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How many multiples of 4 are there between 12 and 96

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How many multiples of 4 are there between 12 and 96 [#permalink]

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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

My answer was 21 and that's incorrect.
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Re: Totally basic [#permalink]

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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
24-3+1 = 22

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Re: Totally basic [#permalink]

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New post 26 May 2010, 12:39
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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: Totally basic [#permalink]

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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: \(\frac{96-12}{4}+1=22\).

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: Totally basic [#permalink]

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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: Totally basic [#permalink]

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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: \(\frac{96-12}{4}+1=22\).

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 three-digit numbers are divisible by 7?

Generally we do--->999-100= 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 three-digit numbers are divisible by 7?

Generally we do--->999-100= 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 three-digit numbers are divisible by 7?

Generally we do--->999-100= 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 one inclusive we only subtract two extremes, for Eg

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 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: Totally basic [#permalink]

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New post 08 Jan 2012, 19:07
You forgot to add 1. It goes like this - (96-12)/4 + 1 = 22
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Re: Totally basic [#permalink]

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marcusaurelius wrote:
I have one concern regarding this How many multiples and how many integers?

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 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 three-digit numbers are divisible by 7?

Generally we do--->999-100= 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 three-digit numbers are divisible by 7?

Generally we do--->999-100= 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 one inclusive we only subtract two extremes, for Eg

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 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 (994-105)/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 (999-100)/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 [#permalink]

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New post 31 Oct 2013, 20:46
Why add one to the final result? I can count from 12-96 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 [#permalink]

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Stoneface wrote:
Why add one to the final result? I can count from 12-96 by four and come up with 22 that way, but I want to know the logic behind it.


Set of consecutive multiples of 4 is an evenly spaced set (arithmetic progression).

If the first term of arithmetic progression is \(a_1\) and the common difference of successive members is \(d\), then the \(n_{th}\) term of the sequence is given by:

\(a_ n=a_1+d(n-1)\) --> \(n=\frac{a_n-a_1}{d} + 1\).

Hope it helps.
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Re: How many multiples of 4 are there between 12 and 96 [#permalink]

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New post 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+(n-1)d
96=12+(n-1)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 [#permalink]

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New post 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+(n-1)D to find the nth term
12+(n-1)4=96
4n-4=84
n=22

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How many multiples of 4 are there between 12 and 96 [#permalink]

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New post 06 Oct 2014, 23:47
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: \(\frac{96-12}{4}+1=22\).

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.


Hi Bunuel,

can i use this formula for not inclusive? But when i use this formula I got different result.
Last multiple of 7 - First Multiple of 7/ 7 - 1

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Re: How many multiples of 4 are there between 12 and 96 [#permalink]

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New post 07 Oct 2014, 01:12
vietnammba 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: \(\frac{96-12}{4}+1=22\).

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.


Hi Bunuel,

can i use this formula for not inclusive? But when i use this formula I got different result.
Last multiple of 7 - First Multiple of 7/ 7 - 1


There are examples with "not inclusive" cases above. Please be more specific.
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How many multiples of 4 are there between 12 and 96 [#permalink]

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New post 07 Oct 2014, 01:24
# of multiples of x in the range = (Last multiple of x in the range - First multiple of x in the range)/x - 1.
Is this formula correct to calculate no of non inclusive multiple of an integer?

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Re: How many multiples of 4 are there between 12 and 96 [#permalink]

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New post 07 Oct 2014, 01:27
vietnammba 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.
Is this formula correct to calculate non inclusive multiple of an integer?


Yes. Again, examples in my post above have not inclusive ranges!
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How many multiples of 4 are there between 12 and 96 [#permalink]

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New post 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, [(990-11)/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 [#permalink]

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New post 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, [(990-11)/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. :shock:
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Re: How many multiples of 4 are there between 12 and 96 [#permalink]

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New post 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, [(990-11)/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. :shock:





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+(n-1)D
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Re: How many multiples of 4 are there between 12 and 96 [#permalink]

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New post 12 Aug 2016, 08:22
Applying some high school math here =>
A(N) = A + (N-1)D
So => 96=4+(N-1)4
=> N=22
Smash that B
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Re: How many multiples of 4 are there between 12 and 96   [#permalink] 12 Aug 2016, 08:22

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How many multiples of 4 are there between 12 and 96

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