If p, x, and y are positive integers, y is odd, and p = x2 + y2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.

(2) x – y = 3

I got lost in the explanation involving the quadratic formula. Can someone please explain.

the answer is a. is it.

I tried looking this problem up, but I couldn't find it. Is this from the challenge archive? If so, are you sure it's from Nov 3? And also, what year? My hunch is that the answer is E, but I haven't fully solved it. Some of these challenge problems can take longer than 2 minutes.

statement 1 - when p is divided by 8, the remainder is 5.

now, when a square of odd nos. is divided by 8, the remainder is always 1. (9/8 =1. 25/8 =1). Secondly if x is divisible by 4, its square must be divisble by 8. And if this is the case, when p is divided by 8, the remainder should be 1. but, when is divided by 8, the remainder is 5. Therefore, x is not divisble by 4. statement 1 is sufficient.

statement 2 - x-y = 3. there can be multiple values for x and y, so this is not sufficient.

Hence, answer A.

Still waiting for a reply - it you'd like us to make the explanation as written more clear to you, we have to be able to read that explanation. :) Please let us know (a) whether this is from the challenge problem archive or some other source and (b) if from the challenge problem archive, Nov 3 of what year?
Thanks!

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

Hi,

Apologies for the delayed response. Also, question is from Oct 27, not Nov 3. Anyway, here's the explaination I can't quite understand:

Statement (1): SUFFICIENT. We know that p is odd. We know from the problem stem that y is odd, which means that y2 is odd. Therefore, x2 must be even (because O = E + O), so x must be even.

However from all of this, we can infer something else—specifically, that x is NOT a multiple of 4. Here’s why:

y2 = (y2 — 1) + 1, which using the quadratic property, yields: y = (y+1)(y—1) +1. Because y is odd, then (y+1)(y—1) is even times even, which is a multiple of 4. However we ALSO know that either y+1 or y—1 is a multiple of 4, because they are consecutive multiples of 2. Therefore, y2 is 1 greater than a multiple of 8. (You can confirm this by thinking about all squared odd numbers: 12 = 1, 32 = 9, 52 = 25, 72 = 49, etc.)

Do NOT worry too much about this stuff - this is way harder than what the test typically expects!

p is odd because when p is divided by 8 (an even number), the remainder is 5 (an odd number).

y is odd because the question stem says so.

y^2 is odd because any odd * any odd = odd

p = x^2 + y^2
odd = ? + odd

odd + even = odd
odd + odd = even

Therefore, x^2 must be even. Therefore, x must be even. Good so far? (If not, test any of the above with real numbers.)

I can represent y^2 as: y^2 + 1 - 1, right? (+1 and -1 just equals zero, so I'm back to y^2.)

Rearrange to y^2 - 1 + 1. (y^2 - 1) is one of the special quadratic identities, and it can be written as: (y+1)(y-1). (If you don't remember this, go back and review quadratic identities; you have to know these for the test!)

So I can replace y^2 -1 with (y+1)(y-1) to get:
y^2 = (y+1)(y-1) +1
I'm just re-writing y^2 in a different form, that's all.

If y is odd, then y+1 is even. If y is odd, then y-1 is also even. So I've got:
y^2 = (y+1)(y-1) +1
odd = (even)(even) +1
Which makes sense - and even number +1 = an odd number.

If anything above doesn't make sense, stop now and test it with real numbers.

the two even integers in the equation above are consecutive, so one is a multiple of 2 and the other is a multiple of both 2 and 4. Think of consecutive even integers: 2, 4, 6, 8, 10, 12, 14, 16, etc. All of them are divisible by 2, and every other one is divisible by 4.

Multiplying the two even numbers together will give us a new even number that is also divisible by 4. Try this out. If the two numbers are 2 and 4, the product is 8, which is divisible by 4. If the two numbers are 4 and 6, the product is 24, which is divisible by 4. Whatever the two numbers are, the product will always be a multiple of 4, because one of the two numbers will always be a multiple of 4.

And we can take this a step further. One of the two numbers will always be divisible by 2 and the other will always be divisible by 4, so the product will always be divisible by 2*4 = 8. (Again, try with real numbers if you're not sure.)

So back to our equation:
y^2 = (y+1)(y-1) +1
odd = (even)(even) +1
odd = (even multiple of 8) + 1

So whatever y^2 is, it will always be [(an even multiple of 8) + 1]. y^2 also has to represent a squared odd integer. So, the first (even multiple of 8 + 1) = 9. Is that the square of an odd integer? Yes. So 3 is a possible value for y. The next (even multiple of 8 + 1) = 17. Is that the square of an odd integer? No. Ignore it. The next (even multiple of 8 + 1) = 25. Is that the square of an odd integer? Yes. So 5 is a possible value for y. Noticing a pattern? Try the next two on your own to see what happens.

Now, back to our original equation:
p = x^2 + y^2
odd = even + (even multiple of 8 + 1)

Statement 1 says that, when p is divided by 8, the remainder is 5. Hmm. When y^2 is divided by 8, the remainder is only 1. Therefore, when the x^2 part of the equation is divided by 8, the remainder has to be 4, in order for the two remainders on the right-hand side to add up to 5 (the total remainder on the left-hand side).

Can something that is divisible by 4 have a remainder of 4? No - never. (Try some real numbers if you're not sure.) So x^2 is not divisible by 4. And if x^2 is not divisible by 4, then x also cannot be divisible by 4. (Try some real numbers again if you're not sure why this is true.)

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT