Let's address your student's concern by demonstrating that systematically finding factor pairs for a number will indeed find all of the factors, including those that might involve multiplying more than two numbers together.

Step 1: Choose a Number

Let's choose the number 60 as an example. We will find all of its factors by finding its factor pairs.

Step 2: Find Factor Pairs for 60

To find all factors systematically, we look for pairs of numbers that multiply to 60:

1. 1 × 60 = 60
2. 2 × 30 = 60
3. 3 × 20 = 60
4. 4 × 15 = 60
5. 5 × 12 = 60
6. 6 × 10 = 60
7. (7 is not a factor since 60 ÷ 7 is not an integer)
8. 8 and 9 are not factors.
9. 10 × 6 = 60 (already listed above with 6 × 10)

Now, we can stop here since we have already found all the pairs. Notice that we only need to go up to the square root of 60 (approximately 7.75). So, checking up to 7 or 8 is sufficient.

From the above, the factors of 60 are:
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Step 3: Addressing the Student's Concern

The student might think that if we consider only pairs, we could miss combinations involving three or more numbers. For example, the student might think:

• "What about 2 × 3 × 10 = 60?"

Explanation of Why No Factors Are Missed

When you multiply 2 × 3 × 10, you are using three numbers, but two of them (2 and 3) multiply to make 6. So, the expression 2 × 3 × 10 is equivalent to 6 × 10. We have already identified 6 and 10 as factors through our factor pairs.

The important point here is that any multiplication of more than two factors can always be reduced to the multiplication of two factors (their product). For example:

• 2 × 3 × 10 = (2 × 3) × 10 = 6 × 10

Thus, this combination is already accounted for in our list of factor pairs.

General Explanation

For any number $N$, if we systematically find all factor pairs $(a, b)$ such that $a \times b = N$, we will find all factors:

1. Pairs Cover All Possible Combinations: Every factor can be part of a factor pair. If $c$ is a factor, then there is some other number $d$ such that $c \times d = N$. Therefore, all factors appear in at least one pair.
2. Multiplication of Multiple Factors Reduces to Pairs: Any product involving more than two factors, like $a \times b \times c$, can always be grouped into pairs: $(a \times b) \times c$. Since we already have all possible pairs, we do not miss any factors by focusing on pairs.

Conclusion

By finding all factor pairs, you systematically account for all possible factors of a number. Even if a product involves more than two numbers, the multiplication can always be reduced to two factors, and thus it is already captured in our method of finding factor pairs. This shows that finding factor pairs will indeed include all factors of a number.

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Would you like further clarification or details on this topic?

5 Related Questions:

1. How can we use factorization to find the greatest common divisor (GCD) of two numbers?
2. What is the prime factorization of a number, and how does it relate to finding all factors?
3. How do we find the least common multiple (LCM) using the factors of a number?
4. Why do we only need to check up to the square root of a number when finding factors?
5. Can the method of finding factor pairs be extended to polynomials or algebraic expressions?

Tip: When finding factors, remember that every divisor has a corresponding divisor pair. This helps ensure you find all possible factors!