Answer
Verified
381.3k+ views
Hint: First of all use the formula of exponents given s ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ and write ${{7}^{63}}$ as ${{\left( {{7}^{3}} \right)}^{21}}$. Find the cube of 7 inside the bracket and write 343 = 344 – 1. Now, use the binomial expansion of the expression
${{\left( x+y \right)}^{n}}$ given as ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$ and substitute x = 344 and y = -1. Write the last term -1 as – (344 – 343) and accordingly write the remainder of the expression when it is divided by 344.
Complete step by step solution:
Here we have been asked to find the remainder of the expression ${{7}^{63}}$ when it is divided by 344. We need to simplify the dividend ${{7}^{63}}$ by using the binomial expansion formula.
Now, using the formula of exponents given as ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ we can write ${{7}^{63}}$ equal to ${{\left( {{7}^{3}} \right)}^{21}}$, so cubing 7 inside the bracket we get,
$\Rightarrow {{7}^{63}}={{\left( 343 \right)}^{21}}$
The base number in the R.H.S can be written as 343 = 344 – 1, so we have,
$\Rightarrow {{7}^{63}}={{\left( 344-1 \right)}^{21}}$
We know that the expansion formula of the binomial expression ${{\left( x+y \right)}^{n}}$ is given as ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$, so substituting x = 344 and y = -1 we get,
\[\begin{align}
& \Rightarrow {{\left( 344-1 \right)}^{21}}={}^{21}{{C}_{0}}{{\left( 344 \right)}^{21}}{{\left( -1 \right)}^{0}}+{}^{21}{{C}_{1}}{{\left( 344 \right)}^{21-1}}{{\left( -1 \right)}^{1}}+....+{}^{21}{{C}_{21}}{{\left( 344 \right)}^{0}}{{\left( -1 \right)}^{21}} \\
& \Rightarrow {{\left( 344-1 \right)}^{21}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-1 \\
& \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-1 \\
\end{align}\]
We can write the last term which is – 1 as – (344 – 343) so we get,
\[\begin{align}
& \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-\left( 344-343 \right) \\
& \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-344+343 \\
\end{align}\]
Clearly we can see that all the terms contain 344 except the last term which is 343 so when we divide the expression by 344 then all the terms will get divided except the last term. Therefore, the remainder of the expression $\left( \dfrac{{{7}^{63}}}{344} \right)$ will be equal to 343.
So, the correct answer is “Option b”.
Note: Note that we have written the last term –1 as –344 + 343 because if we leave it as –1 and will divide the expansion with 344 then the remainder will turn out to be negative which is not possible according to Euclid's division algorithm. So do not think that the last term is –1 so the remainder will be 1 as it will be a wrong approach.
${{\left( x+y \right)}^{n}}$ given as ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$ and substitute x = 344 and y = -1. Write the last term -1 as – (344 – 343) and accordingly write the remainder of the expression when it is divided by 344.
Complete step by step solution:
Here we have been asked to find the remainder of the expression ${{7}^{63}}$ when it is divided by 344. We need to simplify the dividend ${{7}^{63}}$ by using the binomial expansion formula.
Now, using the formula of exponents given as ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ we can write ${{7}^{63}}$ equal to ${{\left( {{7}^{3}} \right)}^{21}}$, so cubing 7 inside the bracket we get,
$\Rightarrow {{7}^{63}}={{\left( 343 \right)}^{21}}$
The base number in the R.H.S can be written as 343 = 344 – 1, so we have,
$\Rightarrow {{7}^{63}}={{\left( 344-1 \right)}^{21}}$
We know that the expansion formula of the binomial expression ${{\left( x+y \right)}^{n}}$ is given as ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$, so substituting x = 344 and y = -1 we get,
\[\begin{align}
& \Rightarrow {{\left( 344-1 \right)}^{21}}={}^{21}{{C}_{0}}{{\left( 344 \right)}^{21}}{{\left( -1 \right)}^{0}}+{}^{21}{{C}_{1}}{{\left( 344 \right)}^{21-1}}{{\left( -1 \right)}^{1}}+....+{}^{21}{{C}_{21}}{{\left( 344 \right)}^{0}}{{\left( -1 \right)}^{21}} \\
& \Rightarrow {{\left( 344-1 \right)}^{21}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-1 \\
& \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-1 \\
\end{align}\]
We can write the last term which is – 1 as – (344 – 343) so we get,
\[\begin{align}
& \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-\left( 344-343 \right) \\
& \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-344+343 \\
\end{align}\]
Clearly we can see that all the terms contain 344 except the last term which is 343 so when we divide the expression by 344 then all the terms will get divided except the last term. Therefore, the remainder of the expression $\left( \dfrac{{{7}^{63}}}{344} \right)$ will be equal to 343.
So, the correct answer is “Option b”.
Note: Note that we have written the last term –1 as –344 + 343 because if we leave it as –1 and will divide the expansion with 344 then the remainder will turn out to be negative which is not possible according to Euclid's division algorithm. So do not think that the last term is –1 so the remainder will be 1 as it will be a wrong approach.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
What percentage of the solar systems mass is found class 8 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE