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M07-14

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Official Solution:

What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 or by 35?

A. 884
B. 890
C. 892
D. 910
E. 945


# of multiples of 11 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(990-11)}{11}+1=90\);

# of multiples of 35 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(980-35)}{35}+1=28\);

# of multiples of both 11 and 35 is 2 (\(11*35=385\) and 770);

So, # of multiples of 11 or 35 in the given range is \(90+28-2=116\). Thus numbers which are not divisible by either of them is \(1000-116=884\).


Answer: A
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Re: M07-14 [#permalink]

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New post 20 Sep 2014, 17:20
Hi Bunuel,
\
Might be a very basic question.I took lot of time to solve this mainly i took time to find the highest mulitple of 35 < 1000.
is there a fast way to do this? How do we quickly find out 980 is the highest multiple of 35 < 1000,

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shankar245 wrote:
Hi Bunuel,
\
Might be a very basic question.I took lot of time to solve this mainly i took time to find the highest mulitple of 35 < 1000.
is there a fast way to do this? How do we quickly find out 980 is the highest multiple of 35 < 1000,


I did this way: 35*30 = 1,050 (easy to spot this, I suppose). 1,050 - 70 = 980.

Hope it helps.
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Re: M07-14 [#permalink]

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New post 20 Sep 2014, 18:00
Okay got it, so a sort of approximation.You replied pretty fast too!

Much apprecited

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Hi all, Bunuel,

What about the following solution?:
1000/11=90,xx so we have 90 multiples of 11 in 1000
1000/35=28,xx so we have 28 multiples of 35 in 1000
Now, because we counted a few multiples of both 11 and 35 we need to subtract them:
1000/(35*11)=2,xx so we need to subtract 2 multiples of both 11 and 35.

Altogether:
90+28-2=116 so we have 1000-116=884 integers that are NOT divisible by 11 or 35.

My true question is this: can we use this technique outside of numbers from [1 to X], with intervals: for example: "what is the number of integers between 754 and 1000 that are not divisible by 11 or 35?". You can't directly use the same technique and calculate 246/11, 246/35 etc. but I wonder if there is a way of adapting the above technique. Maybe by finding only one end of the range (say 990 for the number 11) instead of two (990 and 759 for the number 11)?

I hope this is not too confusing, but I feel there is a way of doing these exercises faster...

Thanks anyways!!

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Re: M07-14 [#permalink]

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New post 26 Mar 2015, 02:23
Bunuel wrote:
Official Solution:


# of multiples of 11 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(990-11)}{11}+1=90\);

# of multiples of 35 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(980-35)}{35}+1=28\);

# of multiples of both 11 and 35 is 2 (\(11*35=385\) and 770);

So, # of multiples of 11 or 35 in the given range is \(90+28-2=116\). Thus numbers which are not divisible by either of them is \(1000-116=884\).


Answer: A


Great approach and easy to remember!

But wouldn't it be less cumbersome if we simplify it to (Last Multiple / Multiple)? Or are their any cases where this does not work?

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New post 26 Mar 2015, 04:03
germangoesgmat wrote:
Bunuel wrote:
Official Solution:


# of multiples of 11 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(990-11)}{11}+1=90\);

# of multiples of 35 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(980-35)}{35}+1=28\);

# of multiples of both 11 and 35 is 2 (\(11*35=385\) and 770);

So, # of multiples of 11 or 35 in the given range is \(90+28-2=116\). Thus numbers which are not divisible by either of them is \(1000-116=884\).


Answer: A


Great approach and easy to remember!

But wouldn't it be less cumbersome if we simplify it to (Last Multiple / Multiple)? Or are their any cases where this does not work?


The formula used in the solution is correct for all cases. So, I'd advice to stick with it.
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New post 19 Apr 2015, 14:50
Bunuel wrote:
What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 or by 35?

A. 884
B. 890
C. 892
D. 910
E. 945



Hi, anyone can tell me, the fastest way to find the last numbers each divisible by 35 and 11?

Thank you very much.

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Re: M07-14 [#permalink]

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New post 05 Jul 2015, 11:43
chukychuk wrote:
Hi all, Bunuel,

What about the following solution?:
1000/11=90,xx so we have 90 multiples of 11 in 1000
1000/35=28,xx so we have 28 multiples of 35 in 1000
Now, because we counted a few multiples of both 11 and 35 we need to subtract them:
1000/(35*11)=2,xx so we need to subtract 2 multiples of both 11 and 35.

Altogether:
90+28-2=116 so we have 1000-116=884 integers that are NOT divisible by 11 or 35.

My true question is this: can we use this technique outside of numbers from [1 to X], with intervals: for example: "what is the number of integers between 754 and 1000 that are not divisible by 11 or 35?". You can't directly use the same technique and calculate 246/11, 246/35 etc. but I wonder if there is a way of adapting the above technique. Maybe by finding only one end of the range (say 990 for the number 11) instead of two (990 and 759 for the number 11)?

I hope this is not too confusing, but I feel there is a way of doing these exercises faster...

Thanks anyways!!


Bumping up for I'd like to highlight how benefitial it is to analyze the answer choices, or at least know what they are. Had the answer choices been 882, 884, and 885, the formula would have been the best method. Otherwise, estimation worked fine. That's what I did.

Thanks,

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Re: M07-14 [#permalink]

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New post 26 Nov 2015, 14:01
i tried to plug in numbers to see how many options I have
last multiple of 11 is 990, thus there must be 90 multiples of 11 only. D and E - eliminated.
now, 35: maximum multiple of 35 is 980. and 980 is the 28th multiple of 35.
together we have ~118 multiples, but since in these multiples are included multiples of both 35 and 11, then there should be less than 118 but more than 110 numbers that are divisible by 35 or 11. B and C eliminated, and remained with A.

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Re: M07-14 [#permalink]

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What if there were more than 2 common multiples of 11 and 35? would there be an easier way of finding the number of common multiples?

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Re: M07-14 [#permalink]

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New post 12 Jul 2016, 13:02
intogamer wrote:
What if there were more than 2 common multiples of 11 and 35? would there be an easier way of finding the number of common multiples?


We can find first common multiple by taking LCM of two numbers under consideration (here 11 and 35) and later find the multiples of LCM within given range to find the required number.

If the LCM is a smaller number , then we can use the process (that has been used to find multiples of 11 and 35 ) again to find number of multiples of LCM.

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Re: M07-14 [#permalink]

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New post 23 Oct 2016, 15:57
Hi! If the question asked : What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 AND by 35? then the only difference in arriving at the answer would be to find out the multiples of 11 & 35 and subtract that from 1,000, correct?..Thank you in advance.

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New post 23 Oct 2016, 22:53
Conquergmat5 wrote:
Hi! If the question asked : What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 AND by 35? then the only difference in arriving at the answer would be to find out the multiples of 11 & 35 and subtract that from 1,000, correct?..Thank you in advance.


Yes. In this case the answer would be 1000 - 2 = 998.
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New post 15 Nov 2016, 01:24
Hi,
I got the answer but just wanted to clarify something.
why am I not being able to apply De morgan's law in this question ?
If A= divisible by 11 and B= divisible by 35

Then the question asks A'union B'
which by demargan's law = Complement of ( A intersection B )= 1000-2

Please help ?

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New post 22 Nov 2016, 16:46
wouldn't it be 1001-116? because it is inclusive

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M07-14 [#permalink]

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New post 29 Nov 2016, 10:41
Bunuel wrote:
Official Solution:

What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 or by 35?

A. 884
B. 890
C. 892
D. 910
E. 945


# of multiples of 11 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(990-11)}{11}+1=90\);

# of multiples of 35 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(980-35)}{35}+1=28\);

# of multiples of both 11 and 35 is 2 (\(11*35=385\) and 770);

So, # of multiples of 11 or 35 in the given range is \(90+28-2=116\). Thus numbers which are not divisible by either of them is \(1000-116=884\).


Answer: A



Hi Bunuel,
I had come across a similar question "How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?". In here we find # of multiples of 13 and 12, and eliminate multiples of both i.e 12*13 = 156. I understand since 156 has been counted both in # of 13's and # of 12's we delete 2 multiples from the answer.

But in the above problem both 11*35 = 385 and 2*11*35 = 770 must be counted twice in each of multiples of 11 and 13. So my doubt is why didn't we delete 4 multiples (385 and 770 are counted both in 11 multiples and 13 multiples) and instead did only 2 multiples (385 and 770).

Please help me understand!!!
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Re: M07-14 [#permalink]

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New post 10 Dec 2016, 04:28
PerseveranceWins wrote:
intogamer wrote:
What if there were more than 2 common multiples of 11 and 35? would there be an easier way of finding the number of common multiples?


We can find first common multiple by taking LCM of two numbers under consideration (here 11 and 35) and later find the multiples of LCM within given range to find the required number.

If the LCM is a smaller number , then we can use the process (that has been used to find multiples of 11 and 35 ) again to find number of multiples of LCM.


Easier approach?
Observation! And intuition. I guess!
You already calculated how many multiples of 11 are present between 1 and 1000- it's 90.
So how many times would you come across a 35 (or a multiple of 35) when you run the multiplication table of 11 up to 90?
11 x 35. one.
11 x 70. two.
11 x 105. nope.
That's two.
Or if you do it the other way...
You already calculated how many multiples of 35 are present between 1 and 1000- it's 28.
How many 11s in 28? 11 and 22. That's two again.
It would take about the same time as the LCM approach.
LCM of 11 and 35? 385.
How many 385s in 1000? 2.
I guess "EASY" depends on the approach one's used to.
Nevertheless, our quest for an easier approach has to stop somewhere!
At the same time, I could, just for fun, dare anyone to come up with an EASIER approach :P

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Re: M07-14 [#permalink]

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New post 10 Dec 2016, 04:44
Sash143 wrote:
Bunuel wrote:
Official Solution:

What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 or by 35?

A. 884
B. 890
C. 892
D. 910
E. 945


# of multiples of 11 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(990-11)}{11}+1=90\);

# of multiples of 35 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(980-35)}{35}+1=28\);

# of multiples of both 11 and 35 is 2 (\(11*35=385\) and 770);

So, # of multiples of 11 or 35 in the given range is \(90+28-2=116\). Thus numbers which are not divisible by either of them is \(1000-116=884\).


Answer: A



Hi Bunuel,
I had come across a similar question "How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?". In here we find # of multiples of 13 and 12, and eliminate multiples of both i.e 12*13 = 156. I understand since 156 has been counted both in # of 13's and # of 12's we delete 2 multiples from the answer.

But in the above problem both 11*35 = 385 and 2*11*35 = 770 must be counted twice in each of multiples of 11 and 13. So my doubt is why didn't we delete 4 multiples (385 and 770 are counted both in 11 multiples and 13 multiples) and instead did only 2 multiples (385 and 770).

Please help me understand!!!


There is one critical aspect of your question ("How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?") that is different from the question discussed in this forum.

If you can't spot it yet, its "but not both".

Your question asks us not to consider multiples of both 12 and 13.

The question discussed in this forum has no such restrictions. We are merely removing repetitions of multiples of both 11 and 35.

Before I sign off, here's the cliched "Hope it helps!"

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