Sets: Union of Sets
If $$W$$ is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3, multiples of 2, or multiples of both, then $$W$$ contains how many numbers?
Incorrect.
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Incorrect.
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Incorrect.
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Incorrect.
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Finally, subtract the eight multiples of 6 from the sum of the multiples of 2 and 3:
>$$\text{Total}= 17 + 25 - 8$$
>>$$= 42 - 8$$
>>$$= 34$$
Therefore, there are a total of 34 multiples of 2, 3, or 6.
Correct.
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Multiples of 3
Step 1. Subtract the extreme multiples of 3 within the range (The greatest is 99, the smallest is 51):
>$$99 - 51 = 48$$
Step 2. Divide by 3:
>$$\frac{48}{3} = 16$$
Step 3. Add 1:
>$$16 + 1 = 17$$.
So there are 17 multiples of 3 within the range: examples are 51, 54, 57, 60, etc.
Multiples of 2
Step 1. Subtract the extreme multiples of 2 within the range (The greatest is 98, the smallest is 50):
>$$98 - 50 = 48$$
Step 2. Divide by 2:
>$$\frac{48}{2} = 24$$.
Step 3. Add 1:
>$$24 + 1 = 25$$.
So there are 25 multiples of 2 within the range: examples are 50, 52, 54, 56, 58, 60 etc.
Add the 17 multiples of 3 and the 25 multiples of 2: $$25+17=42$$. However, by adding the multiples of 2 and the multiples of 3, we are effectively counting several numbers twice: for example, 54 and 60 are parts of both the lists above. So we can't just take $$25+17=42$$.
Find the number of multiples of 6 (which are the double counted, as 6 is divisible by both 2 and 3), and subtract it from 42.
Multiples of 6
Step 1. Subtract the extreme multiples of 6 within the range (The greatest is 96, the smallest is 54):
>$$96 - 54 = 42$$
Step 2. Divide by 6:
>$$\frac{42}{6} = 7$$
Step 3. Add 1:
>$$7 + 1 = 8$$.
So there are eight multiples of 6 within the range: we counted eight numbers twice.
26
32
33
34
35
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