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ps- number of integers

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ps- number of integers [#permalink] New post 06 Oct 2008, 16:06
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what is the total number of integers between 100 and 200 that are divisible by 3?

ans:33
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Re: ps- number of integers [#permalink] New post 07 Oct 2008, 00:00
My approach:

3 * 33 = 99
3 * 66 = 198

Hence, number of integers between 100 and 200 and divisible by 3 will be 66-33 = 33.
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Re: ps- number of integers [#permalink] New post 07 Oct 2008, 05:59
Thank you both. I don't really understand neither of these two answers.

Why you multiply 3 x 33 and then 3 x 66?
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Re: ps- number of integers [#permalink] New post 07 Oct 2008, 06:02
can you please explain your reasoning behind this? Why 198-102/3 and not 200-100/3?

ritula wrote:
(198-102)/3 +1=33

andreasonlinegr wrote:
what is the total number of integers between 100 and 200 that are divisible by 3?

ans:33
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Re: ps- number of integers [#permalink] New post 07 Oct 2008, 10:18
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[quote="andreasonlinegr"]can you please explain your reasoning behind this? Why 198-102/3 and not 200-100/3?

the first and the last multiples of 3 in the given range

however you can still use 200-100/3 = 33.3333 ( the number of multiples cant be a fraction thus still answer is 33.

if the result is 33.9 still only 33 multiples of 3 are there
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Re: ps- number of integers [#permalink] New post 08 Oct 2008, 01:45
andreasonlinegr wrote:
Thank you both. I don't really understand neither of these two answers.

Why you multiply 3 x 33 and then 3 x 66?


The first number that will be in the range of 100-200 and divisible by 3 will be 3*34. Similarly, the last such number will be 3*66.

In orde to take into account the first number, I should consider the number before 100 that is divisible by 3 and that is why 3*33.
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Re: ps- number of integers [#permalink] New post 08 Oct 2008, 13:47
yezz wrote:
andreasonlinegr wrote:
can you please explain your reasoning behind this? Why 198-102/3 and not 200-100/3?

the first and the last multiples of 3 in the given range

however you can still use 200-100/3 = 33.3333 ( the number of multiples cant be a fraction thus still answer is 33.

if the result is 33.9 still only 33 multiples of 3 are there


Dumb question: Why take multiples, subtract them and then divide by 3? Is there a formula for that?
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Re: ps- number of integers [#permalink] New post 08 Oct 2008, 14:08
piper wrote:
yezz wrote:
andreasonlinegr wrote:
can you please explain your reasoning behind this? Why 198-102/3 and not 200-100/3?

the first and the last multiples of 3 in the given range

however you can still use 200-100/3 = 33.3333 ( the number of multiples cant be a fraction thus still answer is 33.

if the result is 33.9 still only 33 multiples of 3 are there


Dumb question: Why take multiples, subtract them and then divide by 3? Is there a formula for that?


every 3 consecutive intigers has a multiple of 3 ..... right??

when you subtract the higher and lower values in the given range you get the total no on intigers in that range......right

then when u devide that number by 3 ( from every 3 consecutive initgers there is a multiple of 3 above ) you will know how many ( groups of 3 consecutives are there and hence you will no the number of multiples of 3)

hope i am clear
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Re: ps- number of integers [#permalink] New post 08 Oct 2008, 14:34
yezz wrote:
every 3 consecutive intigers has a multiple of 3 ..... right??

when you subtract the higher and lower values in the given range you get the total no on intigers in that range......right

then when u devide that number by 3 ( from every 3 consecutive initgers there is a multiple of 3 above ) you will know how many ( groups of 3 consecutives are there and hence you will no the number of multiples of 3)

hope i am clear


Wow, beautifully explained! Thank you.
Re: ps- number of integers   [#permalink] 08 Oct 2008, 14:34
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