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If y>x & x^2 + y^2 = 5101, the value of yx is
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06 Apr 2014, 19:47
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If y>x & \(x^2 + y^2 = 5101\), the value of yx is A: 39 B: 259 C: 547 D: 1 E: 408 I managed to get the answer, took above 2 minutes. Want to know if this can be solved algebraically?
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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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09 Apr 2014, 08:04
ankushbassi wrote: Bunuel,Can you please suggest a shortcut time saving approach for such questions.trying and fixing values take lots of time. Let me give you one that comes to my mind. Focusing on big picture gets you to the answer fairly quickly and without many calculations. Note that sum of squares of two numbers is 5101. So the largest a number can be is around 70  75 since 70^2 = 4900. So the difference of the two numbers cannot be a three digit number at all. Hence we are down to two options: 1 or 39. Checking for 1 is quite easy. If the numbers are consecutive, their squares will be very similar. This means that their squares will be around 2500 since that is approximately the half of 5101. So the numbers could be 50 and 51. All you need to do is check whether 51^2 = 2601 (i.e. 5101  2500). Since it is, so the difference between y and x is 1 and the answer is (D)
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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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06 Apr 2014, 21:04
PareshGmat wrote: If y>x & \(x^2 + y^2 = 5101\), the value of yx is
A: 39
B: 259
C: 547
D: 1
E: 408
I managed to get the answer, took above 2 minutes. Want to know if this can be solved algebraically? write down : nos> 0 1 2 3 4 5 6 7 8 9 Squares(only their unit digit like unit digit of square of 9 is 81 is 1 )> 0 1 4 9 6 5 6 9 4 1 now,observe that \(x^2 + y^2 = 5101\) so unit digit is 1 in above combinations we have 0+1=1 5+6=1 thus, we are asked XY : we get 10=1 65=1. hence D. Hope it is clear Cheers, ~Pegasus.



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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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06 Apr 2014, 21:29
PareshGmat wrote: If y>x & \(x^2 + y^2 = 5101\), the value of yx is
A: 39
B: 259
C: 547
D: 1
E: 408
I managed to get the answer, took above 2 minutes. Want to know if this can be solved algebraically? x^2 + y^2 = 5101 = 2500 + 2601 Hence x = 25 and y = 26 so y  x = 1
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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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06 Apr 2014, 21:54
PerfectScores wrote: PareshGmat wrote: If y>x & \(x^2 + y^2 = 5101\), the value of yx is
A: 39
B: 259
C: 547
D: 1
E: 408
I managed to get the answer, took above 2 minutes. Want to know if this can be solved algebraically? x^2 + y^2 = 5101 = 2500 + 2601 Hence x = 25 and y = 26 so y  x = 1 I did in the same way Answer is correct; correcting the typo \(50^2 = 2500\) \(51^2 = 2601\) x = 50 y = 51
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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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06 Apr 2014, 22:03
PareshGmat wrote: PerfectScores wrote: PareshGmat wrote: If y>x & \(x^2 + y^2 = 5101\), the value of yx is
A: 39
B: 259
C: 547
D: 1
E: 408
I managed to get the answer, took above 2 minutes. Want to know if this can be solved algebraically? x^2 + y^2 = 5101 = 2500 + 2601 Hence x = 25 and y = 26 so y  x = 1 I did in the same way Answer is correct; correcting the typo \(50^2 = 2500\) \(51^2 = 2601\) x = 50 y = 51 The sum of two squares should be 5101. The units digit should be 1. Now just find the units digit of squares from 0  9: 0 1 4 9 6 5 6 9 4 1  Only 0 + 1 will give you 1. Hence you have few options to try out and eventually you will come to 50^2 + 51^2 (try 10, 20, 30, 40 and 50)
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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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08 Apr 2014, 21:23
Bunuel,Can you please suggest a shortcut time saving approach for such questions.trying and fixing values take lots of time.



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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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09 Apr 2014, 08:51
I did it this way to eliminate some choice first.
\((yx)^2+2xy=5101\) \(\rightarrow xy=\frac{[5101(yx)^2]}{2}\)
We have \(xy=\) A. 1790 B. \(<0\) C. \(<0\) D. 2550 E. \(<0\)
Only with choice D, we find \(x=50,y=51\) (for in choice A 179 is a prime number).



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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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09 Apr 2014, 10:55
x^2 +y^2 = 5101 hence, 5101 is sum of two squares. Take the midpoint of 5101 and check for the immediate perfect square lower to it. 50^2 = 2500 is the closest perfect square lower to 5101/2 Now among the option, B, C and E are out of question as the difference is 3 digits and square of them is much higher than 5101. so, xy can be 1 or 39, given 50^2 = 2500, we add one to 50 and check if its square is 2601. and 51^2 = 2601, so answer is 1.



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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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16 Jul 2014, 12:15
Is there anyone who solved this in an algebraic way?



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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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16 Jul 2014, 19:04
ronr34 wrote: Is there anyone who solved this in an algebraic way? I was waiting for the same. Seems that this may hardly be solved algebraically
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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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16 Jul 2014, 19:49
PareshGmat wrote: ronr34 wrote: Is there anyone who solved this in an algebraic way? I was waiting for the same. Seems that this may hardly be solved algebraically This is how I solved it. Since sum of square is 5101. That means one of number's square must come 1 as unit digit. so one of number's unit is either 1 or 9. Either way other number's unit digit will be 0. only one of the answer choice have unit digit as 1, that is the answer "D"



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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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16 Jul 2014, 22:08
PareshGmat wrote: ronr34 wrote: Is there anyone who solved this in an algebraic way? I was waiting for the same. Seems that this may hardly be solved algebraically Note that you have only one equation and two unknowns. There are a lot of other constraints that are hard to handle algebraically. Not every GMAT question is meant for an algebra solution. There are some which are meant to be solved using reasoning.
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Re: If y>x & x^2 + y^2 = 5101, the value of yx is
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