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Question

Answers

${{10}^{10}}\cdot \left( {{10}^{10}}+1 \right)\left( {{10}^{10}}+2 \right)$ is divided by $6$ _____.

(a) 2

(b) 4

(c) 0

(d) 6

Answer

Verified

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Hint: For finding the remainder just consider the base of the number leaving its exponent part and evaluate the remainder. After that proceed to identifying a recurring pattern in the remainder to obtain the final result.

Complete step-by-step answer:

First of all, we would expand the expression given in the question:

$\begin{align}

& {{10}^{10}}\cdot \left( {{10}^{10}}+1 \right)\left( {{10}^{10}}+2 \right)=\left( {{10}^{20}}+{{10}^{10}} \right)\left( {{10}^{10}}+2 \right) \\

& ={{10}^{20+10}}+2\cdot {{10}^{20}}+{{10}^{10+10}}+2\cdot {{10}^{10}} \\

& ={{10}^{30}}+3\times {{10}^{20}}+2\times {{10}^{10}} \\

\end{align}$

Now we would individually calculate the remainder when divided by 6 for the above three terms and then finally add all the evaluated remainder to obtain the final answer.

Considering ${{10}^{30}}$ and dividing it by 6 to obtain the remainder,

$\begin{align}

& {{R}_{1}}=\dfrac{{{10}^{30}}}{6} \\

& {{R}_{1}}=\dfrac{{{10}^{29}}\times 10}{6} \\

& {{R}_{1}}={{10}^{29}}\cdot \dfrac{10}{6} \\

& \therefore {{R}_{1}}=4 \\

\end{align}$

Hence, from the above formulation we obtained that when any exponent of 10 is divided by 6 we get 4 as remainder.

So, for a number having 10 as base and any exponent n we can establish a recurring relation as,

$R\left[ \dfrac{{{10}^{n}}}{6} \right]=4$

Now proceeding to the next term, we get

${{R}_{2}}=\dfrac{3\times {{10}^{20}}}{6}$

Operating in the similar manner as we did for the previous term, we get

\[\begin{align}

& {{R}_{2}}=\dfrac{3\times {{10}^{20}}}{6} \\

& {{R}_{2}}=\dfrac{{{10}^{19}}\cdot \left( 3\times 10 \right)}{6} \\

& {{R}_{2}}={{10}^{19}}\cdot \dfrac{30}{6} \\

& \therefore {{R}_{2}}=0 \\

\end{align}\]

At last, the remainder for last term would be

$\begin{align}

& {{R}_{3}}=\dfrac{2\times {{10}^{10}}}{6} \\

& {{R}_{3}}=\dfrac{{{10}^{9}}\left( 2\times 10 \right)}{6} \\

& {{R}_{3}}={{10}^{9}}\cdot \dfrac{20}{6} \\

& \therefore {{R}_{3}}=2 \\

\end{align}$

Therefore, the sum total of all the reminders would give us the desired remainder.

$\begin{align}

& S={{R}_{1}}+{{R}_{2}}+{{R}_{3}} \\

& S=4+0+2 \\

& S=6 \\

\end{align}$

So, the final remainder is obtained by dividing the sum of remainder S with 6.

$\begin{align}

& R=\dfrac{S}{6} \\

& R=\dfrac{6}{6} \\

& \therefore R=0 \\

\end{align}$

Hence, the correct option is (c).

Note: The key step helpful in solving such questions is the correct formation of fraction through which remainder can be evaluated easily. A common mistake in this problem include the expansion of expression because ${{a}^{m}}\times {{a}^{m}}={{a}^{m+m}}={{a}^{2m}}$ which is wrongly evaluated by some students as ${{a}^{m}}\times {{a}^{m}}={{a}^{{{m}^{m}}}}={{a}^{{{m}^{2}}}}$.

Complete step-by-step answer:

First of all, we would expand the expression given in the question:

$\begin{align}

& {{10}^{10}}\cdot \left( {{10}^{10}}+1 \right)\left( {{10}^{10}}+2 \right)=\left( {{10}^{20}}+{{10}^{10}} \right)\left( {{10}^{10}}+2 \right) \\

& ={{10}^{20+10}}+2\cdot {{10}^{20}}+{{10}^{10+10}}+2\cdot {{10}^{10}} \\

& ={{10}^{30}}+3\times {{10}^{20}}+2\times {{10}^{10}} \\

\end{align}$

Now we would individually calculate the remainder when divided by 6 for the above three terms and then finally add all the evaluated remainder to obtain the final answer.

Considering ${{10}^{30}}$ and dividing it by 6 to obtain the remainder,

$\begin{align}

& {{R}_{1}}=\dfrac{{{10}^{30}}}{6} \\

& {{R}_{1}}=\dfrac{{{10}^{29}}\times 10}{6} \\

& {{R}_{1}}={{10}^{29}}\cdot \dfrac{10}{6} \\

& \therefore {{R}_{1}}=4 \\

\end{align}$

Hence, from the above formulation we obtained that when any exponent of 10 is divided by 6 we get 4 as remainder.

So, for a number having 10 as base and any exponent n we can establish a recurring relation as,

$R\left[ \dfrac{{{10}^{n}}}{6} \right]=4$

Now proceeding to the next term, we get

${{R}_{2}}=\dfrac{3\times {{10}^{20}}}{6}$

Operating in the similar manner as we did for the previous term, we get

\[\begin{align}

& {{R}_{2}}=\dfrac{3\times {{10}^{20}}}{6} \\

& {{R}_{2}}=\dfrac{{{10}^{19}}\cdot \left( 3\times 10 \right)}{6} \\

& {{R}_{2}}={{10}^{19}}\cdot \dfrac{30}{6} \\

& \therefore {{R}_{2}}=0 \\

\end{align}\]

At last, the remainder for last term would be

$\begin{align}

& {{R}_{3}}=\dfrac{2\times {{10}^{10}}}{6} \\

& {{R}_{3}}=\dfrac{{{10}^{9}}\left( 2\times 10 \right)}{6} \\

& {{R}_{3}}={{10}^{9}}\cdot \dfrac{20}{6} \\

& \therefore {{R}_{3}}=2 \\

\end{align}$

Therefore, the sum total of all the reminders would give us the desired remainder.

$\begin{align}

& S={{R}_{1}}+{{R}_{2}}+{{R}_{3}} \\

& S=4+0+2 \\

& S=6 \\

\end{align}$

So, the final remainder is obtained by dividing the sum of remainder S with 6.

$\begin{align}

& R=\dfrac{S}{6} \\

& R=\dfrac{6}{6} \\

& \therefore R=0 \\

\end{align}$

Hence, the correct option is (c).

Note: The key step helpful in solving such questions is the correct formation of fraction through which remainder can be evaluated easily. A common mistake in this problem include the expansion of expression because ${{a}^{m}}\times {{a}^{m}}={{a}^{m+m}}={{a}^{2m}}$ which is wrongly evaluated by some students as ${{a}^{m}}\times {{a}^{m}}={{a}^{{{m}^{m}}}}={{a}^{{{m}^{2}}}}$.

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