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The best approach

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by rivers96 » Fri May 02, 2008 2:46 pm
Hi, Does anybody know which is the best approach to this problem?

[url=https://www.postimage.org/image.php?v=aV1bbca0][img]https://www.postimage.org/aV1bbca0.jpg[/img][/url]


thanks a lot,

Rivers
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Source: — Data Sufficiency |

by tmmyc » Mon May 05, 2008 8:48 pm
For this problem, you will need to be comfortable with how odds and evens interact when added and multiplied.

(1)
Assume c is even, then this statement holds true if d is even or odd: 2*(2+1) = 6 Even
Assume c is odd, then this statement holds true if d is odd: 3*(3+1) = 12 Even
Insufficient, since c can be either even or odd.

(2)
Assume c is even, then this statement holds true if d is even or odd: (2+2)(1+4) = 20 Even
Assume c is odd, then this statement holds true if d is even: (1+2)(2+4) = 18 Even
Insufficient, since c can be either even or odd.

Together
Assume c is even, then these two statements hold true if d is even or odd:
2*(2+1) = 6 Even
(2+2)(1+4) = 20 Even

Assume c is odd, then
(1) is true only if d is odd, but not even
3*(3+1) = 12 Even
3*(3+2) = 15 Odd

(2) is true only if d is even, but not odd
(1+2)(2+4) = 18 Even
(1+2)(2+3) = 15 Odd

Since d cannot be both even and odd at the same time, this case does not work and c cannot be odd. Only the first case works so c must be even. Sufficient.
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