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How many multiples of 4 are there between 12 and 96, inclusive?

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

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26 May 2010, 11:21
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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

Problem Solving
Question: 11
Category: Arithmetic Properties of numbers
Page: 63
Difficulty: 550

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

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26 May 2010, 11:55
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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: How many multiples of 4 are there between 12 and 96, inclusive?  [#permalink]

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02 Feb 2011, 03:20
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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: How many multiples of 4 are there between 12 and 96, inclusive?  [#permalink]

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31 Oct 2013, 19:46
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1
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, inclusive?  [#permalink]

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01 Nov 2013, 00:20
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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, inclusive?  [#permalink]

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30 Dec 2013, 21:56
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My approach would be this way:
12/4 = 3;
96/4 = 24;

Since both are inclusive, I will go with (Last - First + 1) concept:
24 - 3 + 1
= 22;

Ans is (B)
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Re: How many multiples of 4 are there between 12 and 96, inclusive?  [#permalink]

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23 Mar 2015, 21:30
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Hi All,

This is an example of a "fencepost" problem, since you have to include the "posts" (the numbers 12 and 96) in your calculation. There are a couple of different ways to approach these types of prompts, depending on what you're given and how comfortable you are with the 'technical aspects' of the math.

Here, we have a pretty easy situation: we're asked for all the multiples of 4 between 12 and 96, INCLUSIVE (meaning we have to include the 12 and the 96).

(4)(25) = 100, so there are 25 positive multiples of 4 when dealing with all positive integers from 1 to 100.

(4)(24) = 96, so there are 24 positive multiples of 4 when dealing with all positive integers from 1 to 96, INCLUSIVE.

We now have to remove the multiples of 4 that DO NOT fit the given range....

There are two: 4 and 8

24 - 2 = 22 total multiples of 4

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

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19 Mar 2018, 15:10
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

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

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31 Oct 2018, 13:52
Is there a difference in the equation, when the numbers are not inclusive?
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Re: How many multiples of 4 are there between 12 and 96, inclusive?  [#permalink]

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31 Oct 2018, 18:07
Hi TestTaker9,

To answer your immediate question: YES - the equation would change IF we were NOT including 12 and 96. Since both of those numbers are multiples of 4, we would have to subtract 2 from the total. As an alternative, you could recalculate using the lower-most and upper-most multiples of 4 that would be in the range of what you were looking for. In your hypothetical, that would be 16 and 92.

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

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16 May 2019, 19:02
Total multiples of 4 to 96 =96/4=24
4 has two multiples up to 12 which are 4 & 8.
So, multiples of 4 for 12 to 96 inclusive= 24-2=22 (Ans. B)

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

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18 May 2019, 14:36
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Since all the numbers will be in ap
So the first number will be 12 and last is 96.
An=a+(n-1)d
96=12+(n-1)4
21=n-1
N=22

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

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18 May 2019, 14:37
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Since all the numbers will be in ap
So the first number will be 12 and last is 96.
An=a+(n-1)d
96=12+(n-1)4
21=n-1
N=22

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

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18 Aug 2019, 23:47
TestTaker9 wrote:
Is there a difference in the equation, when the numbers are not inclusive?

Yes there will be a difference then.
The range would be 16 to 92
and in between there are 76 numbers out of which 19 will be divisble by 4 and +1, 20 would be the answer.
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Re: How many multiples of 4 are there between 12 and 96, inclusive?   [#permalink] 18 Aug 2019, 23:47