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The product of two positive integers is 96. The sum of the two number

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kiran120680
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The product of two positive integers is 96. The sum of the two number [#permalink] The product of two positive integers is 96. The sum of the two number   17 Apr 2019, 05:06
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The product of two positive integers is 96. The sum of the two number is 22. Which of the following could be the difference of the two integers?


A. 2
B. 5
C. 6
D. 8
E. 10
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E
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lucajava
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The product of two positive integers is 96. The sum of the two number [#permalink] The product of two positive integers is 96. The sum of the two number   17 Apr 2019, 10:34
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\(ab = 96 = 2^5 * 3\)

\(a + b = 22\)

A few attempts will lead you to the right answer. In this case, \(96 = 16 * 6\) and \(16 + 6 = 22\). Hence, \(16 - 6 = 10\).
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sahibr
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Re: The product of two positive integers is 96. The sum of the two number [#permalink] The product of two positive integers is 96. The sum of the two number   17 Apr 2019, 17:31
Let the two integers be x and y.

Hence from the question we know that,

xy=96

x+y=22
Or y=22-x

Hence xy = 96
can be written as
x(22-x)=96

=>22x-x(sq)=96

=>x(sq)-22x+96=0

=>x(sq)-16x-6x+96=0

=>x(x-16)-6(x-16)=0

=>(x-16)(x-6)=0

=>x=16 or x=6

Hence, y=16 or y=6 respectively

Difference between the two integers[x-y or y-x] gives us 16-6=10

Hence, option E.
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Re: The product of two positive integers is 96. The sum of the two number [#permalink] The product of two positive integers is 96. The sum of the two number   19 Apr 2019, 10:16
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kiran120680 wrote:
The product of two positive integers is 96. The sum of the two number is 22. Which of the following could be the difference of the two integers?


A. 2
B. 5
C. 6
D. 8
E. 10


We can create the equations in which x = one number and y = the other number:

xy = 96

and

x + y = 22

x = 22 - y

Substituting, we have:

(22 - y)y = 96

22y - y^2 = 96

y^2 - 22y + 96 = 0

(y - 16)(y - 6) = 0

So y = 16 or 6

When y = 16, x = 6, or when y = 6, x = 16.

So the difference could be 10.

Alternate Solution:

Letting x and y denote the numbers, we are given that x + y = 22 and xy = 96. Let’s find (x + y)^2:

(x + y)^2 = 484

x^2 + 2xy + y^2 = 484

Let’s subtract 4xy = 384 from each side of this equation:

x^2 + 2xy + y^2 - 4xy = 484 - 384

x^2 - 2xy + y^2 = 100

(x - y)^2 = 100

x - y = 10 or x - y = -10

Answer: E
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Hasini22
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Re: The product of two positive integers is 96. The sum of the two number [#permalink] The product of two positive integers is 96. The sum of the two number   07 Dec 2023, 17:30
Products of 96
(1*96,2*48,3*32,4*24,6*16)
Only adding 6 and 16 gives 22.
Therefore the difference is 10

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Re: The product of two positive integers is 96. The sum of the two number [#permalink] The product of two positive integers is 96. The sum of the two number   08 Dec 2023, 19:37
i solved it on the basis of logic
Stmt 1 - xy = 96 --Which means either x or y is even
Stmt 2 - x + y = 22 ---Since sum is even therefore both x and y even.

Now unit digit of 96 is 6. Therefore look for number that satisfy these conditions. You will arrive at 16,6 in a jiffy
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Re: The product of two positive integers is 96. The sum of the two number [#permalink]
08 Dec 2023, 19:37
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