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

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

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

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).

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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!!!

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

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.

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

MAGOOSH OFFICIAL SOLUTION:

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

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

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?

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.

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

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

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-

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