What least number must be subtracted from each of the numbers `7`, `17` and `47` so that the remainders are in continued proportion ?

Answer

To solve the problem of finding the least number that must be subtracted from the numbers 7, 17, and 47 so that the remainders are in continued proportion, we can follow these steps:

### Step-by-Step Solution:

1. **Define the variable**: Let the least number to be subtracted be \( x \).

2. **Write the remainders**: The remainders after subtracting \( x \) from each number will be:
- From 7: \( 7 - x \)
- From 17: \( 17 - x \)
- From 47: \( 47 - x \)

3. **Set up the proportion**: According to the property of continued proportion, we have:
\[ \frac{7 - x}{17 - x} = \frac{17 - x}{47 - x} \]

4. **Cross-multiply**: This gives us the equation:
\[ (7 - x)(47 - x) = (17 - x)(17 - x) \]

5. **Expand both sides**:
- Left side:
\[ 7 \cdot 47 - 7x - 47x + x^2 = 329 - 54x + x^2 \]
- Right side:
\[ (17 - x)^2 = 17^2 - 2 \cdot 17 \cdot x + x^2 = 289 - 34x + x^2 \]

6. **Set the equation**: Now we have:
\[ 329 - 54x + x^2 = 289 - 34x + x^2 \]

7. **Cancel \( x^2 \) from both sides**:
\[ 329 - 54x = 289 - 34x \]

8. **Rearrange the equation**: Move all terms involving \( x \) to one side and constant terms to the other:
\[ 329 - 289 = 54x - 34x \]
\[ 40 = 20x \]

9. **Solve for \( x \)**:
\[ x = \frac{40}{20} = 2 \]

### Final Answer:
The least number that must be subtracted from each of the numbers 7, 17, and 47 is **2**.