How many of the factors of 210 are odd numbers greater than 1?

1. Translate the problem requirements: We need to find how many factors of 210 are both odd (not divisible by 2) and greater than 1. This means we're looking for divisors of 210 that contain no factor of 2.
2. Find the prime factorization of 210: Break down 210 into its prime components to understand its structure and identify which factors will be odd.
3. Identify the odd prime factors: Separate the even prime factor (2) from the odd prime factors, since odd factors can only come from combinations of odd primes.
4. Generate all combinations of odd prime factors: Systematically list all possible products of the odd prime factors to find every odd factor greater than 1.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in everyday language. We want to count how many numbers can divide 210 evenly, but with two specific conditions:

• The factor must be odd - this means it cannot be divisible by 2
• The factor must be greater than 1 - this excludes the number 1 itself

Think of it this way: if we have a number that divides 210 evenly, and that number contains no factors of 2, then it's an odd factor. We need to find all such numbers bigger than 1.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical requirements

2. Find the prime factorization of 210

To understand what factors 210 has, we need to break it down into its basic building blocks - the prime numbers that multiply together to make 210.

Let's work step by step:

• \(210 \div 2 = 105\) (so 2 is a prime factor)
• \(105 \div 3 = 35\) (so 3 is a prime factor)
• \(35 \div 5 = 7\) (so 5 is a prime factor)
• 7 is already prime

Therefore: \(210 = 2 \times 3 \times 5 \times 7\)

This prime factorization tells us exactly what "ingredients" we have to work with when building factors of 210.

3. Identify the odd prime factors

Now here's the key insight: for a factor to be odd, it cannot contain the prime factor 2. Looking at our factorization \(210 = 2 \times 3 \times 5 \times 7\), we can separate this into:

• Even prime factor: 2
• Odd prime factors: 3, 5, and 7

Any odd factor of 210 must be built using only the odd prime factors: 3, 5, and 7. We cannot use the factor of 2 at all, because that would make our result even.

Process Skill: INFER - Recognizing that odd factors can only come from odd prime components

4. Generate all combinations of odd prime factors

Now we systematically find every way to combine the odd primes 3, 5, and 7. Think of this like choosing ingredients from a menu - we can pick none, one, two, or all three of these primes:

Using one prime at a time:

• 3
• 5
• 7

Using two primes at a time:

• \(3 \times 5 = 15\)
• \(3 \times 7 = 21\)
• \(5 \times 7 = 35\)

Using all three primes:

• \(3 \times 5 \times 7 = 105\)

Let's verify these are all factors of 210:

• \(210 \div 3 = 70\) ✓
• \(210 \div 5 = 42\) ✓
• \(210 \div 7 = 30\) ✓
• \(210 \div 15 = 14\) ✓
• \(210 \div 21 = 10\) ✓
• \(210 \div 35 = 6\) ✓
• \(210 \div 105 = 2\) ✓

All seven numbers are indeed factors of 210, all are odd, and all are greater than 1.

Process Skill: CONSIDER ALL CASES - Systematically examining every possible combination to ensure we don't miss any factors

5. Final Answer

We found exactly seven odd factors of 210 that are greater than 1:

3, 5, 7, 15, 21, 35, and 105

The answer is E. Seven

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Missing the "greater than 1" constraint

Students often focus solely on finding "odd factors" and forget that the question specifically excludes 1. Since 1 is technically an odd factor of every number, failing to apply this constraint would lead to counting 1 as a valid answer, resulting in 8 factors instead of 7.

Faltering Point 2: Misunderstanding what makes a factor "odd"

Students may think they need to check each factor individually for oddness rather than recognizing the key insight: odd factors can only be formed using odd prime factors. This leads to a much more complicated approach of finding all factors first, then filtering for odd ones.

Faltering Point 3: Confusing "factors" with "prime factors"

Some students might think the question is asking only for odd prime factors of 210, which would give them just 3 (the primes 3, 5, and 7) instead of all possible odd factors including their combinations.

Errors while executing the approach

Faltering Point 1: Incomplete systematic enumeration

When generating combinations of the odd primes {3, 5, 7}, students often miss some combinations. Common omissions include forgetting two-factor combinations like \(3 \times 7 = 21\) or \(5 \times 7 = 35\), leading to an undercount of the total odd factors.

Faltering Point 2: Arithmetic errors in multiplication

Simple calculation mistakes when computing products like \(3 \times 5 = 15\), \(3 \times 7 = 21\), \(5 \times 7 = 35\), or \(3 \times 5 \times 7 = 105\) can lead to incorrect factor values, making verification against 210 fail and causing confusion about which factors are valid.

Errors while selecting the answer

Faltering Point 1: Including the number 1 in the final count

Even if students correctly identify all seven odd factors (3, 5, 7, 15, 21, 35, 105), they might add 1 to their list at the last moment, thinking "1 is odd and divides everything," forgetting the "greater than 1" requirement and selecting answer choice with 8 factors (if available) or getting confused about their count.