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Divisibility DS

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by theachiever » Mon Dec 24, 2012 10:17 pm
If R is an integer, is R evenly divisible by 3?
(1) 2R is evenly divisible by 3
(2) 3R is evenly divisible by 3
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Source: — Data Sufficiency |
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by puneetkhurana2000 » Mon Dec 24, 2012 10:27 pm
Statement 1) 2R is evenly divisible by 3, so 2R can be 3,6,9,12,15...so acceptable values of R are 3,6,9... and all these are evenly divisible by 3. Sufficient!!!

Statement 2) 3R is evenly divisible by 3, so 3R can be 3,6,9,12,15...so acceptable values of R are 1,2,3,4,5... and some of these(3,6..) are evenly divisible by 3 and some(1,2,5..) are not. Not Sufficient!!!

Answer A.
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by tabsang » Tue Dec 25, 2012 1:58 am
theachiever wrote:If R is an integer, is R evenly divisible by 3?
(1) 2R is evenly divisible by 3
(2) 3R is evenly divisible by 3
Statement 1: 2R is evenly divisible by 3
Let's examine this a little closely:
2 x R is evenly divisible by 3
Since 2 is definitely not divisible by 3, for the condition to be true, R has to be divisible by 3.
SUFFICIENT.

Statement 2: 3R is evenly divisible by 3
This says:
3 x R is evenly divisible by 3
Since 3 is definitely divisible by 3, for the condition to be true, R may or may not be divisible by 3.
INSUFFICIENT.

Thus, the answer is [spoiler](A)[/spoiler]

Cheers,
Taz

P.S.: If this cleared your doubt, please take a moment to hit the "Thank" button. :)
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by tabsang » Tue Dec 25, 2012 2:08 am
puneetkhurana2000 wrote:Statement 1) 2R is evenly divisible by 3, so 2R can be 3,6,9,12,15...so acceptable values of R are 3,6,9...
Hey Puneet,

As per your explanation, if
2R can be 3,6,9,12,15...
... so according to that, the first acceptable value of R would be 3/2 right (for 2R to be 3)?

Now since R is an integer and 2R is completely divisible by 3 (as per statement 1), the values of 2R would be 6,12,18,... as opposed to 3,6,9,12,15...

Cheers,
Taz
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by puneetkhurana2000 » Tue Dec 25, 2012 6:23 am
puneetkhurana2000 wrote:
Statement 1) 2R is evenly divisible by 3, so 2R can be 3,6,9,12,15...so acceptable values of R are 3,6,9...
Hey Puneet,

Taz wrote:
As per your explanation, if
Quote:
2R can be 3,6,9,12,15...
... so according to that, the first acceptable value of R would be 3/2 right (for 2R to be 3)?

Now since R is an integer and 2R is completely divisible by 3 (as per statement 1), the values of 2R would be 6,12,18,... as opposed to 3,6,9,12,15...

Cheers,
Taz
It says the acceptable values of R (not 2R) are 3,6,9...

Hope this makes sense!!!
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