Find the cube of the following numbers using Nikhilam Sutra.
$
\left( {\text{a}} \right){\text{ 103}} \\
\left( {\text{b}} \right){\text{ 105}} \\
\left( {\text{c}} \right){\text{ 107}} \\
$
Answer
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Hint- Here, we will proceed by finding out the base and the sub-base defined in Nikhilam Sutra method and then, we will find the difference which is another term defined in this method. Then, we will apply the formula i.e., Cube of the number = \[{s^2}\](Number + 2d)/s $ \times $\[3{d^2}\]/\[{d^3}\] where s is the sub-base and d is the difference.
Complete step-by-step solution -
$\left( {\text{a}} \right)$ In the first part, we have to find the cube of the number 103
Since, base is the power of 10 which is nearest to the number whose cube is required
This number is nearest to 100 (which is a power of 10) i.e., Base = 100
Sub-base is the quotient when dividing the number whose cube is required to the base. So, when 103 is divided by 100 we will be getting 1 as the quotient which is the required sub-base.
i.e., Sub-base, s = 1
Difference is obtained by subtracting the product of the base and sub-base from the given number whose cube is required.
i.e., Difference, d = (Number) – (Base)(Sub-base)
Here, d = 103 – (100)(1) = 103 – 100 = 3
As we know that the cube according to Nikhilam Sutra is given by
Cube of the number = \[{s^2}\](Number + 2d)/s $ \times $\[3{d^2}\]/\[{d^3}\] where s is the sub-base and d is the difference
By substituting all the values in the above expression, we get
${\left( {103} \right)^3}$= \[{1^2}\](103 + 2(3))/1 $ \times $\[3{\left( 3 \right)^2}\]/\[{3^3}\]
$ \Rightarrow {\left( {103} \right)^3}$= 109/27/27
As the base is 100 so we will take 2 digits that’s why both the numbers 27 will remain as it is.
$ \Rightarrow {\left( {103} \right)^3}$ = 1092727
Therefore, the cube of the number 103 is 1092727.
$\left( {\text{b}} \right)$ In the second part, we have to find the cube of the number 105
Since, base is the power of 10 which is nearest to the number whose cube is required
This number is nearest to 100 (which is a power of 10) i.e., Base = 100
Sub-base is the quotient when dividing the number whose cube is required to the base. So, when 105 is divided by 100 we will be getting 1 as the quotient which is the required sub-base.
i.e., Sub-base, s = 1
Difference is obtained by subtracting the product of the base and sub-base from the given number whose cube is required.
i.e., Difference, d = (Number) – (Base)(Sub-base)
Here, d = 105 – (100)(1) = 105 – 100 = 5
As we know that the cube according to Nikhilam Sutra is given by
Cube of the number = \[{s^2}\](Number + 2d)/s $ \times $\[3{d^2}\]/\[{d^3}\] where s is the sub-base and d is the difference
By substituting all the values in the above expression, we get
${\left( {105} \right)^3}$= \[{1^2}\](105 + 2(5))/1 $ \times $\[3{\left( 5 \right)^2}\]/\[{5^3}\]
$ \Rightarrow {\left( {105} \right)^3}$= 115/75/125
As the base is 100 so we will take 2 digits that’s why the digit 1 from 125 will get added to 75.
$ \Rightarrow {\left( {105} \right)^3}$= 115/76/25 = 1157625
Therefore, the cube of the number 105 is 1157625.
$\left( {\text{c}} \right)$ In the third part, we have to find the cube of the number 107
Since, base is the power of 10 which is nearest to the number whose cube is required
This number is nearest to 100 (which is a power of 10) i.e., Base = 100
Sub-base is the quotient when dividing the number whose cube is required to the base. So, when 107 is divided by 100 we will be getting 1 as the quotient which is the required sub-base.
i.e., Sub-base, s = 1
Difference is obtained by subtracting the product of the base and sub-base from the given number whose cube is required.
i.e., Difference, d = (Number) – (Base)(Sub-base)
Here, d = 107 – (100)(1) = 107 – 100 = 7
As we know that the cube according to Nikhilam Sutra is given by
Cube of the number = \[{s^2}\](Number + 2d)/s $ \times $\[3{d^2}\]/\[{d^3}\] where s is the sub-base and d is the difference
By substituting all the values in the above expression, we get
${\left( {107} \right)^3}$= \[{1^2}\](107 + 2(7))/1 $ \times $\[3{\left( 7 \right)^2}\]/\[{7^3}\]
$ \Rightarrow {\left( {107} \right)^3}$= 121/147/343
As the base is 100 so we will take 2 digits that’s why the digit 3 from 343 will get added to 147 and we will get 121/150/43 and then, the digit 1 from 150 will get added to 121.
$ \Rightarrow {\left( {107} \right)^3}$= 122/50/43 = 1225043
Therefore, the cube of the number 107 is 1225043.
Note- In this particular the base defined was 100 for all three numbers 103,105 and 107 whose cubes were required. If we have to find the cube of the number corresponding to which the base will be 10. For that problem, we will make sure that in the final representation of the cube (i.e., x/y/z where x, y and z are three numbers) all the numbers except the first one are 1-digit numbers.
Complete step-by-step solution -
$\left( {\text{a}} \right)$ In the first part, we have to find the cube of the number 103
Since, base is the power of 10 which is nearest to the number whose cube is required
This number is nearest to 100 (which is a power of 10) i.e., Base = 100
Sub-base is the quotient when dividing the number whose cube is required to the base. So, when 103 is divided by 100 we will be getting 1 as the quotient which is the required sub-base.
i.e., Sub-base, s = 1
Difference is obtained by subtracting the product of the base and sub-base from the given number whose cube is required.
i.e., Difference, d = (Number) – (Base)(Sub-base)
Here, d = 103 – (100)(1) = 103 – 100 = 3
As we know that the cube according to Nikhilam Sutra is given by
Cube of the number = \[{s^2}\](Number + 2d)/s $ \times $\[3{d^2}\]/\[{d^3}\] where s is the sub-base and d is the difference
By substituting all the values in the above expression, we get
${\left( {103} \right)^3}$= \[{1^2}\](103 + 2(3))/1 $ \times $\[3{\left( 3 \right)^2}\]/\[{3^3}\]
$ \Rightarrow {\left( {103} \right)^3}$= 109/27/27
As the base is 100 so we will take 2 digits that’s why both the numbers 27 will remain as it is.
$ \Rightarrow {\left( {103} \right)^3}$ = 1092727
Therefore, the cube of the number 103 is 1092727.
$\left( {\text{b}} \right)$ In the second part, we have to find the cube of the number 105
Since, base is the power of 10 which is nearest to the number whose cube is required
This number is nearest to 100 (which is a power of 10) i.e., Base = 100
Sub-base is the quotient when dividing the number whose cube is required to the base. So, when 105 is divided by 100 we will be getting 1 as the quotient which is the required sub-base.
i.e., Sub-base, s = 1
Difference is obtained by subtracting the product of the base and sub-base from the given number whose cube is required.
i.e., Difference, d = (Number) – (Base)(Sub-base)
Here, d = 105 – (100)(1) = 105 – 100 = 5
As we know that the cube according to Nikhilam Sutra is given by
Cube of the number = \[{s^2}\](Number + 2d)/s $ \times $\[3{d^2}\]/\[{d^3}\] where s is the sub-base and d is the difference
By substituting all the values in the above expression, we get
${\left( {105} \right)^3}$= \[{1^2}\](105 + 2(5))/1 $ \times $\[3{\left( 5 \right)^2}\]/\[{5^3}\]
$ \Rightarrow {\left( {105} \right)^3}$= 115/75/125
As the base is 100 so we will take 2 digits that’s why the digit 1 from 125 will get added to 75.
$ \Rightarrow {\left( {105} \right)^3}$= 115/76/25 = 1157625
Therefore, the cube of the number 105 is 1157625.
$\left( {\text{c}} \right)$ In the third part, we have to find the cube of the number 107
Since, base is the power of 10 which is nearest to the number whose cube is required
This number is nearest to 100 (which is a power of 10) i.e., Base = 100
Sub-base is the quotient when dividing the number whose cube is required to the base. So, when 107 is divided by 100 we will be getting 1 as the quotient which is the required sub-base.
i.e., Sub-base, s = 1
Difference is obtained by subtracting the product of the base and sub-base from the given number whose cube is required.
i.e., Difference, d = (Number) – (Base)(Sub-base)
Here, d = 107 – (100)(1) = 107 – 100 = 7
As we know that the cube according to Nikhilam Sutra is given by
Cube of the number = \[{s^2}\](Number + 2d)/s $ \times $\[3{d^2}\]/\[{d^3}\] where s is the sub-base and d is the difference
By substituting all the values in the above expression, we get
${\left( {107} \right)^3}$= \[{1^2}\](107 + 2(7))/1 $ \times $\[3{\left( 7 \right)^2}\]/\[{7^3}\]
$ \Rightarrow {\left( {107} \right)^3}$= 121/147/343
As the base is 100 so we will take 2 digits that’s why the digit 3 from 343 will get added to 147 and we will get 121/150/43 and then, the digit 1 from 150 will get added to 121.
$ \Rightarrow {\left( {107} \right)^3}$= 122/50/43 = 1225043
Therefore, the cube of the number 107 is 1225043.
Note- In this particular the base defined was 100 for all three numbers 103,105 and 107 whose cubes were required. If we have to find the cube of the number corresponding to which the base will be 10. For that problem, we will make sure that in the final representation of the cube (i.e., x/y/z where x, y and z are three numbers) all the numbers except the first one are 1-digit numbers.
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