The product of two numbers is 7168 and their H.C.F is 16. Find the sum of all possible numbers-

Options

A. 640
B. 860
C. 460
D. Data inadequate

Answer

To solve the problem step by step, we will follow the given information and use the relationship between the product of two numbers, their H.C.F., and their L.C.M.

### Step 1: Understand the relationship
We know that for any two numbers \( x \) and \( y \):
\[ x \times y = \text{H.C.F.}(x, y) \times \text{L.C.M.}(x, y) \]

### Step 2: Define the numbers
Let the two numbers be \( x = 16a \) and \( y = 16b \), where \( a \) and \( b \) are co-prime (since their H.C.F. is 1).

### Step 3: Set up the equation
Given that the product \( x \times y = 7168 \), we can substitute:
\[ (16a) \times (16b) = 7168 \]
This simplifies to:
\[ 256ab = 7168 \]

### Step 4: Solve for \( ab \)
Now, we divide both sides by 256:
\[ ab = \frac{7168}{256} = 28 \]

### Step 5: Find pairs of \( (a, b) \)
Next, we need to find pairs of integers \( (a, b) \) such that \( ab = 28 \). The pairs of factors of 28 are:
- \( (1, 28) \)
- \( (2, 14) \)
- \( (4, 7) \)

### Step 6: Calculate the corresponding numbers
Now we will calculate the corresponding numbers \( x \) and \( y \) for each pair:
1. For \( (1, 28) \):
- \( x = 16 \times 1 = 16 \)
- \( y = 16 \times 28 = 448 \)
- Sum = \( 16 + 448 = 464 \)

2. For \( (2, 14) \):
- \( x = 16 \times 2 = 32 \)
- \( y = 16 \times 14 = 224 \)
- Sum = \( 32 + 224 = 256 \)

3. For \( (4, 7) \):
- \( x = 16 \times 4 = 64 \)
- \( y = 16 \times 7 = 112 \)
- Sum = \( 64 + 112 = 176 \)

### Step 7: Calculate the total sum of all possible numbers
Now we add all the possible sums:
\[ 464 + 256 + 176 = 896 \]

### Final Answer
The sum of all possible numbers is \( 896 \).

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