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If a , b, and c are integers

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If a , b, and c are integers

by logitech » Sun Dec 07, 2008 3:53 pm
If a , b, and c are integers and

ab^2/c is a positive even integer, which of the following must
be true?


I. ab is even
II. ab > 0
III. c is even
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by earth@work » Sun Dec 07, 2008 4:49 pm
IMO: I only

ab^2/c is a positive even integer
I. ab is even : since ab^2 is even integer the either a or b or both have to be even implies ab is even, both a & b cannot be odd
either case ab will be even...true always
II. ab > 0 not necessary true, ab^2/c positive implies a,b^2 & c are all positive but we cannot say about b...it cud be positive or negative
III. c is even...not necessarily true, c cud be odd if a is odd as ab^2 is an integer.
if b is even : a cud be even leads to c even; OR a is odd leads to c odd or even
if b is odd, then a is even & c cud be even or odd
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by logitech » Sun Dec 07, 2008 5:00 pm
Good work!

OA I only
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by pbanavara » Sun Dec 07, 2008 5:53 pm
well .. why not I and II ?

ab has to be positive because .. if b is negative .. a can be negative and c can be negative. Am I missing something ?

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by hwiya320 » Sun Dec 07, 2008 5:59 pm
pbanavara wrote:well .. why not I and II ?

ab has to be positive because .. if b is negative .. a can be negative and c can be negative. Am I missing something ?

- pradeep
you just answered yourself.

The question is what MUST be true.

if A was negative and B was positive, then ab < 0,

but if C was negative, the statement is true again that the value is positive.

Therefore, it does NOT have to be true at all times.
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by pbanavara » Sun Dec 07, 2008 6:01 pm
hwiya320 wrote:
pbanavara wrote:well .. why not I and II ?

ab has to be positive because .. if b is negative .. a can be negative and c can be negative. Am I missing something ?

- pradeep
you just answered yourself.

The question is what MUST be true.

if A was negative and B was positive, then ab < 0,

but if C was negative, the statement is true again that the value is positive.

Therefore, it does NOT have to be true at all times.
Oh man .. missed that MUST clause .. danka
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by bsandhyav » Thu Dec 11, 2008 4:02 am
A little confused here.....

must be = should necessarily be true ??
or
must be = could be true??
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by niraj_a » Thu Dec 11, 2008 4:12 am
I only

i did the MGMAT may (dunno why it clicked)

ab^2 = c * even

(note how doing this doesn't violate the rule that c could be 0, because the question doesn't test for that here).

so for RHS to be correct, LHS must be even as well. II is out since b's sign can be hidden coz of the square. c doesn't have to be even for RHS to be even, because anything multiplied by even will be even.
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by Scott@TargetTestPrep » Wed Dec 13, 2017 3:58 pm
logitech wrote:If a, b, and c are integers and a*b^2/c is a positive even integer, which of the following must be true?

I. ab is even
II. ab > 0
III. c is even

A. I only
B. II only
C. I and II
D. I and III
E. I, II, and III
We are given that (a*b^2)/c is a positive even integer. Therefore, a*b^2 must be even. (If a*b^2 is odd, (a*b^2)/c can't ever be even.)

Now recall that the product of an even number and any integer is even, so either a or b, or both, must be even. Thus we see that ab must be an even integer. However, ab DOES NOT have to be greater than zero, since a could be -2 and b could be 1. Finally, we see that c does not have to be even, since a could be -2, b could be 1, and c = -1. Thus, only Roman numeral I must be true.

Answer: A

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