What is the remainder when `3^(444)+4^(333)` is divided by 5 ?

Answer

To find the remainder when \( 3^{444} + 4^{333} \) is divided by 5, we can use properties of modular arithmetic and the cyclicity of powers.

### Step 1: Find the cyclicity of \( 3^n \) modulo 5
We start by calculating the first few powers of 3 modulo 5:

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

After \( 3^4 \), the pattern repeats every 4 terms. Thus, the cyclicity of \( 3^n \) modulo 5 is 4.

### Step 2: Determine \( 444 \mod 4 \)
Next, we find \( 444 \mod 4 \) to determine which power of 3 we need:

\[ 444 \div 4 = 111 \quad \text{(exactly, with a remainder of 0)} \]

Thus, \( 444 \mod 4 = 0 \).

### Step 3: Calculate \( 3^{444} \mod 5 \)
Since \( 444 \mod 4 = 0 \), we have:

\[ 3^{444} \equiv 3^0 \equiv 1 \mod 5 \]

### Step 4: Find the cyclicity of \( 4^n \) modulo 5
Now, we calculate the first few powers of 4 modulo 5:

- \( 4^1 \mod 5 = 4 \)
- \( 4^2 \mod 5 = 16 \mod 5 = 1 \)

The cyclicity of \( 4^n \) modulo 5 is 2.

### Step 5: Determine \( 333 \mod 2 \)
Next, we find \( 333 \mod 2 \):

\[ 333 \div 2 = 166 \quad \text{(with a remainder of 1)} \]

Thus, \( 333 \mod 2 = 1 \).

### Step 6: Calculate \( 4^{333} \mod 5 \)
Since \( 333 \mod 2 = 1 \), we have:

\[ 4^{333} \equiv 4^1 \equiv 4 \mod 5 \]

### Step 7: Combine results
Now we combine the results of \( 3^{444} \) and \( 4^{333} \):

\[ 3^{444} + 4^{333} \equiv 1 + 4 \equiv 5 \mod 5 \]

### Step 8: Find the remainder
Since \( 5 \mod 5 = 0 \), the remainder when \( 3^{444} + 4^{333} \) is divided by 5 is:

\[ \text{Remainder} = 0 \]