What is the remainder when `3^(257)` is divided by 80

Answer

To find the remainder when \( 3^{257} \) is divided by 80, we can follow these steps:

### Step 1: Rewrite the exponent
We can express \( 3^{257} \) as:
\[ 3^{257} = 3 \cdot 3^{256} \]

### Step 2: Simplify \( 3^{256} \)
Next, we can express \( 256 \) in terms of powers of \( 4 \):
\[ 3^{256} = 3^{4 \cdot 64} = (3^4)^{64} \]
Calculating \( 3^4 \):
\[ 3^4 = 81 \]
Thus, we have:
\[ 3^{256} = 81^{64} \]

### Step 3: Rewrite \( 81^{64} \)
We can express \( 81 \) in a different form:
\[ 81 = 80 + 1 \]
So, we can rewrite \( 81^{64} \) as:
\[ 81^{64} = (80 + 1)^{64} \]

### Step 4: Expand using the Binomial Theorem
Using the Binomial Theorem, we can expand \( (80 + 1)^{64} \):
\[ (80 + 1)^{64} = \sum_{r=0}^{64} \binom{64}{r} 80^r \cdot 1^{64-r} \]
This expansion gives us:
\[ = \binom{64}{0} 80^0 + \binom{64}{1} 80^1 + \binom{64}{2} 80^2 + \ldots + \binom{64}{64} 80^{64} \]

### Step 5: Identify terms divisible by 80
In this expansion, all terms where \( r \geq 1 \) will be multiples of \( 80 \). The only term that is not a multiple of \( 80 \) is the first term:
\[ \binom{64}{0} 80^0 = 1 \]

### Step 6: Combine with the factor of 3
Now, substituting back, we have:
\[ 3^{257} = 3 \cdot (80 + 1)^{64} = 3 \cdot \left( \text{(multiple of 80)} + 1 \right) \]
Thus:
\[ 3^{257} = 3 \cdot \text{(multiple of 80)} + 3 \]

### Step 7: Find the remainder
When we divide \( 3^{257} \) by \( 80 \), the term \( 3 \cdot \text{(multiple of 80)} \) contributes nothing to the remainder, leaving us with:
\[ \text{Remainder} = 3 \]

### Final Answer
Therefore, the remainder when \( 3^{257} \) is divided by \( 80 \) is:
\[ \boxed{3} \]