How many prime numbers between 1 and 100 are factors of 7{,}150?

1. Translate the problem requirements: We need to find how many prime numbers between 1 and 100 can divide 7,150 evenly (are factors of 7,150). This means finding the prime factorization of 7,150 and counting how many of those prime factors fall between 1 and 100.
2. Find the prime factorization of 7,150: Break down 7,150 into its constituent prime factors by systematically dividing by small primes.
3. Identify which prime factors fall in the given range: Check which of the prime factors from the factorization are between 1 and 100.
4. Count the qualifying prime factors: Tally up the prime factors that meet our criteria to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this question is really asking us. We need to find prime numbers that are also factors of 7,150.

Think of it this way: if I have a number like 12, its factors are all the numbers that divide evenly into 12. For example, 1, 2, 3, 4, 6, and 12 are all factors of 12 because they divide evenly with no remainder.

Now, among those factors, some might be prime numbers. A prime number is a number greater than 1 that has exactly two factors: 1 and itself. For example, 2 and 3 are prime factors of 12.

So our job is to find all the prime numbers between 1 and 100 that divide evenly into 7,150.

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

2. Find the prime factorization of 7,150

The best way to find which prime numbers are factors of 7,150 is to break down 7,150 into its basic building blocks - its prime factors.

Let's start by dividing 7,150 by small prime numbers:

Step 1: Check if 2 divides 7,150
\(7,150 ÷ 2 = 3,575\) (yes, it divides evenly)

Step 2: Check if 2 divides 3,575
3,575 is odd, so 2 doesn't divide it. Let's try 3.

Step 3: Check if 3 divides 3,575
Sum of digits: 3 + 5 + 7 + 5 = 20. Since 20 is not divisible by 3, neither is 3,575. Let's try 5.

Step 4: Check if 5 divides 3,575
3,575 ends in 5, so it's divisible by 5: \(3,575 ÷ 5 = 715\)

Step 5: Check if 5 divides 715
715 ends in 5, so: \(715 ÷ 5 = 143\)

Step 6: Factor 143
Let's try dividing by small primes. \(143 ÷ 11 = 13\) (both 11 and 13 are prime)

So our complete factorization is:
\(7,150 = 2 × 5 × 5 × 11 × 13 = 2 × 5^2 × 11 × 13\)

3. Identify which prime factors fall in the given range

Now we need to check which of our prime factors are between 1 and 100.

Our prime factors from the factorization are: 2, 5, 11, and 13

Let's check each one:

• 2: Is 2 between 1 and 100? Yes ✓
• 5: Is 5 between 1 and 100? Yes ✓
• 11: Is 11 between 1 and 100? Yes ✓
• 13: Is 13 between 1 and 100? Yes ✓

All of our prime factors fall within the range of 1 to 100.

4. Count the qualifying prime factors

Let's count up our prime factors that meet the criteria:

Prime factors of 7,150 that are between 1 and 100:

1. 2
2. 5
3. 11
4. 13

That gives us 4 prime numbers.

Final Answer

We found that \(7,150 = 2 × 5^2 × 11 × 13\), which means it has exactly 4 prime factors (2, 5, 11, and 13) that are between 1 and 100.

The answer is D. Four

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "factors" means: Students might confuse "factors of 7,150" with "multiples of primes" or think they need to find primes that add up to something related to 7,150. The question asks for prime numbers that divide evenly into 7,150, but students may not clearly understand this relationship.

2. Overlooking the constraint "between 1 and 100": Students might focus only on finding all prime factors of 7,150 without paying attention to the range limitation. They could devise an approach that finds all prime factors regardless of the given range constraint.

3. Confusing prime factors with all factors: Students might think they need to find ALL factors of 7,150 that happen to be prime, rather than understanding that they need to find the prime factorization first and then check which prime factors fall in the given range.

Errors while executing the approach

1. Arithmetic errors during prime factorization: Students commonly make division errors when breaking down 7,150. For example, they might incorrectly calculate 3,575 ÷ 5 or make errors when checking divisibility rules (like incorrectly adding digits to test divisibility by 3).

2. Incomplete factorization: Students might stop the factorization process too early, missing some prime factors. For instance, they might correctly find 2 and 5 but fail to completely factor 143 into 11 × 13, stopping when they reach a number that seems difficult to factor.

3. Missing repeated prime factors: Students might count the prime factor 5 twice because it appears as \(5^2\) in the factorization, not realizing that 5 is still just one unique prime factor, even though it appears with an exponent greater than 1.

Errors while selecting the answer

1. Double-counting repeated primes: If students found the factorization \(2 × 5^2 × 11 × 13\), they might count 5 twice because of the exponent, leading them to select "Five" instead of "Four" as the final answer.

2. Forgetting to check the range constraint at the end: Students might correctly find all prime factors but forget to verify that each one falls between 1 and 100 before counting them for the final answer, though in this case all factors do fall in range.