If M is the least common multiple of 90, 196, and 300, which of the following is NOT a factor...

1. Translate the problem requirements: We need to find M (the least common multiple of 90, 196, and 300), then determine which answer choice is NOT a factor of M. This means four answer choices will divide M evenly, but one will not.
2. Break down each number into prime factors: Convert 90, 196, and 300 into their prime factorizations to see the building blocks clearly.
3. Construct the LCM using highest powers: For each prime that appears, take the highest power from any of the three numbers to build M.
4. Test each answer choice for divisibility: Check whether each answer choice's prime factorization can be formed using the primes available in M, identifying the one that cannot.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked to find. We need to find M, which is the least common multiple (LCM) of three numbers: 90, 196, and 300.

Think of the LCM as the smallest number that all three of these numbers can divide into evenly. For example, if we had numbers 4 and 6, their LCM would be 12 because 12 is the smallest number that both 4 and 6 divide into without remainders.

Once we find M, we need to check which of the five answer choices is NOT a factor of M. A factor is a number that divides evenly into another number. So four of our answer choices will divide evenly into M, but one will not.

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

2. Break down each number into prime factors

To find the LCM systematically, we need to break each number down to its most basic building blocks - its prime factors. Think of this like taking apart a LEGO structure to see what individual pieces were used to build it.

Let's break down each number:

For 90:
\(90 = 2 \times 45\)
\(45 = 3 \times 15\)
\(15 = 3 \times 5\)
So \(90 = 2 \times 3^2 \times 5\)

For 196:
\(196 = 4 \times 49\)
\(4 = 2^2\)
\(49 = 7^2\)
So \(196 = 2^2 \times 7^2\)

For 300:
\(300 = 3 \times 100\)
\(100 = 4 \times 25 = 2^2 \times 5^2\)
So \(300 = 2^2 \times 3 \times 5^2\)

Now we can see the prime factorizations clearly:

• \(90 = 2^1 \times 3^2 \times 5^1\)
• \(196 = 2^2 \times 7^2\)
• \(300 = 2^2 \times 3^1 \times 5^2\)

3. Construct the LCM using highest powers

To build the LCM, imagine we need enough of each type of prime 'ingredient' to satisfy all three numbers. We take the highest power of each prime that appears in any of the factorizations.

Looking at our prime factors:

• For prime 2: We see \(2^1\) (in 90), \(2^2\) (in 196), and \(2^2\) (in 300). The highest power is \(2^2\)
• For prime 3: We see \(3^2\) (in 90), no 3 (in 196), and \(3^1\) (in 300). The highest power is \(3^2\)
• For prime 5: We see \(5^1\) (in 90), no 5 (in 196), and \(5^2\) (in 300). The highest power is \(5^2\)
• For prime 7: We see no 7 (in 90), \(7^2\) (in 196), and no 7 (in 300). The highest power is \(7^2\)

Therefore: \(M = 2^2 \times 3^2 \times 5^2 \times 7^2\)

Let's calculate this step by step:
\(M = 4 \times 9 \times 25 \times 49\)
\(M = 36 \times 25 \times 49\)
\(M = 900 \times 49\)
\(M = 44,100\)

4. Test each answer choice for divisibility

Now we need to check which answer choice is NOT a factor of \(M = 44,100\). For a number to be a factor of M, its prime factorization must only use primes that are available in M, and not use higher powers than what M contains.

Remember: \(M = 2^2 \times 3^2 \times 5^2 \times 7^2\), so any factor can only use up to \(2^2\) for the prime 2, up to \(3^2\) for the prime 3, up to \(5^2\) for the prime 5, and up to \(7^2\) for the prime 7.

Choice A: 600
\(600 = 6 \times 100 = 2 \times 3 \times 10^2 = 2 \times 3 \times (2 \times 5)^2 = 2 \times 3 \times 2^2 \times 5^2 = 2^3 \times 3^1 \times 5^2\)

This needs \(2^3\), but M only has \(2^2\). Since 600 requires more factors of 2 than M contains, 600 cannot be a factor of M.

Choice B: 700
\(700 = 7 \times 100 = 7 \times 10^2 = 7 \times (2 \times 5)^2 = 7^1 \times 2^2 \times 5^2\)
This fits within M's available primes: \(2^2\) ✓, \(5^2\) ✓, \(7^1\) ✓

Choice C: 900
\(900 = 9 \times 100 = 3^2 \times 10^2 = 3^2 \times (2 \times 5)^2 = 2^2 \times 3^2 \times 5^2\)
This fits within M's available primes: \(2^2\) ✓, \(3^2\) ✓, \(5^2\) ✓

Choice D: 2,100
\(2,100 = 21 \times 100 = 3 \times 7 \times 10^2 = 3^1 \times 7^1 \times (2 \times 5)^2 = 2^2 \times 3^1 \times 5^2 \times 7^1\)
This fits within M's available primes: \(2^2\) ✓, \(3^1\) ✓, \(5^2\) ✓, \(7^1\) ✓

Choice E: 4,900
\(4,900 = 49 \times 100 = 7^2 \times 10^2 = 7^2 \times (2 \times 5)^2 = 2^2 \times 5^2 \times 7^2\)
This fits within M's available primes: \(2^2\) ✓, \(5^2\) ✓, \(7^2\) ✓

Process Skill: APPLY CONSTRAINTS - Using the limitation that factors cannot exceed the prime powers available in M

Final Answer

Choice A (600) is NOT a factor of M because it requires \(2^3 = 8\) factors of 2, but M only contains \(2^2 = 4\) factors of 2. All other choices can be formed using the prime factors available in \(M = 2^2 \times 3^2 \times 5^2 \times 7^2 = 44,100\).

The answer is A.

Common Faltering Points

Errors while devising the approach

1. Confusing LCM with GCD: Students might mistakenly think they need to find the Greatest Common Divisor (GCD) instead of the Least Common Multiple (LCM). This would lead them to look for common factors rather than building up to include all prime factors.

2. Misunderstanding what "NOT a factor" means: Students might get confused about whether they're looking for something that IS a factor or something that is NOT a factor, especially since four of the five choices will actually be factors.

3. Attempting to calculate LCM through listing multiples: Instead of using prime factorization, students might try to find the LCM by listing out multiples of each number (90, 180, 270... and 196, 392, 588... and 300, 600, 900...), which becomes unwieldy and error-prone with larger numbers.

Errors while executing the approach

1. Prime factorization errors: Students commonly make mistakes when breaking numbers into prime factors. For example, they might incorrectly factor 196 as \(2^3 \times 7\) instead of \(2^2 \times 7^2\), or miss that \(300 = 2^2 \times 3^1 \times 5^2\) and write it as \(2^3 \times 3 \times 5\).

2. Incorrectly applying the "highest power" rule: When constructing the LCM, students might take the lowest power instead of the highest power of each prime, or they might add the exponents instead of taking the maximum. For instance, seeing \(2^1\), \(2^2\), \(2^2\) they might use \(2^1\) or try \(2^5\) instead of \(2^2\).

3. Arithmetic computation errors: When calculating \(M = 2^2 \times 3^2 \times 5^2 \times 7^2 = 4 \times 9 \times 25 \times 49\), students often make multiplication errors, especially in the final steps like \(900 \times 49 = 44,100\).

Errors while selecting the answer

1. Stopping at the first choice that works: Students might check choice A (600), find that it doesn't divide 44,100 evenly, and immediately select it without verifying that the other choices actually DO work as factors of M.

2. Misreading the question requirement: Even after correctly computing everything, students might select a choice that IS a factor instead of the one that is NOT a factor, especially under time pressure or if they lose track of what the question is asking for.