To solve the problem step by step, we need to find the least number \( x \) that meets the given conditions.
### Step 1: Find the LCM of the numbers
We need to find the least common multiple (LCM) of the numbers 12, 15, 18, 20, and 27.
- **Prime factorization**:
- \( 12 = 2^2 \times 3^1 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 18 = 2^1 \times 3^2 \)
- \( 20 = 2^2 \times 5^1 \)
- \( 27 = 3^3 \)
- **LCM Calculation**:
- Take the highest power of each prime:
- For \( 2 \): \( 2^2 \)
- For \( 3 \): \( 3^3 \)
- For \( 5 \): \( 5^1 \)
Thus, the LCM is:
\[
LCM = 2^2 \times 3^3 \times 5^1 = 4 \times 27 \times 5 = 540
\]
### Step 2: Adjust for the remainder
Since \( x \) leaves a remainder of 2 when divided by 12, 15, 18, 20, and 27, we can express \( x \) as:
\[
x = 540m + 2
\]
where \( m \) is a positive integer.
### Step 3: Ensure divisibility by 23
Next, we need \( x \) to be divisible by 23:
\[
540m + 2 \equiv 0 \mod 23
\]
Calculating \( 540 \mod 23 \):
\[
540 \div 23 \approx 23.478 \quad \Rightarrow \quad 23 \times 23 = 529 \quad \Rightarrow \quad 540 - 529 = 11
\]
So, \( 540 \equiv 11 \mod 23 \).
Now we have:
\[
11m + 2 \equiv 0 \mod 23 \quad \Rightarrow \quad 11m \equiv -2 \mod 23 \quad \Rightarrow \quad 11m \equiv 21 \mod 23
\]
### Step 4: Solve for \( m \)
To solve \( 11m \equiv 21 \mod 23 \), we need the multiplicative inverse of 11 modulo 23. Testing values, we find:
\[
11 \times 21 \equiv 1 \mod 23
\]
Thus, the inverse is \( 21 \). Multiply both sides by 21:
\[
m \equiv 21 \times 21 \mod 23 \quad \Rightarrow \quad m \equiv 441 \mod 23
\]
Calculating \( 441 \mod 23 \):
\[
441 \div 23 \approx 19.173 \quad \Rightarrow \quad 23 \times 19 = 437 \quad \Rightarrow \quad 441 - 437 = 4
\]
So, \( m \equiv 4 \mod 23 \).
### Step 5: Calculate \( x \)
Substituting \( m = 4 \) into the equation for \( x \):
\[
x = 540 \times 4 + 2 = 2160 + 2 = 2162
\]
### Step 6: Find the sum of the digits of \( x \)
Now, we calculate the sum of the digits of \( 2162 \):
\[
2 + 1 + 6 + 2 = 11
\]
### Step 7: Calculate the quotient
Finally, we divide \( x \) by the sum of its digits:
\[
\frac{2162}{11} = 196
\]
Thus, the quotient is:
\[
\boxed{196}
\]
