Divisibility Problem: Finding Number of Students Divisible by Both 6 and 4

Q: Why can't I just multiply 6 × 4 = 24 to get the answer?

That works here by luck, but multiplying doesn't always give the LCM! For example, LCM(6,9) = 18, not 54. Always use prime factorization: find the highest power of each prime factor.

Q: How do I find LCM using prime factorization?

Break each number into prime factors: 6 = $ 2 \times 3 $ and 4 = $ 2^2 $. Then take the highest power of each prime: $ 2^2 \times 3^1 = 12 $.

Q: What does 'exact division' mean in word problems?

Exact division means no remainders - every student gets placed in a group with no one left over. This tells you the total must be divisible by the group size.

LCM Applications with Range Constraints

A math teacher divides the class into groups to solve exercises together.

On Tuesday, he divides them into groups of 6.

On Wednesday, he divides them into groups of 4.

The division is exact and no student is left without a group.

The number of students in a class is more than 21 but less than 31.

How many students are in the class?

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Step-by-step solution

To solve this problem, we'll follow these steps:

Calculate the LCM of 6 and 4.
Find the number between 21 and 31 that is a multiple of this LCM.

Step 1: Calculate the LCM of 6 and 4.
- The prime factorization of 6 is $2 \times 3$ .
- The prime factorization of 4 is $2^2$ .
- The LCM is derived by taking the highest power of each prime number involved: $LCM(6, 4) = 2^2 \times 3 = 12$ .

Step 2: Identify the multiples of 12 within the range (21, 31).
- Multiples of 12 are: $12, 24, 36, \ldots$
- In the range between 21 and 31, the only multiple of 12 is 24.

Therefore, the number of students in the class is $24$ .

Final Answer

$24$

Key Points to Remember

Essential concepts to master this topic

Concept: Find LCM when number must be divisible by multiple values
Method: LCM(6,4) = $2^2 \times 3 = 12$ using highest prime powers
Verify: Check 24 ÷ 6 = 4 and 24 ÷ 4 = 6 with no remainders ✓

Common Mistakes

Avoid these frequent errors

Finding GCD instead of LCM
Don't find the Greatest Common Divisor (GCD = 2) when you need divisibility by both numbers! GCD finds shared factors, but you need the smallest number divisible by both. Always find LCM for 'divisible by both' problems.

Practice Quiz

Test your knowledge with interactive questions

Will a number divisible by 6 necessarily be divisible by 3?

FAQ

Everything you need to know about this question

Why can't I just multiply 6 × 4 = 24 to get the answer?

That works here by luck, but multiplying doesn't always give the LCM! For example, LCM(6,9) = 18, not 54. Always use prime factorization: find the highest power of each prime factor.

How do I find LCM using prime factorization?

Break each number into prime factors: 6 = $2 \times 3$ and 4 = $2^2$ . Then take the highest power of each prime: $2^2 \times 3^1 = 12$ .

What if there are multiple multiples in the range?

List all multiples of the LCM and check which ones fit your constraints. For LCM = 12: multiples are 12, 24, 36... Only 24 falls between 21 and 31.

How do I check my answer is correct?

Verify both conditions: 24 ÷ 6 = 4 (exact groups of 6) and 24 ÷ 4 = 6 (exact groups of 4). Also confirm 21 < 24 < 31. All conditions satisfied!

What does 'exact division' mean in word problems?

Exact division means no remainders - every student gets placed in a group with no one left over. This tells you the total must be divisible by the group size.

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