Smithtown High School is holding a lottery to raise money.

I selected 6 and quickly realized my mistake of over-counting when I did that.

fluke's approach is good and that is what I used although I over-counted by not finding out the count of numbers divisible by 6,14,21 and 42

I have another method to solve this:
select nearest multiple of 2,3,7 to 200. i.e 210.

\(=210 (1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{7})\)
\(=60\)

Hand pick numbers not divisible by 2,3,or 7 between 201 and 210.
205 ans 209 i.e. only 2 numbers.

Therefore total numbers between 0-200
(C) 60-2 = 58

You can also do this question by quick approximation.

1, 2, 3, ... 200 - here there are 100 odd numbers and 100 even. Ignore the even numbers since they are all divisible by 2. We are left with 100 odd numbers.

1, 3, 5, 7, 9, 11, 13, 15, 17, 19... 199

Every group of 3 will have a multiple of 3. We will be able to make 33 groups of 3 numbers each so we will have 33 multiples of 3 here. Remove these multiples of 3. We are left with 67 numbers.

The multiples of 7 in 1 to 200 are: 7, 14, 21, ... 196 (28 multiples)
Half of them will be even so ignore them. You are left with 14. About a third of them will be multiples of 3 so you will have about 9 multiples of 7 which will still be in the leftover 67 numbers. Remove them to get 67 - 9 = 58 numbers.

Answer (C)

Let’s first determine the number of multiples of 2, 3, and 7.

Number of multiples of 2:

(200 - 2)/2 + 1 = 100

Number of multiples of 3:

(198 - 3)/3 + 1 = 66

Number of multiples of 7:

(196 - 7)/7 + 1 = 28

Some of the above outcomes double-count certain numbers. For example, all multiples of 6 were double-counted because they are multiples of both 2 and of 3. Similarly, multiples of 14 were double-counted as multiples of both 2 and 7. Multiples of 21 were double-counted as multiples of both 3 and 7. So we will have to subtract those double-counted numbers from the total.
Number of multiples of 2 and 3, i.e., 6:

(198 - 6)/6 + 1 = 33

Number of multiples of 2 and 7, i.e., 14:

(196 - 14)/14 + 1 = 14

Number of multiples of 3 and 7, i.e., 21:

(189 - 21)/21 + 1 = 9

Finally, we need to determine of the number of multiples of 2 and 3 and 7, i.e., 42.
(168 - 42)/42 + 1 = 4

We subtracted these 4 numbers twice during the correction for double-counting, so we will have to add them back in.

So the total number of possible winners are 100 + 66 + 28 - 33 - 14 - 9 + 4 = 142. Therefore, 200 - 142 = 58 tickets win none of the prizes.

Alternate Solution:

Since exactly half of 200 tickets are divisible by 2, 200 - 100 = 100 tickets do not win a t-shirt.

There are (198 - 3)/3 + 1 = 66 tickets that win gift certificates. Of these 66 tickets, exactly half are divisible by 2 and the remaining half do not win t-shirts. Of the 100 tickets that do not win a t-shirt, 33 win gift certificates; therefore 100 - 33 = 67 tickets win neither gift certificates nor t-shirts.

Among the 67 tickets that win neither gift certificates nor t-shirts, there are some that win movie passes and these tickets are multiples of 7 that are neither even nor divisible by 3. These tickets are 7 x 1 = 7, 7 x 5 = 35, 7 x 7 = 49, 7 x 11 = 77, 7 x 13 = 91, 7 x 17 = 119, 7 x 19 = 133, 7 x 23 = 161 and 7 x 25 = 175. If we subtract these tickets from the 67 tickets that win neither gift certificates nor t-shirts, we will obtain the number of tickets that win nothing, i.e. 67 - 9 = 58.

Answer: C

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