When the number `3^(98)` is divided by 5, then remainder is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the remainder when \( 3^{98} \) is divided by 5, we can use modular arithmetic and properties of exponents. Here’s a step-by-step solution:

### Step 1: Identify the pattern in powers of 3 modulo 5
First, we will calculate the first few powers of 3 modulo 5 to identify any patterns.

- \( 3^1 \mod 5 = 3 \)
- \( 3^2 \mod 5 = 9 \mod 5 = 4 \)
- \( 3^3 \mod 5 = 12 \mod 5 = 2 \)
- \( 3^4 \mod 5 = 6 \mod 5 = 1 \)

### Step 2: Observe the cycle
From the calculations, we see that the powers of 3 modulo 5 repeat every 4 terms:
- \( 3^1 \equiv 3 \)
- \( 3^2 \equiv 4 \)
- \( 3^3 \equiv 2 \)
- \( 3^4 \equiv 1 \)

### Step 3: Determine the exponent modulo 4
Since the pattern repeats every 4 terms, we can find \( 3^{98} \mod 5 \) by first reducing the exponent modulo 4:
\[ 98 \mod 4 = 2 \]
This means \( 3^{98} \equiv 3^2 \mod 5 \).

### Step 4: Calculate \( 3^2 \mod 5 \)
Now, we can use our earlier calculation:
\[ 3^2 \mod 5 = 4 \]

### Conclusion
Thus, the remainder when \( 3^{98} \) is divided by 5 is \( \boxed{4} \).