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If \(\mathrm{b}\) is a positive integer and \(\mathrm{b}^4\) is divisible by \(81\), which of the following could be the remainder when \(\mathrm{b}\) is divided by \(9\)?
Let's break down what this problem is asking in everyday language. We have a positive integer b, and when we raise it to the fourth power (\(\mathrm{b}^4\)), that result can be divided by 81 with no remainder left over. Given this constraint, we want to find out what remainder is possible when we divide b itself by 9.
Think of it like this: if I tell you that when you take a number, multiply it by itself four times, and then divide by 81, you get a whole number with nothing left over, what can you tell me about what's left over when I divide the original number by 9?
Process Skill: TRANSLATE - Converting the divisibility language into a clear constraint we can work with
Now let's think about what 81 really is. If we break it down into its most basic building blocks (prime factors), we get:
\(81 = 3 \times 3 \times 3 \times 3 = 3^4\)
So when the problem says "\(\mathrm{b}^4\) is divisible by 81," it's really saying "\(\mathrm{b}^4\) is divisible by \(3^4\)." This means when we completely break down \(\mathrm{b}^4\) into its prime factors, we need to find at least four 3's among those factors.
Technical notation: If \(81 | \mathrm{b}^4\), then \(3^4 | \mathrm{b}^4\)
Here's the key insight: if \(\mathrm{b}^4\) needs to have at least four factors of 3, how many factors of 3 does b itself need to have?
Let's say b has exactly n factors of 3. When we raise b to the fourth power, we get \(\mathrm{b}^4\), which will have 4n factors of 3 (because we're multiplying b by itself 4 times).
For \(\mathrm{b}^4\) to be divisible by \(3^4\), we need: \(4\mathrm{n} \geq 4\)
This means: \(\mathrm{n} \geq 1\)
So b must have at least one factor of 3, which means b must be divisible by 3.
Process Skill: INFER - Drawing the non-obvious conclusion about what the constraint on \(\mathrm{b}^4\) tells us about b itself
Since we've established that b must be a multiple of 3, let's see what remainders are possible when we divide multiples of 3 by 9.
The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...
Let's find the remainder when each is divided by 9:
We can see a pattern: when multiples of 3 are divided by 9, the only possible remainders are 0, 3, and 6.
Now let's check each answer choice:
Process Skill: APPLY CONSTRAINTS - Using our deduced constraint that b must be a multiple of 3 to eliminate impossible answer choices
The answer is D. 6.
To verify: if b leaves remainder 6 when divided by 9, then b could be 6, 15, 24, 33, etc. Let's check with \(\mathrm{b} = 6\):
\(\mathrm{b}^4 = 6^4 = 1296\)
\(1296 \div 81 = 16\) with remainder 0 ✓
This confirms that when b has remainder 6 upon division by 9, the condition that \(\mathrm{b}^4\) is divisible by 81 can be satisfied.
Students often fail to recognize that if \(\mathrm{b}^4\) is divisible by 81, this places constraints on b itself. They might try to work directly with the remainders without first establishing what type of number b must be. This leads to testing all answer choices without the crucial insight that b must be divisible by 3.
Many students see "divisible by 81" and don't think to break down 81 into its prime factors (\(3^4\)). Without this step, they miss the key insight about how many factors of 3 are needed in \(\mathrm{b}^4\), and consequently in b itself. This makes the problem much harder to solve systematically.
Some students might incorrectly think that since we want remainders when b is divided by 9, they should focus on what happens when 9 divides into various numbers, rather than understanding that the constraint "\(\mathrm{b}^4\) divisible by 81" limits what b can be in the first place.
When determining how many factors of 3 are needed in b, students might set up the inequality incorrectly. If b has n factors of 3, then \(\mathrm{b}^4\) has 4n factors of 3. For divisibility by \(3^4\), we need \(4\mathrm{n} \geq 4\), so \(\mathrm{n} \geq 1\). Students sometimes write this as \(\mathrm{n} \geq 4\) or make other arithmetic errors in this crucial step.
When systematically checking what remainders multiples of 3 can have when divided by 9, students often make basic division errors. For example, they might incorrectly calculate that \(15 \div 9\) gives remainder 3 instead of remainder 6, leading to wrong conclusions about possible remainders.
Students might correctly identify that only remainders 0, 3, and 6 are possible for multiples of 3 divided by 9, but then choose 3 or 0 from the mental shortcut that "smaller remainders are more common" rather than checking which of these actually appears in the answer choices. Since only 6 appears among the choices, they need to recognize that D is the only possible answer.