Let x be the smallest 5-digit number such that when it is divided by 5,6,7 and 21, it leaves the same remainder 4. What is the sum of the digits of x?

Options

A. 11
B. 13
C. 10
D. 17

Answer

To solve the problem, we need to find the smallest 5-digit number \( x \) such that when \( x \) is divided by 5, 6, 7, and 21, it leaves a remainder of 4.

### Step-by-Step Solution:

1. **Understanding the Condition**:
We need \( x \) to leave a remainder of 4 when divided by 5, 6, 7, and 21. This can be expressed mathematically as:
\[ x \equiv 4 \ (\text{mod} \ 5) \]
\[ x \equiv 4 \ (\text{mod} \ 6) \]
\[ x \equiv 4 \ (\text{mod} \ 7) \]
\[ x \equiv 4 \ (\text{mod} \ 21) \]

2. **Finding the Common Modulus**:
Since all these conditions require the same remainder, we can simplify our calculations by finding the least common multiple (LCM) of the divisors (5, 6, 7, and 21).
- The prime factorization of the numbers is:
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 7 = 7^1 \)
- \( 21 = 3^1 \times 7^1 \)

The LCM is obtained by taking the highest power of each prime:
\[ \text{LCM}(5, 6, 7, 21) = 2^1 \times 3^1 \times 5^1 \times 7^1 = 210 \]

3. **Setting Up the Equation**:
Since \( x \equiv 4 \ (\text{mod} \ 210) \), we can express \( x \) as:
\[ x = 210k + 4 \]
where \( k \) is a non-negative integer.

4. **Finding the Smallest 5-Digit Number**:
We need \( x \) to be at least 10,000:
\[ 210k + 4 \geq 10000 \]
Subtracting 4 from both sides:
\[ 210k \geq 9996 \]
Dividing both sides by 210:
\[ k \geq \frac{9996}{210} \approx 47.2 \]
Since \( k \) must be an integer, we take \( k = 48 \).

5. **Calculating \( x \)**:
Now substituting \( k = 48 \) back into the equation for \( x \):
\[ x = 210 \times 48 + 4 = 10080 + 4 = 10084 \]

6. **Finding the Sum of the Digits of \( x \)**:
The digits of \( 10084 \) are 1, 0, 0, 8, and 4. Therefore, the sum of the digits is:
\[ 1 + 0 + 0 + 8 + 4 = 13 \]

### Final Answer:
The sum of the digits of \( x \) is **13**.