Find the least number which when divided by 24,32 and 36 leaves the remainders 19,27 and 31, respectively.

Answer

To find the least number which, when divided by 24, 32, and 36, leaves the remainders 19, 27, and 31 respectively, we can follow these steps:

### Step 1: Set Up the Equations
We can express the conditions given in the problem as equations:
- Let the least number be \( x \).
- According to the problem:
- \( x \equiv 19 \mod 24 \)
- \( x \equiv 27 \mod 32 \)
- \( x \equiv 31 \mod 36 \)

### Step 2: Rewrite the Equations
We can rewrite these congruences in terms of \( x \):
- From \( x \equiv 19 \mod 24 \), we can express \( x \) as:
\[ x = 24k + 19 \quad \text{(for some integer } k\text{)} \]
- From \( x \equiv 27 \mod 32 \):
\[ x = 32m + 27 \quad \text{(for some integer } m\text{)} \]
- From \( x \equiv 31 \mod 36 \):
\[ x = 36n + 31 \quad \text{(for some integer } n\text{)} \]

### Step 3: Solve the First Two Equations
We can set the first two expressions for \( x \) equal to each other:
\[ 24k + 19 = 32m + 27 \]
Rearranging gives:
\[ 24k - 32m = 8 \]
Dividing through by 8:
\[ 3k - 4m = 1 \]

### Step 4: Find Integer Solutions
We can find integer solutions for \( k \) and \( m \). Rearranging gives:
\[ 3k = 4m + 1 \implies k = \frac{4m + 1}{3} \]
To ensure \( k \) is an integer, \( 4m + 1 \) must be divisible by 3. Testing values of \( m \):
- If \( m = 1 \): \( k = \frac{4(1) + 1}{3} = \frac{5}{3} \) (not an integer)
- If \( m = 2 \): \( k = \frac{4(2) + 1}{3} = \frac{9}{3} = 3 \) (integer)

So, \( k = 3 \) and \( m = 2 \) is a solution.

### Step 5: General Solution for \( k \) and \( m \)
The general solution can be expressed as:
\[ k = 3 + 4t \quad \text{and} \quad m = 2 + 3t \quad \text{(for some integer } t\text{)} \]

### Step 6: Substitute Back to Find \( x \)
Substituting \( k \) back into the equation for \( x \):
\[ x = 24(3 + 4t) + 19 = 72 + 96t + 19 = 91 + 96t \]

### Step 7: Solve the Third Equation
Now we substitute \( x = 91 + 96t \) into the third equation \( x \equiv 31 \mod 36 \):
\[ 91 + 96t \equiv 31 \mod 36 \]
Calculating \( 91 \mod 36 \):
\[ 91 \div 36 = 2 \quad \text{(remainder } 19\text{)} \]
Thus, we have:
\[ 19 + 24t \equiv 31 \mod 36 \]
This simplifies to:
\[ 24t \equiv 12 \mod 36 \]
Dividing through by 12:
\[ 2t \equiv 1 \mod 3 \]
The solution is \( t \equiv 2 \mod 3 \). Thus, \( t = 2 + 3s \) for integer \( s \).

### Step 8: Substitute Back to Find \( x \)
Substituting \( t \) back into \( x \):
\[ x = 91 + 96(2 + 3s) = 91 + 192 + 288s = 283 + 288s \]
The least value occurs when \( s = 0 \):
\[ x = 283 \]

### Final Answer
The least number which, when divided by 24, 32, and 36, leaves the remainders 19, 27, and 31 respectively is **283**.