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# How many integers between 100 and 150 inclusive, can be

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How many integers between 100 and 150 inclusive, can be [#permalink]

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25 Oct 2005, 21:59
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many integers between 100 and 150 inclusive, can be evenly divided by neither 3 nor 5?

a) 33
b) 28
c) 27
d) 26
e) 24

Last edited by sudhagar on 26 Oct 2005, 17:06, edited 2 times in total.
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26 Oct 2005, 09:09
arsaboo wrote:
Are you sur ethe options are correct

Yes, the options are correct. I have edited for couple of typo error in the question part though.
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26 Oct 2005, 09:48
sudhagar wrote:
How many integers between 10 and 150 inclusive, can be evenly divided by neither 3 nor 5?

a) 33
b) 28
c) 27
d) 26
e) 24

All div by 3 -
12, 15, ... 150 => A=47 be the total number of numbers divisible by 3

All div by 5 -
10, 15, ... 150 => B=29 be the total number of numbers divisible by 5

Numbers divisible by both 3 and 5 in common -
15, 30, 45, ... 150 => A n B = 10

So total numbers divisible by 5 and 3 = A + B - (A n B) = 66

All others -> 141 - 66 = 75 but this is not in the list of choices!!

IMO, btw 10 and 150, there is definitely more than what is in the given answer choices!

I just tried out, if it is 100 and 150, then, by the same method, I get -
51 - (17+11-3) = 26

Last edited by gsr on 26 Oct 2005, 17:47, edited 1 time in total.
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26 Oct 2005, 10:16
heheh... really dont know wether my thinking is correct..
I got to 66 as well using two strategies... the first one is the same as GSR. The second is:

prime factorization of all 10 is 2*5. The next integer divisable by 5 would be 15... thus 3*5... next one 4*5... etc till 30*5 = 150.
The number of integers divisble by 5 from 10 to 100 inclusive then is:
30 - 2 + 1 = 29

Same for 3.. so the first number after 11 to be divisable by 3 is 12. that is 2*2*3 or 4*3... next one is 5*3 ... etc till 50*3 = 150.
The number of integers divisible by 3 from 10 to 100 inclusive then is:
50 - 4 + 1 = 47

Now we add these two numbers 29 + 47 = 76. BUT, we have to substract 1 because 3*5 and 5*3 has been counted in both. so 75.

Total integers between 10 and 150 inclusive is 150 - 10 + 1 = 141
So total integers NOT divisible by 5 nor 3 is 141 - 75 = 66

NOW , since the question is asking for evenly divisable integers, and the prime factorization demonstrated that the integers are "consecutive", we can assume that every second number will result an even integer. Thus 66/2 = 33 -> A

Could that be what they are asking for?
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26 Oct 2005, 17:08
I am extremely sorry guys. I didn't realize that I have typed 10 instead of 100, even during my first edit

I really apologize for straining your brains unnecessarily. I have edited the question and it should be correct now. I would post the OA by end of today.
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26 Oct 2005, 18:01
150-102 = 48, 48/3 = 16 +1 (to count both 102 and 150) = 17

50/5 = 10 +1 (to include 100 and 150) = 11

105, 120, 135, 150 are shared

28 - 4 = 24

E

is this not right?
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26 Oct 2005, 18:05
Nevermind, its 26
Its asking for how many numbers are NOT divisible by either..
so 50 - 24 = 26

D
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26 Oct 2005, 19:17
OA is D.

Gsr has found it irrespective of me posting an typo question. I apologize again, as this would have wasted some of your precious prep time .
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27 Oct 2005, 14:29
Numbers between 100 and 150 inclusive = 51 (includes 100 AND 150, or does the inclusive only apply to the 150??)
Numbers divisible by 5 = (150 - 100)/5 + 1 = 11
Numbers divisible by 3 = (150-102)/3 + 1 = 17
Numbers divisible by 3&5 = 105, 120, 135, 150 = 4

51 - (11+17-4) = 27

Answer choice C, am I missing something here?
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27 Oct 2005, 16:18
Yes i didn't realize that....your explanation makes sense! i change my answer to 27.
Sudhagar, is OA really D? If so, can you please explain why?

Thanks
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28 Oct 2005, 07:33
lol, I saw this on the forum yesterday and posted a note saying that I thought the answer should be C (27). I got home and cracked open my Princeton Review book for some practice and there was the same problem. I worked it out again and got 27. OA was indeed C (27) in the Princeton Review book.
28 Oct 2005, 07:33
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