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Veritas Prep GMAT Instructor V
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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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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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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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$$[m]$$[/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) x-y=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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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.

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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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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I am a little confused about solving the problem:

Stmnt 2: x-y =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.

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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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aliasjit wrote:
I am a little confused about solving the problem:

Stmnt 2: x-y =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.

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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1
VeritasPrepKarishma wrote:
aliasjit wrote:
I am a little confused about solving the problem:

Stmnt 2: x-y =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.

y is odd, yes, but why must y be 1?

Responding to a pm:
Quote:
because as per statement 2 if x-y =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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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(a-n) + 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:
x-y = 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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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) X-Y=3
X-odd=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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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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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.

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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$$y$$ 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

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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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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.

Can someone tell me where I went wrong? I did:
p=8q+5=x^2+y^2
then x^2 = 8q + 5 - y^2
8q is even, 5 is odd, y^2 is odd (since y is odd)
so x^2 = even + odd - odd = even
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From the statement one we know that p=8t+5
I listed possible values for p:
5, 13, 21, 29, 37, 45... from the list I found only 45 as a sum of two squares (6^2+3^2) and chose answer A.

But I was not sure 45 is the only number that satisfies p=x^2+y^2. How to know whether a number other than 45 satisfies p=x^2+y^2 and p=8t+5 requirement
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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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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?

Your analysis till now is fine but it is incomplete. We do get that x is even but we also get that x is a multiple of 2 but not 4 as explained in the post above: http://gmatclub.com/forum/if-p-x-and-y- ... ml#p837890

Is the following solution correct?

x2=8q+5+y2

Since x is an integer the right hand side must also be an integer. Since right hand side is even it has to be a perfect even square to equal the left side.
Y cannot assume just any value it has to assume only those which will make the right side an even perfect square. So x must be even and and a multiple of 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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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.

Buneul, Is this a GMAT style question? I need an expert opinion like yours in this regard.
As I understand, GMAT is a test of critical thinking and less of math. The complex substitution method above and peculiar properties of 8 cannot be tested in GMAT, in my opinion. Can you please clarify?

Regards. Re: If p, x, and y are positive integers, y is odd, and p = x^2   [#permalink] 23 May 2019, 01:35

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