The problem asks for the least number divisible by 10, 18, 21, and 30. To find the least such number, we need to determine the Least Common Multiple (LCM) of these numbers.

Step 1: Prime factorization of each number:

• $10 = 2 \times 5$
• $18 = 2 \times 3^2$
• $21 = 3 \times 7$
• $30 = 2 \times 3 \times 5$

Step 2: Find the LCM by taking the highest power of each prime factor:

• The prime factors involved are 2, 3, 5, and 7.
• The highest power of 2 is $2^1$.
• The highest power of 3 is $3^2$.
• The highest power of 5 is $5^1$.
• The highest power of 7 is $7^1$.

Step 3: Multiply these together to find the LCM: $LCM = 2^1 \times 3^2 \times 5^1 \times 7^1 = 2 \times 9 \times 5 \times 7 = 630$

Therefore, the least number that is divisible by 10, 18, 21, and 30 is 630.

Would you like further clarification or details on this process?

Here are 5 related questions for further practice:

1. How do you calculate the LCM using the prime factorization method?
2. What is the greatest common divisor (GCD) of 10, 18, 21, and 30?
3. How would the LCM change if we added 15 to the list of numbers?
4. How is the LCM of two numbers different from the LCM of multiple numbers?
5. Can you solve for the LCM of the numbers 12, 24, 36, and 72?

Tip: A great way to practice LCM problems is by first breaking down the numbers into their prime factorizations and then selecting the highest power of each factor.