In this question, we are given y is odd
and p=x^2+y^2. we want to chk if x is divisible by 4.
From 1 we get
p=8z+5 where z=0,1,2......
put p=5
we get x=2 or 1 and y =1 or 2. But since y is odd, x=2,y=1
If you chk putting all other values no one satisfy the equation.
Hence suff.

From 2 we get x=y+3
put in original
we get
p=y^2+9+6y+y^2
p=2y^2+6y+9
we dont know the value of y, so no clue as x is divisible by 4 or not.

Hence the ans is A.

jimmiejaz wrote:In this question, we are given y is odd
and p=x^2+y^2. we want to chk if x is divisible by 4.
From 1 we get
p=8z+5 where z=0,1,2......
put p=5
we get x=2 or 1 and y =1 or 2. But since y is odd, x=2,y=1
If you chk putting all other values no one satisfy the equation.
Hence suff.

There are actually many values that sastify the rule, but in all of them x is not a multiple of 4.

e.g., we could pick y = 5 and x = 6 (since that would mean p = 61, which is 7*8 + 5).

So, while I've managed to convince myself that we always get a "NO" answer to the question, I still can't find an algebraic solution (at least not one findable in under 5 minutes).

yeah...i too tested values of p = 13 and p = 21 and found that A was NO for both of these.

srisl11 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

Please help
OA A

This is the solution given in the forum...
Answer
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.)

Since y2 divided by 8 yields a remainder of 1, from statement (1) we need x2 divided by 8 to yield a remainder of 4. This means that x must be even, but NOT be a multiple of 4, because if it were a multiple of 4, then x2 would be a multiple of 16, and therefore a multiple of 8.

Statement (2): INSUFFICIENT. We know that x is even. However, we can come up with two different examples which lead to different answers. For example, if x = 8 and y = 5, then x is a multiple of 4, but if x = 6 and y = 3, then x is not a multiple of 4.

The answer is A: Statement (1) is sufficient to answer the question, but statement (2) is insufficient.

srisl11 wrote:

srisl11 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

Please help
OA A

This is the solution given in the forum...
Answer
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.)

Since y2 divided by 8 yields a remainder of 1, from statement (1) we need x2 divided by 8 to yield a remainder of 4. This means that x must be even, but NOT be a multiple of 4, because if it were a multiple of 4, then x2 would be a multiple of 16, and therefore a multiple of 8.

Statement (2): INSUFFICIENT. We know that x is even. However, we can come up with two different examples which lead to different answers. For example, if x = 8 and y = 5, then x is a multiple of 4, but if x = 6 and y = 3, then x is not a multiple of 4.

The answer is A: Statement (1) is sufficient to answer the question, but statement (2) is insufficient.

That's a great proof, but the chances of anyone coming up with that in under 5 minutes seems mighty slim to me. Gogo picking numbers!

Square of an odd no when divided by 8 leaves remainder 1.

Number 1, 3, 5, 7, 9, 11
Square 1, 9, 25, 49, 81, 121

y is odd => y^2 divided by 8 leaves remainder 1.
=> y^2 divided by 4 leaves remainder 1.

p = x^2 + y^2
when p is divided by 8, the remainder is 5
=> x^2 when divided by 8 remainder 4 (as y^2 divided by 8 leaves 1)
=> x is divisible by 4

x is even; y is odd..
y is 5, x = 2 - x not divisible by 4
y is 7, x = 4 - x is divisible by 4 INCONSISTENT ANSWERS...

A

austin wrote:Square of an odd no when divided by 8 leaves remainder 1.

Number 1, 3, 5, 7, 9, 11
Square 1, 9, 25, 49, 81, 121

y is odd => y^2 divided by 8 leaves remainder 1.
=> y^2 divided by 4 leaves remainder 1.

p = x^2 + y^2
when p is divided by 8, the remainder is 5
=> x^2 when divided by 8 remainder 4 (as y^2 divided by 8 leaves 1)
=> x is divisible by 4

x is even; y is odd..
y is 5, x = 2 - x not divisible by 4
y is 7, x = 4 - x is divisible by 4 INCONSISTENT ANSWERS...

A

Excellent work austin !!!

Thank you Stuart and Austin

austin wrote:Square of an odd no when divided by 8 leaves remainder 1.

Number 1, 3, 5, 7, 9, 11
Square 1, 9, 25, 49, 81, 121

y is odd => y^2 divided by 8 leaves remainder 1.
=> y^2 divided by 4 leaves remainder 1.

p = x^2 + y^2
when p is divided by 8, the remainder is 5
=> x^2 when divided by 8 remainder 4 (as y^2 divided by 8 leaves 1)
=> x is divisible by 4
x is even; y is odd..
y is 5, x = 2 - x not divisible by 4
y is 7, x = 4 - x is divisible by 4 INCONSISTENT ANSWERS...

A

didn't understand on how the bolded conclusion was derived.

if x^2/ 8 gives a reminder of 4; how does that implies that x is divisbile of 4?

plz explain. TIA

y is an odd number. square of an odd number divided by 8 leaves remainder 1.

p = x^2 + y^2
when p is divided by 8, the remainder is 5
When y^2 is divided by 8, the remainder is 1 (from above)
=> when x^2 is divided by 8, the remanider is 4
=> x^2 is divisible by 4 => x is divisible by 4

The remainder when p is divided by 8 = the remainder when X^2 is divided by 8 + the remainder when y^2 is divided by 8...

Hope its clear now...

austin wrote:Square of an odd no when divided by 8 leaves remainder 1.

Austin, is this a formula?
I tried different numbers and found this true. But just want to know this is a formula or this is what you found out by yourself. Thanks.

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.

how is it that we know p is odd? i understand the problem stem says y is odd, but no where does the problem say p is odd?

thanks

Spring: You get it with practice.. the more you see numbers it just happens (trust me on this...)..

heffsz:
"how is it that we know p is odd"???
When p is divided by 8, the remainder is 5 => p has to be odd...

??