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If p, x, and y are positive integers, y is odd, and p = x^2
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21 May 2020, 15:49
VeritasKarishma wrote: netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 Such questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibilityandremaindersifyou.htmlComing to your question, First thing that comes to mind is if y is odd, \(y^2\) is also odd. If \(y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1\) Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1. Stmnt 1: When p is divided by 8, the remainder is 5. When y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4. \(x^2 = 8a + 4\) (i.e. we can make 'a' groups of 8 and 4 will be leftover) \(x^2 = 4(2a+1)\) This implies \(x = 2*\sqrt{Odd Number}\)because (2a+1) is an odd number. Square root of an odd number will also be odd. Therefore, we can say that x is not divisible by 4. Sufficient. Stmnt 2: x  y = 3 Since y is odd, we can say that x will be even (Even  Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient. Answer (A). Hi Experts  chetan2u, VeritasKarishma, Bunuel, nick1816, GMATBustersJust wondering why doesn't my method come to the same conclusion as yours ? i thought A was INS P = odd obviously ( i listed out values of P like 5  13  21  29  37  45  53  61 ..... and they were all odd) P = \(x^{2}\) + \(y^{2}\) We know Y = odd. Thus \(y^{2}\) will be odd as well Thus \(x^{2}\) = P  \(y^{2}\) \(x^{2}\) = Odd  \(Odd^{2}\) \(x^{2}\) = Even ( as odd  odd = Even) Thus x = \(\sqrt[]{Even}\) or x = Even Given x is even  x can be 2 or 4 or 6 or 8 Thus INS Please let me know why doesn't this strategy work



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If p, x, and y are positive integers, y is odd, and p = x^2
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22 May 2020, 07:31
 first of all always write completely no selfmade abbreviation.
It took me more than 30 sec to conclude INS = insufficient.
 Also u said p is odd obviously. you should write as per statement 1 p is odd.
quoting p is odd without writing St 1 is not right, it seems u r concluding p is odd without St1.
 u have taken p as an odd integer, it is not just an odd integer, it is a specific type of odd integer, the specific info u have missed.
p = 5, 13, 21... not p can be any odd integer.
so u got x = even is right, but x is not any even integer it is even integer which is not multiple of 2 ( you missed this because u took p as any odd number) x cannot be equal to 4 or 8... these are the flaws in your solution, for correct solution, refer VeritasKarishma's solution.
Happy learning jabhatta@umail.iu.edu wrote: VeritasKarishma wrote: netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 Such questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibilityandremaindersifyou.htmlComing to your question, First thing that comes to mind is if y is odd, \(y^2\) is also odd. If \(y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1\) Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1. Stmnt 1: When p is divided by 8, the remainder is 5. When y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4. \(x^2 = 8a + 4\) (i.e. we can make 'a' groups of 8 and 4 will be leftover) \(x^2 = 4(2a+1)\) This implies \(x = 2*\sqrt{Odd Number}\)because (2a+1) is an odd number. Square root of an odd number will also be odd. Therefore, we can say that x is not divisible by 4. Sufficient. Stmnt 2: x  y = 3 Since y is odd, we can say that x will be even (Even  Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient. Answer (A). Hi Experts  chetan2u, VeritasKarishma, Bunuel, nick1816, GMATBustersJust wondering why doesn't my method come to the same conclusion as yours ? i thought A was INS P = odd obviously ( i listed out values of P like 5  13  21  29  37  45  53  61 ..... and they were all odd) P = \(x^{2}\) + \(y^{2}\) We know Y = odd. Thus \(y^{2}\) will be odd as well Thus \(x^{2}\) = P  \(y^{2}\) \(x^{2}\) = Odd  \(Odd^{2}\) \(x^{2}\) = Even ( as odd  odd = Even) Thus x = \(\sqrt[]{Even}\) or x = Even Given x is even  x can be 2 or 4 or 6 or 8 Thus INS Please let me know why doesn't this strategy work
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Re: If p, x, and y are positive integers, y is odd, and p = x^2
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25 May 2020, 15:23
jabhatta@umail.iu.edu wrote: VeritasKarishma wrote: netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 Answer (A). Just wondering why doesn't my method come to the same conclusion as yours ? i thought A was INS P = odd obviously ( i listed out values of P like 5  13  21  29  37  45  53  61 ..... and they were all odd) P = \(x^{2}\) + \(y^{2}\) We know Y = odd. Thus \(y^{2}\) will be odd as well Thus \(x^{2}\) = P  \(y^{2}\) \(x^{2}\) = Odd  \(Odd^{2}\) \(x^{2}\) = Even ( as odd  odd = Even) Thus x = \(\sqrt[]{Even}\) or x = Even Given x is even  x can be 2 or 4 or 6 or 8 Thus INS Please let me know why doesn't this strategy work a couple things you missed analyzing in statement 1: 1. Statement 1 can be written as \(x^{2}\) + \(y^{2}\) = 8m + 5 you establish that \(x^{2}\) is even and \(y^{2}\) is odd. 2. you seemed to not evaluate any further here. You can actually use statement 1 to further test and determine what you know. 3. so to fill in the gaps  I would have continued to assess what values of x can be in the statement. The only one that you care to eliminate is 2. 4. I personally filled in values of m starting with 0 since i need to find the smallest value that x could be. m can't be negative because you know any number \(n^{2}\) produces a non negative number and will create a negative if less than 0. 5. when you try to plug 0 in, you realize you cannot and that the smallest value of m is 1 in order for the formula to work. so when m = 1, x = 4. this is the minimum value of m and also then the minimum value of x. x is divisible by 4.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2
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28 May 2020, 11:20
VeritasKarishma wrote: Eshaninan wrote: Am i missing something, why cant we take stmt 2 as follows: squaring xy=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4 The question is: "Is x divisible by 4?" not "Is p divisible by 4?" x is even since y is odd. We don't know whether x is divisible by only 2 or 4 as well. Regarding Statement 2  If x  y = 3... squaring both the sides we get x^2+y^2  2xy = 9... since x^2+y^2 = p we can say that p  2xy = 9 or p = 9 + 2xythis implies that p = odd + even number hence p = odd since p is odd it can not be divisible by 4. Hence, sufficient and correct choice should be D. Is the thought process correct, if not, why?



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Re: If p, x, and y are positive integers, y is odd, and p = x^2
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05 Jul 2020, 12:47
chauhandeepak8 wrote: VeritasKarishma wrote: Eshaninan wrote: Am i missing something, why cant we take stmt 2 as follows: squaring xy=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4 The question is: "Is x divisible by 4?" not "Is p divisible by 4?" x is even since y is odd. We don't know whether x is divisible by only 2 or 4 as well. Regarding Statement 2  If x  y = 3... squaring both the sides we get x^2+y^2  2xy = 9... since x^2+y^2 = p we can say that p  2xy = 9 or p = 9 + 2xythis implies that p = odd + even number hence p = odd since p is odd it can not be divisible by 4. Hence, sufficient and correct choice should be D. Is the thought process correct, if not, why? The question is if x is divisible by 4 and not if p is divisible by 4. Based on your thought process since p is odd and it is given that y is odd and p = x^2 + y^2, we know that x^2 should be even. If x = 2, then x is not divisible by 4 and if x = 4, then it is divisible by 4. Hence not sufficient.



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Re: If p, x, and y are positive integers, y is odd, and p = x^2
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16 Jul 2020, 11:22
netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3 I think it was just easier to use various numbers Statement II is insufficient alone. For Statement I, I used various numbers for Option A and noted that x is not divisible by 4 Given Statement 1: p when divided by 8 leaves a remainder of 5 p = 8q + 5 Numbers that work here for p are 5, 13, 21, 29, 37, 45, 53 etc., Also, p = x^2 + y^2, y is odd So of p = 5 > y should be 1 since when y = 3, y^2 = 9. Therefore, p = 2^2 + 1^2 > x = 2 [Not div. by 4] Similarly, when p = 13 > p = 2^2 + 3^2 > x = 2[Not div. by 4] p = 21 > Can't think of numbers here p = 29 > p = 2^2 + 5^2 > x = 2[Not div by 4] p = 37 > p = 6^2 + 1^2 > x = 6[Not div by 4] p = 45 > p = 6^2 + 3^2 > x = 6[Not div by 4] p = 53 > p = 2^2 + 7^2 > x = 2[Not div by 4] Since x is not div by 4 for any of the situations above, it is safe to conclude that A is sufficient !!!
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Re: If p, x, and y are positive integers, y is odd, and p = x^2
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