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How many positive integers less than 5,000 are evenly divisi [#permalink]
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12 Jul 2011, 22:20
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How many positive integers less than 5,000 are evenly divisible by neither 15 nor 21? A. 4,514 B. 4,475 C. 4,521 D. 4,428 E. 4,349
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Re: Integers less than 5,000 are not divisible by 15 or 21 ? [#permalink]
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12 Jul 2011, 23:00
integers less than 5000 divisible by 15 5000/15 =333.something , so 333
integers less than 5000 divisible by 21 5000/21 = 238.## , so 238
we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 47 =524
now subtract that from 4999. 4999 524 = 4475 answer B.



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Re: Integers less than 5,000 are not divisible by 15 or 21 ? [#permalink]
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04 Apr 2014, 06:10
Aj85 wrote: integers less than 5000 divisible by 15 5000/15 =333.something , so 333
integers less than 5000 divisible by 21 5000/21 = 238.## , so 238
we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 47 =524
now subtract that from 4999. 4999 524 = 4475 answer B. Question here, why don't you multiply 47*2 I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this? Thanks Cheers J



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Re: Integers less than 5,000 are not divisible by 15 or 21 ? [#permalink]
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04 Apr 2014, 08:12
jlgdr wrote: Aj85 wrote: integers less than 5000 divisible by 15 5000/15 =333.something , so 333
integers less than 5000 divisible by 21 5000/21 = 238.## , so 238
we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 47 =524
now subtract that from 4999. 4999 524 = 4475 answer B. Question here, why don't you multiply 47*2 I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this? Thanks Cheers J 47 numbers are counted TWICE, we need them to be counted ONCE, so that's why are subtracting 47 once. Similar questions to practice: whatisthenumberofintegersfrom1to1000inclusive126153.htmlwhatisthetotalnumberofpositiveintegersthatareless128104.htmlhowmanyintegersfrom1to200inclusivearedivisib109333.htmlhowmanyevenintegersnwhere100n200aredivisib103779.htmlHope it helps.
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Re: Integers less than 5,000 are not divisible by 15 or 21 ? [#permalink]
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04 Apr 2014, 21:57
Bunuel wrote: jlgdr wrote: Aj85 wrote: integers less than 5000 divisible by 15 5000/15 =333.something , so 333
integers less than 5000 divisible by 21 5000/21 = 238.## , so 238
we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 47 =524
now subtract that from 4999. 4999 524 = 4475 answer B. Question here, why don't you multiply 47*2 I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this? Thanks Cheers J 47 numbers are counted TWICE, we need them to be counted ONCE, so that's why are subtracting 47 once. Similar questions to practice: whatisthenumberofintegersfrom1to1000inclusive126153.htmlwhatisthetotalnumberofpositiveintegersthatareless128104.htmlhowmanyintegersfrom1to200inclusivearedivisib109333.htmlhowmanyevenintegersnwhere100n200aredivisib103779.htmlHope it helps. Hi Bunuel, Is there a shorter way around this question?This approach took me 5 minutes!



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Re: How many positive integers less than 5,000 are evenly divisi [#permalink]
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04 Apr 2014, 21:59
Option B.(Took 5 minutes to solve with the following approach!) All nos. from \(1\) to \(5000\) since we're given POSITIVE integers.\(=4999\) Multiples of \(15\) in this range\(=4995/15=333\) Multiples of \(21=4998/21=238\) Multiples of both \(15\) and \(21\)=multiples of \(105=47\) Now \(answer=4999333238+47=4475\)



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Re: How many positive integers less than 5,000 are evenly divisi [#permalink]
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16 Apr 2014, 23:53
AKG1593 wrote: Option B.(Took 5 minutes to solve with the following approach!) All nos. from \(1\) to \(5000\) since we're given POSITIVE integers.\(=4999\) Multiples of \(15\) in this range\(=4995/15=333\) Multiples of \(21=4998/21=238\) Multiples of both \(15\) and \(21\)=multiples of \(105=47\) Now \(answer=4999333238+47=4475\) This seems to be one of those killer problems that aren't necessarily conceptually difficult, but unless you have (what for me is) very highpowered arithmetic (long division) skills you're going to be in a time crunch here and in danger of compromising your quant section on test day. Practice, practice, practice those fundamentals! One trick, because the ACs are so close together, is that you want to be careful to account for the fact that "less than 5000" accounts for only 4,999 integers, not including 5000. (Turns out on this problem it's not of such great concern, but on many other problems, it definitely is!) For this question, huge waste of about 2030 seconds was the initial rephrase: "How many integers.. divisible by NEITHER 15 or 21." Don't know why but that tripped me up. Do they mean all of the numbers not divisible by 105 (Neither 15 AND 21, I saw the language trick there and ruled this out but not after wasting some time on math). What were are therefore looking for is Opposite: 4999 (minus) all the numbers that 15 or 21 go into (minus) overlap between multiples of 15 and 21. Let the long division begin!



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Re: How many positive integers less than 5,000 are evenly divisi [#permalink]
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21 Nov 2014, 19:11
hey Bunuel, What does evenly divisible number means? I thought the quotient for such division should be multiple of 2. Thanks, Amit Bunuel wrote: jlgdr wrote: Aj85 wrote: integers less than 5000 divisible by 15 5000/15 =333.something , so 333
integers less than 5000 divisible by 21 5000/21 = 238.## , so 238
we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 47 =524
now subtract that from 4999. 4999 524 = 4475 answer B. Question here, why don't you multiply 47*2 I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this? Thanks Cheers J 47 numbers are counted TWICE, we need them to be counted ONCE, so that's why are subtracting 47 once. Similar questions to practice: whatisthenumberofintegersfrom1to1000inclusive126153.htmlwhatisthetotalnumberofpositiveintegersthatareless128104.htmlhowmanyintegersfrom1to200inclusivearedivisib109333.htmlhowmanyevenintegersnwhere100n200aredivisib103779.htmlHope it helps.



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Re: How many positive integers less than 5,000 are evenly divisi [#permalink]
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22 Nov 2014, 06:36



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Re: How many positive integers less than 5,000 are evenly divisi [#permalink]
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21 Feb 2017, 13:49
Bunuel wrote: jlgdr wrote: Aj85 wrote: integers less than 5000 divisible by 15 5000/15 =333.something , so 333
integers less than 5000 divisible by 21 5000/21 = 238.## , so 238
we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 47 =524
now subtract that from 4999. 4999 524 = 4475 answer B. Question here, why don't you multiply 47*2 I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this? Thanks Cheers J 47 numbers are counted TWICE, we need them to be counted ONCE, so that's why are subtracting 47 once. Similar questions to practice: http://gmatclub.com/forum/whatisthen ... 26153.htmlhttp://gmatclub.com/forum/whatisthet ... 28104.htmlhttp://gmatclub.com/forum/howmanyinte ... 09333.htmlhttp://gmatclub.com/forum/howmanyeven ... 03779.htmlHope it helps. In questions like these do we always have to take an LCM ?



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Re: How many positive integers less than 5,000 are evenly divisi [#permalink]
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03 Apr 2018, 00:38
I’m sorry for my ignorance, but why do we take the LCM of 15 and 21 to remove double count? I multiplied 15*21= 315 and divided 5000/315= 15 times to get the overlap. I know it is wrong now, but can someone explain it to me please?
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Re: How many positive integers less than 5,000 are evenly divisi [#permalink]
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03 Apr 2018, 03:41
Juliaz wrote: I’m sorry for my ignorance, but why do we take the LCM of 15 and 21 to remove double count? I multiplied 15*21= 315 and divided 5000/315= 15 times to get the overlap. I know it is wrong now, but can someone explain it to me please?
Posted from my mobile device The numbers which are divisible by both 15 and 21 are numbers which are divisible by the least common multiple of 15 and 21, which is 105: 105, 210, 315, 420, ... If you simply multiply you are loosing many numbers, for example, 105, 210, 420, ...
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Re: How many positive integers less than 5,000 are evenly divisi
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