How many different positive prime factors of 5^(20) + 5^(17) are there?

1. Translate the problem requirements: We need to find how many different positive prime factors exist in the expression \(5^{20} + 5^{17}\). A prime factor is a prime number that divides the expression evenly.
2. Factor out the common term: Identify and extract the highest common factor from both terms to simplify the expression into a more manageable form.
3. Analyze the remaining factor: Examine the simplified expression to identify what additional factors beyond the obvious ones are present.
4. Identify all prime factors: Systematically determine each distinct prime number that divides the factored expression and count them.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We have the expression \(5^{20} + 5^{17}\), and we need to find how many different positive prime factors this expression has.

A prime factor is simply a prime number (like 2, 3, 5, 7, 11, etc.) that divides our expression evenly. For example, if our expression equals 60, then the prime factors would be 2, 3, and 5 because \(60 = 2^2 \times 3 \times 5\).

So our goal is to break down \(5^{20} + 5^{17}\) into its prime components and count how many different primes appear.

Process Skill: TRANSLATE - Converting the problem language into a clear mathematical objective

2. Factor out the common term

Looking at our expression \(5^{20} + 5^{17}\), I notice both terms contain powers of 5. Let me think about what they have in common.

\(5^{20}\) means 5 multiplied by itself 20 times
\(5^{17}\) means 5 multiplied by itself 17 times

Since both terms contain at least 17 factors of 5, I can factor out \(5^{17}\) from both terms:

\(5^{20} + 5^{17} = 5^{17} \times 5^3 + 5^{17} \times 1\)
                \(= 5^{17} \times (5^3 + 1)\)
                \(= 5^{17} \times (125 + 1)\)
                \(= 5^{17} \times 126\)

This is much easier to work with! Now I need to find the prime factors of \(5^{17} \times 126\).

Process Skill: SIMPLIFY - Factoring out common terms to make the problem more manageable

3. Analyze the remaining factor

Now I have \(5^{17} \times 126\). Let me find the prime factors step by step.

From \(5^{17}\): This gives us the prime factor 5 (appearing 17 times, but we only count distinct primes).

From 126: I need to break this down into prime factors. Let me do this systematically:
126 ÷ 2 = 63 (so 2 is a prime factor)
63 ÷ 3 = 21 (so 3 is a prime factor)
21 ÷ 3 = 7 (3 appears again, but we already counted it)
7 ÷ 7 = 1 (so 7 is a prime factor)

Therefore: \(126 = 2 \times 3^2 \times 7\)

So our complete factorization is:
\(5^{20} + 5^{17} = 5^{17} \times 2 \times 3^2 \times 7\)

4. Identify all prime factors

Looking at our factorization \(5^{17} \times 2 \times 3^2 \times 7\), I can identify all the distinct prime factors:

1. The prime 5 (from the \(5^{17}\) term)
2. The prime 2 (from factoring 126)
3. The prime 3 (from factoring 126)
4. The prime 7 (from factoring 126)

Let me verify: \(5^{17} \times 2 \times 3^2 \times 7 = 5^{17} \times 2 \times 9 \times 7 = 5^{17} \times 126\) ✓

Therefore, there are exactly 4 different positive prime factors.

Final Answer

The expression \(5^{20} + 5^{17}\) has exactly four different positive prime factors: 2, 3, 5, and 7.

The answer is D. Four

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "prime factors" means
Students might confuse "prime factors" with "total number of prime factors including repetitions" or "prime factorization." For example, if they see \(5^{17} \times 2 \times 3^2 \times 7\), they might count 5 seventeen times plus 2 once plus 3 twice plus 7 once, giving them 21 total factors instead of recognizing we only want the 4 distinct primes: 2, 3, 5, and 7.

2. Not recognizing the factoring opportunity
Students might attempt to calculate \(5^{20} + 5^{17}\) directly instead of factoring out the common term \(5^{17}\). Since \(5^{20}\) and \(5^{17}\) are enormous numbers, trying to work with them directly would make the problem nearly impossible to solve by hand and lead students to think the problem requires complex calculations.

3. Forgetting that \(5^{17}\) contributes the prime factor 5
After factoring to get \(5^{17} \times 126\), students might focus only on finding the prime factors of 126 (which are 2, 3, and 7) and forget that the \(5^{17}\) term contributes 5 as an additional distinct prime factor, leading them to conclude there are only 3 prime factors instead of 4.

Errors while executing the approach

1. Arithmetic errors when calculating \(5^3\)
When factoring out \(5^{17}\), students need to compute \(5^3 = 125\). Some might incorrectly calculate this as 75 or 150, leading to the wrong value for \((5^3 + 1)\) and subsequently incorrect prime factorization of the remaining term.

2. Incomplete or incorrect prime factorization of 126
Students might make errors in the step-by-step factorization: \(126 = 2 \times 63 = 2 \times 3 \times 21 = 2 \times 3 \times 3 \times 7 = 2 \times 3^2 \times 7\). Common mistakes include stopping early (like \(126 = 6 \times 21\)) without fully reducing to primes, or making division errors along the way.

3. Computational errors when verifying \(5^3 + 1 = 126\)
Students might miscalculate \(125 + 1 = 126\), or make errors when checking their work, leading them to doubt their factoring approach or proceed with incorrect values.

Errors while selecting the answer

1. Counting repeated prime factors instead of distinct primes
Even after correctly finding the factorization \(5^{17} \times 2 \times 3^2 \times 7\), students might count the exponents and think there are more than 4 factors because 3 appears twice \((3^2)\) or because 5 appears 17 times, leading them to select "E. More than four" instead of the correct answer "D. Four."

2. Double-counting or missing prime factors in the final tally
Students might correctly identify the primes 2, 3, 5, and 7 but then make a counting error in their final step – either missing one of the primes or accidentally counting one twice, leading them to select "C. Three" or "E. More than four" instead of "D. Four."