A two digit number is four times the sum and three times the product of its digits. Find the number:

Options

A. A)12
B. B)22
C. C)14
D. D)24

Answer

To find the two-digit number that is four times the sum and three times the product of its digits, we can follow these steps:

### Step 1: Define the digits
Let the two-digit number be represented as \( N \), where:
- \( A \) is the tens digit,
- \( B \) is the units digit.

Thus, the number can be expressed as:
\[ N = 10A + B \]

### Step 2: Set up the equations
According to the problem:
1. The number \( N \) is four times the sum of its digits:
\[ N = 4(A + B) \]
2. The number \( N \) is also three times the product of its digits:
\[ N = 3(A \times B) \]

### Step 3: Substitute \( N \)
From the first equation, we can substitute \( N \):
\[ 10A + B = 4(A + B) \]
Expanding this gives:
\[ 10A + B = 4A + 4B \]

### Step 4: Rearrange the equation
Rearranging the equation to isolate terms involving \( A \) and \( B \):
\[ 10A - 4A = 4B - B \]
This simplifies to:
\[ 6A = 3B \]
Dividing both sides by 3 gives:
\[ 2A = B \]
Thus, we can express \( B \) in terms of \( A \):
\[ B = 2A \]

### Step 5: Substitute \( B \) into the second equation
Now, substitute \( B = 2A \) into the second equation:
\[ 10A + 2A = 3(A \times 2A) \]
This simplifies to:
\[ 12A = 6A^2 \]

### Step 6: Rearrange and factor
Rearranging gives:
\[ 6A^2 - 12A = 0 \]
Factoring out \( 6A \):
\[ 6A(A - 2) = 0 \]

### Step 7: Solve for \( A \)
Setting each factor to zero gives:
1. \( 6A = 0 \) → \( A = 0 \) (not valid for a two-digit number)
2. \( A - 2 = 0 \) → \( A = 2 \)

### Step 8: Find \( B \)
Using \( A = 2 \) to find \( B \):
\[ B = 2A = 2 \times 2 = 4 \]

### Step 9: Form the two-digit number
Now, substituting \( A \) and \( B \) back into the expression for \( N \):
\[ N = 10A + B = 10(2) + 4 = 20 + 4 = 24 \]

### Conclusion
The two-digit number is \( 24 \).

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