Smithtown High School is holding a lottery to raise money.

Question

Smithtown High School is holding a lottery to raise money. The tickets are assigned numbers from 1 to 200. Tickets with numbers divisible by 2 win t-shirts, tickets with numbers divisible by 3 win gift certificates, and tickets with numbers divisible by 7 win movie passes. How many tickets win none of the prizes?

A. 6
B. 52
C. 58
D. 142
E. 194

A. 6
B. 52
C. 58
D. 142
E. 194

fluke · Accepted Answer

Total numbers divisible by:

3->200/3=66.66; Ignore the decimal part=66
2->200/2=100;
7->200/7=28.Something; Ignore decimal part=28
3*2->200/6=33
3*7->200/21=9
2*7->200/14=14
3*2*7->200/42=4

Thus, total numbers that can win prize= 66+100+28-33-9-14+4=142  A \cup B \cup C=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C) ]
Non-winnable numbers=200-142=58

Ans: "C"

subhashghosh · Answer

Obviously, the number of T-shirts not divisible by 2,3 or 7 will not win any prize

So # of T-shirts that are multiples of 2 = p

200 = 2 + (p-1)2 => p = 100

# of T-shirts that are multiples of 3 = q

198 = 3 + (q - 1)3

=> q = 66 + 1 = 67

# of T-shirts that are multiples of 7 = r

196 = 7 + (r-1)7

=> r = 28 + 1 = 29

Now we have to find out the common multiples of 2 and 3 = 6

198 = 6 + (n - 1)6

=> n = 33 + 1 = 34

and 2 and 7 = 14

196 = 14 + (n-1)14

=> n = 14 + 1 = 15

and 3 and 7 = 21

=> 189 = 21 + (n-1)21

=> n = 10

and 2*3*7 = 42

168 = 42 + (n-1)42

=> n = 5

So Total # that can win prize = 100 + 67 + 29 - 34 - 15 - 10 + 5

= 201 - 59 = 142

SO T-shirts that won't win prize = 200 - 142 = 58

Ansswer - C

puneetj · Answer

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

puneetj · Answer

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

PiyushK · Answer

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

KarishmaB · Answer

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)

ScottTargetTestPrep · Answer

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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Smithtown High School is holding a lottery to raise money.

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