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represents the remainder of 3x/2, which of the following

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represents the remainder of 3x/2, which of the following  [#permalink] New post 12 Sep 2003, 03:04
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A
B
C
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E

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[x] represents the remainder of 3x/2, which of the following
expression is equal to 1:
a. [2x] b. [3x]+1 c. [2x] +1


I dont have the official answer to this one..
i m having a hard time interpreting this.


prat

Last edited by Praetorian on 13 Sep 2003, 20:07, edited 1 time in total.
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 [#permalink] New post 12 Sep 2003, 03:29
Interesting question.

[X] represents remainder of number 3x/2
Now, question asks which of the following is equal to 1.
Now, 3[x] + 1 represents remainder of number: 3(3x)/2 + 1 {replacing value of [x]}
=>i.e. remainder of 9x/2 + 1
9x/2 will have remainder either 1 or 0
Thus, 9x/2 + 1 will have only 1 as remainder.
Why?
when 9x/2 has remainder 0 => 9x/2 + 1 has remainder 1
9x/2 has remainder 1=> 9x/2 + 1 has remainder 0

For choice A, number is 0.
choice B, number is 0 or 1.
choice C, number is 1 or 3.
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 [#permalink] New post 13 Sep 2003, 18:24
I guess answer would be [2x]+1 , but is not among the choices :(


IMO all others are different than 1
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 [#permalink] New post 13 Sep 2003, 18:49
MartinMag wrote:
I guess answer would be [2x]+1 , but is not among the choices :(


IMO all others are different than 1


Martin

your answer choice is there...but can you explain please?

thanks
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 [#permalink] New post 13 Sep 2003, 19:56
praetorian123 wrote:
MartinMag wrote:
I guess answer would be [2x]+1 , but is not among the choices :(


IMO all others are different than 1


Martin

your answer choice is there...but can you explain please?

thanks


2[x]+1 would throw a different answer than [2x]+1


[2x]+1 is not there :?
CEO
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 [#permalink] New post 13 Sep 2003, 20:10
MartinMag wrote:
praetorian123 wrote:
MartinMag wrote:
I guess answer would be [2x]+1 , but is not among the choices :(


IMO all others are different than 1


Martin

your answer choice is there...but can you explain please?

thanks


2[x]+1 would throw a different answer than [2x]+1


[2x]+1 is not there :?


Hey martin

Its right there :)

no, actually my question had a typo...

Sorry about that guys!

[2x] will have a remainder of 0...

so.. obviously we need [2x] + 1..

thanks
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 [#permalink] New post 13 Sep 2003, 20:23
answer is c I guess... Although depending on what x is, b could be valid too...

case a:
[2x] is the remainder of 3(2x)/2. The fraction resolves itself quite nicely to 3x, so there is no remainder. [2x]=0.

case b:
[3x] is the remainder of 3(3x)/2. This fraction may or may not have a remainder, depending on what x is. If x=2, then [3x]+1 = 1. If x=7, then [3x]+1=1+1=2

case c:
looking at (a) above, [2x]=0, so [2x]+1=1.

definitely c, and maybe b depending on the value of x...

oops, just read the answers but don't understand... did you fix your typo in the question prat?

regards,
Maniaque
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 [#permalink] New post 13 Sep 2003, 20:32
Maniaque27 wrote:

definitely c, and maybe b depending on the value of x...

regards,
Maniaque



Hey

yes, i did correct the typo

The very fact that you can have 1 or 2 as remainder makes B invalid..

We are looking for the best answer..remember :)

thanks
  [#permalink] 13 Sep 2003, 20:32
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represents the remainder of 3x/2, which of the following

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