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# In the x-y plane, point (p, q) is a lattice point if both p

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Manager
Joined: 06 Oct 2015
Posts: 86
Re: In the x-y plane, point (p, q) is a lattice point if both p  [#permalink]

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10 Sep 2016, 02:16
Awesome. That helps me a lot. Thank you.
Manager
Joined: 23 Aug 2017
Posts: 117
Schools: ISB '21 (A)
Re: In the x-y plane, point (p, q) is a lattice point if both p  [#permalink]

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20 Jun 2019, 09:20
1

Here's how I approached the problem:

The eqn of the circle is : (x+2)^2 + (y-1)^2 =36
So for all points inside the circle the expression becomes (x+2)^2 + (y-1)^2 <36
We need to find integer solutions(x,y) for this
So both (x+2) and (y-1) can range from min -5 to max 5 ...We need to find the no of suitable combinations...For each value of (x+2) and (y-1) we get unique values for x and y...So lets say x+2=a and y-1=b
a^2 + b^2 < 36
For a=0 : b= -5 to 5....11 sets of (a,b)
for a=1/-1: b= -5 to 5....11*2=22 sets
for a=2/-2: b= -5 to 5....11*2=22 sets
for a=3/-3: b= -5 to 5....11*2=22 sets (since 3^2 + 5^2 < 36)
for a=4/-4: b= -4 to 4......9*2=18 sets (since 4^2 +5^2 > 36....but 4^2 + 3^2 < 36)
for a=5/-5: b= -3 to 3......7*2=14 sets (since 5^2 +4^2 > 36...but 5^2 + 3^2 < 36)

So total no of solutions = 11+22+22+22+18+14 = 109

Thanks
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Re: In the x-y plane, point (p, q) is a lattice point if both p  [#permalink]

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20 Jun 2019, 21:04
Debashis Roy wrote:

Here's how I approached the problem:

The eqn of the circle is : (x+2)^2 + (y-1)^2 =36
So for all points inside the circle the expression becomes (x+2)^2 + (y-1)^2 <36
We need to find integer solutions(x,y) for this
So both (x+2) and (y-1) can range from min -5 to max 5 ...We need to find the no of suitable combinations...For each value of (x+2) and (y-1) we get unique values for x and y...So lets say x+2=a and y-1=b
a^2 + b^2 < 36
For a=0 : b= -5 to 5....11 sets of (a,b)
for a=1/-1: b= -5 to 5....11*2=22 sets
for a=2/-2: b= -5 to 5....11*2=22 sets
for a=3/-3: b= -5 to 5....11*2=22 sets (since 3^2 + 5^2 < 36)
for a=4/-4: b= -4 to 4......9*2=18 sets (since 4^2 +5^2 > 36....but 4^2 + 3^2 < 36)
for a=5/-5: b= -3 to 3......7*2=14 sets (since 5^2 +4^2 > 36...but 5^2 + 3^2 < 36)

So total no of solutions = 11+22+22+22+18+14 = 109

Thanks

Yes, this is perfectly fine too. Though use of symmetry can reduce some effort.
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Karishma
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Joined: 23 Aug 2017
Posts: 117
Schools: ISB '21 (A)
Re: In the x-y plane, point (p, q) is a lattice point if both p  [#permalink]

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20 Jun 2019, 21:34
Yep definitely ur solution is more concise
Manager
Joined: 07 May 2018
Posts: 56
Re: In the x-y plane, point (p, q) is a lattice point if both p  [#permalink]

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24 Jun 2019, 13:15
krazyvshank wrote:
mikemcgarry wrote:
jakeqs wrote:
Not sure if this works but seems the easiest way would be to find area of the circle=36Pi and subtract the 4 lattice points on the circle. Maybe it only works in this example and not always.

Dear jakeqs,
As I expressed to tvrs09 above, what you are suggesting is an approximation method. Estimation is a perfectly fine strategy for the GMAT Quant.
If the answers are spread out, then estimation has a better chance of determining a unique answer. In what you suggested here, it was a totally lucky fluke that it wound up with the exact answer. Often, estimation does not produce an exact answer, but that's OK. Again, if the answer choices are widely space apart, then estimation is often enough to determine a single answer.
Does all this make sense?
Mike

Mike,

This is a beautiful question. I approached this sort of like a permutations problem. We have x^2 + y^2 < 36 for points to lie inside the circle.
Taking integer squares we can consider 0,1,2,3,4,5,-1,-2,-3,-4,-5. 11 in total.
For 0,1,2,3,-1,-2,-3: we can have 11 possible pairs (including itself): 11*7 = 77
For 4,-4: we can't have 5 or -5: 2 * (11-2) = 18
For 5,-5: we can't have 5,-5,4,-4: 2 * (11-4) = 14
77 + 18 + 14 = 109.
Hope this is also an alternate simple method.

Hi krazyvshank,

Could you explain why you did two times the (11-2)
Re: In the x-y plane, point (p, q) is a lattice point if both p   [#permalink] 24 Jun 2019, 13:15

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