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If \(\mathrm{n}\) and \(\mathrm{k}\) are positive integers and \(\mathrm{n + k < 8}\), how many different values of the product \(\mathrm{nk}\) are possible
Let's break down what this problem is asking us to find. We have two positive integers, which means they're counting numbers like 1, 2, 3, 4, and so on. We'll call these numbers n and k.
The constraint is that when we add n and k together, their sum must be less than 8. So \(\mathrm{n + k < 8}\).
Our goal is to find how many different values are possible when we multiply n and k together (the product \(\mathrm{nk}\)), given this constraint.
Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding
Since n and k are positive integers and \(\mathrm{n + k < 8}\), let's systematically list all possible combinations:
When \(\mathrm{n = 1}\): k can be 1, 2, 3, 4, 5, or 6 (since \(\mathrm{1 + k}\) must be less than 8)
When \(\mathrm{n = 2}\): k can be 1, 2, 3, 4, or 5 (since \(\mathrm{2 + k}\) must be less than 8)
When \(\mathrm{n = 3}\): k can be 1, 2, 3, or 4 (since \(\mathrm{3 + k}\) must be less than 8)
When \(\mathrm{n = 4}\): k can be 1, 2, or 3 (since \(\mathrm{4 + k}\) must be less than 8)
When \(\mathrm{n = 5}\): k can be 1 or 2 (since \(\mathrm{5 + k}\) must be less than 8)
When \(\mathrm{n = 6}\): k can only be 1 (since \(\mathrm{6 + k}\) must be less than 8)
When \(\mathrm{n = 7}\) or higher: No valid values of k, since even with \(\mathrm{k = 1}\), we'd have \(\mathrm{n + k ≥ 8}\)
Process Skill: CONSIDER ALL CASES - Systematically examining every possible scenario
Now let's calculate the product \(\mathrm{nk}\) for each valid pair. Note that \(\mathrm{(n,k)}\) and \(\mathrm{(k,n)}\) give the same product, so we only need to consider each product once:
From \(\mathrm{n = 1}\): \(\mathrm{1×1 = 1}\), \(\mathrm{1×2 = 2}\), \(\mathrm{1×3 = 3}\), \(\mathrm{1×4 = 4}\), \(\mathrm{1×5 = 5}\), \(\mathrm{1×6 = 6}\)
From \(\mathrm{n = 2}\): \(\mathrm{2×1 = 2}\) (already counted), \(\mathrm{2×2 = 4}\) (already counted), \(\mathrm{2×3 = 6}\) (already counted), \(\mathrm{2×4 = 8}\), \(\mathrm{2×5 = 10}\)
From \(\mathrm{n = 3}\): \(\mathrm{3×1 = 3}\) (already counted), \(\mathrm{3×2 = 6}\) (already counted), \(\mathrm{3×3 = 9}\), \(\mathrm{3×4 = 12}\)
From \(\mathrm{n = 4}\): \(\mathrm{4×1 = 4}\) (already counted), \(\mathrm{4×2 = 8}\) (already counted), \(\mathrm{4×3 = 12}\) (already counted)
From \(\mathrm{n = 5}\): \(\mathrm{5×1 = 5}\) (already counted), \(\mathrm{5×2 = 10}\) (already counted)
From \(\mathrm{n = 6}\): \(\mathrm{6×1 = 6}\) (already counted)
Let's collect all the unique product values we found:
1, 2, 3, 4, 5, 6, 8, 9, 10, 12
Counting these distinct values: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12
That's exactly 10 different possible values for the product \(\mathrm{nk}\).
Process Skill: SIMPLIFY - Organizing our results to clearly see the final answer
There are 10 different possible values for the product \(\mathrm{nk}\) when n and k are positive integers with \(\mathrm{n + k < 8}\).
This matches answer choice C: 10.
Faltering Point 1: Misinterpreting "positive integers" to include zero. Students might think that n or k can be 0, which would lead them to consider additional pairs like \(\mathrm{(0,1), (0,2)}\), etc. This would result in counting extra product values and arriving at an incorrect answer.
Faltering Point 2: Misunderstanding the constraint "\(\mathrm{n + k < 8}\)" as "\(\mathrm{n + k ≤ 8}\)". This small but critical error would allow students to include pairs where the sum equals 8, such as \(\mathrm{(1,7), (2,6), (3,5)}\), and \(\mathrm{(4,4)}\), leading to additional product values like 7, 15, and 16.
Faltering Point 3: Failing to recognize that the order of n and k doesn't matter for the product. Some students might think they need to count \(\mathrm{(2,3)}\) and \(\mathrm{(3,2)}\) as giving different products, leading to confusion in their systematic approach and potentially missing some valid combinations.
Faltering Point 1: Making arithmetic errors when calculating products. For instance, students might incorrectly calculate \(\mathrm{3×4 = 11}\) instead of 12, or \(\mathrm{2×5 = 12}\) instead of 10, which would lead to an incorrect count of distinct values.
Faltering Point 2: Failing to systematically list all valid pairs. Students might skip certain combinations, particularly when n gets larger. For example, they might forget to consider pairs like \(\mathrm{(5,2)}\) or \(\mathrm{(6,1)}\), missing important product values like 10 and 6.
Faltering Point 3: Double-counting the same product values. When organizing their work, students might list the same product multiple times (like counting both \(\mathrm{2×3=6}\) and \(\mathrm{1×6=6}\) as separate values) without recognizing they represent the same distinct value, leading to an inflated count.
Faltering Point 1: Counting the total number of valid pairs instead of the number of distinct products. Students might correctly find all 21 valid \(\mathrm{(n,k)}\) pairs but mistakenly think this is what the question is asking for, rather than counting the distinct product values.
Faltering Point 2: Including duplicate products in the final count. Even after calculating all products correctly, students might fail to eliminate duplicates in their final list, counting repeated values like 6 (which appears multiple times) as separate distinct values, leading to an answer higher than 10.