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What is the remainder when,
${{10}^{10}}\cdot \left( {{10}^{10}}+1 \right)\left( {{10}^{10}}+2 \right)$ is divided by $6$ _____.
(a) 2
(b) 4
(c) 0
(d) 6

Last updated date: 24th Jul 2024
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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:
 & {{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}} \\

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,
 & {{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 \\

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
 & {{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 \\

At last, the remainder for last term would be
 & {{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 \\

Therefore, the sum total of all the reminders would give us the desired remainder.
 & S={{R}_{1}}+{{R}_{2}}+{{R}_{3}} \\
 & S=4+0+2 \\
 & S=6 \\

So, the final remainder is obtained by dividing the sum of remainder S with 6.
 & R=\dfrac{S}{6} \\
 & R=\dfrac{6}{6} \\
 & \therefore R=0 \\
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 ${{a}^{m}}\times {{a}^{m}}={{a}^{{{m}^{m}}}}={{a}^{{{m}^{2}}}}$.