The tens digit of a three-digit number is 3. If the digits at units and hundreds places are interchanged then the number thus formed is 396 more than the previous one. Also the sum of the units digit and hundreds digit is 14. Then what is the number?

Options

A. 480
B. 539
C. 593
D. 935

Answer

To solve the problem step by step, we can break it down as follows:

### Step 1: Define the three-digit number
Let the three-digit number be represented as \( 100Y + 30 + X \), where:
- \( Y \) is the digit in the hundreds place,
- \( 3 \) is the digit in the tens place,
- \( X \) is the digit in the units place.

### Step 2: Set up the equations based on the problem statement
1. We know that the sum of the unit digit and the hundreds digit is 14:
\[ X + Y = 14 \quad \text{(Equation 1)} \]

2. When the digits at the units and hundreds places are interchanged, the new number formed is:
\[ 100X + 30 + Y \]
According to the problem, this new number is 396 more than the original number:
\[ 100X + 30 + Y = (100Y + 30 + X) + 396 \]

### Step 3: Simplify the second equation
Rearranging the equation gives:
\[ 100X + 30 + Y - 100Y - 30 - X = 396 \]
This simplifies to:
\[ 99X - 99Y = 396 \]
Dividing the entire equation by 99:
\[ X - Y = 4 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( X + Y = 14 \) (Equation 1)
2. \( X - Y = 4 \) (Equation 2)

We can add these two equations:
\[ (X + Y) + (X - Y) = 14 + 4 \]
This simplifies to:
\[ 2X = 18 \]
Thus, we find:
\[ X = 9 \]

### Step 5: Substitute back to find \( Y \)
Now substitute \( X = 9 \) back into Equation 1:
\[ 9 + Y = 14 \]
Solving for \( Y \):
\[ Y = 14 - 9 = 5 \]

### Step 6: Form the original number
Now that we have \( Y = 5 \) and \( X = 9 \), we can form the original number:
\[ 100Y + 30 + X = 100(5) + 30 + 9 = 500 + 30 + 9 = 539 \]

### Final Answer
The original number is **539**.

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