If k is a positive integer and 5^(k) is a factor of the product of the odd integers from 99...

1. Translate the problem requirements: We need to find how many times 5 appears as a factor in the product \(99 \times 101 \times 103 \times \ldots \times 199\) (all odd integers from 99 to 199). The greatest possible k is the total count of factor 5 in this product.
2. Identify which odd numbers contain factors of 5: Among odd integers, only those ending in 5 will be divisible by 5, so we need to count odd multiples of 5 in our range.
3. List the relevant odd multiples of 5: Find all numbers of the form \(5 \times \text{(odd number)}\) that fall between 99 and 199 inclusive.
4. Count factors of 5 in each identified number: For each odd multiple of 5, determine how many times 5 divides into it by checking for higher powers of 5.
5. Sum all factors of 5: Add up the total count of factors of 5 from all the identified numbers to get the maximum value of k.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're really being asked to find. We have a product of all odd integers from 99 to 199, which looks like: \(99 \times 101 \times 103 \times 105 \times \ldots \times 197 \times 199\).

The question asks: what's the greatest possible value of k such that \(5^k\) divides this entire product? In plain English, we're counting how many times we can "pull out" a factor of 5 from this big multiplication.

Think of it this way: if we had \(10 \times 15 \times 25\), we could write this as \((2 \times 5) \times (3 \times 5) \times (5 \times 5) = 2 \times 3 \times 5^4\). So \(k = 4\) in this case because we have four factors of 5 total.

Process Skill: TRANSLATE - Converting the factor question into a counting problem

2. Identify which odd numbers contain factors of 5

Since we're only looking at odd numbers, we need to figure out which odd numbers between 99 and 199 are divisible by 5.

Here's the key insight: an odd number is divisible by 5 if and only if it ends in 5. Why? Because numbers ending in 0 are even, so among numbers divisible by 5, only those ending in 5 can be odd.

So we're looking for odd numbers ending in 5 in our range: 105, 115, 125, 135, 145, 155, 165, 175, 185, 195.

3. List the relevant odd multiples of 5

Let's systematically list all odd multiples of 5 from 99 to 199:

• Start with the first odd multiple of 5 ≥ 99: that's 105
• Continue: 105, 115, 125, 135, 145, 155, 165, 175, 185, 195
• Stop at 195 since the next would be 205, which exceeds 199

So our complete list is: 105, 115, 125, 135, 145, 155, 165, 175, 185, 195
That's 10 numbers total.

4. Count factors of 5 in each identified number

Now comes the crucial part: some of these numbers have more than one factor of 5! We need to count carefully.

Let's check each number:

• \(105 = 21 \times 5\) → one factor of 5
• \(115 = 23 \times 5\) → one factor of 5
• \(125 = 5^3 = 5 \times 5 \times 5\) → three factors of 5!
• \(135 = 27 \times 5\) → one factor of 5
• \(145 = 29 \times 5\) → one factor of 5
• \(155 = 31 \times 5\) → one factor of 5
• \(165 = 33 \times 5\) → one factor of 5
• \(175 = 7 \times 25 = 7 \times 5^2\) → two factors of 5!
• \(185 = 37 \times 5\) → one factor of 5
• \(195 = 39 \times 5\) → one factor of 5

Process Skill: CONSIDER ALL CASES - Checking for higher powers of 5, not just single factors

5. Sum all factors of 5

Now we add up all the factors of 5 we found:

• Most numbers contribute 1 factor each: 105, 115, 135, 145, 155, 165, 185, 195 → 8 factors
• 175 contributes 2 factors
• 125 contributes 3 factors

Total: \(8 + 2 + 3 = 13\) factors of 5

Therefore, the greatest possible value of k is 13.

Final Answer

The greatest possible value of k is 13, which corresponds to answer choice B.

To verify: \(5^{13}\) divides the product of odd integers from 99 to 199, but \(5^{14}\) does not, since we found exactly 13 factors of 5 in the entire product.

Common Faltering Points

Errors while devising the approach

Students often confuse "\(5^k\) is a factor of the product" with "\(5^k\) appears in each number." They might think they need to find k such that every odd number from 99 to 199 is divisible by \(5^k\), rather than understanding that they need to count the total number of factors of 5 across all numbers in the product.

Students may overlook that the problem specifically asks for odd integers from 99 to 199, and accidentally include even numbers like 100, 110, 120, etc. This would lead them to count factors of 5 in numbers like \(100 = 2^2 \times 5^2\) or \(150 = 2 \times 3 \times 5^2\), inflating their final count.

Many students assume each number contributes at most one factor of 5, failing to recognize that some numbers like \(125 = 5^3\) or \(175 = 7 \times 5^2\) contain multiple factors of 5. This leads to undercounting and selecting a smaller answer.

Errors while executing the approach

Students often make errors when breaking down numbers like 125 or 175. For example, they might write \(125 = 25 \times 5\) and count only 2 factors instead of recognizing \(125 = 5^3 = 5 \times 5 \times 5\) with 3 factors. Similarly, for \(175 = 25 \times 7\), they might count only 1 factor instead of 2.

When listing odd multiples of 5 from 99 to 199, students may skip numbers or include incorrect ones. Common errors include forgetting 105 (since 99 isn't divisible by 5), including 199 incorrectly thinking it ends in 5, or miscounting the sequence 105, 115, 125, ..., 195.

Even with correct factor counts for each number, students may make simple addition errors when summing up: 8 (single factors) + 2 (from 175) + 3 (from 125) = 13. They might incorrectly add these as \(8 + 2 + 3 = 12\) or 14.

Errors while selecting the answer

Students might correctly calculate that there are 13 total factors of 5 but then select answer choice A (10) thinking about the count of numbers (10 odd multiples of 5) rather than the total number of factors of 5 (13).

Some students might think "the greatest possible value of k" means they should add 1 to their calculated result, reasoning that if \(5^{13}\) divides the product, then maybe \(5^{14}\) also could. This stems from not understanding that k represents exactly how many factors of 5 exist in the prime factorization.