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Re: PS: Divisible by 4 [#permalink]
14 Aug 2009, 12:46

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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

SOL:

St1: Here we will have to use a peculiar property of number 8. The square of any odd number when divided by 8 will always yield a remainder of 1!!

This means that y^2 MOD 8 = 1 for all y => p MOD 8 = (x^2 + 1) MOD 8 = 5 => x^2 MOD 8 = 4

Now if x is divisible by 4 then x^2 MOD 8 will be zero. And also x cannot be an odd number as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a non-multiple of 4. => SUFFICIENT

St2: x - y = 3 Since y can be any odd number, x could also be either a multiple or a non-multiple of 4. => NOT SUFFICIENT

Re: PS: Divisible by 4 [#permalink]
15 Aug 2009, 12:49

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Very good solution I did not know this property of 8. Kudos to you.

By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works.

Re: PS: Divisible by 4 [#permalink]
16 Dec 2010, 06:39

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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 \(y=odd=2k+1\) --> \(8q+5=x^2+(2k+1)^2\) --> \(x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)\).

So, \(x^2=4(2q+1-k^2-k)\). Now, if \(k=odd\) then \(2q+1-k^2-k=even+odd-odd-odd=odd\) and if \(k=even\) then \(2q+1-k^2-k=even+odd-even-even=odd\), so in any case \(2q+1-k^2-k=odd\) --> \(x^2=4*odd\) --> 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 --> \(x-odd=3\) --> \(x=even\) but not sufficient to say whether it's multiple of 4.

Re: PS: Divisible by 4 [#permalink]
18 Dec 2010, 10:23

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maliyeci wrote:

Very good solution I did not know this property of 8. Kudos to you.

By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works.

So for any odd number n, n^2=1 mod 8

Its not something one shall already know before attacking a question, you may realize properties like this when u start solving a question. Even I didn't know about this property of 8.

I approached the question in following way:

Stmt 1: P/8=(x^2+y^2)/8; using remainder theorem; rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8] if x is divisible by 4, then x^2= 4k*4k= 16K=8*2K is also divisible by 8. now to anaylze rem[y^2/8]; start putting suitable values of y; i.e all odd values starting from 1. for y=1; rem(1/8)=1 for y=3; rem(9/8)=1 for y=5;rem(25/8)=1

so you observe this pattern here. coming back to ques now, as rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]= rem[x^2/8] + 1 =5; this means rem[x^2/8] is not 0; which implies x is not divisible my 8; Sufficient

Stmt2: y being odd can be accept both 3 and 5 as values and we get different results; thus Insufficient

Thus OA is A _________________

The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!

Re: PS: Divisible by 4 [#permalink]
19 Dec 2010, 06:49

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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/divisibility-and-remainders-if-you.html

Coming 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.

Re: PS: Divisible by 4 [#permalink]
16 Jul 2012, 17:19

Am i missing something, why cant we take stmt 2 as follows: squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4

Re: PS: Divisible by 4 [#permalink]
16 Jul 2012, 22:24

Expert's post

Eshaninan wrote:

Am i missing something, why cant we take stmt 2 as follows: squaring x-y=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. _________________

Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
13 Sep 2013, 19:25

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from first statement p = 8j + 5 Put j as 1, 2,3,4,5... p would be 13, 21,29, 37,45... Now in the formula p= x^2+y^2 put 1,3,5,7 as value of y ( as y is odd) to get x. You will notic the possible value of x is 2 which is not divisble by 4.

Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
23 Dec 2013, 22:45

For Statement 1: since p when divided by 8 leaves remainder 5.We obtain the following equation p= 8q+5 We know y is odd. If we write p =x^2+y^2 then we get the eqn: x^2+y^2=8q+5 Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd. Then x^2= odd - y^2 i.e x^2=even ie x= even But it's not sufficient to answer the question whether x is a multiple of 4? By this logic i get E as my answer. Statement 2: is insufficient.

Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
30 Dec 2013, 22:39

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Abheek wrote:

For Statement 1: since p when divided by 8 leaves remainder 5.We obtain the following equation p= 8q+5 We know y is odd. If we write p =x^2+y^2 then we get the eqn: x^2+y^2=8q+5 Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd. Then x^2= odd - y^2 i.e x^2=even ie x= even But it's not sufficient to answer the question whether x is a multiple of 4?

Re: PS: Divisible by 4 [#permalink]
29 Apr 2014, 21:15

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 \(y=odd=2k+1\) --> \(8q+5=x^2+(2k+1)^2\) --> \(x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)\).

So, \(x^2=4(2q+1-k^2-k)\). Now, if \(k=odd\) then \(2q+1-k^2-k=even+odd-odd-odd=odd\) and if \(k=even\) then \(2q+1-k^2-k=even+odd-even-even=odd\), so in any case \(2q+1-k^2-k=odd\) --> \(x^2=4*odd\) --> 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 --> \(x-odd=3\) --> \(x=even\) but not sufficient to say whether it's multiple of 4.

Answer: A.

A) \(8a + 5 = x^2 + y^2\) \(even + odd = x^2 + odd\) \(x^2=even\) therefore x can be 2 ,not divisible by 4. or 4 ,divisble by 4 Hence Insufficient

B) x - y = 3 x - odd = odd x= even but x can be 2 ,not divisible by 4 , or 4 ,divisble by 4 . Hence Insufficient.

Re: PS: Divisible by 4 [#permalink]
30 Apr 2014, 01:36

Expert's post

abid1986 wrote:

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 \(y=odd=2k+1\) --> \(8q+5=x^2+(2k+1)^2\) --> \(x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)\).

So, \(x^2=4(2q+1-k^2-k)\). Now, if \(k=odd\) then \(2q+1-k^2-k=even+odd-odd-odd=odd\) and if \(k=even\) then \(2q+1-k^2-k=even+odd-even-even=odd\), so in any case \(2q+1-k^2-k=odd\) --> \(x^2=4*odd\) --> 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 --> \(x-odd=3\) --> \(x=even\) but not sufficient to say whether it's multiple of 4.

Answer: A.

A) \(8a + 5 = x^2 + y^2\) \(even + odd = x^2 + odd\) \(x^2=even\) therefore x can be 2 ,not divisible by 4. or 4 ,divisble by 4 Hence Insufficient

B) x - y = 3 x - odd = odd x= even but x can be 2 ,not divisible by 4 , or 4 ,divisble by 4 . Hence Insufficient.

Please note that the correct answer is A. _________________

Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
18 Aug 2014, 01:43

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alphonsa wrote:

For statement 1 , wouldn't plugging in values be a better option?

No. When you need to establish something, plugging in values is not fool proof.

Anyway, in this question, how will you plug in values? You cannot assume a value for x since that is what you need to find. You will assume a value for y and a value for p such that they satisfy all conditions. This itself will be quite tricky. Then when you do get a value for x, you will find that it will be even but not divisible by 4. How can you be sure that this will hold for every value of y and p?

When a statement is not sufficient, plugging in values can work - you find two opposite cases - one which answers in yes and the other which answers in no. Then you know that the statement alone is not sufficient. But when the statement is sufficient, it is very hard to prove that it will hold for all possible values using number plugging alone. You need to use logic in that case. _________________

Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
29 Aug 2014, 04:25

Hi Karishma,

Thanks for the explanation to the question. I was just wondering how the answer would change if we change the question stem a little bit. What if the question asks if p (instead of x) is divisible by 4?

In this scenario, statement 1 would be sufficient since if something leaves a remainder of 5, it would leave a remainder of 1 upon division by 4

For statement 2, we know that x = y+3, so x is even. If we square it, it would surely be divisible by 4. Now if a number (y^2, which is odd) non-divisible by 4 is added to a number divisible by 4, the result would surely be not divisible by 4. So statement 2 would also be sufficient.

Is this reasoning correct? just for practicing the concept

gmatclubot

Re: If p, x, and y are positive integers, y is odd, and p = x^2
[#permalink]
29 Aug 2014, 04:25

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