Which of the following is the product of two integers whose

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

Hey guys! I managed to figure a shorter way out.

For the sum of 2 numbers to be odd i.e. 11, we must have a case of one even number and one odd number since -

Even + Odd = Odd

Thus on observing the numbers, first try numbers that have both even and odd factors. Since all numbers are clearly factors of 2, simply try checking for odd numbers.

A. -42 (One number must be positive and one must be negative)
7 x 6, 3 x 14, 2 x 21
B. -28
7 x 4 (Won't fulfill the condition when you consider any one number to be negative) , 2 x 14
C. 12
6 x 2, 4 x 3, 12 x 1 (won't fulfill the condition)
D. 16
Not considering since I know it has all even factors
E. 32
Not considering since I know it has all even factors

Answer: A

-42..
took some time..would love to see a short cut..

14*3=42 14-3=11

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?

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.

neat..same as my method but its not short enough took a valuabe 50 seconds ..cause i started with answer choice E...

Bunuel , Any neat solution for this one please?

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.

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.
_________________

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.

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.

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

[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.

For the sum of two integers to be positive 11, the two integers could both be positive or one of it is positive and the other one is negative. Since the sum of the two integers is positive 11 , then the negative one must be smaller one of the two integers.

-3 + 14 = 11 and (-3)(14)= -42

how did you take a*b=42 , this was not given in the question
The only information given was a+b = 11

Took me 3 minutes. But I saw this:

x + y = -11 --> odd + even = odd

Therefore,
odd x even = even...all of the choices are even.

Which odd/even numbers add to -11?
(-6) + (-5)
(-4) + (-7)
(-2) + (-9)
(-8) + (-3)
(-14) + (3) YES <-- only one that gives us a solution in the list.
_________________

As per question we need to find the value of x (11-x) = 11x - x^2

we need to check all the values from Option A to E. Fortunately Option A provides us with the solutions with prime factorization of equation -x^2 + 11x + 42 = 0, (x+14) (x-3) = 0.

No other options satisfies the equation.

Hope it helps

Correct option : A

we all know formula with quadratic Equation for its roots (x-a)(x-b)

- a\({x^2}\) + bx + c

- (-b + \(\sqrt{{{b^2} - 4ac}}\)) / 2

- the only option gives Integer when substitute is (-42) rest all are fractions or decimals