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How many positive integers less than \(\mathrm{100}\) are neither multiples of \(\mathrm{2}\) or \(\mathrm{3}\)?
Let's break down what we're looking for in everyday language. We want to count positive integers from 1 to 99 that are NOT multiples of 2 AND NOT multiples of 3.
In other words, we're looking for numbers that can't be divided evenly by 2 (so they're odd numbers) AND can't be divided evenly by 3 either.
For example, the number 5 qualifies because \(5 \div 2 = 2.5\) (not a whole number) and \(5 \div 3 = 1.67...\) (not a whole number).
The number 6 does NOT qualify because \(6 \div 2 = 3\) (whole number), so 6 is a multiple of 2.
Process Skill: TRANSLATE - Converting the phrase "neither multiples of 2 or 3" into mathematical understanding
We start by counting all positive integers less than 100.
Since we want positive integers less than 100, we're looking at the numbers: 1, 2, 3, 4, ..., 99.
That gives us a total of 99 numbers to start with.
Now we need to remove the numbers we DON'T want. But we have to be careful not to remove the same number twice!
Let's think about this step by step:
Step 3a: Count multiples of 2 from 1 to 99
Multiples of 2 are: 2, 4, 6, 8, ..., 98
To count these: \(98 \div 2 = 49\) multiples of 2
Step 3b: Count multiples of 3 from 1 to 99
Multiples of 3 are: 3, 6, 9, 12, ..., 99
To count these: \(99 \div 3 = 33\) multiples of 3
Step 3c: Count multiples of BOTH 2 and 3 (multiples of 6)
Here's the tricky part! Some numbers are multiples of both 2 and 3. For example, 6 is divisible by both 2 and 3.
A number that's a multiple of both 2 and 3 must be a multiple of 6.
Multiples of 6 are: 6, 12, 18, 24, ..., 96
To count these: \(96 \div 6 = 16\) multiples of 6
Process Skill: CONSIDER ALL CASES - Recognizing that some numbers are counted in both categories
Now we apply the inclusion-exclusion principle in plain English:
Numbers we want = Total numbers - Multiples of 2 - Multiples of 3 + Multiples of both 2 and 3
Why do we ADD BACK the multiples of both 2 and 3? Because when we subtracted multiples of 2 AND subtracted multiples of 3, we accidentally subtracted the numbers that are multiples of both twice! So we need to add them back once.
Let's calculate:
• Total positive integers less than 100: 99
• Subtract multiples of 2: \(99 - 49 = 50\)
• Subtract multiples of 3: \(50 - 33 = 17\)
• Add back multiples of both (multiples of 6): \(17 + 16 = 33\)
Therefore, there are 33 positive integers less than 100 that are neither multiples of 2 nor multiples of 3.
The answer is 33, which corresponds to choice D. 33.
To verify this makes sense: we started with 99 numbers, removed about half (the even numbers), removed about one-third of what remained (multiples of 3), but added back some we double-counted. Getting 33 as our final count is reasonable.
1. Misinterpreting "neither...nor" as "not both"
Students often confuse "neither multiples of 2 nor 3" with "not multiples of both 2 and 3". The question asks for numbers that are NOT multiples of 2 AND NOT multiples of 3 (like 5, 7, 11). However, students might incorrectly think it's asking for numbers that aren't multiples of both 2 and 3 simultaneously, which would include numbers like 4 (multiple of 2 only) or 9 (multiple of 3 only).
2. Not recognizing the need for inclusion-exclusion principle
Some students might try to simply subtract multiples of 2 and multiples of 3 from the total without realizing they're double-counting numbers that are multiples of both 2 and 3 (i.e., multiples of 6). This leads them to think: \(99 - 49 - 33 = 17\), missing the critical step of adding back the overlap.
3. Attempting to list numbers individually instead of using systematic counting
Students might try to manually list all numbers that aren't multiples of 2 or 3 (like 1, 5, 7, 11, 13, 17, 19, 23...), which is time-consuming and prone to errors, rather than using the more efficient inclusion-exclusion approach.
1. Incorrectly calculating the number of multiples within the range
When finding multiples of 2 less than 100, students might incorrectly calculate \(100 \div 2 = 50\) instead of recognizing that since we want numbers less than 100, the largest multiple of 2 is 98, giving us \(98 \div 2 = 49\) multiples. Similar errors can occur with multiples of 3 and 6.
2. Arithmetic errors in the inclusion-exclusion calculation
Even with the correct approach, students may make simple arithmetic mistakes in the final calculation: \(99 - 49 - 33 + 16\). Common errors include forgetting to add back the 16, or making mistakes in the subtraction steps (like getting \(99 - 49 = 51\) instead of 50).
3. Using wrong LCM for overlap calculation
Some students might incorrectly calculate the overlap of multiples of 2 and 3. Instead of recognizing that numbers divisible by both 2 and 3 must be divisible by \(\mathrm{LCM}(2,3) = 6\), they might use \(2 \times 3 = 6\) without understanding why, or worse, try to count the overlap differently.
No likely faltering points - once students correctly execute the inclusion-exclusion calculation and arrive at 33, the answer selection is straightforward as it directly matches option D.