If `N=3^(14)+3^(13)-12`, then what is the largest prime factor of N ?

Options

A. 11
B. 79
C. 13
D. 73

Answer

To solve the problem, we need to find the largest prime factor of \( N = 3^{14} + 3^{13} - 12 \).

### Step-by-Step Solution:

1. **Factor out common terms**:
\[ N = 3^{14} + 3^{13} - 12 \]
We can factor out \( 3^{13} \) from the first two terms:
\[ N = 3^{13}(3 + 1) - 12 \]
This simplifies to:
\[ N = 3^{13} \cdot 4 - 12 \]

2. **Rewrite the expression**:
Now we can rewrite \( N \):
\[ N = 4 \cdot 3^{13} - 12 \]

3. **Factor out 4**:
We can factor out 4 from the expression:
\[ N = 4(3^{13} - 3) \]

4. **Simplify the expression inside the parentheses**:
We can factor \( 3 \) out of \( 3^{13} - 3 \):
\[ N = 4 \cdot 3(3^{12} - 1) \]

5. **Use the difference of squares**:
The term \( 3^{12} - 1 \) can be factored using the difference of squares:
\[ 3^{12} - 1 = (3^6 - 1)(3^6 + 1) \]

6. **Further factor \( 3^6 - 1 \)**:
We can again use the difference of squares on \( 3^6 - 1 \):
\[ 3^6 - 1 = (3^3 - 1)(3^3 + 1) \]

7. **Calculate \( 3^3 - 1 \) and \( 3^3 + 1 \)**:
\[ 3^3 - 1 = 27 - 1 = 26 \]
\[ 3^3 + 1 = 27 + 1 = 28 \]

8. **Combine all factors**:
Now we can write \( N \) as:
\[ N = 4 \cdot 3 \cdot (3^6 - 1)(3^6 + 1) = 4 \cdot 3 \cdot (26)(28)(3^6 + 1) \]

9. **Identify prime factors**:
- The prime factorization of \( 4 \) is \( 2^2 \).
- The prime factorization of \( 3 \) is \( 3 \).
- The prime factorization of \( 26 \) is \( 2 \cdot 13 \).
- The prime factorization of \( 28 \) is \( 2^2 \cdot 7 \).
- The term \( 3^6 + 1 \) needs to be evaluated separately, but it is guaranteed to be larger than the previous factors.

10. **Evaluate \( 3^6 + 1 \)**:
\[ 3^6 = 729 \implies 3^6 + 1 = 730 \]
The prime factorization of \( 730 \) is \( 2 \cdot 5 \cdot 73 \).

11. **List all prime factors**:
The prime factors of \( N \) are \( 2, 3, 5, 7, 13, 73 \).

12. **Identify the largest prime factor**:
The largest prime factor of \( N \) is:
\[ \boxed{73} \]