If b is a positive integer and b^4 is divisible by 81, which of the following could be the remainder...

1. Translate the problem requirements: We need to find which remainder is possible when b is divided by 9, given that b is a positive integer and \(\mathrm{b}^4\) is divisible by 81. This means 81 divides \(\mathrm{b}^4\) evenly, and we want to know what's left over when b is divided by 9.
2. Break down the divisibility condition using prime factorization: Since \(81 = 3^4\), the condition "\(\mathrm{b}^4\) is divisible by 81" means \(\mathrm{b}^4\) must contain at least four factors of 3, which tells us something about how many factors of 3 that b itself must have.
3. Determine the minimum prime factor requirement for b: For \(\mathrm{b}^4\) to have at least four factors of 3, b must have at least one factor of 3, meaning b must be divisible by 3.
4. Analyze possible remainders when multiples of 3 are divided by 9: Since b must be a multiple of 3, we can write \(\mathrm{b} = 3\mathrm{k}\) for some positive integer k. When we divide \(3\mathrm{k}\) by 9, we get different remainders depending on the value of k.
5. Test each answer choice against our constraint: Check which of the given options (1, 2, 4, 6, 8) can actually occur as remainders when a multiple of 3 is divided by 9.

Execution of Strategic Approach

1. Translate the problem requirements

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

2. Break down the divisibility condition using prime factorization

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 \(3^4 | \mathrm{b}^4\)

3. Determine the minimum prime factor requirement for b

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

4. Analyze possible remainders when multiples of 3 are divided by 9

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:

• \(3 \div 9 = 0\) remainder 3
• \(6 \div 9 = 0\) remainder 6
• \(9 \div 9 = 1\) remainder 0
• \(12 \div 9 = 1\) remainder 3
• \(15 \div 9 = 1\) remainder 6
• \(18 \div 9 = 2\) remainder 0
• \(21 \div 9 = 2\) remainder 3
• \(24 \div 9 = 2\) remainder 6

We can see a pattern: when multiples of 3 are divided by 9, the only possible remainders are 0, 3, and 6.

5. Test each answer choice against our constraint

Now let's check each answer choice:

1. Remainder 1: Can a multiple of 3 have remainder 1 when divided by 9? No, as we saw above.
2. Remainder 2: Can a multiple of 3 have remainder 2 when divided by 9? No.
3. Remainder 4: Can a multiple of 3 have remainder 4 when divided by 9? No.
4. Remainder 6: Can a multiple of 3 have remainder 6 when divided by 9? Yes! We saw examples like 6, 15, 24, etc.
5. Remainder 8: Can a multiple of 3 have remainder 8 when divided by 9? No.

Process Skill: APPLY CONSTRAINTS - Using our deduced constraint that b must be a multiple of 3 to eliminate impossible answer choices

4. Final Answer

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.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the relationship between \(\mathrm{b}^4\) and b

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.

2. Not recognizing the need for prime factorization

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.

3. Confusing the direction of the constraint

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.

Errors while executing the approach

1. Incorrect calculation of required factors of 3 in b

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.

2. Arithmetic errors when finding remainders

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.

Errors while selecting the answer

1. Selecting a remainder that seems plausible without verification

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.