To solve the problem of finding the least number that is exactly divisible by 18, 30, and 45 without any remainder, we need to find the Least Common Multiple (LCM) of these three numbers. Here is the step-by-step solution:
### Step 1: Prime Factorization
First, we need to find the prime factorizations of each of the numbers.
- 18
[tex]\( 18 = 2 \times 3^2 \)[/tex]
- 30
[tex]\( 30 = 2 \times 3 \times 5 \)[/tex]
- 45
[tex]\( 45 = 3^2 \times 5 \)[/tex]
### Step 2: Identify the Highest Powers of Each Prime Factor
Next, we identify the highest powers of all the prime factors that appear in the factorizations of these numbers.
- The prime factor 2:
- In [tex]\(18\)[/tex], [tex]\(2 = 2^1\)[/tex]
- In [tex]\(30\)[/tex], [tex]\(2 = 2^1\)[/tex]
- In [tex]\(45\)[/tex], [tex]\(2\)[/tex] does not appear.
- The highest power of 2 is [tex]\(2^1\)[/tex].
- The prime factor 3:
- In [tex]\(18\)[/tex], [tex]\(3^2\)[/tex]
- In [tex]\(30\)[/tex], [tex]\(3^1\)[/tex]
- In [tex]\(45\)[/tex], [tex]\(3^2\)[/tex]
- The highest power of 3 is [tex]\(3^2\)[/tex].
- The prime factor 5:
- In [tex]\(18\)[/tex], [tex]\(5\)[/tex] does not appear.
- In [tex]\(30\)[/tex], [tex]\(5^1\)[/tex]
- In [tex]\(45\)[/tex], [tex]\(5^1\)[/tex]
- The highest power of 5 is [tex]\(5^1\)[/tex].
### Step 3: Multiply These Highest Powers Together
The LCM is given by multiplying the highest powers of all prime factors together.
[tex]\[
\text{LCM} = 2^1 \times 3^2 \times 5^1
\][/tex]
Calculate the product:
[tex]\[
2^1 = 2
\][/tex]
[tex]\[
3^2 = 9
\][/tex]
[tex]\[
5^1 = 5
\][/tex]
[tex]\[
\text{LCM} = 2 \times 9 \times 5
\][/tex]
[tex]\[
\text{LCM} = 2 \times 9 = 18
\][/tex]
[tex]\[
\text{LCM} = 18 \times 5 = 90
\][/tex]
Therefore, the least number that is exactly divisible by 18, 30, and 45 without any remainder is [tex]\( \boxed{90} \)[/tex].
