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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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01 Mar 2015, 19:45
Lucky2783 wrote: are there more tricks like this , ODD^2 MOD 8 = 1 ?
thanks Lucky There can be innumerable properties of numbers and you can not say which one will be useful and which one not on exam day  hence an exercise in learning all such properties is wasted effort. Instead try to figure out how to arrive at them on your own when required. For example, here y is odd and you need to check divisibility of y^2 by 8. y = 2a+1 y^2 = (2a+1)^2 = 4a^2 + 4a + 1 = 4a(a+1) + 1 Now, notice that one of a and (a+1) must be even so 4a(a+1) has to be divisible by 8. This means y^2 will always leave remainder 1 when divided by 8. Focus on developing these inference skills and they will serve you much better on the exam.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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29 Apr 2015, 22:46
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 Statement 1 : When p is divided by 8, the remainder is 5. This implies that P is odd ( Only odd numbers can leave odd remainders with Even numbers ) Therefore X must be even > Remainder(x^2+y^2)/8 = 5 > Remainder(x^2/8) + Remainder(y^2/8)= 5 > Remainder(x^2/8) + 1 = 5 > Remainder(x^2/8) = 4 Now find value of x by substituting even numbers 2,4,6,... The numbers satisfying above equation are 2,6,10...... which aren't multiples of 4. SUFFICIENT STATEMENT B x – y = 3 X is even ( Given Y is odd ) it can be any even number NOT SUFFICIENT Hence A



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If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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25 Oct 2015, 04:49
\([m]\)[/m] 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 x,y,p are integers >0 ,y is odd (1) p=8a+5 means p is an odd number: \(x^2\)=odd\(odd^2\)=even, as we are delaing only with integers, x is equal to 0,4,16,36,64,100 etc and is divisible by 4 (0 is also an even number and divisible by 4) SUFFICIENT (2) xy=3 means x=odd+odd=even, x=0 is divisible by 4, x=6 is not divisible by 4, two different answers,so NOT SUFFICIENT btw. the difference between (1) and (2) when \(x^2\)=even must be divisible by 4, but when x=even it could be divisible by 4.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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22 Nov 2015, 02:43
Can't we think of it in a simpler way, without using much properties.
Statement 1 says that "When p is divided by 8, the remainder is 5", therefore we can infer that p is an odd number. now since it is given that y is odd therefore square of y will also be odd, by this we come to know that x has to be an even number. P(ODD)=x(EVEN)+y(ODD)
Now, we take the smallest value of a positive even integer for x, i.e. 2. 2^2=4, therefore YES. By rule, square of any any even integer is always divisible by 4. Therefore statement 1 is sufficient
Statement 2 states that x – y = 3, if X=4 and Y=1, then Yes If X=6 and Y=3, then No
hence insufficient.
Answer:A
Im not a pro so might end up doing something wrong. Please do correct me if i make any mistake here.
Thanks



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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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26 Nov 2015, 06:58
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. 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 The question is actually asking whether when x=4m?(m;positive integer), p=(4m)^2+odd^2=odd? There are 3 variables (p,x,y) but only 2 equations are given by the conditions, so there is high chance (D) will be the answer For condition 1, p=8q+5=odd(q: positive integer) p=5,13,21......., When y=1,3,5.... and x=2,6,10.... the answer is always 'no' so the condition is sufficient For condition 2, the answer becomes 'yes' for x=4,y=1, but 'no' for x=6, y=3, so this is insufficient and the answer becomes (A). For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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27 Apr 2017, 04:59
I am a little confused about solving the problem:
Stmnt 2: xy =3
we know y is odd. and if Y is odd as per problem statement y cannot be anything but 1 as x and y both are positive integers.
Therefore x =4 and is divisible by 4.
Please help me understand where I am going wrong with this.



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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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27 Apr 2017, 05:07
aliasjit wrote: I am a little confused about solving the problem:
Stmnt 2: xy =3
we know y is odd. and if Y is odd as per problem statement y cannot be anything but 1 as x and y both are positive integers.
Therefore x =4 and is divisible by 4.
Please help me understand where I am going wrong with this. y is odd, yes, but why must y be 1?
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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03 May 2017, 04:31
VeritasPrepKarishma wrote: aliasjit wrote: I am a little confused about solving the problem:
Stmnt 2: xy =3
we know y is odd. and if Y is odd as per problem statement y cannot be anything but 1 as x and y both are positive integers.
Therefore x =4 and is divisible by 4.
Please help me understand where I am going wrong with this. y is odd, yes, but why must y be 1? Responding to a pm: Quote: because as per statement 2 if xy =3 and x and y are both +ve integers could y be anything but 1 But as per statement 2, we do not know that x must be 4. x must be even since y is odd and difference between x and y is odd. But will it be 4, we do not know.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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19 Jun 2017, 18:33
y is odd => y = 2n+1 y^2 = 4*even + 1 or 8n + 1; where n can be any positive integer Statement 1: P = 8a + 5 8a + 5 = x^2 + 8n + 1 x^2 = 8(an) + 4 = 4(2a – 2n + 1) = 4(odd) x = 2(odd) ; x = 2, 6, 10, 14…… Hence X is not divisible by 4 SUFFICIENT Statement 2: xy = 3; x = 3 – odd; x = even. X can be 2, 4, 6, 8; NOT SUFFICIENT
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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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18 Aug 2017, 21:09
p=x²+y² Y=odd
question is is x/4=integer 1) p/8 gives a reminder of 5 or p=q8+5(q is the quotient) P can be =5, 13,18,23,28,34,39,44....... let us check out options by plugin in above nos p=5 ie 5=x²+y² now y can only be 1 and x=2. which is not divisible by 4 13=x²+y²=(2)²+(3)² again not divisible by 4 18=(3)²+(3)² again 3 is not divisible by 4 23= (>4<5)²+(3)² again not divisible by 4 Statement is sufficient 2) XY=3 Xodd=odd X has to be even. It can be 2 which is not divisible by 4 or 4 which is divisible by 4 Statement is not sufficient



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If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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20 Aug 2017, 09:40
siddharthmk wrote: I did this question this way. I found it simple.
1. p=x^2+y^2 y is odd p div 8 gives remainder 5. A number which gives remainder 5 when divided by 8 is odd.
so (x^2 + y^2)/8 = oddnumber (x^2 + y^2) = 8 * oddnumber (this is an even number without doubt)
x^2 + y^2 is even. Since y is odd to get x^2+y^2 even x must also be odd.
X is an odd number not divisible by 4
Option A: 1 alone is sufficient Hello All, A quick observation regarding text highlighted above: I believe we can NOT write like this. Lets say p = x^2+y^2 = 5 or 13 or 21 or 29 so on. Then we can Not write 5/8 = odd number or 13/8 = odd number Isn't? Rather we can write 13/8 = 1 + 5/8 i.e. Dividend/Divisor = Quotient + Reminder/Divisor . Any comments maths expert Bunel/Karishma ?



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Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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04 Sep 2017, 04:13
Bunuel wrote: nonameee wrote: Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?
Thank you. 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 > \(p=8q+5=x^2+y^2\) > as given that \(y=odd=2k+1\) > \(8q+5=x^2+(2k+1)^2\) > \(x^2=8q+44k^24k=4(2q+1k^2k)\). So, \(x^2=4(2q+1k^2k)\). Now, if \(k=odd\) then \(2q+1k^2k=even+oddoddodd=odd\) and if \(k=even\) then \(2q+1k^2k=even+oddeveneven=odd\), so in any case \(2q+1k^2k=odd\) > \(x^2=4*odd\) > in order \(x\) to be multiple of 4 \(x^2\) must be multiple of 16 but as we see it's not, so \(x\) is not multiple of 4. Sufficient. (2) x – y = 3 > \(xodd=3\) > \(x=even\) but not sufficient to say whether it's multiple of 4. Answer: A. For statement 1 , can we not directly say that since the rem is 5 when divided by 8 ,p wll be odd. Since y is already odd => x is odd . hence not divisible by 4?



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If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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15 Nov 2017, 11:00
\(y\) is odd , Let \(y = 2k + 1\) (any odd number can be expressed like this)
Statement 1: Now \(p % 8 = 5\) => \((x^2 + y^2 => x^2 + (2k + 1)^2)\) % 8 = 5 => \((x^2 + 4k^2 + 4k + 1)\) % 8 = 5 => \((x^2 + 4k^2 + 4k 4 + 4 + 1)\) % 8 = 5 (since 5 is remainder) \((x^2 + 4k^2 + 4k  4)\) % 8 = 0 \((x^2 + 4(k^2 + 4k  1))\) % 8 = 0 now if \(k = odd => k^2 + 4k  1\) is odd now if \(k = even => k^2 + 4k  1\) is still odd so, \((x^2 + 4 * odd)\) % 8 = 0 if \(x\) were mulitple of 4, then \(x = 4m\) => \((16m^2 + 4 * odd)\) % 8=> this won't be zero as \(16m^2\) is divisible by 8, but \(4*odd\) is not. so x cannot be 4m (or multiple of 4)
sufficient
Statement 2: Clearly not sufficient
Answer (A)




If p, x, and y are positive integers, y is odd, and p = x^2
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