Question

The remainder when \[{{27}^{10}}+{{7}^{51}}\] is divided by \[10\], is equal to?

Answer 1

Hint: In order to find the remainder, we will be converting \[{{27}^{10}}\] into the powers of \[7\]then to the powers of \[81\]by applying the power rule i.e. \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\] and we will be checking the remainder in each of the case. Then we will be applying the same formula to the \[{{7}^{51}}\] and write it in the powers of \[343\] and check out the remainder when divided by \[10\]. At last, we will be summing the remainders from both the cases and check out the remainder when divided by \[10\] and this will be our required answer.

Complete step by step answer:
Now let us learn about exponents and powers. The general form of an exponent is \[{{a}^{n}}\] where \[a\] is called as the base and \[n\] is the exponent. The exponents follow the multiplication law, the division law and the negative power law. The value of any base to the power zero is always one.
Now let us find the remainder when \[{{27}^{10}}+{{7}^{51}}\] is divided by \[10\].
Consider \[{{27}^{10}}\],
\[27\] when divided by \[10\], we get the remainder as \[7\].
So now \[{{27}^{10}}\] can be expressed as \[{{7}^{10}}\]
\[{{7}^{10}}\] can also be written as \[{{\left( {{7}^{2}} \right)}^{5}}={{49}^{5}}\] using the formula \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\].
Now when we divide \[49\] by \[10\], we get the remainder as \[9\].
Now we can write \[{{49}^{5}}\] as \[{{9}^{5}}\] .
Now we can express this as \[{{\left( {{9}^{2}} \right)}^{2}}\times 9={{81}^{2}}\times 9\].
Now when we divide \[81\] by \[10\], we obtain \[1\] as the remainder.
\[\Rightarrow 1\times 9=9\]
From the above considerations, we can conclude that \[{{27}^{10}}\] when divided by \[10\], we get the remainder as \[9\].
Now let us consider \[{{7}^{51}}\]
It can be written as \[{{7}^{51}}={{\left( {{7}^{3}} \right)}^{17}}\]
And \[{{\left( {{7}^{3}} \right)}^{17}}={{343}^{17}}\]
Now when \[343\] is divided by \[10\], we obtain \[3\] as the remainder.
So \[{{343}^{17}}={{3}^{17}}\]
Now we can simplify it further as \[{{\left( {{3}^{4}} \right)}^{4}}\times 3\]
\[\Rightarrow {{81}^{4}}\times 3\]
\[\Rightarrow 1\times 3=3\]
We can conclude that \[{{7}^{51}}\] when divided by \[10\] leaves \[3\] as the remainder.
Now let us add the remainders from both the cases. We get,
\[\Rightarrow 9+3=12\]
Now when \[12\] is divided by \[10\], we get the remainder as \[2\].
\[\therefore \] The remainder when \[{{27}^{10}}+{{7}^{51}}\] is divided by \[10\] is \[2\].

Note: We can use exponents and powers in our everyday life while expressing the distance between the sun and the earth. We can use exponents whenever the value to be expressed is large as exponents make it easier.