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# xy and yx are a pair of two digit positive integers with reversed digi

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Director
Joined: 07 Dec 2014
Posts: 923
xy and yx are a pair of two digit positive integers with reversed digi [#permalink]

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11 Nov 2017, 17:11
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65% (hard)

Question Stats:

44% (01:29) correct 56% (01:26) wrong based on 32 sessions

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xy and yx are a pair of two digit positive integers with reversed digits. For how many such pairs does yx-xy=y^2-x^2?

A. 2
B. 3
C. 4
D. 5
E. 6
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Aug 2009
Posts: 5720
xy and yx are a pair of two digit positive integers with reversed digi [#permalink]

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11 Nov 2017, 19:53
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gracie wrote:
xy and yx are a pair of two digit positive integers with reversed digits. For how many such pairs does yx-xy=y^2-x^2?

A. 2
B. 3
C. 4
D. 5
E. 6

Hi..

=$$yx-xy=y^2-x^2...........10y+x-10x-y+(x-y)(x+y)=0.......9(x-y)+(x-y)(x+y)=0.. (y-x)(9-(x+y)=0$$
Either x=y or x+y=9.... But pair means x and y should be different otherwise pair would consist of just one number example 22
So numbers with sum of digits as 9 are
18 and 81
27 and 72
36 and 63
45 and 54

Since pair of numbers is asked 18 and 81 will be same as 81 and 18.

C
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xy and yx are a pair of two digit positive integers with reversed digi [#permalink]

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11 Nov 2017, 20:19
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gracie wrote:
xy and yx are a pair of two digit positive integers with reversed digits. For how many such pairs does yx-xy=y^2-x^2?

A. 2
B. 3
C. 4
D. 5
E. 6

$$yx-xy=y^2-x^2$$

$$=>10y+x-10x-y=(y-x)(y+x) => 9y-9x-(y-x)(y+x)=0$$

or $$(y-x)[9-(y+x)]=0 =>$$ either $$y=x$$ or $$y+x=9$$

Now it is given that $$xy$$ & $$yx$$ are a pair of two digit integers, hence $$y$$ is not equal to $$x$$ because in that case there will be only a single no and not a pair

so $$y+x=9$$ the pairs possible are

(1,8), (2,7), (3,6), (4,5) = $$4$$.........[here (4,5) & (5,4) represent the same pair]

Option C
xy and yx are a pair of two digit positive integers with reversed digi   [#permalink] 11 Nov 2017, 20:19
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# xy and yx are a pair of two digit positive integers with reversed digi

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