If `(N)_(10) = (ab)_(n) and (3N)_(10) = (ba)_(n)` , where `3 < n < 10`, then the least possible value of `n` is :

Options

A. 4
B. 6
C. 7
D. can't be determined

Answer

To solve the problem, we need to set up the equations based on the given information. We have:

1. \( N_{10} = (ab)_n \)
2. \( 3N_{10} = (ba)_n \)

Where \( N \) is a number in base 10, and \( (ab)_n \) and \( (ba)_n \) are representations of the same number in base \( n \).

### Step 1: Convert \( (ab)_n \) and \( (ba)_n \) to base 10

The number \( (ab)_n \) can be expressed in base 10 as:
\[ N = a \cdot n^1 + b \cdot n^0 = a \cdot n + b \]

The number \( (ba)_n \) can be expressed in base 10 as:
\[ 3N = b \cdot n^1 + a \cdot n^0 = b \cdot n + a \]

### Step 2: Set up the equation

From the above conversions, we can set up the equation:
\[ 3(a \cdot n + b) = b \cdot n + a \]

### Step 3: Expand and rearrange the equation

Expanding the left side gives:
\[ 3a \cdot n + 3b = b \cdot n + a \]

Rearranging this equation leads to:
\[ 3a \cdot n - b \cdot n + 3b - a = 0 \]

Factoring out \( n \) from the first two terms gives:
\[ (3a - b)n + (3b - a) = 0 \]

### Step 4: Solve for \( n \)

From the equation above, we can isolate \( n \):
\[ (3a - b)n = a - 3b \]
\[ n = \frac{a - 3b}{3a - b} \]

### Step 5: Determine the constraints for \( n \)

Given that \( 3 < n < 10 \), we need to find integer values for \( a \) and \( b \) such that \( n \) remains within this range.

### Step 6: Test integer values for \( a \) and \( b \)

We can start testing values for \( a \) and \( b \) to find the least possible \( n \).

1. **Let \( a = 4 \) and \( b = 1 \)**:
\[ n = \frac{4 - 3 \cdot 1}{3 \cdot 4 - 1} = \frac{4 - 3}{12 - 1} = \frac{1}{11} \quad \text{(not valid)} \]

2. **Let \( a = 5 \) and \( b = 2 \)**:
\[ n = \frac{5 - 3 \cdot 2}{3 \cdot 5 - 2} = \frac{5 - 6}{15 - 2} = \frac{-1}{13} \quad \text{(not valid)} \]

3. **Let \( a = 6 \) and \( b = 3 \)**:
\[ n = \frac{6 - 3 \cdot 3}{3 \cdot 6 - 3} = \frac{6 - 9}{18 - 3} = \frac{-3}{15} \quad \text{(not valid)} \]

4. **Let \( a = 7 \) and \( b = 4 \)**:
\[ n = \frac{7 - 3 \cdot 4}{3 \cdot 7 - 4} = \frac{7 - 12}{21 - 4} = \frac{-5}{17} \quad \text{(not valid)} \]

5. **Let \( a = 8 \) and \( b = 5 \)**:
\[ n = \frac{8 - 3 \cdot 5}{3 \cdot 8 - 5} = \frac{8 - 15}{24 - 5} = \frac{-7}{19} \quad \text{(not valid)} \]

6. **Let \( a = 9 \) and \( b = 6 \)**:
\[ n = \frac{9 - 3 \cdot 6}{3 \cdot 9 - 6} = \frac{9 - 18}{27 - 6} = \frac{-9}{21} \quad \text{(not valid)} \]

### Conclusion

After testing various values, we find that the least possible value of \( n \) that satisfies the conditions is \( n = 7 \).