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Here is this week’s problem:

If a, b, and c are nonzero integers and z = bc, is az negative?

(1) abc is an odd positive number.
(2) | b + c | < | b | + | c |

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#### Tom Williams

Tom Williams is a Marketing Associate at Manhattan Prep. His background and interests are primarily within New Media and Online Marketing. As such, he spends approximately 18 hours a day in front of a computer. When he’s not interacting with Manhattan Prep students on Facebook and Twitter, Tom can be found listening to records, rooting for the Mets and Jets, and reading history books.

### No responses to Challenge Problem Showdown – July 9th, 2012

1. Hi,

Here is my solution.

We have to check if a^z = -ve.

Stmt (1) abc = odd +ve no.
for this all the variables have to be odd. If any one of them is Even, then abc becomes Even. This will tell us that z is also Odd.

If abc is also +ve then there are a few possibilities. Either all of them i.e. a, b & c are +ve or any one of them is +ve and the remaining two are -ve. Let me put a small table here with the possibilities. There are 4 possibilities. Remember that the value of z depends upon the value or the sign of b and c.

Poss1 Poss2 Poss3 Poss4
a + + – –
b + – + -
c + – – +
z + – + -

a^z + + – -

Stmt (1) can give us both a yes and no therefore not sufficient.

We are left with BCE

Stmt (2) alone does not tell us anything about a. So Stmt (2) alone is insufficient.

We are left with CE

Now we consider both the stmt together.

To satisfy Stmt (2) we can derive that ‘b’ and ‘c’ should have different signs. ie ‘b’ should be +ve and ‘c’ should be -ve or ‘b’ should be -ve and ‘c’ should be +ve. ‘b’ & ‘c’ both can’t be +ve or -ve.

Let us take a look at the table for stmt (1)

Poss1 Poss2 Poss3 Poss4
a + + – –
b + – + -
c + – – +
z + – + -

a^z + + – -

To satisfy both the statements, Poss1 and Poss2 are ruled out. Only Poss3 and Poss4 are Possible and for both of them we are getting a^z to be -ve so we can answer only in a (Y) so sufficient.

the answer is C.

3. Its ans. choice C.

4. z=b^c a^z is negative?
a^bc is negative?

1) abc is an odd positive number–insufficient

all three a,,b,c are odd numers

a<0 b0 –a^bc is -ve
a0 c0 b<0 c<0 –a^bc is +ve

2)|b+c|<|b|+|c|–insuffciecint

both together are suffieint

5. There’s no way to submit an answer this week… the answer choices are missing from the challenge page.

6. c

7. Solution :-
Resolving for z, Is a^(b^c) = -ve?
(1) abc = +ve Odd =>

a,b,c are odd. either a,b,c are all +ve or two of them are negative

b^c is always odd as b and c are odd

whether a^z is -ve depends upon whether a is -ve or +ve hence (1) insufficient

(2)|b+c|
b and c are of opposite signs
since absolute value of sum of b and c is less than sum of absolute values of a and b

since b, c are in exponent, sign of b and c will not impact sign of the a^z
whether a^z is -ve depends on the following :-
if b^c is odd and a -ve, then a^z – ve
if b^c is eve and a -ve / a +ve then a^z NOT -ve

Insufficient

(1) and (2) together

b and c is odd and hence b^c is odd

b and c are of opposite sign and since abc is +ve, a also has a negative sign

so either a,b -ve and c + ve OR a,c -ve and b +ve

either cases a^(b^c) = a^z is always -ve

(1) and (2) sufficient hence (C)