What is the smallest number whch when divided by 6, 18, 24 leaves a remainder of 2, 14 and 20 respectively ?

Answer

To solve the problem of finding the smallest number that leaves specific remainders when divided by given divisors, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( x \) such that:
- \( x \mod 6 = 2 \)
- \( x \mod 18 = 14 \)
- \( x \mod 24 = 20 \)

### Step 2: Adjust the equations
We can rewrite the conditions based on the remainders:
- From \( x \mod 6 = 2 \), we can express this as \( x = 6k + 2 \) for some integer \( k \).
- From \( x \mod 18 = 14 \), we can express this as \( x = 18m + 14 \) for some integer \( m \).
- From \( x \mod 24 = 20 \), we can express this as \( x = 24n + 20 \) for some integer \( n \).

### Step 3: Find the common difference
Next, we observe the differences between the divisors and their respective remainders:
- For \( 6 \) and \( 2 \): \( 6 - 2 = 4 \)
- For \( 18 \) and \( 14 \): \( 18 - 14 = 4 \)
- For \( 24 \) and \( 20 \): \( 24 - 20 = 4 \)

This shows that the common difference is \( 4 \).

### Step 4: Calculate the LCM of the divisors
Now we find the least common multiple (LCM) of the divisors \( 6, 18, \) and \( 24 \):
- The prime factorization gives us:
- \( 6 = 2 \times 3 \)
- \( 18 = 2 \times 3^2 \)
- \( 24 = 2^3 \times 3 \)
- The LCM is taken by taking the highest power of each prime:
- LCM = \( 2^3 \times 3^2 = 72 \)

### Step 5: Adjust the LCM with the common difference
Now, we need to adjust the LCM by subtracting the common difference \( 4 \):
- Calculate \( 72 - 4 = 68 \).

### Step 6: Verify the solution
We need to check if \( 68 \) satisfies all the original conditions:
- \( 68 \mod 6 = 2 \) (Correct)
- \( 68 \mod 18 = 14 \) (Correct)
- \( 68 \mod 24 = 20 \) (Correct)

### Conclusion
Thus, the smallest number that meets all the conditions is \( \boxed{68} \).

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