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How many multiples of 4 are there between 12 and 96, inclusive?
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26 May 2010, 12:21
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How many multiples of 4 are there between 12 and 96, inclusive? A. 21 B. 22 C. 23 D. 24 E. 25
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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26 May 2010, 12:55
marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive?
21 22 23 24 25
My answer was 21 and that's incorrect. \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\). In the original case: \(\frac{9612}{4}+1=22\). If the question were: how many multiples of 5 are there between 7 and 35, not inclusive? Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is 5; \(\frac{30(5)}{5}+1=8\). OR:How many multiples of 7 are there between 28 and 1, not inclusive? Last multiple of 7 IN the range is 7; First multiple of 7 IN the range is 21; \(\frac{7(21)}{7}+1=3\). Hope it helps.
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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26 May 2010, 12:37
marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive?
21 22 23 24 25
My answer was 21 and that's incorrect. 12 is the 3rd multiple of 4 96 is the 24th multiple of 4 243+1 = 22




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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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26 May 2010, 12:39
22 multiples of 4 between 12 and 96 inclusive. from 4 * 3 upto 4 *24, (3,4,...,24). Hence, 22 multiples !
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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02 Feb 2011, 04:20
Thanks Bunuel. One day, after my GMAT is over, and that day will come soon, you should bake a cake for you
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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08 Jan 2012, 18:49
Bunuel wrote: marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive?
21 22 23 24 25
My answer was 21 and that's incorrect. \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\). In the original case: \(\frac{9612}{4}+1=22\). If the question were: how many multiples of 5 are there between 7 and 35, not inclusive? Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is 5; \(\frac{30(5)}{5}+1=8\). OR:How many multiples of 7 are there between 28 and 1, not inclusive? Last multiple of 7 IN the range is 7; First multiple of 7 IN the range is 21; \(\frac{7(21)}{7}+1=3\). Hope it helps. I have one concern regarding this How many multiples and how many integers? How many of the threedigit numbers are divisible by 7? Generally we do>999100= 899 so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1 but the answer is only 128,How is that possible????? As per our knowledge we know that >If the question says both inclusive we add +1 at the last, For eg How many of the threedigit numbers are divisible by 7? Generally we do>999100= 899 so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1 >If the question says both not inclusive we add 1 at the last, For eg How many of the threedigit numbers are divisible by 7? Generally we do>999100= 899 so 899/7=128 so we need to include both digits so we add 1 to total so it is 1281 >If the question says one inclusive we only subtract two extremes, for Eg How many of the threedigit numbers are divisible by 7? Generally we do>999100= 899 so 899/7=128, so 128 is the answer..... Anything wrong with my concept???????Please explain i am totally confused with this............... If anything wrong,please explain how to with these kind of problems, Like How many integers between two numbers? and How many integers between two multiples?
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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08 Jan 2012, 19:07
You forgot to add 1. It goes like this  (9612)/4 + 1 = 22
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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03 Sep 2012, 16:33
marcusaurelius wrote: I have one concern regarding this How many multiples and how many integers?
How many of the threedigit numbers are divisible by 7?
Generally we do>999100= 899
so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1
but the answer is only 128,How is that possible?????
As per our knowledge we know that
>If the question says both inclusive we add +1 at the last, For eg
How many of the threedigit numbers are divisible by 7?
Generally we do>999100= 899
so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1
>If the question says both not inclusive we add 1 at the last, For eg
How many of the threedigit numbers are divisible by 7?
Generally we do>999100= 899
so 899/7=128 so we need to include both digits so we add 1 to total so it is 1281
>If the question says one inclusive we only subtract two extremes, for Eg
How many of the threedigit numbers are divisible by 7?
Generally we do>999100= 899
so 899/7=128, so 128 is the answer.....
Anything wrong with my concept???????Please explain i am totally confused with this...............
If anything wrong,please explain how to with these kind of problems, Like How many integers between two numbers? and How many integers between two multiples? three digit numbers divisible by 7 or multiples of 7 are from 105 to 994. Now if u do (994105)/7 + 1 { as both are inclusive } you get 127 + 1 = 128 as the answer... the problem in your approach is that neither 100 nor 999 are multiples of 7 and hence you need to solve (999100)/7 with both NOT INCLUSIVE and you get the correct answer 128
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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31 Oct 2013, 20:46
Why add one to the final result? I can count from 1296 by four and come up with 22 that way, but I want to know the logic behind it.



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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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15 Nov 2013, 21:37
[quote="marcusaurelius"]How many multiples of 4 are there between 12 and 96, inclusive?
A. 21 B. 22 C. 23 D. 24 E. 25
12,16....96
applying AP, an=a+(n1)d 96=12+(n1)4 by solving this equation,the result is
n = 22 thats is the answer



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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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10 Dec 2013, 02:41
marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive?
A. 21 B. 22 C. 23 D. 24 E. 25
My answer was 21 and that's incorrect. We can use AP fourmula here, A+(n1)D to find the nth term 12+(n1)4=96 4n4=84 n=22



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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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21 Apr 2016, 17:41
Bunuel wrote: \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).
I know this is a really old thread but isn't this an unnecessary step for this formula? I've seen you use it on similar problems including a couple on the GMAT Club tests and I don't understand why you subtract by the first multiple. For example multiples of 11 in 1000, [(99011)/11] + 1 is the same as (990/11) so why add the additional subtraction and addition steps?
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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22 Apr 2016, 11:03
redfield wrote: Bunuel wrote: \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).
I know this is a really old thread but isn't this an unnecessary step for this formula? I've seen you use it on similar problems including a couple on the GMAT Club tests and I don't understand why you subtract by the first multiple. For example multiples of 11 in 1000, [(99011)/11] + 1 is the same as (990/11) so why add the additional subtraction and addition steps? Bunuel's approach is perfect... Here you are trying to calculate the numbers divisible by 11 from 0  1000 But say in a situation like  Quote: Find the total no of numbers between 900  990 , (both inclusive) which are divisible by 11 Here your formula will not work . Actually we have 2 sets  1. 0  899 2. 900  990 We are interested in the 2nd set So, U can go this way ( Bunnels approach) \(Total \ no \ of \ Numbers \ divisible \ by \ 11 \ up \ to \ 990  Total \ no \ of \ Numbers \ divisible \ by \ 11 \ up \ to 899\) + \(The \ number \ 990\ itself \ (\ which \ is \ divisible \ by\ 11)\) Hope I am clear, please feel free to revert in case of any doubt.
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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22 Apr 2016, 11:11
Abhishek009 wrote: redfield wrote: Bunuel wrote: \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).
I know this is a really old thread but isn't this an unnecessary step for this formula? I've seen you use it on similar problems including a couple on the GMAT Club tests and I don't understand why you subtract by the first multiple. For example multiples of 11 in 1000, [(99011)/11] + 1 is the same as (990/11) so why add the additional subtraction and addition steps? Bunuel's approach is perfect... Here you are trying to calculate the numbers divisible by 11 from 0  1000 But say in a situation like  Quote: Find the total no of numbers between 900  990 , (both inclusive) which are divisible by 11 Here your formula will not work . Actually we have 2 sets  1. 0  899 2. 900  990 We are interested in the 2nd set So, U can go this way ( Bunnels approach) \(Total \ no \ of \ Numbers \ divisible \ by \ 11 \ up \ to \ 990  Total \ no \ of \ Numbers \ divisible \ by \ 11 \ up \ to 899\) + \(The \ number \ 990\ itself \ (\ which \ is \ divisible \ by\ 11)\) Hope I am clear, please feel free to revert in case of any doubt. I feel compelled to say this But this ain't no formula actually This is just an extension of the basic AP series formula => An = A+(n1)D Regards Stone Cold
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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01 Mar 2017, 04:21
Bunuel wrote: marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive?
21 22 23 24 25
My answer was 21 and that's incorrect. \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\). In the original case: \(\frac{9612}{4}+1=22\). If the question were: how many multiples of 5 are there between 7 and 35, not inclusive? Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is 5; \(\frac{30(5)}{5}+1=8\). OR:How many multiples of 7 are there between 28 and 1, not inclusive? Last multiple of 7 IN the range is 7; First multiple of 7 IN the range is 21; \(\frac{7(21)}{7}+1=3\). Hope it helps.
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File comment: Please refer to attachment. There are 7 multiples of 5 i.e. 5, 5, 10, 15, 20, 25, 30. Then according to answer you did they are 8 multiples. How? please explain.
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How many multiples of 4 are there between 12 and 96, inclusive?
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01 Mar 2017, 04:24
madannadam9 wrote: Bunuel wrote: marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive?
21 22 23 24 25
My answer was 21 and that's incorrect. \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\). In the original case: \(\frac{9612}{4}+1=22\). If the question were: how many multiples of 5 are there between 7 and 35, not inclusive? Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is 5; \(\frac{30(5)}{5}+1=8\). OR:How many multiples of 7 are there between 28 and 1, not inclusive? Last multiple of 7 IN the range is 7; First multiple of 7 IN the range is 21; \(\frac{7(21)}{7}+1=3\). Hope it helps. You missed 0, which is a multiple of every integer.
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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19 Mar 2018, 16:10
marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive?
A. 21 B. 22 C. 23 D. 24 E. 25 We can determine the number of multiples of 4 from 12 to 96, inclusive, by using the following formula: (largest multiple of 4  smallest multiple of 4)/4 + 1 (96  12)/4 + 1 =84/4 + 1 = 21 + 1 = 22 Answer: B
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Re: How many multiples of 4 are there between 12 and 96, inclusive?
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07 Aug 2018, 05:23
Bunuel wrote: marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive?
21 22 23 24 25
My answer was 21 and that's incorrect. \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\). In the original case: \(\frac{9612}{4}+1=22\). If the question were: how many multiples of 5 are there between 7 and 35, not inclusive? Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is 5; \(\frac{30(5)}{5}+1=8\). OR:How many multiples of 7 are there between 28 and 1, not inclusive? Last multiple of 7 IN the range is 7; First multiple of 7 IN the range is 21; \(\frac{7(21)}{7}+1=3\). Hope it helps. May I know why the addition of 1 is constant for inclusive as well as non inclusive ?? Posted from my mobile device




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