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505-555 Level|   Multiples and Factors|                  
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Bunuel
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Bunuel
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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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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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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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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

Final Answer:
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marcusaurelius
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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Is there a difference in the equation, when the numbers are not inclusive?
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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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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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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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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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TestTaker9
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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Multiples of 4 are there between 12 and 96, inclusive: This means that there will be an A.P. with a common difference of '4'.

96 = 12 + (n-1) 4

=> 96 - 12 + 4= 4n
=> 4n = 88

=> n = 22

Answer D
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Bunuel
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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.

  • \(Number \ of \ multiples \ of \ x \ in \ the \ range = \)

    \(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

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,

We add 1 even if it's not inclusive? As shown in your extra examples.

Thank you :)

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How many multiples of 4 are there between 12 and 96, inclusive?
4* (3) = 12
4* (24) = 96

=24-3 +1 =22 (B)
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Bunuel
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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.

  • \(Number \ of \ multiples \ of \ x \ in \ the \ range = \)

    \(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

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.


This formula is helpful for many questions regarding sets or sums of sets with consecutively spaced integers. I'm just taking a moment to explain where the formula comes from, because it's something the GMAT tests in a sneaky way.

This value of 'last number' - 'first number' is called the 'range' of a set. That can basically be thought of as the 'distance' from the smallest number to the largest.

Students often have a sense that we should divide that distance by the number we're counting by (in the original question, '4').

The thing that often gets forgotten is the '+1.' Why do we have that plus one?

Well, what do we solve for when we divide 'the range' by 'the number we're counting by?'

For instance, the range from 12 to 96 is 84. If we divide that 84 by the 'four' we're counting by, we get 21. What does that 21 'signify' for this problem???

It tells you how many 'jumps' of four you make to get from 12 to 96.

12 + (4 + 4 + 4 + 4 .... +4) {21 times} = 96

Visualized this might be seen as:


12 _(+4)_ 16 _(+4)_ 20 _(+4)_ 24 .... 92 _(+4)_ 96

So we have 21 of those '+4s'... that is 21 gaps BETWEEN the numbers in the set. In order to have 21 gaps between numbers, you must have 22 numbers total! That's why we must add 1.

POINT IS: Do not mix up the 'range' with the 'number of numbers' (and it's very easy to do when you 'cut' the range into pieces, as we do in this problem).

(Random aside, I often see people try to count up how many multiples of [the number] there are in a range of the first '10' numbers, say, and then multiply that by how many '10s' there are in the whole set. This won't often work---some '10' will have more multiples of that number than other '10s.'

For instance 0-10 has three multiples of 3. 11-20 has three multiples of 3. But 21-30 has *four* multiples of 3. Stuff like that will happen often.
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Stoneface
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.
8*12=96
3*4=12
.'. 3*8=24
(1-12) 3 times 4 so, less 2 times .
Remainder 24-2=22

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Bunuel
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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.

  • \(Number \ of \ multiples \ of \ x \ in \ the \ range = \)

    \(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

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 would like to have a small understanding in the concept below.
How many multiples of 7 are there between -28 and -1, not inclusive?
Why are we not considering 0 as multiple of 7?
Last multiple of 7 IN the range is 0;
First multiple of 7 IN the range is -21;

\(\frac{0-(-21)}{7}+1=4\).
Hence 4 multiples -21,-14,-7,0 in the given range.
0 is multiple of every number right?
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