Find the remainder when $100^{100}$ is divided by $18$.

Answer 1

To find the remainder when \( 100^{100} \) is divided by 18, we will start by reducing the base modulo 18.

First, calculate \( 100 \mod 18 \):
\[
100 \div 18 = 5 \quad \text{with a remainder of } 10
\]
Thus,
\[
100 \equiv 10 \pmod{18}
\]

Next, we need to calculate \( 100^{100} \mod 18 \), which is equivalent to finding \( 10^{100} \mod 18 \).

We observe a pattern in the powers of 10 modulo 18. Start by calculating the lower powers of 10 modulo 18:
\[
10^1 \equiv 10 \pmod{18}
\]
\[
10^2 = 100 \equiv 10 \pmod{18}
\]

Both \( 10^1 \) and \( 10^2 \) when taken modulo 18 give a remainder of 10. We see that:
\[
10^k \mod 18 = 10 \quad \text{for any positive integer } k
\]

To confirm and finalize, use induction. Assume \( 10^n \equiv 10 \pmod{18} \). Then:
\[
10^{n+1} = 10^n \cdot 10 \equiv 10 \cdot 10 \equiv 100 \equiv 10 \pmod{18}
\]

Thus, the pattern holds true for \( n+1 \) whenever it holds for \( n \), starting from the base cases.

Therefore,
\[
10^{100} \equiv 10 \pmod{18}
\]

Consequently,
\[
100^{100} \equiv 10 \pmod{18}
\]

Thus, the remainder when \( 100^{100} \) is divided by 18 is \(\boxed{10}\).