Question

What will be the remainder if \[{2^{100}}\] is divided by 17?

Answer 1

Hint: We are given a number with base and power. Now we need not to find the power of 2 upto 100. We just need to find power of 2 such that it is near to any multiple of 17. Such that the remainder is possibly 1 so that any power of 1 is always 1. That will help in solving the problem easily.

Complete step by step solution:
Given that \[{2^{100}}\] is divided by 17.
We know that \[{2^{100}}\] can be written as \[{2^{88}} \times {2^{12}}\]
We know that \[{2^8} = 256\]
And at the same time \[17 \times 15 = 255\]
So on dividing we get 1 as remainder so we get,
\[ =\dfrac{{{2^{88}} \times {2^{12}}}}{{17}}\]
We know that \[{a^{mn}} = {\left( {{a^m}} \right)^n}\]
So we can rewrite above expression as
\[ =\dfrac{{{{\left( {{2^8}} \right)}^{11}} \times {2^{12}}}}{{17}}\]
Now on dividing 256 by 17 we get 1 as remainder. So
\[ = {1^{11}} \times {2^{12}}\]
We know that 1 if raised to any power is always 1.
\[ = 1 \times {2^{12}}\]
\[ = {2^{12}}\]
This is the correct answer.
So, the correct answer is “ \[{2^{12}}\] ”.

Note: Here note that if we have an idea for any higher multiple of 17 that dividing higher power of 2 gives 1 as remainder can also be taken. That will make the calculations easier. Whereas the method mentioned above is also correct. Note that when we face problems like this such that divided by 18 or divided by 7 we need to find or convert the power such that the divisor divides the number such that the remainder is 1 only.