2nd term of the series is 9.
So, the unit digit of \[{3^{58}}\] is 9.
Therefore, the unit digit of $\left( {{7^{95}} - {3^{58}}} \right)$ is \[3 - 9 = 13 - 9\] (borrow from previous digit of 3) = 4.
Note: For every number, the unit digits of the powers of the numbers will be repeated in a series after some period. This is called cycling of the unit digits. Using this we can find the units of numbers with bigger powers.
Complete step-by-step Solution:
We are given $\left( {{7^{95}} - {3^{58}}} \right)$ and we have to find its unit digit.
Consider \[{7^{95}}\] .
We know that
$
{7^1} = 7 \\
{7^2} = 49 \\
{7^3} = 343 \\
{7^4} = 2401 \\
{7^5} = 16807 \\
{7^6} = 117649 \\
{7^7} = 823543 \\
{7^8} = 5764801 \\
$
The unit digit of 7 power 1 is 7, 7 power 2 is 9, 7 power 3 is 3, 7 power 4 is 1, 7 power 5 is 7, 7 power 6 is 9, 7 power 7 is 3, 7 power 8 is 1.
As we can observe, the series of the unit digits of powers of 7 is 7, 9, 3, 1.
After every 4 powers the unit digits get repeated.
So divide 95 by 4.
$\dfrac{{95}}{4}$ Gives remainder 3. This means that 95 is 3rd in the series of unit digits as (92+3). And the unit digit of the 3rd term is 3.
So, the unit digit of \[{7^{95}}\] is 3.
Consider \[{3^{58}}\] .
We know that
$
{3^1} = 3 \\
{3^2} = 9 \\
{3^3} = 27 \\
{3^4} = 81 \\
{3^5} = 243 \\
{3^6} = 729 \\
{3^7} = 2187 \\
{3^8} = 6561 \\
$
The unit digit of 3 power 1 is 3, 3 power 2 is 9, 3 power 3 is 7, 3 power 4 is 1, 3 power 5 is 3, 3 power 6 is 9, 3 power 7 is 7, 3 power 8 is 1.
As we can observe, the series of the unit digits of powers of 3 is 3, 9, 7, 1.
After every 4 powers the unit digits get repeated here also.
So divide 58 by 4.
$\dfrac{{58}}{4}$ Gives remainder 2. This means that 58 is 2nd in the series of unit digits as (56+2). And the unit digit of the 2nd term of the series is 9.
So, the unit digit of \[{3^{58}}\] is 9.
Therefore, the unit digit of $\left( {{7^{95}} - {3^{58}}} \right)$ is \[3 - 9 = 13 - 9\] (borrow from previous digit of 3) = 4.
Note: For every number, the unit digits of the powers of the numbers will be repeated in a series after some period. This is called cycling of the unit digits. Using this we can find the units of numbers with bigger powers.
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