When $10^{100}$ is divided by 7, the remainder is ?
To find the remainder when \(10^{100}\) is divided by 7, we use Fermat's Little Theorem. This theorem states that for a prime number \(p\) and an integer \(a\) not divisible by \(p\):
\[ a^{p-1} \equiv 1 \pmod{p} \]In this case, \(a = 10\) and \(p = 7\).
Step 1: Apply Fermat's Little Theorem
According to the theorem, with \(a=10\) and \(p=7\), we have:
\[ 10^{7-1} \equiv 10^6 \equiv 1 \pmod{7} \]Step 2: Rewrite the exponent
We express the exponent 100 in terms of 6:
\[ 100 = 6 \times 16 + 4 \]This allows us to rewrite \(10^{100}\) as:
\[ 10^{100} = 10^{6 \times 16 + 4} = (10^6)^{16} \times 10^4 \]Step 3: Simplify using the theorem
Since \(10^6 \equiv 1 \pmod{7}\), we can substitute this into our expression:
\[ (10^6)^{16} \equiv 1^{16} \equiv 1 \pmod{7} \]Therefore, \(10^{100}\) simplifies to:
\[ 10^{100} \equiv 1 \times 10^4 \equiv 10^4 \pmod{7} \]Step 4: Calculate the remainder of \(10^4 \mod 7\)
First, let's find the remainder of \(10^2\) when divided by 7:
\[ 10^2 = 100 \]Dividing 100 by 7 gives 14 with a remainder of 2:
\[ 100 \div 7 = 14 \, \text{remainder } 2 \]So, we have:
\[ 10^2 \equiv 2 \pmod{7} \]Now we can find \(10^4 \pmod{7}\):
\[ 10^4 = (10^2)^2 \equiv 2^2 \equiv 4 \pmod{7} \]Final Answer: The remainder when \(10^{100}\) is divided by 7 is 4.