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# In the x-y coordinate plane, the distance between (m, n) and (3, 4) is

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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In the x-y coordinate plane, the distance between (m, n) and (3, 4) is [#permalink]

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12 Dec 2016, 23:55
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Question Stats:

51% (01:32) correct 49% (01:55) wrong based on 47 sessions

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In the x-y coordinate plane, the distance between (m, n) and (3, 4) is$$√26$$. If m and n are integers, how many possible cases are there?
A. 5 B. 6 C. 7 D. 8 E. 9
[Reveal] Spoiler: OA

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4898
GPA: 3.82
Re: In the x-y coordinate plane, the distance between (m, n) and (3, 4) is [#permalink]

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15 Dec 2016, 00:15
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==> From $$(m-3)^2+(n-4)^2$$=(√26)2, you get (m-3, n-4)=(±1, ±5) or (±5, ±1), so there are total of 8 numbers. Therefore, the answer is D.

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Joined: 12 Nov 2016
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Re: In the x-y coordinate plane, the distance between (m, n) and (3, 4) is [#permalink]

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20 Feb 2017, 18:48
I don't understand this question
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 806
Re: In the x-y coordinate plane, the distance between (m, n) and (3, 4) is [#permalink]

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20 Feb 2017, 22:18
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Nunuboy1994 wrote:
I don't understand this question

• The distance between any two points in a coordiante plane is given by -
o $$(x_1-x_2)^2 + (y_1-y_2)^2 = d^2$$
o where $$d$$ is the distance between the points ($$x_1,y_1$$) and ($$x_2,y_2$$)
• In the question, we are given that the distance between ($$m,n$$) and ($$3,4$$) is $$√26$$
• Using the above formula, we can write -
o $$(m-3)^2 + (n-4)^2 = (√26)^2$$
o $$(m-3)^2 + (n-4)^2 = 26$$
• Since m and n are integers, the value of $$(m-3)^2$$ and $$(n-4)^2$$ will also be integers.
o We can write $$(m-3)^2 + (n-4)^2 = 25 + 1$$
o Now there are a number of possibilities - either
 $$(m-3)^2 = 25$$ and $$(n-4)^2 = 1$$
 which gives us $$m-3 =$$ +$$5$$ or -$$5$$ and $$(n-4) =$$+$$1$$or -$$1$$
• Thus from here we will get 2 possibilities of m and 2 possibilities of n. And the total number of cases possible are $$2 *2 = 4$$
OR
• There is also a possibility that $$(m-3)^2 = 1$$ and $$(n-4)^2 = 5$$
o which gives us $$m-3 =$$ +$$1$$ or -$$1$$ and $$n-4 =$$ +$$5$$ or -$$5$$
• Thus, from here also there are 2 possibilities each for m and n. And the total number of cases possible are $$2 * 2 = 4$$
• Therefore, the total number of cases possible for ($$m, n$$) are $$4 + 4 = 8$$.

Thanks,
Saquib
Quant Expert
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Re: In the x-y coordinate plane, the distance between (m, n) and (3, 4) is   [#permalink] 20 Feb 2017, 22:18
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