Let x be the least number which on being divided by 8, 12, 15, 24, 25 and 40 leaves a remainder of 7 in each case. What will be the remainder when x is divided by 29?

Options

A. 18
B. 27
C. 19
D. 20

Answer

To solve the problem, we need to find the least number \( x \) that leaves a remainder of 7 when divided by the numbers 8, 12, 15, 24, 25, and 40.

### Step 1: Understand the condition
Since \( x \) leaves a remainder of 7 when divided by these numbers, we can express \( x \) in the form:
\[ x = k \cdot \text{lcm}(8, 12, 15, 24, 25, 40) + 7 \]
where \( k \) is a non-negative integer and lcm denotes the least common multiple.

### Step 2: Calculate the least common multiple (LCM)
To find the LCM of the numbers, we first find the prime factorization of each number:
- \( 8 = 2^3 \)
- \( 12 = 2^2 \cdot 3^1 \)
- \( 15 = 3^1 \cdot 5^1 \)
- \( 24 = 2^3 \cdot 3^1 \)
- \( 25 = 5^2 \)
- \( 40 = 2^3 \cdot 5^1 \)

Now, we take the highest power of each prime that appears in these factorizations:
- The highest power of \( 2 \) is \( 2^3 \) (from 8, 24, and 40).
- The highest power of \( 3 \) is \( 3^1 \) (from 12, 15, and 24).
- The highest power of \( 5 \) is \( 5^2 \) (from 25).

Thus, the LCM is:
\[ \text{lcm}(8, 12, 15, 24, 25, 40) = 2^3 \cdot 3^1 \cdot 5^2 = 8 \cdot 3 \cdot 25 = 600 \]

### Step 3: Express \( x \)
Now substituting the LCM back into the equation for \( x \):
\[ x = k \cdot 600 + 7 \]

### Step 4: Find the least \( x \)
To find the least positive \( x \), we set \( k = 1 \):
\[ x = 1 \cdot 600 + 7 = 607 \]

### Step 5: Find the remainder when \( x \) is divided by 29
Now we need to find \( 607 \mod 29 \):
\[ 607 \div 29 \approx 20.93 \quad \text{(take the integer part, which is 20)} \]
Calculating \( 20 \cdot 29 = 580 \).

Now, subtract this from 607:
\[ 607 - 580 = 27 \]

Thus, the remainder when \( x \) is divided by 29 is:
\[ \boxed{27} \]