From the consecutive integers -{10} to 10 inclusive, 20 integers are randomly chosen…..

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Question: From the consecutive integers (-10) to (10) inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

Options:

• A. ((-10)^{20})
• B. ((-10)^{10})
• C. 0
• D. (-(10)^{19})
• E. (-(10)^{20})

Solution

1. Translate the problem requirements: We need to understand what “least possible value” means for a product of 20 integers, where each integer can be any value from -10 to 10 (including repetitions)
2. Analyze the impact of zero on the product: Determine whether including zero in our selection would give us the minimum possible product value
3. Explore negative versus positive products: Compare the magnitude of products when we have different numbers of negative factors, focusing on odd vs even counts of negative numbers
4. Optimize the magnitude through strategic integer selection: Choose the specific integers that will give us the largest absolute value while maintaining the desired sign

Execution of Strategic Approach

1. Translate the problem requirements Let’s break down what this problem is really asking. We need to pick 20 integers, and each pick can be any number from -10, -9, -8, …, 0, 1, 2, …, 9, 10. We can pick the same number multiple times – that’s what “repetitions allowed” means.

The key phrase is “least possible value” of the product. This means we want the most negative result possible. Think of it this way: if I multiply a bunch of numbers together, I get either a positive result, a negative result, or zero. Among all possible results, we want the one that’s farthest to the left on the number line.

Process Skill: TRANSLATE – Converting “least possible value” to “most negative result possible”

2. Analyze the impact of zero on the product Here’s a simple question: what happens if we include zero in our product?

Let’s say we pick: 5, -3, 7, 0, -2, and 15 more numbers. No matter what those other 15 numbers are, our final product will be (5 \times (-3) \times 7 \times 0 \times (-2) \times \ldots = 0).

Any product that includes zero equals zero. So if we include zero among our 20 numbers, our product will be 0.

But can we do better than 0? Can we get a negative result? Yes! If we avoid zero entirely and pick our numbers carefully, we can get a negative product. Since any negative number is less than 0, getting a negative product would give us a smaller (more negative) result than 0.

Conclusion: We should avoid picking 0 to get the least possible value.

3. Explore negative versus positive products Now let’s think about when products are positive versus negative. This follows a simple rule that most students know:

• Even number of negative factors → positive product
• Odd number of negative factors → negative product

Since we want the least (most negative) possible value, we want a negative product. This means we need an odd number of negative integers among our 20 picks.

Let’s think about this concretely. If we pick 19 copies of ((-10)) and 1 copy of ((+10)), we get: ((-10) \times (-10) \times (-10) \times \ldots \times (-10) \times (+10))

That’s 19 negative factors and 1 positive factor. 19 is odd, so our result will be negative. Good!

The calculation gives us: ((-10)^{19} \times (+10) = (-10)^{19} \times 10 = -10^{20})

What if we tried 20 copies of ((-10))? Then we’d have ((-10)^{20} = +10^{20}), which is positive. That’s larger than our negative result, so it’s not what we want.

4. Optimize the magnitude through strategic integer selection Now we know we want a negative product with the largest possible absolute value. We established that we need an odd number of negative factors.

To maximize the absolute value, we should pick the numbers with the largest absolute values available. The largest absolute value in our range is 10, which appears as both +10 and -10.

Here’s our optimal strategy:

• Pick ((-10)) nineteen times (odd number of negative factors ensures negative result)
• Pick ((+10)) one time (to maximize magnitude without changing the sign)

This gives us: ((-10)^{19} \times (+10) = -10^{19} \times 10 = -10^{20})

We can write this as (-(10^{20})), which matches answer choice E: (-10^{20}).

Process Skill: CONSIDER ALL CASES – We systematically considered zero, positive products, and negative products to find the optimal case

Common Faltering Points

Errors while devising the approach

• Misunderstanding “least possible value”: Students often confuse “least” with “smallest magnitude” rather than “most negative.” They might think the answer should be close to zero or have the smallest absolute value, when actually we want the most negative number possible.
• Overlooking the impact of zero: Students may not immediately realize that including zero in the product automatically makes the entire product equal to zero. They might spend time trying to optimize around zero instead of recognizing that avoiding zero entirely can lead to a more negative (and therefore smaller) result.
• Forgetting the sign rule for products: Students may not systematically consider that an odd number of negative factors produces a negative product while an even number produces a positive product. Without this insight, they can’t strategically choose the right combination of positive and negative numbers.

Errors while executing the approach

• Arithmetic errors with negative exponents: Students frequently make sign errors when calculating products like ((-10)^{19} \times 10). They might incorrectly write this as ((-10)^{20}) instead of recognizing it equals (-10^{20}), leading to a positive result instead of the correct negative result.
• Confusion between ((-10)^{20}) and (-(10)^{20}): Students often struggle with the difference between these expressions. ((-10)^{20}) equals positive (10^{20}) because we’re raising a negative number to an even power, while (-(10)^{20}) equals negative (10^{20}) because we’re taking the negative of (10^{20}).

Errors while selecting the answer

• Mixing up answer choice notation: After correctly calculating (-10^{20}), students may select choice A: ((-10)^{20}) instead of choice E: (-(10)^{20}). The similar-looking notation can cause confusion, especially since ((-10)^{20} = +10^{20}) while (-(10)^{20} = -10^{20}).
• Second-guessing the negative result: Students might doubt their negative answer because the magnitude (10^{20}) seems extremely large. They may incorrectly choose a smaller magnitude option like (-(10)^{19}) thinking it’s more reasonable, not realizing that we specifically want to maximize the magnitude while keeping the result negative.