Smallest 3-digit number divisible by 6, 8, and 12

To find the smallest 3-digit number that is divisible by 6, 8, and 12, we need to find the least common multiple (LCM) of these numbers. The LCM is the smallest number that is divisible by all the given numbers.

Step 1: Prime factorization of the numbers
Let's start by finding the prime factorization of each number:

- Prime factorization of 6: 2 * 3
- Prime factorization of 8: 2 * 2 * 2
- Prime factorization of 12: 2 * 2 * 3

Step 2: Determine the highest power of each prime factor
Next, we need to determine the highest power of each prime factor among the given numbers. This is done by taking the maximum count of each prime factor:

- Highest power of 2: 2^3 = 8
- Highest power of 3: 3^1 = 3

Step 3: Calculate the LCM
Finally, we can calculate the LCM by multiplying the highest powers of the prime factors:

LCM = 2^3 * 3^1 = 8 * 3 = 24

Step 4: Finding the smallest 3-digit number divisible by 6, 8, and 12
Now that we have the LCM, we need to find the smallest 3-digit number that is divisible by 24. The smallest 3-digit number is 100, so we can start from there and check if it is divisible by 24:

- 100 ÷ 24 = 4 remainder 4

Since 100 divided by 24 leaves a remainder of 4, it is not divisible by 24. We need to find the next multiple of 24 greater than 100.

- 24 * 5 = 120

Therefore, the smallest 3-digit number divisible by 6, 8, and 12 is 120.

Summary:
- Prime factorization of 6: 2 * 3
- Prime factorization of 8: 2 * 2 * 2
- Prime factorization of 12: 2 * 2 * 3
- Highest power of 2: 2^3 = 8
- Highest power of 3: 3^1 = 3
- LCM = 2^3 * 3^1 = 8 * 3 = 24
- Smallest 3-digit number divisible by 24 is 120.

Answer is 120 which is divisible by 6 8 12