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by shashank.ism » Mon Feb 08, 2010 7:00 am
How many even integers n, where 100 < n < 200, are divisible neither by seven nor by nine?

a) 40
b) 37
c) 39
d) 38
e) 41
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by harsh.champ » Mon Feb 08, 2010 7:36 am
shashank.ism wrote:How many even integers n, where 100 < n < 200, are divisible neither by seven nor by nine?

a) 40
b) 37
c) 39
d) 38
e) 41
Since 7 and 9 are primes,we have to keep in mind not to double count divisors of 63(126 and 189)

Hence,for 7 it starts with 105 and ends at 196 . [196-105 = 91 hence,13 divisors]
For 9 , we start at 108 and end at 198 [ 198 - 108 = 90 ,hence 10 divisors]

[spoiler]Ans. is 49(no. of even integers) - (13 + 10 - 2) = 38 D.[/spoiler]
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by ajith » Mon Feb 08, 2010 10:43 am
shashank.ism wrote:How many even integers n, where 100 < n < 200, are divisible neither by seven nor by nine?

a) 40
b) 37
c) 39
d) 38
e) 41
100 < n < 200,
No of Even integers divisible by 9 = 198-108/18 = 5
No of even integers divisible by 7 = (196-112)/14 = 6
No of even integers divisible by 9 and 7 = 1 [126]

Total no of even integers divisible by either 9 or 7 = 5+6-1 =10
Total no of even integers divisible by neither 9 nor 7 = 48 -10 = 38
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by pkvermaastra » Wed Dec 18, 2013 11:37 pm
Why have u done 196 -112 and not 196-105 .I have confusion in this .
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by theCodeToGMAT » Thu Dec 19, 2013 12:44 am
Can some one point out my mistake, i tried the question and got a different answer from the posts above.

Thanks!

Total Even numbers = 49

Even Multiple of "7" = 112 --> 196 ==> terms = 7
Even Multiple of "9" = 108 --> 198 ==> terms = 6
Even Multiples of 7x9 = 63 => 63X2 = 126 .. So, 1

So, Total = 49 - (7+6 - 1) = 37
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by GMATinsight » Sat Nov 05, 2016 9:12 pm
shashank.ism wrote:How many even integers n, where 100 < n < 200, are divisible neither by seven nor by nine?

a) 40
b) 37
c) 39
d) 38
e) 41
Answer: option C
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by Jay@ManhattanReview » Wed Dec 07, 2016 12:49 am
theCodeToGMAT wrote:Can some one point out my mistake, i tried the question and got a different answer from the posts above.

Thanks!

Total Even numbers = 49

Even Multiple of "7" = 112 --> 196 ==> terms = 7 numbers
Even Multiple of "9" = 108 --> 198 ==> terms = 6 numbers
Even Multiples of 7x9 = 63 => 63X2 = 126 .. So, 1 number

So, Total = 49 - (7+6 - 1) = 37
You are correct Rahul,

100 < n < 200 => There are 99 numbers: 49 even and 50 odd.

No of Even integers divisible by 9 = 198-108/(2*9) = 6
No of even integers divisible by 7 = (196-112)/(2*7) = 7
No of even integers divisible by 9 and 7 = 1 [2*9*7=126]

Total no of even integers divisible by either 9 or 7 = 6+7-1 = 12 numbers
Total no of even integers divisible by neither 9 nor 7 = 49 -12 = 37 numbers

Hope this helps!

-Jay

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by [email protected] » Wed Dec 07, 2016 10:20 am
Hi All,

In these types of questions, there's nothing wrong with approaching the prompt with some 'brute force' math. Since the answer choices to this question are consecutive integers, if you 'missed' any of the possibilities because you were trying to come up with an 'elegant' approach (or if you did math 'in your head'), then you'd end up choosing one of the wrong answers (and you wouldn't even know it). There are a lot of details to deal with here, so you have to make sure to take the proper notes.

We're asked for the number of EVEN INTEGERS between 100 and 200 (NOT inclusive) that are NOT divisible by 7 nor by 9.

There are 100 numbers from 101 to 200; half are even (50) and half are odd (50). However, we're not supposed to include 200 in the list, so we have 49 total even numbers to think about. Rather than list all of the numbers that we're looking for, I'm going to list the numbers that we're NOT looking for (the ones that ARE divisible by 7 or 9 or both).

EVEN Multiples of 7 in this range:
112
126
140
154
168
182
196

EVEN Multiples of 9 in this range:
108
126
144
162
180
198

If you place these 2 lists side-by-side, you'll notice that 1 number "overlaps": 126 - but you're not allowed to count that number 'twice.' Thus there are 7 + 6 - 1 = 12 numbers that don't fit what we're looking for.

49 - 12 = 37

Final Answer: B

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by Matt@VeritasPrep » Thu Dec 08, 2016 7:46 pm
One thing to be careful of here is how close together the answers are. It's very easy to miss one number somewhere and to come out with the wrong result, so be strategic on test day and either commit to taking the time to work through the cases (if you don't know the general approach) or just decide this one isn't worth the time and risk and make an educated guess.
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by Jay@ManhattanReview » Thu Dec 08, 2016 9:50 pm
I do agree with both of you. Yes, since the options differ by one number, and the range 100 < n < 200 is not too wide, it is wise to spend 20 second more and be assured.

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