To find the least number which leaves a remainder of 6 when divided by 8, 12, or 15, we can follow these steps:
### Step 1: Define the problem
Let the number we are looking for be \( x \). According to the problem, when \( x \) is divided by 8, 12, or 15, it leaves a remainder of 6. This can be expressed mathematically as:
- \( x \equiv 6 \mod 8 \)
- \( x \equiv 6 \mod 12 \)
- \( x \equiv 6 \mod 15 \)
### Step 2: Rewrite the equations
From the above congruences, we can express \( x \) in terms of the divisors:
- \( x = 8k + 6 \) for some integer \( k \)
- \( x = 12m + 6 \) for some integer \( m \)
- \( x = 15n + 6 \) for some integer \( n \)
### Step 3: Simplify the equations
We can subtract 6 from each equation to focus on the multiples of the divisors:
- \( x - 6 = 8k \)
- \( x - 6 = 12m \)
- \( x - 6 = 15n \)
This implies that \( x - 6 \) is a common multiple of 8, 12, and 15.
### Step 4: Find the Least Common Multiple (LCM)
To find the least number that is divisible by 8, 12, and 15, we need to calculate the LCM of these numbers.
**Prime factorization:**
- \( 8 = 2^3 \)
- \( 12 = 2^2 \times 3^1 \)
- \( 15 = 3^1 \times 5^1 \)
**Determine the highest powers of each prime factor:**
- For \( 2 \): highest power is \( 2^3 \) (from 8)
- For \( 3 \): highest power is \( 3^1 \) (from both 12 and 15)
- For \( 5 \): highest power is \( 5^1 \) (from 15)
**Calculate the LCM:**
\[
\text{LCM} = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5
\]
Calculating this step-by-step:
- \( 8 \times 3 = 24 \)
- \( 24 \times 5 = 120 \)
Thus, the LCM of 8, 12, and 15 is 120.
### Step 5: Solve for \( x \)
Since \( x - 6 \) is the LCM, we have:
\[
x - 6 = 120
\]
Adding 6 to both sides gives:
\[
x = 120 + 6 = 126
\]
### Conclusion
The least number which leaves a remainder of 6 when divided by 8, 12, or 15 is **126**.
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