What is the smallest positive integer n such that the product 140n is the square of an integer?

1. Translate the problem requirements: We need to find the smallest positive integer n such that when we multiply 140 by n, the result is a perfect square (can be expressed as some integer squared).
2. Break down 140 into prime factors: Identify the basic building blocks of 140 to understand what's missing for a perfect square.
3. Analyze perfect square requirements: Determine what additional factors n must provide so that all prime factors appear an even number of times.
4. Find the minimal n and verify: Calculate the smallest n that satisfies our requirements and confirm it works.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in simple terms. We have the number 140, and we need to multiply it by some positive integer n. The result of this multiplication (140n) needs to be a "perfect square" - meaning it can be written as some whole number multiplied by itself.

For example, 36 is a perfect square because \(36 = 6 \times 6 = 6^2\). Similarly, 100 is a perfect square because \(100 = 10 \times 10 = 10^2\).

So we're asking: what's the smallest positive number n that makes 140n equal to some integer squared?

Process Skill: TRANSLATE - Converting the mathematical language into concrete understanding

2. Break down 140 into prime factors

To understand what we're working with, let's break down 140 into its most basic building blocks - its prime factors. Think of this like taking apart a Lego structure to see what individual pieces we have.

Let's factor 140 step by step:

• 140 = 2 × 70
• 70 = 2 × 35
• 35 = 5 × 7

So we can write: \(140 = 2 \times 2 \times 5 \times 7\)

In mathematical notation: \(140 = 2^2 \times 5^1 \times 7^1\)

3. Analyze perfect square requirements

Here's the key insight: for any number to be a perfect square, each of its prime factors must appear an even number of times. This is because when we square a number, every prime factor gets doubled.

For example, if we have \(6^2 = (2 \times 3)^2 = 2^2 \times 3^2\), each prime factor appears an even number of times (2 times each).

Looking at our factorization of \(140 = 2^2 \times 5^1 \times 7^1\):

• The factor 2 appears 2 times (even) ✓
• The factor 5 appears 1 time (odd) ✗
• The factor 7 appears 1 time (odd) ✗

For 140n to be a perfect square, we need n to provide one more factor of 5 and one more factor of 7 to make their counts even.

Therefore, the smallest n must be: \(n = 5 \times 7 = 35\)

Process Skill: INFER - Drawing the non-obvious conclusion about even powers in perfect squares

4. Find the minimal n and verify

We determined that n = 35. Let's verify this works:

\(140n = 140 \times 35 = 4,900\)

Let's check if 4,900 is indeed a perfect square:

\(140 \times 35 = (2^2 \times 5^1 \times 7^1) \times (5^1 \times 7^1) = 2^2 \times 5^2 \times 7^2\)

Now all prime factors appear an even number of times! This means:

\(4,900 = (2^1 \times 5^1 \times 7^1)^2 = (2 \times 5 \times 7)^2 = 70^2\)

Indeed, \(70^2 = 4,900\), confirming our answer.

Let's also verify that 35 is the smallest such n by checking that no smaller option works:

• n = 5: gives us \(140 \times 5 = 700 = 2^2 \times 5^2 \times 7^1\) (7 still appears an odd number of times)
• n = 7: gives us \(140 \times 7 = 980 = 2^2 \times 5^1 \times 7^2\) (5 still appears an odd number of times)

Final Answer

The smallest positive integer n such that 140n is a perfect square is 35.

This corresponds to answer choice D. 35

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding what makes a perfect square
Students may think that for 140n to be a perfect square, they just need to find a value of n that makes 140n equal to some familiar perfect square like 144 or 169, rather than understanding the fundamental requirement that all prime factors must appear an even number of times. This leads to random trial-and-error instead of systematic analysis.

Faltering Point 2: Attempting to solve without prime factorization
Students may try to work backwards from the answer choices by multiplying 140 by each option and then checking if the result is a perfect square through estimation or square root calculation. While this might work, it's inefficient and doesn't build understanding of the underlying mathematical principle.

Errors while executing the approach

Faltering Point 1: Incorrect prime factorization of 140
Students may make arithmetic errors when breaking down 140 into prime factors. Common mistakes include writing 140 = 2 × 5 × 14 (forgetting that 14 isn't prime) or 140 = 4 × 35 (using 4 instead of breaking it down to \(2^2\)). These errors lead to incorrect identification of which prime factors need to be "completed" to even powers.

Faltering Point 2: Misapplying the "even powers" rule
Even when students understand that prime factors need even powers, they may incorrectly think they need to make ALL powers even by adding factors, rather than recognizing that \(2^2\) already has an even power and doesn't need adjustment. This could lead them to include an extra factor of 2 in their answer.

Errors while selecting the answer

Faltering Point 1: Calculation errors in verification
Students may correctly determine that n = 35 but then make arithmetic mistakes when verifying that \(140 \times 35 = 4,900\) or that \(4,900 = 70^2\). Simple multiplication errors could make them doubt their correct answer and choose a different option.