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If p, x, and y are positive integers, y is odd, and p = x^2 02 May 2022, 08:43
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ArjunKumar25
This is the first time, I am posting on GMAT club because for the first time I feel I found the best way to solve a question that has not been posted amongst the top replies. Please give me kudos.

1) P=8A+5
8A will be even and, odd=even+odd
Therefore P is odd.

Now, We use the information from the question stem
P=x+y
Since P is odd, and
p(odd)=x+y(odd)
therefore X is even.
Sufficient

2) X-Y = 3
X is either even and Y is odd
or
X is odd and Y is even.
Insufficient
Show more

Your approach for each statement is not right. For statement 1, concluding that X is even is not enough to deem the statement sufficient because if x = 2 then x is not divisible by 4 but if x = 4 then x is divisible by 4. Please review KarishmaB approach, which demonstrates that \(x=2*\sqrt{Odd Number}\), and therefore, x is never divisible by 4.

For statement 2, the question stem provides that y is odd. A correct way to show that it is insufficient is to test x = 6 and y = 3 > x is not divisible by 4, and x = 4 and y = 1 > x is divisible by 4.
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If p, x, and y are positive integers, y is odd, and p = x^2 27 Jul 2022, 20:50
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Bunuel
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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.
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Hi everyone,

"in order \(x\) to be multiple of 4 \(x^2\) must be multiple of 16 "

Can we apply the same rule to other numbers?
i.e. in order \(x\) to be multiple of 5 \(x^2\) must be multiple of 25

But the reverse isn't true right?

Thank you
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If p, x, and y are positive integers, y is odd, and p = x^2 28 Jul 2022, 01:54
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MyndNathamon
Bunuel
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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.

Hi everyone,

"in order \(x\) to be multiple of 4 \(x^2\) must be multiple of 16 "

Can we apply the same rule to other numbers?
i.e. in order \(x\) to be multiple of 5 \(x^2\) must be multiple of 25

But the reverse isn't true right?

Thank you
Show more

We got that \(x^2=4*odd\) and are asked to find out whether x is a multiple of 4:

    - If \(x\) were a multiple of 4, then \(x^2\) would have been a multiple of 16 but \(x^2=4*odd\) shows that it is not. Therefore, \(x\) is not a multiple of 4.

Now, if we were asked to find whether integer x is a multiple of 5 and we were also given that x^2 is NOT a multiple of 25, then we could say that x is NOT a multiple of 5 because if it were than x^2 would have been a multiple of 25.

Does this make sense?
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If p, x, and y are positive integers, y is odd, and p = x^2 25 Oct 2022, 08:58
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KarishmaB
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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: https://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.

Answer (A).
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Dear KarishmaB, EXPERT,

Stmt2: x - y = 3
We know y >=1 and y is odd so y can be 1,3,5----
x = 3 + odd = even and since y minimum value is 1, x=4 ---> divisible by 4

Hence Option D is the answer IMO.
Where am I going wrong?
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If p, x, and y are positive integers, y is odd, and p = x^2 31 Aug 2024, 05:49
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METHOD 1 :
By algebraic formula : (x-y)^2 = x^2 + y^2 - 2xy

We know x-y = 3
Therefore (x-y)^2 = 9 which is an odd number
2xy will always be an even number

Substituting in the above equation:
odd = x^2 + y^2 - even
x^2 + y^2 = odd - even = odd number

hence, p = x^2 + y^2 is not divisible by 4.

METHOD 2 :
x-y = 3
y is odd. Which means x is even.
So x^2 is even and y^2 is odd. x^2 + y^2 = even + odd = odd.
hence, p = x^2 + y^2 is not divisible by 4.

So why can't statement B alone be sufficient to answer the question?
In that case, option D, where both A and B individually is sufficient should be the answer.
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If p, x, and y are positive integers, y is odd, and p = x^2 23 Oct 2024, 13:16
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mr1234
METHOD 1 :
By algebraic formula : (x-y)^2 = x^2 + y^2 - 2xy

We know x-y = 3
Therefore (x-y)^2 = 9 which is an odd number
2xy will always be an even number

Substituting in the above equation:
odd = x^2 + y^2 - even
x^2 + y^2 = odd - even = odd number

hence, p = x^2 + y^2 is not divisible by 4.

METHOD 2 :
x-y = 3
y is odd. Which means x is even.
So x^2 is even and y^2 is odd. x^2 + y^2 = even + odd = odd.
hence, p = x^2 + y^2 is not divisible by 4.

So why can't statement B alone be sufficient to answer the question?
In that case, option D, where both A and B individually is sufficient should be the answer.
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The question asks whether x is divisible by 4, not p.
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If p, x, and y are positive integers, y is odd, and p = x^2 14 Jan 2025, 12:23
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netcaesar
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
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Statement 1:
1. As power doesn’t affect the nature of odd or even, we can write p=x+y. where y=Odd and P remain 5 when divided by 8 it refers to p=odd
So Odd= even+odd; so it is clear that x=even.
2. Y[sup]2[/sup] mod 8 remain 1.
p≡ 5(mod 8)
x[sup]2[/sup]+y[sup]2[/sup]≡5 (mod8)
x[sup]2[/sup]+1≡5
x[sup]2[/sup]≡4 (mod 8)
· If x=2 2[sup]2[/sup]= 4≡ 4 (mod 8)
· X=4 4[sup]2[/sup]=16≡ 0 (mod 8)
· X=6 6[sup]2[/sup]= 36 ≡ 4 (mod 8)
So x= 2 or 6 which is not divisible by 4. Statement 1 is sufficient
Statement 2:
x – y = 3
x- odd=3; even-odd=odd. That all statement 2 says. So not sufficient information is given.
Ans: A
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If p, x, and y are positive integers, y is odd, and p = x^2 20 Jul 2025, 09:49
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Hi,
I had a doubt about statement 2. Since it is given that x and y are positive integers, and x-y = 3 => x = 3+y. We know that x is even and x > 3. Since x and y are positive integers, x >= 4, and hence x is divisible by 4.
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If p, x, and y are positive integers, y is odd, and p = x^2 20 Jul 2025, 10:14
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ledzep10
Hi,
I had a doubt about statement 2. Since it is given that x and y are positive integers, and x-y = 3 => x = 3+y. We know that x is even and x > 3. Since x and y are positive integers, x >= 4, and hence x is divisible by 4.
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Just because x is even and greater than 3 doesn’t mean it must be divisible by 4. For example, x = 6 satisfies all given conditions but is not divisible by 4.
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If p, x, and y are positive integers, y is odd, and p = x^2 Updated on: 11 Aug 2025, 00:22
Originally posted by gmatchile on 11 Aug 2025, 00:17.
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p, x, and y are positive integers, and is odd? 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 Let's try 1) This tells us that p is not even, that is, p is odd. If p is odd and y is odd, y^2 is also odd, then x must necessarily be even, since (even)^2 is even. So y^2 (odd) + x^2 (even) = odd = p Then the smallest positive even integer is 2 and 2^2 = 4, the smallest odd number is 1 and 1^2 = 1. Therefore, 4 + 1 = 5, and 5 divided by 8 gives a remainder of 5. Therefore, we cannot guarantee that x is divisible by 4. 1) It's sufficient. Let's analyze 2) x - y = 3 If x is odd, x must necessarily be even. The smallest positive odd integer is 1, so x would have to be 4. If y = 3, x would have to be 6, and 6 is not divisible by 4. 2) It's not sufficient. Answer A
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