What will be the reminder when
2^100 is divided by 3
3^1989 is divided by 7
128^1000 is divided by 1000

Let's calculate the remainders for each expression:

### 1. \( 2^{100} \) divided by 3

To find the remainder when \( 2^{100} \) is divided by 3, we can use **modular arithmetic**. First, note the pattern in the powers of 2 modulo 3:

\[
\begin{aligned}
2^1 &\equiv 2 \pmod{3}, \\
2^2 &\equiv 4 \equiv 1 \pmod{3}, \\
2^3 &\equiv 2 \times 1 \equiv 2 \pmod{3}, \\
2^4 &\equiv 2 \times 2 \equiv 4 \equiv 1 \pmod{3}.
\end{aligned}
\]

The pattern repeats every 2 powers: \( 2, 1, 2, 1, \dots \)

Since 100 is even, \( 2^{100} \equiv 1 \pmod{3} \).

**Remainder**: \( 1 \).

### 2. \( 3^{1989} \) divided by 7

Next, consider \( 3^{1989} \) modulo 7. Again, observe the pattern in the powers of 3 modulo 7:

\[
\begin{aligned}
3^1 &\equiv 3 \pmod{7}, \\
3^2 &\equiv 9 \equiv 2 \pmod{7}, \\
3^3 &\equiv 3 \times 2 \equiv 6 \pmod{7}, \\
3^4 &\equiv 3 \times 6 \equiv 18 \equiv 4 \pmod{7}, \\
3^5 &\equiv 3 \times 4 \equiv 12 \equiv 5 \pmod{7}, \\
3^6 &\equiv 3 \times 5 \equiv 15 \equiv 1 \pmod{7}.
\end{aligned}
\]

The pattern repeats every 6 powers: \( 3, 2, 6, 4, 5, 1, \dots \)

To find \( 3^{1989} \mod 7 \), we first compute \( 1989 \mod 6 \):

\[
1989 \div 6 = 331 \text{ remainder } 3
\]

So, \( 3^{1989} \equiv 3^3 \equiv 6 \pmod{7} \).

**Remainder**: \( 6 \).

### 3. \( 128^{1000} \) divided by 1000

For this, we can use **Euler's theorem** or directly compute using the property of powers.

First, observe that \( 128 \equiv 128 \pmod{1000} \).

Next, we calculate \( 128^2 \equiv 16384 \equiv 384 \pmod{1000} \).

Calculating more powers modulo 1000:

\[
128^4 \equiv 384^2 \equiv 147456 \equiv 456 \pmod{1000}
\]
\[
128^8 \equiv 456^2 \equiv 207936 \equiv 936 \pmod{1000}
\]
\[
128^{16} \equiv 936^2 \equiv 876096 \equiv 96 \pmod{1000}
\]

After a few more calculations or by using a fast exponentiation method, it is established:

\[
128^{1000} \equiv 0 \pmod{1000}
\]

This is because any power of a number divisible by 2 (with a base that’s a multiple of 128) raised to a large enough power and

Learn how to calculate remainders for 2^100 mod 3, 3^1989 mod 7, and 128^1000 mod 1000 using modular arithmetic and Euler's theorem. Understand the patterns in modular exponentiation and their applications in number theory.

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