Question

How do you find the remainder of ${{7}^{2002}}$ divided by $101$?

Answer 1

Hint: To find the remainder of${{7}^{2002}}$when divided by$101$, we will use Fermat's little theorem. Fermat's little theorem states that if p is a prime number, then for any integer a, the number ${{a}^{p}}-a$ is an integer multiple of p. It is also expressed as:
$\Rightarrow {{a}^{p-1}}\equiv 1\left( \bmod p \right)$
Here the (MOD) means modulo.The modulo (or mod) is the remainder after dividing one number another. Fermat's little theorem is a fundamental theorem in elementary number theory, which helps to compute powers of integers modulo prime numbers.

Complete step by step answer:
Now we have been given that ${{7}^{2002}}$, if we compare this from Fermat's little theorem then, there is$7$ and p is $101$.
Now we will find out the remainder of the given question step by step by using the Fermat's theorem.
Now we can write the given ${{7}^{2002}}$ as by Fermat's theorem,
$\Rightarrow {{7}^{101-1}}\equiv 1\left( \bmod 101 \right)$
Now subtracting one from $101$ , then we get
$\Rightarrow {{7}^{100}}\equiv 1\left( \bmod 101 \right)$
Now, multiply the power of above equation by $20$ on both sides of the equation, then we get
$\Rightarrow {{\left( {{7}^{100}} \right)}^{20}}\equiv {{1}^{20}}\left( \bmod 101 \right)$
Now we know if we put any power on one it will be one, it means ${{1}^{a}}=1$ for a is greater than equal to one. Now we know that${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ , so the ${{\left( {{7}^{100}} \right)}^{20}}={{7}^{100\times 20}}={{7}^{2000}}$
Now putting this we get,
$\Rightarrow {{7}^{2000}}\equiv 1\left( \bmod 101 \right)$
Now we will multiply the both side of the equation with${{7}^{2}}$, then we get
$\Rightarrow {{7}^{2000}}\centerdot {{7}^{2}}\equiv 1\cdot {{7}^{2}}\left( \bmod 101 \right)$
Since we know very well that ${{7}^{2}}$ is equal to $49$. And we also know that ${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$ , now we will use this formula to solve
$\Rightarrow {{7}^{2000}}\cdot {{7}^{2}}={{7}^{2000+2}}={{7}^{2002}}$ .
Now putting these values, we get
$\Rightarrow {{7}^{2002}}\equiv 49\left( \bmod 101 \right)$
Now by using Fermat's little theorem we get the remainder of the given question which is $49$.

Hence the remainder of ${{7}^{2002}}$ when divided by $101$ is $49$.

Note: whenever we solve these types of questions, we need to know that when we divide any term or any polynomial then the obtained remainder is always positive. We are done that the remainder is$49$. Indeed, if you denote the remainder by r then$0\le r\le 100$and$r\equiv 49\left( \bmod 101 \right)$. This means that $101|r-49$ and since$-49\le r-49<52$ , then you get $r-49=0$