Find the least number which when divided by 12, 15, 18 and 27 leaves remainders 8, 11, 14 and 23 respectively.

Answer

To find the least number which, when divided by 12, 15, 18, and 27, leaves remainders of 8, 11, 14, and 23 respectively, we can follow these steps:

### Step 1: Set Up the Equations
We can express the problem in terms of congruences:
- Let the number be \( x \).
- We have the following congruences:
- \( x \equiv 8 \mod 12 \)
- \( x \equiv 11 \mod 15 \)
- \( x \equiv 14 \mod 18 \)
- \( x \equiv 23 \mod 27 \)

### Step 2: Rewrite the Congruences
To simplify the calculations, we can rewrite each congruence:
- From \( x \equiv 8 \mod 12 \), we can write \( x = 12k + 8 \) for some integer \( k \).
- From \( x \equiv 11 \mod 15 \), we can write \( x = 15m + 11 \) for some integer \( m \).
- From \( x \equiv 14 \mod 18 \), we can write \( x = 18n + 14 \) for some integer \( n \).
- From \( x \equiv 23 \mod 27 \), we can write \( x = 27p + 23 \) for some integer \( p \).

### Step 3: Find the Least Common Multiple (LCM)
To solve these congruences, we need to find the least common multiple (LCM) of the divisors:
- The prime factorization of the numbers is:
- \( 12 = 2^2 \times 3^1 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 18 = 2^1 \times 3^2 \)
- \( 27 = 3^3 \)

Now, we take the highest power of each prime:
- \( 2^2 \) from 12
- \( 3^3 \) from 27
- \( 5^1 \) from 15

Thus, the LCM is:
\[ \text{LCM} = 2^2 \times 3^3 \times 5^1 = 4 \times 27 \times 5 = 540 \]

### Step 4: Adjust for the Remainders
Since all the remainders are less than the respective divisors, we can find the least number by subtracting the common difference from the LCM:
- The common difference is 4 (since \( 12 - 8 = 4 \), \( 15 - 11 = 4 \), \( 18 - 14 = 4 \), \( 27 - 23 = 4 \)).
- Therefore, the least number is:
\[ x = 540 - 4 = 536 \]

### Final Answer
The least number which, when divided by 12, 15, 18, and 27, leaves remainders 8, 11, 14, and 23 respectively is **536**.