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How many of the factors of \(\mathrm{210}\) are odd numbers greater than \(\mathrm{1}\)?
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:
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
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:
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.
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:
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
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:
Using two primes at a time:
Using all three primes:
Let's verify these are all factors of 210:
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
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
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.
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.
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.