How many different positive integers are factors of 441?

1. Translate the problem requirements: We need to find how many different positive integers divide evenly into 441 (i.e., how many positive divisors 441 has)
2. Break down 441 into prime factors: Express 441 as a product of prime numbers to reveal its underlying structure
3. Apply the divisor counting principle: Use the prime factorization to systematically count all possible combinations that form factors
4. Verify with the answer choices: Check our result against the given options to confirm our calculation

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're actually looking for. When we ask "how many positive integers are factors of 441," we're asking: how many different positive whole numbers can divide evenly into 441?

Think of it this way: if you have 441 objects and you want to arrange them into equal groups, how many different group sizes are possible? Each possible group size is a factor of 441.

For example, if 21 divides evenly into 441, then 21 is a factor. If 15 doesn't divide evenly, then 15 is not a factor.

Process Skill: TRANSLATE - Converting the question about "factors" into the concrete concept of "numbers that divide evenly"

2. Break down 441 into prime factors

To find all factors systematically, we need to break 441 down into its building blocks - the prime numbers that multiply together to make 441.

Let's start dividing 441 by small prime numbers:
• Is 441 divisible by 2? No, it's odd.
• Is 441 divisible by 3? Let's check: \(4 + 4 + 1 = 9\), and 9 is divisible by 3, so yes!

\(441 \div 3 = 147\)

Now let's factor 147:
• Is 147 divisible by 3? \(1 + 4 + 7 = 12\), and 12 is divisible by 3, so yes!

\(147 \div 3 = 49\)

Now we have 49. We recognize that \(49 = 7 \times 7 = 7^2\)

So putting it all together: \(441 = 3 \times 3 \times 7 \times 7 = 3^2 \times 7^2\)

3. Apply the divisor counting principle

Now here's the key insight: every factor of 441 must be made up of some combination of these prime factors.

Since \(441 = 3^2 \times 7^2\), any factor of 441 must have the form \(3^a \times 7^b\) where:
• a can be 0, 1, or 2 (we can use zero, one, or two 3's)
• b can be 0, 1, or 2 (we can use zero, one, or two 7's)

Let's list all the possibilities:
• \(3^0 \times 7^0 = 1 \times 1 = 1\)
• \(3^1 \times 7^0 = 3 \times 1 = 3\)
• \(3^2 \times 7^0 = 9 \times 1 = 9\)
• \(3^0 \times 7^1 = 1 \times 7 = 7\)
• \(3^1 \times 7^1 = 3 \times 7 = 21\)
• \(3^2 \times 7^1 = 9 \times 7 = 63\)
• \(3^0 \times 7^2 = 1 \times 49 = 49\)
• \(3^1 \times 7^2 = 3 \times 49 = 147\)
• \(3^2 \times 7^2 = 9 \times 49 = 441\)

So our factors are: 1, 3, 7, 9, 21, 49, 63, 147, 441

Counting these up: that's 9 different factors.

Alternatively, we can use the counting principle directly: since a has 3 choices (0, 1, 2) and b has 3 choices (0, 1, 2), the total number of factors is \(3 \times 3 = 9\).

4. Verify with the answer choices

Our calculation shows that 441 has 9 positive factors.

Looking at the answer choices:

1. 4
2. 6
3. 7
4. 9 ✓
5. 11

Our answer of 9 matches choice D.

Final Answer

The number of different positive integers that are factors of 441 is 9.

The answer is D.

Common Faltering Points

Errors while devising the approach

1. Confusing "factors" with "prime factors"

Students often misinterpret the question and think they need to find only the prime factors of 441 (which would be just 3 and 7). They miss that the question asks for ALL positive integer factors, including composite numbers like 9, 21, 49, etc. This leads them to count only 2 factors instead of all 9.

2. Attempting to find factors by trial division instead of using prime factorization

Some students try to find factors by testing divisibility with every number from 1 to 441, which is time-consuming and error-prone. They may miss factors or double-count, especially for a number as large as 441. The systematic approach using prime factorization and the divisor formula is much more reliable.

Errors while executing the approach

1. Incorrect prime factorization of 441

Students may make errors when breaking down 441 into prime factors. Common mistakes include: incorrectly concluding that 441 is divisible by 2 (it's odd), making arithmetic errors when dividing (like getting \(441 \div 3 = 143\) instead of 147), or failing to recognize that \(49 = 7^2\). This leads to an incorrect prime factorization and wrong factor count.

2. Misapplying the divisor counting formula

Even with the correct prime factorization \(441 = 3^2 \times 7^2\), students may incorrectly apply the formula. They might add the exponents instead of multiplying: \((2 + 1) + (2 + 1) = 6\), or forget to add 1 to each exponent: \(2 \times 2 = 4\). The correct application is \((2 + 1) \times (2 + 1) = 3 \times 3 = 9\).

3. Incomplete enumeration of factors

When listing all factors manually, students often forget to include 1 and 441 itself, or miss some of the middle factors like 21 or 63. They might also accidentally list the same factor twice, leading to an incorrect count.

Errors while selecting the answer

No likely faltering points - the calculation directly gives the number of factors, and there are no unit conversions or additional interpretations needed for the final answer selection.