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Which of the following is the product of two integers whose

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Which of the following is the product of two integers whose [#permalink]

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New post 29 Mar 2008, 08:09
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Which of the following is the product of two integers whose sum is 11?

(A) -42
(B) -28
(C) 12
(D) 26
(E) 32
[Reveal] Spoiler: OA
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Re: Which of the following is the product of two integers whose [#permalink]

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New post 29 Mar 2008, 08:20
-42..
took some time..would love to see a short cut..

14*3=42 14-3=11
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New post 29 Mar 2008, 08:36
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I have managed to solve it, but I am wondering whether there is a more efficient process.

I've done as below:

x + y = 11
x= 11-y


xy=(11-y)y=11y-y^2

Then, I plugged in with:
y=1 -> xy=11-1=10
y=2 -> xy=22-4=18
y=3 -> xy=33-9=24
y=4 -> xy=44-16=28
y=5 -> xy=55-25=30
y=6 -> xy=66-36=30
y=7 -> xy=77-49=28

Then I thought, that xy must be negative, then started to plug-in negative numbers:
y=-1 -> xy=-11-1=-12
y=-2 -> xy=-22-4=-26
y=-3 -> xy=-33-9=-42

Bingo! But this takes far too long. Can anyone help?
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New post 29 Mar 2008, 08:43
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Yeah, there doesn't seem to be an easy way to just plug stuff in and solve algebraically. My method may shed some light though:

let the two numbers be x and y, and N be the answer

With the given information, I know that
xy=N and that x+y = 11

which gives:
x(11-x)=N
11x-x^2=N
0 = x^2 - 11x + N
0=(x ?)(x ?)

At this point, I had the following thoughts:

(A) -42 : to get -42, it must be (x + ?)(x - ?) so how can a positive and negative number add to -11 and yet still multiply together for -42?
(B) -28 : similar thought process to (A) if you didn't start by evaluating A first
(C) 12: (x + ?)(x + ?) or (x - ?)(x - ?)
(D) 26
(E) 32


Truthfully, this is one of those questions where going through the entire calculation may not be necessary and just trying out numbers would be quicker.
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New post 29 Mar 2008, 09:27
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a+b=11

a*b=42

a*(11-a) = 42

a^2 - 11a + 42 = 0

(a-14)(a+3) --> a=14 is a valid solution. therefore, b=-3 .... a+b = 14-3 = 11 and a*b=(14)*(-3) = -42
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New post 29 Mar 2008, 09:33
neat..same as my method but its not short enough took a valuabe 50 seconds ..cause i started with answer choice E...


pmenon wrote:
a+b=11

a*b=42

a*(11-a) = 42

a^2 - 11a + 42 = 0

(a-14)(a+3) --> a=14 is a valid solution. therefore, b=-3 .... a+b = 14-3 = 11 and a*b=(14)*(-3) = -42
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Re: Which of the following is the product of two integers whose [#permalink]

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Re: Which of the following is the product of two integers whose [#permalink]

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New post 29 Jan 2015, 10:09
Bunuel , Any neat solution for this one please?
:(
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Re: Which of the following is the product of two integers whose [#permalink]

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New post 30 Dec 2015, 20:38
A+B=11
A=1 B=10 -> AB=10 no
A=2 B=9 -> 18 no
A=3 B=8 - no
A=4 B=7 - NO
A=5 B=6 - NO
A=7 B=4 - NO
A=8 B=3 - NO
A=9 B=2 - NO
A=10 B=1 - N0
A=11 B=0 - NO
A=12 B=-1 - NO
A=13 B=-2 - NO
A=14 B=-3 -> AB=-42. WE HAVE SUCH AN ANSWER, THUS A IS THE CORRECT ONE.
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Which of the following is the product of two integers whose [#permalink]

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New post 03 Mar 2016, 10:17
Answer A.

It helps to break down the answer choices into (prime) factors and adding the (prime) factors to see whether the sum might equal 11.

\(-42\) = \(2 * 21 * -1\) = \(2 * 7 * 3 * -1\)
Adding the factors together yields 11 --> \(2 + 7 + 3 -1 = 11\)
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Which of the following is the product of two integers whose [#permalink]

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New post 03 Mar 2016, 10:26
Expert's post
saiesta wrote:
Answer A.

It helps to break down the answer choices into (prime) factors and adding the (prime) factors to see whether the sum might equal 11.

\(-42\) = \(2 * 21 * -1\) = \(2 * 7 * 3 * -1\)
Adding the factors together yields 11 --> \(2 + 7 + 3 -1 = 11\)


-1 is NOT a prime factor. Prime factors as per the definition are positive integers. You were lucky with your approach.

The simplest way is to realize that x+y=11 for which xy = options given. When you see option (A), it should trigger the observation that you can even use negative integers.

Factors of 42 : 1,2,3,6,7,14,21,42 (and their negative counterparts). Now see that -3 and 14 are in this group and 14+(-3)=11 (14*-3=-42). Thus A is the correct answer.

Hope this helps.
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Re: Which of the following is the product of two integers whose [#permalink]

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New post 12 Mar 2016, 04:40
lumone wrote:
Which of the following is the product of two integers whose sum is 11?

(A) -42
(B) -28
(C) 12
(D) 26
(E) 32



I solved this Question like this,

Let the two integers are x,y
x+y=11 (Given)
xy=? (Needed)

instead of solving this algebraically, Test the Answer choices

A. -42
Do the factorization : (-1,42)----> There sum is not 11--eliminate
(-2,21)---->There sum is not 11--eliminate
(-3,14)-----> there sum is 11 Bingo!!!!

So, my answer is A...

As the answer is in A, it took me very less time to answer the question. but i think this method is be simple and efficient.
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Re: Which of the following is the product of two integers whose [#permalink]

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New post 13 Mar 2016, 06:54
lumone wrote:
Which of the following is the product of two integers whose sum is 11?

(A) -42
(B) -28
(C) 12
(D) 26
(E) 32


I just plugged in the values. If both the integers are positive, then their product will be positive. With positive options, we cannot plug in any values.
Therefore, one integer is positive and another is negative.
(12,1) or (13,2) or (14,3) and so on.
(14, -3) fits the answer.
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Re: Which of the following is the product of two integers whose [#permalink]

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New post 07 May 2016, 02:05
hard one
two number can not both positive because after pick some positive number we can not have the result

5*6=30
4*7=28

one of them must be nagative

42= 2*3*7
we need factorization.

so 2*7 and 3 is ok.

A
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Re: Which of the following is the product of two integers whose [#permalink]

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New post 07 May 2016, 03:41
[quote="lumone"]Which of the following is the product of two integers whose sum is 11?

(A) -42
(B) -28
(C) 12
(D) 26
(E) 32

got it right after putting values.
any shortcut.
Re: Which of the following is the product of two integers whose   [#permalink] 07 May 2016, 03:41
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