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Re: Set A comprises all 3-digit numbers that are multiples of 6. Set B com [#permalink]
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FCOCGALVAN wrote:

Hi!

Just one question,

Multiples of 4:
First term = 100
Last Term = 1000

Why are you using 1000 if its just 3 digit integers?

Kind regardS!


FCOCGALVAN

My bad!
I have made the necessary changes. Thanks
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Re: Set A comprises all 3-digit numbers that are multiples of 6. Set B com [#permalink]
Why can't we consider multiples of 24 that are 3 digit numbers ?

First term: 24*5 = 120
Last term: 24*41 = 984

Number of terms that would be common in both lists of 6 and 4 multiples gives 37 (41-5+1) here. Which does not seem to give the right answer.
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Re: Set A comprises all 3-digit numbers that are multiples of 6. Set B com [#permalink]
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Vavi1234 wrote:
Why can't we consider multiples of 24 that are 3 digit numbers ?

First term: 24*5 = 120
Last term: 24*41 = 984

Number of terms that would be common in both lists of 6 and 4 multiples gives 37 (41-5+1) here. Which does not seem to give the right answer.


Should not be a multiple of 8!
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Re: Set A comprises all 3-digit numbers that are multiples of 6. Set B com [#permalink]
why (225+1)/2 for finding divisible by 8.... or there is another way to find divisible by 8....
because i dont understand what you did for finding no. which are divisible by 8...
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Re: Set A comprises all 3-digit numbers that are multiples of 6. Set B com [#permalink]
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void wrote:
why (225+1)/2 for finding divisible by 8.... or there is another way to find divisible by 8....
because i dont understand what you did for finding no. which are divisible by 8...


No, there is not another way.

Those are integers rules you should know as second skin to perform well on the test

see here https://gre.myprepclub.com/forum/gre-quant ... tml#p51913

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Re: Set A comprises all 3-digit numbers that are multiples of 6. Set B com [#permalink]
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void wrote:
why (225+1)/2 for finding divisible by 8.... or there is another way to find divisible by 8....
because i dont understand what you did for finding no. which are divisible by 8...


We have 225 numbers which are divisible by 4, we need to find the numbers which are divisible by 8 here!
You know, every second number in the sequence would be divisible by 8 i.e. half of the numbers would be divisible by 8

Since we have the first term and last term divisible by 4, we will have 1 more term in the sequence which is divisible by only 4 than only 8.

Example: 4, 8, 12, 16, 20
How many numbers are divisible by 4 only: 3
How many of them are divisible by just 8: 2

\(3 = \frac{(5+1)}{2}\)

Same logic goes for a question something like this: How many terms in the sequence are odd?
1, 2, 3, 4, 5, 6, 7, 8, 9

Last term and First term are both odd - so we will have 1 more odd term than the total number of even terms
odd terms = \((9+1)/2 = 5\)
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Re: Set A comprises all 3-digit numbers that are multiples of 6. Set B com [#permalink]
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