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If \(\mathrm{n} = 9! - 6^4\), which of the following is the greatest integer \(\mathrm{k}\) such that \(3^\mathrm{k}\) is a factor of \(\mathrm{n}\)?
Let's break this down in plain English first. We have a number n that equals \(\mathrm{9!}\) minus \(\mathrm{6^4}\). We want to find the biggest value of k where \(\mathrm{3^k}\) divides evenly into n.
Think of it this way: if \(\mathrm{3^k}\) is a factor of n, that means we can divide n by \(\mathrm{3^k}\) and get a whole number with no remainder. We want the largest possible k for which this works.
So our job is to find the highest power of 3 that "goes into" the result when we subtract \(\mathrm{6^4}\) from \(\mathrm{9!}\).
Process Skill: TRANSLATE - Converting the mathematical language "greatest integer k such that \(\mathrm{3^k}\) is a factor of n" into understandable terms
Now let's figure out how many factors of 3 are hiding in each piece separately.
For \(\mathrm{9!}\): We need to count how many times 3 appears as a factor in \(\mathrm{9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1}\)
Let's look at each number:
- 3 contributes one factor of 3
- \(\mathrm{6 = 2 × 3}\) contributes one factor of 3
- \(\mathrm{9 = 3 × 3}\) contributes two factors of 3
So \(\mathrm{9!}\) has \(\mathrm{1 + 1 + 2 = 4}\) factors of 3 total.
This means \(\mathrm{9! = 3^4 × \text{(some number not divisible by 3)}}\)
For \(\mathrm{6^4}\): Since \(\mathrm{6 = 2 × 3}\), we have \(\mathrm{6^4 = (2 × 3)^4 = 2^4 × 3^4}\)
So \(\mathrm{6^4}\) has exactly 4 factors of 3.
This means \(\mathrm{6^4 = 3^4 × 2^4 = 3^4 × 16}\)
Now let's write both pieces in a form that clearly shows their factors of 3.
For \(\mathrm{9!}\): We know \(\mathrm{9! = 3^4 × m}\), where m is some integer not divisible by 3
To find m: \(\mathrm{9! = 362,880}\) and \(\mathrm{3^4 = 81}\), so \(\mathrm{m = 362,880 ÷ 81 = 4,480}\)
We can verify: 4,480 is not divisible by 3 (\(\mathrm{4 + 4 + 8 + 0 = 16}\), and 16 is not divisible by 3)
For \(\mathrm{6^4}\): We already found \(\mathrm{6^4 = 3^4 × 16}\)
So our expression becomes:
\(\mathrm{n = 9! - 6^4 = 3^4 × 4,480 - 3^4 × 16}\)
Since both terms have \(\mathrm{3^4}\) as a factor, we can factor it out:
\(\mathrm{n = 3^4 × 4,480 - 3^4 × 16}\)
\(\mathrm{n = 3^4 × (4,480 - 16)}\)
\(\mathrm{n = 3^4 × 4,464}\)
Now we need to check if 4,464 has any additional factors of 3:
\(\mathrm{4,464 ÷ 3 = 1,488}\)
\(\mathrm{1,488 ÷ 3 = 496}\)
\(\mathrm{496 ÷ 3 = 165.33...}\) (not a whole number)
So \(\mathrm{4,464 = 3^2 × 496}\), and 496 is not divisible by 3.
Therefore: \(\mathrm{n = 3^4 × 3^2 × 496 = 3^6 × 496}\)
Since 496 is not divisible by 3, the highest power of 3 that divides n is \(\mathrm{3^6}\).
Process Skill: MANIPULATE - Factoring out common terms and systematically checking for additional factors
The greatest integer k such that \(\mathrm{3^k}\) is a factor of n is \(\mathrm{k = 6}\).
Looking at our answer choices, this corresponds to choice D.
1. Misunderstanding what "factor" means in the context
Students might confuse "\(\mathrm{3^k}\) is a factor of n" with "n is a factor of \(\mathrm{3^k}\)" or think they need to find how many times they can multiply \(\mathrm{3^k}\) to get n, rather than understanding they need to find the highest power of 3 that divides evenly into n.
2. Attempting to calculate n directly first
Many students will try to compute \(\mathrm{9! = 362,880}\) and \(\mathrm{6^4 = 1,296}\), then subtract to get \(\mathrm{n = 361,584}\), and then try to factor this large number. This approach is much more time-consuming and error-prone compared to working with prime factorizations from the start.
3. Forgetting to use the subtraction structure strategically
Students might calculate the powers of 3 in \(\mathrm{9!}\) and \(\mathrm{6^4}\) separately but fail to recognize that since both terms have the same power of 3 (\(\mathrm{3^4}\)), they can factor this out to simplify the problem significantly.
1. Incorrectly counting factors of 3 in \(\mathrm{9!}\)
When finding factors of 3 in \(\mathrm{9!}\), students often miss that \(\mathrm{9 = 3^2}\) contributes TWO factors of 3, not just one. They might count only 3, 6, and 9 as contributing one factor each, getting 3 total instead of the correct 4 factors.
2. Arithmetic errors when checking divisibility
When determining if 4,464 has additional factors of 3, students might make division errors or stop checking too early. For example, they might incorrectly calculate \(\mathrm{4,464 ÷ 3}\) or fail to continue dividing 1,488 by 3 to find the second additional factor.
3. Incorrectly handling the factorization of \(\mathrm{6^4}\)
Students might write \(\mathrm{6^4}\) as having 4 factors of 6 rather than recognizing that \(\mathrm{6 = 2 × 3}\), so \(\mathrm{6^4 = (2 × 3)^4 = 2^4 × 3^4}\), leading to wrong counts of factors of 3.
1. Adding instead of using the highest power
After finding that \(\mathrm{9!}\) has 4 factors of 3 and \(\mathrm{6^4}\) has 4 factors of 3, students might incorrectly think the answer is \(\mathrm{4 + 4 = 8}\), not recognizing that the subtraction and factoring process changes how the powers combine.
2. Forgetting to account for additional factors in the difference
Students might correctly factor out \(\mathrm{3^4}\) from both terms but then forget to check whether \(\mathrm{(4,480 - 16) = 4,464}\) contains additional factors of 3, stopping at \(\mathrm{k = 4}\) instead of the correct \(\mathrm{k = 6}\).