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xy and yx are reversed two digit integers. If the sum of the digits eq

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xy and yx are reversed two digit integers. If the sum of the digits eq  [#permalink]

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New post 31 Mar 2019, 16:57
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A
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E

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xy and yx are reversed two digit integers. If the sum of the digits equals the difference between the squares of the digits, what could be the product of the digits?

A. 8
B. 12
C. 15
D.18
E. 24
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Re: xy and yx are reversed two digit integers. If the sum of the digits eq  [#permalink]

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New post 31 Mar 2019, 17:35
Just calculate each option. For B 1+2=3 and 2^2-1=4-1=3.

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Re: xy and yx are reversed two digit integers. If the sum of the digits eq  [#permalink]

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New post 31 Mar 2019, 17:52
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If the sum of the digits equals the difference between the squares of the digits.
=> (x + y) = (x^2 - y^2)
=> (x + y) = (x + y)(x - y)
=>x-y = 1
So, XY = 21, 32, 43, 54, 65, 76, 87, 98

From the above possible numbers, the only product in options is 12 for number 43.

Answer: B
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Re: xy and yx are reversed two digit integers. If the sum of the digits eq  [#permalink]

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New post 31 Mar 2019, 20:20
Lowkya wrote:
If the sum of the digits equals the difference between the squares of the digits.
=> (x + y) = (x^2 - y^2)
=> (x + y) = (x + y)(x - y)
=>x-y = 1
So, XY = 21, 32, 43, 54, 65, 76, 87, 98

From the above possible numbers, the only product in options is 12 for number 43.



Answer: B




Can u explain how did u arrive at B? It is asking product of the digits

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Re: xy and yx are reversed two digit integers. If the sum of the digits eq  [#permalink]

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New post 01 Apr 2019, 08:40
1
Shef08 wrote:
Lowkya wrote:
If the sum of the digits equals the difference between the squares of the digits.
=> (x + y) = (x^2 - y^2)
=> (x + y) = (x + y)(x - y)
=>x-y = 1
So, XY = 21, 32, 43, 54, 65, 76, 87, 98

From the above possible numbers, the only product in options is 12 for number 43.



Answer: B




Can u explain how did u arrive at B? It is asking product of the digits

Posted from my mobile device


Sure.
Yes, it is asking for the product of digits.

From the above solution you can see that possible numbers are XY = 21, 32, 43, 54, 65, 76, 87, 98.
=> Possible products are - 2, 6, 12,20, 30, 42, 56, 72.


Only 12 is in the options. So, B.

Hope this helps.

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Re: xy and yx are reversed two digit integers. If the sum of the digits eq  [#permalink]

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New post 03 Apr 2019, 18:28
gracie wrote:
xy and yx are reversed two digit integers. If the sum of the digits equals the difference between the squares of the digits, what could be the product of the digits?

A. 8
B. 12
C. 15
D.18
E. 24


Without loss of generality of the problem, we can assume that x > y. So we have:

x + y = x^2 - y^2

x + y = (x + y)(x - y)

Dividing both sides of the equation by (x + y) (since it won’t be 0), we have:

1 = x - y

The only way we can get (x - y) to equal 1 is if x is one more than y. Therefore, we see that x and y must be consecutive integers. Of the answer choices, we see that only 12 can be the product of two consecutive integers (12 = 3 x 4), so 12 is the correct answer.

Answer: B
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Re: xy and yx are reversed two digit integers. If the sum of the digits eq   [#permalink] 03 Apr 2019, 18:28
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