The digit of a 3-digit number in Base 4 get reversed when it is converted into Base 3. How many such numbers exist?

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find how many 3-digit numbers in base 4 get reversed when converted to base 3. Let's denote a 3-digit number in base 4 as \( ABC_4 \), where \( A \), \( B \), and \( C \) are the digits in base 4.

### Step 1: Understand the representation in base 4
A 3-digit number in base 4 can be expressed as:
\[ N = A \cdot 4^2 + B \cdot 4^1 + C \cdot 4^0 = 16A + 4B + C \]
where \( A, B, C \) can take values from \( 0 \) to \( 3 \) (since they are base 4 digits).

### Step 2: Understand the representation in base 3
When this number is reversed in base 3, it becomes \( CBA_3 \). The value of this number in base 3 can be expressed as:
\[ M = C \cdot 3^2 + B \cdot 3^1 + A \cdot 3^0 = 9C + 3B + A \]

### Step 3: Set up the equation
According to the problem, these two representations are equal:
\[ 16A + 4B + C = 9C + 3B + A \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ 16A + 4B + C - 9C - 3B - A = 0 \]
\[ 15A + B - 8C = 0 \]
This simplifies to:
\[ 15A + B = 8C \]

### Step 5: Solve for possible values of \( A, B, C \)
Since \( A, B, C \) are digits in base 4, we have:
- \( A \) can be \( 1, 2, 3 \) (it cannot be \( 0 \) as it's a 3-digit number).
- \( B \) and \( C \) can be \( 0, 1, 2, 3 \).

We can now test the possible values of \( A \) to find valid combinations of \( B \) and \( C \).

#### Case 1: \( A = 1 \)
\[ 15(1) + B = 8C \implies 15 + B = 8C \implies B = 8C - 15 \]
- If \( C = 2 \): \( B = 8(2) - 15 = 16 - 15 = 1 \) (valid)
- If \( C = 3 \): \( B = 8(3) - 15 = 24 - 15 = 9 \) (invalid)

So, one valid combination is \( (A, B, C) = (1, 1, 2) \).

#### Case 2: \( A = 2 \)
\[ 15(2) + B = 8C \implies 30 + B = 8C \implies B = 8C - 30 \]
- If \( C = 4 \): \( B = 8(4) - 30 = 32 - 30 = 2 \) (invalid since \( C \) can only be \( 0, 1, 2, 3 \))

No valid combinations.

#### Case 3: \( A = 3 \)
\[ 15(3) + B = 8C \implies 45 + B = 8C \implies B = 8C - 45 \]
- If \( C = 6 \): \( B = 8(6) - 45 = 48 - 45 = 3 \) (invalid since \( C \) can only be \( 0, 1, 2, 3 \))

No valid combinations.

### Conclusion
The only valid combination we found is \( (A, B, C) = (1, 1, 2) \), which corresponds to the number \( 112_4 \).

Thus, there is only **one** such number that exists.

**Final Answer**: 1