To find the least number that is divisible by all numbers from 1 to 5, we'll need to determine the Least Common Multiple (LCM) of these numbers.
Step-by-step solution:
1. Identify the prime factors of each number:
- [tex]\(1\)[/tex] (Note: [tex]\(1\)[/tex] is not included in prime factorization as it doesn't affect LCM)
- [tex]\(2\)[/tex] is already a prime number ([tex]\(2\)[/tex])
- [tex]\(3\)[/tex] is already a prime number ([tex]\(3\)[/tex])
- [tex]\(4\)[/tex] can be expressed as [tex]\(2^2\)[/tex]
- [tex]\(5\)[/tex] is already a prime number ([tex]\(5\)[/tex])
2. List the highest power of each prime factor appearing in the factorizations:
- For [tex]\(2\)[/tex], the highest power is [tex]\(2^2\)[/tex] (from [tex]\(4\)[/tex])
- For [tex]\(3\)[/tex], the highest power is [tex]\(3\)[/tex] (from [tex]\(3\)[/tex])
- For [tex]\(5\)[/tex], the highest power is [tex]\(5\)[/tex] (from [tex]\(5\)[/tex])
3. Multiply these highest powers together to find the LCM:
- [tex]\(2^2 = 4\)[/tex]
- [tex]\(3\)[/tex]
- [tex]\(5\)[/tex]
Combining these, we get:
[tex]\[
\text{LCM} = 2^2 \times 3 \times 5
\][/tex]
4. Calculate the value:
[tex]\[
\text{LCM} = 4 \times 3 \times 5
\][/tex]
Multiplying these step by step:
- [tex]\(4 \times 3 = 12\)[/tex]
- [tex]\(12 \times 5 = 60\)[/tex]
Therefore, the least number that is divisible by all numbers from 1 to 5 is [tex]\( \boxed{60} \)[/tex].
