To solve the problem, we need to find the least number \( x \) that meets the following criteria:
1. When divided by 8, 9, 12, 14, and 36, it leaves a remainder of 4.
2. It is divisible by 11.
### Step 1: Find the LCM of the divisors
We first need to find the least common multiple (LCM) of the numbers 8, 9, 12, 14, and 36.
- **Prime factorization**:
- \( 8 = 2^3 \)
- \( 9 = 3^2 \)
- \( 12 = 2^2 \times 3^1 \)
- \( 14 = 2^1 \times 7^1 \)
- \( 36 = 2^2 \times 3^2 \)
- **LCM calculation**:
- Take the highest power of each prime:
- For 2: \( 2^3 \)
- For 3: \( 3^2 \)
- For 7: \( 7^1 \)
Thus, the LCM is:
\[
\text{LCM} = 2^3 \times 3^2 \times 7^1 = 8 \times 9 \times 7 = 504
\]
### Step 2: Formulate the equation for \( x \)
Since \( x \) leaves a remainder of 4 when divided by 8, 9, 12, 14, and 36, we can express \( x \) as:
\[
x = 504k + 4
\]
for some integer \( k \).
### Step 3: Ensure \( x \) is divisible by 11
We need \( x \) to be divisible by 11:
\[
504k + 4 \equiv 0 \mod{11}
\]
First, calculate \( 504 \mod 11 \):
\[
504 \div 11 = 45 \quad \text{(integer part)}
\]
\[
504 - (45 \times 11) = 504 - 495 = 9
\]
Thus, \( 504 \equiv 9 \mod{11} \).
Now, substitute this back into our equation:
\[
9k + 4 \equiv 0 \mod{11}
\]
This simplifies to:
\[
9k \equiv -4 \mod{11}
\]
Since \(-4 \equiv 7 \mod{11}\), we have:
\[
9k \equiv 7 \mod{11}
\]
### Step 4: Solve for \( k \)
To solve for \( k \), we can test integer values:
- For \( k = 1 \):
\[
9 \times 1 \equiv 9 \mod{11} \quad \text{(not a solution)}
\]
- For \( k = 2 \):
\[
9 \times 2 \equiv 18 \equiv 7 \mod{11} \quad \text{(solution found)}
\]
### Step 5: Calculate \( x \)
Now substitute \( k = 2 \) back into the equation for \( x \):
\[
x = 504 \times 2 + 4 = 1008 + 4 = 1012
\]
### Step 6: Find the sum of the digits of \( x \)
Now, we need to find the sum of the digits of \( 1012 \):
\[
1 + 0 + 1 + 2 = 4
\]
### Final Answer
The sum of the digits of \( x \) is **4**.
