Let x be the least number which when divided by 8, 12, 20, 28, 35 leaves a remainder of 5 in each case. What is the sum of digits of x?

Options

A. 11
B. 14
C. 15
D. 17

Answer

To solve the problem, we need to find the least number \( x \) that leaves a remainder of 5 when divided by 8, 12, 20, 28, and 35.

### Step-by-step Solution:

1. **Understand the Remainder Condition**:
Since \( x \) leaves a remainder of 5 when divided by these numbers, we can express \( x \) as:
\[ x = k \cdot \text{LCM}(8, 12, 20, 28, 35) + 5 \]
where \( k \) is some integer.

2. **Calculate the LCM**:
We need to find the least common multiple (LCM) of the numbers 8, 12, 20, 28, and 35.
- The prime factorization of each number is:
- \( 8 = 2^3 \)
- \( 12 = 2^2 \times 3 \)
- \( 20 = 2^2 \times 5 \)
- \( 28 = 2^2 \times 7 \)
- \( 35 = 5 \times 7 \)

- To find the LCM, we take the highest power of each prime:
- For \( 2 \): \( 2^3 \) (from 8)
- For \( 3 \): \( 3^1 \) (from 12)
- For \( 5 \): \( 5^1 \) (from 20 and 35)
- For \( 7 \): \( 7^1 \) (from 28 and 35)

- Therefore, the LCM is:
\[ \text{LCM} = 2^3 \times 3^1 \times 5^1 \times 7^1 = 8 \times 3 \times 5 \times 7 \]

3. **Calculate the LCM Value**:
- First calculate \( 8 \times 3 = 24 \)
- Then \( 24 \times 5 = 120 \)
- Finally \( 120 \times 7 = 840 \)

So, \( \text{LCM}(8, 12, 20, 28, 35) = 840 \).

4. **Substitute Back to Find \( x \)**:
Now, substituting back into our equation for \( x \):
\[ x = k \cdot 840 + 5 \]
To find the least number, we take \( k = 1 \):
\[ x = 1 \cdot 840 + 5 = 845 \]

5. **Sum of the Digits of \( x \)**:
Now, we need to find the sum of the digits of \( 845 \):
- The digits are 8, 4, and 5.
- Therefore, the sum is:
\[ 8 + 4 + 5 = 17 \]

### Final Answer:
The sum of the digits of \( x \) is \( 17 \).