The least number that must be subtracted from 1294 so that the remainder when divided by 9, 11 and 13 will leave in each case the same remainder 6, is :

Options

A. 2
B. 3
C. 1
D. 4

Answer

To solve the problem, we need to find the least number that must be subtracted from 1294 so that when the result is divided by 9, 11, and 13, it leaves a remainder of 6 in each case.

### Step-by-step Solution:

1. **Understand the Problem**: We need to find a number \( x \) such that when we subtract \( x \) from 1294, the result leaves a remainder of 6 when divided by 9, 11, and 13.

2. **Set Up the Equation**: If we denote the number we subtract as \( x \), then we want:
\[ (1294 - x) \mod 9 = 6 \]
\[ (1294 - x) \mod 11 = 6 \]
\[ (1294 - x) \mod 13 = 6 \]

3. **Find the LCM of the Divisors**: To find a common condition for all three moduli (9, 11, and 13), we calculate the least common multiple (LCM) of these numbers.
- The LCM of 9, 11, and 13 is:
\[ \text{LCM}(9, 11, 13) = 9 \times 11 \times 13 = 1287 \]

4. **Adjust for the Remainder**: Since we want the result to leave a remainder of 6, we can express this as:
\[ 1294 - x = 1287k + 6 \]
for some integer \( k \).

5. **Rearranging the Equation**: Rearranging gives us:
\[ x = 1294 - (1287k + 6) = 1288 - 1287k \]

6. **Finding the Least \( x \)**: To find the least value of \( x \), we can set \( k = 1 \):
\[ x = 1288 - 1287 \times 1 = 1 \]

7. **Conclusion**: The least number that must be subtracted from 1294 is \( \boxed{1} \).