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(1) r^3+r^2 is even --> both even and odd values of r satisfy this statement: odd^3+odd^2=odd+odd=even and even^3+even^2=even+even=even. Not sufficient.

(2) 3r is divisible by 6 --> \frac{3r}{6}=integer --> \frac{r}{2}=integer --> r=2*integer --> r=even. Sufficient.

Re: DS - odd/even problem [#permalink]
20 Nov 2007, 00:59

yuefei wrote:

Is integer R even?

1) R3 + R2 is even 2) 3R is divisible by 6

Since (R3+R2) is even(It is given in first condition)
now R2 is always is even(R*2)
for (R3+R2) to be Even, R3 must be even bcos only even + even = Even.
for this R must be even bcos EvenR * 3 = even.
therefore condition one alone is sufficient.
now from i condition 2 it is clear the R is even.
therfore 2 alone itself is sufficient.

Re: DS - odd/even problem (m04q06) [#permalink]
10 Nov 2010, 08:03

In statement #2, what would happen if zero is considered? Zero is an integer, however it is neither odd nor even. Am I thinking too much for the purpose of GMAT?

Also if I encounter this question on GMAT exam, should I consider zero at all?

Re: DS - odd/even problem (m04q06) [#permalink]
10 Nov 2010, 20:33

peterwang wrote:

In statement #2, what would happen if zero is considered? Zero is an integer, however it is neither odd nor even. Am I thinking too much for the purpose of GMAT?

Also if I encounter this question on GMAT exam, should I consider zero at all?

Thanks

zero is a non-negative and non-positive EVEN number for GMAT....

Re: DS - odd/even problem (m04q06) [#permalink]
14 Nov 2011, 07:19

Statement 1) R^2 (1 + R) is Even.

This can be Even*Even OR Odd*Even, which results in an Even number. Hence insufficient.

Statement 2) 3R/ 6 is even. 3R will be multiple of 6. R will take all those values that are even,and when multiplied by 3 will give a multiple of 6. Therefore, R is even. Sufficient.

Re: DS - odd/even problem (m04q06) [#permalink]
14 Nov 2012, 08:08

for 1: R = 1 => R cube + R sqr = 2 (even) R = 2 => R cube + R sqr = 12 (even) R = 3 => R cube + R sqr = 36 (even) R = 4 => R cube + R sqr = 82 (even)

so we get all values(even + odd) of R for which the eqn is even. NOT SUFFICIENT

for 2: R = 1 => 3R = 3 => is div by 6 (no) R = 2 => 3R = 6 => is div by 6 (yes) R = 3 => 3R = 9 => is div by 6 (no) R = 4 => 3R = 12 => is div by 6 (yes)

thus it gives us "is div by 6 -yes" of eqn for even values of R. Hence B is SUFFICIENT.