Finding the Smallest Number Divisible by 24, 48, and 60

The problem asks for the smallest number that is completely divisible by 24, 48, and 60. This is equivalent to finding the Least Common Multiple (LCM) of these three numbers.

Calculating the LCM using Prime Factorization

We will use the prime factorization method to find the LCM.

1. Find the prime factors of each number:

   • $24 = 2 \times 12 = 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1$
   • $48 = 2 \times 24 = 2 \times (2^3 \times 3^1) = 2^4 \times 3^1$
   • $60 = 2 \times 30 = 2 \times 2 \times 15 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3^1 \times 5^1$

2. Identify the highest power of each prime factor present in any of the factorizations:

   • The prime factors involved are 2, 3, and 5.
   • Highest power of 2: $2^4$ (from the factorization of 48)
   • Highest power of 3: $3^1$ (present in all factorizations)
   • Highest power of 5: $5^1$ (from the factorization of 60)

3. Multiply these highest powers together to find the LCM:

   LCM = $2^4 \times 3^1 \times 5^1$

   LCM = $16 \times 3 \times 5$

   LCM = $48 \times 5$

   LCM = $240$

Therefore, the smallest number that is completely divisible by 24, 48, and 60 is 240.