If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x,...

1. Translate the problem requirements: We need to find the highest power of 5 that divides \(150!\) (the product of integers 1 through 150). This means counting how many times 5 appears as a factor in this multiplication.
2. Count direct multiples of 5: Identify how many numbers from 1 to 150 are divisible by 5, as each contributes at least one factor of 5.
3. Count additional factors from higher powers of 5: Some numbers contribute more than one factor of 5 (like \(25 = 5^2\), \(125 = 5^3\)), so we need to count these extra contributions systematically.
4. Sum all factor contributions: Add up all the factors of 5 from multiples of \(5^1\), \(5^2\), \(5^3\), etc. until the power of 5 exceeds 150.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking. We have \(x = 1 \times 2 \times 3 \times 4 \times \ldots \times 150\), which is what mathematicians call "150 factorial" or \(150!\). The question wants to know: what's the highest power of 5 that divides evenly into this huge product?

Think of it this way: if we could factor this giant multiplication completely, how many 5's would we find? For example, if we had a smaller case like \(1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10\), we'd find the number 5 contributes one factor of 5, and the number \(10 = 2 \times 5\) contributes another factor of 5, giving us \(5^2\) total.

So we need to systematically count every factor of 5 hidden in the numbers from 1 to 150.

Process Skill: TRANSLATE - Converting the factorial notation and "factor" language into a concrete counting problem

2. Count direct multiples of 5

First, let's count how many numbers from 1 to 150 are divisible by 5. These are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ..., 145, 150.

To count these systematically: we want multiples of 5 up to 150. The largest multiple of 5 that doesn't exceed 150 is 150 itself. So we have \(5 \times 1\), \(5 \times 2\), \(5 \times 3\), ..., \(5 \times 30\). That's 30 multiples.

We can verify: \(150 \div 5 = 30\), so there are 30 direct multiples of 5.

Each of these contributes at least one factor of 5 to our product. So far we have at least 30 factors of 5.

3. Count additional factors from higher powers of 5

But wait! Some numbers contribute more than one factor of 5. For example:

• \(25 = 5 \times 5 = 5^2\), so it contributes 2 factors of 5
• \(50 = 2 \times 5^2\), so it contributes 2 factors of 5
• \(125 = 5^3\), so it contributes 3 factors of 5

We already counted one factor of 5 from each of these numbers. Now we need to count the additional factors.

Additional factors from multiples of \(5^2 = 25\):
Multiples of 25 from 1 to 150: 25, 50, 75, 100, 125, 150
How many? \(150 \div 25 = 6\) multiples of 25
Each contributes 1 additional factor of 5 beyond what we already counted.
Additional factors from \(5^2\): 6

Additional factors from multiples of \(5^3 = 125\):
Multiples of 125 from 1 to 150: just 125 itself
How many? \(150 \div 125 = 1\) (since \(150 \div 125 = 1.2\), we take the whole number part)
Additional factors from \(5^3\): 1

Additional factors from multiples of \(5^4 = 625\):
Since \(625 > 150\), there are no multiples of 625 in our range.
Additional factors from \(5^4\): 0

4. Sum all factor contributions

Now we add up all the factors of 5:

• From multiples of \(5^1\): 30 factors
• From multiples of \(5^2\): 6 additional factors
• From multiples of \(5^3\): 1 additional factor
• From multiples of \(5^4\) and higher: 0 additional factors

Total factors of 5: \(30 + 6 + 1 = 37\)

Therefore, \(5^{37}\) is the highest power of 5 that divides \(150!\), so \(y = 37\).

Final Answer

The greatest possible value of y is 37.

Looking at our answer choices:

1. 30
2. 34
3. 36
4. 37
5. 39

Our answer matches choice D) 37. ✓

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding what "\(5^y\) is a factor of x" means
Students often confuse this with simply counting how many numbers from 1 to 150 are divisible by 5. They might think the answer is just \(150 \div 5 = 30\), missing the crucial insight that some numbers contribute multiple factors of 5. The question asks for the highest power of 5 that divides \(150!\), not just how many multiples of 5 exist.

Faltering Point 2: Not recognizing the need to count higher powers of 5
Many students realize they need to count multiples of 5, but fail to recognize that numbers like 25, 50, 75, etc. contain \(5^2\) as a factor and thus contribute more than one factor of 5 to the total product. They stop after counting direct multiples instead of systematically examining \(5^2\), \(5^3\), etc.

Faltering Point 3: Confusion about the factorial notation and product interpretation
Some students might not fully grasp that \(x = 1 \times 2 \times 3 \times \ldots \times 150\) means they need to examine the prime factorization of this entire product. They might attempt to find patterns or use other approaches instead of the systematic factor counting method.

Errors while executing the approach

Faltering Point 1: Arithmetic errors in division calculations
When calculating \(150 \div 5 = 30\), \(150 \div 25 = 6\), \(150 \div 125 = 1.2 \rightarrow 1\), students often make basic division errors or incorrectly handle the floor function (taking only the whole number part). For example, they might round \(150 \div 125 = 1.2\) up to 2 instead of down to 1.

Faltering Point 2: Double-counting factors of 5
Students sometimes add the counts incorrectly, thinking they need to multiply instead of add, or they misunderstand which factors are "additional." For instance, they might count the factor of 5 in 25 twice - once as a multiple of 5 and again as a multiple of 25, instead of understanding that multiples of 25 contribute one additional factor beyond what was already counted.

Faltering Point 3: Stopping the process too early
Some students stop at \(5^2 = 25\) and forget to check \(5^3 = 125\). Since \(125 < 150\), there is indeed one multiple of 125 (which is 125 itself), contributing one additional factor. Missing this step would give an answer of 36 instead of 37.

Errors while selecting the answer

Faltering Point 1: Selecting an intermediate calculation result
Students might select 30 (answer choice A) if they only counted direct multiples of 5, or select 36 (answer choice C) if they forgot to include the contribution from multiples of 125. These represent stopping points in their calculation rather than the final answer.

No likely faltering points for misreading answer format, as the question clearly asks for the value of y and all answer choices are integers matching the expected format.