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

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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


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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 31 Mar 2015, 04:48
Bunuel wrote:
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.



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

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New post 31 Mar 2015, 06:23
1
1
Bunuel wrote:
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.


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

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New post 31 Mar 2015, 06:49
2
AmoyV wrote:
Bunuel wrote:
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.



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


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
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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 31 Mar 2015, 07:54
chetan2u wrote:
AmoyV wrote:
Bunuel wrote:
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.



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


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


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
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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 31 Mar 2015, 10:46
1
Bunuel wrote:
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.


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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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 31 Mar 2015, 18:49
1
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!!!
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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 01 Apr 2015, 04:53
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
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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 01 Apr 2015, 22:39
1
Tmoni26 wrote:
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


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.
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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 02 Apr 2015, 01:10
1
Bunuel wrote:
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.


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
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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 06 Apr 2015, 04:46
Bunuel wrote:
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.


MAGOOSH OFFICIAL SOLUTION:
Attachment:
abcdef_text.PNG
abcdef_text.PNG [ 17.26 KiB | Viewed 4886 times ]

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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 23 Aug 2017, 02:02
Apologies, i'm getting confused with total integers, why do you multiply the numbers? Shouldn't each of this be treated as individual numbers?
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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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Re: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 23 Aug 2017, 02:22
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?
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If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 23 Aug 2017, 02:25
bae wrote:
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.
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If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is  [#permalink]

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New post 23 Aug 2017, 14:18
Bunuel wrote:
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.

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
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