If a, b, c, d, e and f are integers and (ab + cdef)

Question

If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is the maximum number of integers that can be negative?

A. 2
B. 3
C. 4
D. 5
E. 6

Kudos for a correct  solution .

Lucky2783 · Accepted Answer

Case 1: |ab| > |cdef| 
Then we have 1negative in ab and 3 negatives in cdef

Case 2 |ab| <|cdef| 
Then we can have 2negative in ab and 3 negatives in cdef

Eg: 
-3*-2 + 4*-7*-5*-6 <0

Answer 5.

Without knowing magnitudes it is hard to determine but as question asks for maximum value it should be second case . I.e. D

Good question for number properties revision .

chetan2u · Answer

hi  AmoyV ,

maximum will be 5..
you dont require both the multiplicatin to be negative for entire equation to be negative...
any one a or b can be negative to make ab negative and it can still be more(away from 0) than the multiplication of 4 other -ve numbers...
actually by writing minimum required as 1 out of 6,you are actually meaning 5 out of 6 also possible as you will see 5 or 1 will give you same equation..
ans D

AmoyV · Answer

Minimuum should be 1

Maximum should be 4:

1 out of a or b to make the multiplication negative
3 out of c, d, e or f to make the multiplication negative.

Negative+Negative<0

Answer:C

AmoyV · Answer

Crap. Yeah, thanks.

Funny part is I had considered this option when I read the question but messed up the execution. Made a mental note of slowing down. :D

OptimusPrepJanielle · Answer

The number can be negative in following conditions: 1) If ab is a bigger negative number than cdef => Both a & b cannot be negative at a time. 2) If cdef is a bigger negative number than ab => All four c, d, e & f cannot be negative at a time Let one of a & b be negative and let all 4 of c, d, e & f be negative. So, ab will be negative and cdef will be positive but their sum will be negative. Hence maximum 5 can be negative. Hence option (D). -- Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course

sabineodf · Answer

Originally I chose E but now I changed my mind! I ignored that the sum of the numbers had to be under zero, and tried to make the number positive. Now I think the answer is D --

I put in values to try and make it work:

25 * (-1) + (-1 * -2 * -3 * -4) =
-25 + 24 =  -1

That's 5 negative numbers that create a value under 1. one more value were to have been positive, the numbers summed up would not be less than 1. Silly mistake I made originally.

Note to self: READ THE QUESTION!!!

Tmoni26 · Answer

For maximum number of negative values, then

ab must be negative, so either a or b can be negative, we assume a is negative

cdef should also give us a negative value and we can actually have 3 integers negative here to give us an overall negative value

so my answer is 4 for the maximum number of negative integers

b9n920 · Answer

Hi there,

As u said, ab must be negative, does cdef have to be negative?  What if |ab| > |cdef| which makes the expression negative.

So, ab->negative, -> a negative, b positive. And c,d,e,f all negative. And |ab|>|cdef| makes the whole thing negative. So,  5 can be negative. D.

~Binit.

aj13783 · Answer

looking at the eqn: (ab + cdef) < 0

In order for the sum of two terms to be -ve, one term has to be -ve for sure. With this mind, we can build mutliple scenarios. To restrict time take to solve this problem assume the same small integers for all unkowns.

a = -2, b= -2, c= -2, d=-2, e=-2, f= 2

-2*-2 + -2*-2*-2*2 = -12 < 0

It can have a max of 5 -ve integers. Option D

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION: abcdef_text.PNG

bae · Answer

Apologies, i'm getting confused with total integers, why do you multiply the numbers? Shouldn't each of this be treated as individual numbers?

Bunuel · Answer

ab means a*b and cdef means c*d*e*f.

bae · Answer

So should the question rephrased as the products of the integers ... .

Or we can make the assumption that when we find similiar questions during the GMAT exam, we should just treat it as multiple of products?

Bunuel · Answer

If cdef were a 4-digit number it would have been mentioned explicitly. Without that, cdef can only be c*d*e*f since only multiplication sign (*) is usually omitted.

generis · Answer

To achieve negative result here with maximum number of negative integers, we can't use six. An even number of negative factors yields positive. Six negative factors "eat up" all the individual terms.

Five negative factors, however, will work.

Maximize and minimize

1. Maximize the number of negative factors in (cdef): there can be three. Odd number of negative factors = negative product.

2. Maximize negative factors by two more. Make both a and b negative, but minimize their positive product's absolute value.

Make c, d, and e negative, and for ease, keep them small. Let f be the one positive number. Make it large so that |(cdef)| is > (ab).

Result is a small positive number plus a larger negative number, which yields a negative.

3. Let a, b, c, d, and e = -1
Let f = 20

(-1)(-1) + (-1)(-1)(-1)(20) =

1 + (-20) = 
1 - 20 =
-19

4. Alternatively, make all but a or b negative, and one number large because:
-- if one term in (ab) is negative, and 
-- the absolute value of (ab) is greater than (cdef) 
-- the result is a very small negative number (far to the right on the number line) plus a smaller positive number, such that
-- sum of terms with those properties is less than 0.

Let b, c, d, e, and f = -1
Let a = 20

(20)(-1) + (-1)(-1)(-1)(-1) =
-20 + 1 = -19

The maximum number of negative factors possible for this expression is 5

Answer D

PV66 · Answer

It's a very simple one, yet easily prone to a mistake.
 
The equation states that the addition if two numbers needs to be negative. 
Consider cdef first. For the max negative integers, we can have 3 out of 4 to be negative.
Now going to 'ab', this can be either negative or positive for the sum to be negative. If you were to consider 'ab' to be negative, only one of a or b can be negative. However it is also possible that ab is a positive number and lesser in magnitude than cdef. In which case both a and b can be negative. Which leaves us with 2 negative integers for 'ab' and 3 for 'cdef. Hence the max negative integers is 2+3=5.

Cheers,
PV66

Posted from my mobile device

Fdambro294 · Answer

Given: (ab + cdef) < 0

There are 2 Possible Scenarios:

Case 1:
1 Expression is (+)Positive and 1 Expression is (-)Negative

OR

Case 2:
BOTH Expressions are (-)Negative

Start with Case 2:

ab < 0

cdef < 0

Rule: for the Product of Factors to have a (-)Negative Result, there must be an ODD Count of (-)Negative Factors

for ab< 0 ------ we must have 1 (+)Positive Value and 1(-)Negative Value

and

for cdef < 0 ------ we can MAXIMIZE the Count of (-)Negative Values by making 3 Integers (-)Neg.

4 Negative Integers*******

Case 1: 1 Expression is (+)Positive and 1 Expression is (-)Negative

(1)we can first try to make cdef = (+)Pos. ..........and ab = (-)Neg.

Rule: for the Product of Factors to have a (+)Positive Result, there must be an EVEN Count of (-)Negative Factors

we can therefore make c - d - e - f - ALL 4 (-)Negative

and

we can make 1 of ab (+)Positive and the OTHER (-)Negative

5 Possible (-)Negative Integers****

or

(2) we can make ---> ab = (+)Pos. ----- and cdef = (-)Neg.

We can have both Integers A and B = (-)Negative Values and the Result of the Product will be (+)Positive

2 (-)Neg. Values

and

we can have 3 (-)Neg. Values in the expression ----> c*d*e*f

5 Possible (-)Negative Integer Values******

This is the MAXIMUM Possible Count of (-)Neg. Values we can have and the expression still stands TRUE.

If All 6 Values were Negative then:

a * b = (-) * (-) = +Positive Value

c*d*e*f = (-) * (-) * (-) * (-) = + Positive Value

the Expression would NOT hold True

MAXIMUM (-)Negative Integers = 5

-D-

luisdicampo · Answer

Deconstructing the Question

We want the maximum number of negative integers among a, b, c, d, e, f such that

$ab + cdef < 0$

The key is to analyze signs.

$ab$ is positive if $a$ and $b$ have the same sign, and negative if they have opposite signs.

$cdef$ is positive if there are an even number of negatives among $c, d, e, f$, and negative if there are an odd number of negatives.

We want as many negatives as possible while still making the sum negative.

Step-by-step

First test whether all $6$ integers can be negative.

If all six are negative, then

$ab > 0$

because the product of two negatives is positive, and

$cdef > 0$

because the product of four negatives is also positive.

So

$ab + cdef > 0$

This does not satisfy the inequality. So $6$ negatives is impossible.

Now test $5$ negatives.

We only need one example that works.

Let

$a=-1,\ b=-1,\ c=-2,\ d=-2,\ e=-2,\ f=1$

Then there are exactly $5$ negative integers.

Now compute:

$ab = (-1)(-1)=1$

$cdef = (-2)(-2)(-2)(1)=-8$

So

$ab + cdef = 1 + (-8) = -7 < 0$

This works.

Therefore $5$ negatives is possible, and $6$ negatives is impossible.

Answer D

If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is

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