Find the lowest number which gives a remainder 3 when divided by any of the numbers 6, 7 and 8.

Answer

To solve the problem of finding the lowest number that gives a remainder of 3 when divided by 6, 7, and 8, we can follow these steps:

### Step 1: Find the LCM of 6, 7, and 8
To find a number that gives the same remainder when divided by multiple numbers, we first need to find the Least Common Multiple (LCM) of those numbers.

- The prime factorization of the numbers is:
- 6 = 2 × 3
- 7 = 7 (itself, as it's a prime number)
- 8 = 2³

- To find the LCM, we take the highest power of each prime number:
- The highest power of 2 is 2³ (from 8).
- The highest power of 3 is 3¹ (from 6).
- The highest power of 7 is 7¹ (from 7).

Thus, the LCM is:
\[ \text{LCM} = 2^3 \times 3^1 \times 7^1 = 8 \times 3 \times 7 = 168 \]

### Step 2: Add the remainder to the LCM
Since we need a number that gives a remainder of 3 when divided by 6, 7, and 8, we can take the LCM and add 3 to it:
\[ \text{Number} = \text{LCM} + 3 = 168 + 3 = 171 \]

### Step 3: Verify the result
We need to confirm that 171 gives a remainder of 3 when divided by each of the numbers 6, 7, and 8.

- **Dividing by 6:**
\[ 171 \div 6 = 28 \quad \text{Remainder: } 171 - (28 \times 6) = 171 - 168 = 3 \]

- **Dividing by 7:**
\[ 171 \div 7 = 24 \quad \text{Remainder: } 171 - (24 \times 7) = 171 - 168 = 3 \]

- **Dividing by 8:**
\[ 171 \div 8 = 21 \quad \text{Remainder: } 171 - (21 \times 8) = 171 - 168 = 3 \]

Since all divisions confirm that the remainder is indeed 3, we conclude that the lowest number that satisfies the condition is:
\[ \text{Answer: } 171 \]