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

Question

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

KarishmaB · Accepted Answer

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://youtu.be/A5abKfUBFSc 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). This question is discussed in detail here: https://anaprep.com/number-properties-a ... emainders/

samrus98 · Answer

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

ANS: A

Bunuel · Answer

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 ;

Since given that  y=odd=2k+1 , then:

8q+5=x^2+(2k+1)^2 ;
 x^2=8q+4-4k^2-4k ;
 x^2=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 .  Thus,  x^2=4*odd .

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

(2) x – y = 3:

x-odd=3 ;
 x=even .

The above is NOT sufficient to say whether x is multiple of 4.

Answer: A.

siddharthmk · Answer

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

nonameee · Answer

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.

Eshaninan · Answer

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

KarishmaB · Answer

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.

sanjaykvbsingh · Answer

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.

https://gmatclub.com/static/posted_from.png     Posted from    GMAT ToolKit

Abheek · Answer

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.

KarishmaB · Answer

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: if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html#p837890

alphonsa · Answer

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

KarishmaB · Answer

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.

Lucky2783 · Answer

are there more tricks like this , ODD^2 MOD 8 = 1 ?

thanks 
Lucky

KarishmaB · Answer

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.

MathRevolution · Answer

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.

KarishmaB · Answer

You can but does it help you arrive at the answer?
p = x^2 + y^2
p divided by 8 leaves rem 5

If p = 13, p = 2^2 + 3^2 
Also, p leaves 5 rem on division by 8.
x is not divisible by 4

If p = 29, p = 2^2 + 5^2
Also, p leaves 5 rem on division by 8.
x is not divisible by 4

What does this prove? Does it prove that x is ALWAYS not divisible by 4? No. It is not divisible in the two examples we took. But what if it is divisible by 4 in some other cases? How do we establish that x will NEVER be divisible by 4? Only then can we say that our stmnt is sufficient to give us the same answer every time. We will need to do it logically.

SiddharthR · Answer

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  
Similarly, when p = 13 --> p = 2^2 + 3^2 --> x = 2 
p = 21 -> Can't think of numbers here
p = 29 -> p = 2^2 + 5^2 -> x = 2 
p = 37 -> p = 6^2 + 1^2 -> x = 6 
p = 45 -> p = 6^2 + 3^2 -> x = 6 
p = 53 -> p = 2^2 + 7^2 -> x = 2

Since x is not div by 4 for any of the situations above, it is safe to conclude that A is sufficient !!!

vaibhav1221 · Answer

Easy 1 minute approach.

Square of every odd integer when divided by 8 gives a remainder of 1.

Square of every even integer when divided by 8 gives a remainder of either 0 or 4.  (Square of every alternate even number starting from 2, when divided by 8 leave a remainder of 4)

For  \frac{P}{8}  to have a remainder of 5,  x  has to be even.

So, we can write the values of  x^2  and  y^2  as follows
 x^2 = 8r + 4 
 y^2 = 8q + 1

Where  r  and  q  are quotients when  x^2  and  y^2  are divided by 8.

p = x^2 + y^2 = 8r + 4 + 8q + 1

p = x^2 + y^2 = 8(r + q) + 5

From the highlighted portion above and the fact that  x^2  is even, we can conclude that x has to be either 2, 6, 10, 14, 18... so on. None of them are divisible by 4.

Hence, A is sufficient.

B is clearly insufficient.

ArjunKumar25 · Answer

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

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards