Given x and y are positive integers such that y is odd, is x divisible

given Y is odd.
1) x^2 + y^2 = 8k+5
8k+5 , for different values of the constant k gives : 5, 13, 21, 29, 37, 45 ...
x^2 + y^2 = 5,13,21,29,37,45 .....
since y = odd, different values of y give :
y=1 : x^2 = 4, 12, 20, 28, 36, 44, ....
y=3 : x^2 = 4, 12, 20, 28, 36, ..... (leave out the negative values)
y=5 : x^2 = 4, 12, 20, ....
Thus x is not divisible by 4. Suff.

2) x-y = 3, and we know y is odd, say 2k+1.
Thus x = 2k+1+3 = 4 +2k.
Now this gives us x = 4,6,8,10.... So x may or may not be divisible by 4.
Not Suff.

Ans A.

+1 to A.

Putting y = (2n + 1) in y^2, we get (4n^2 + 1 +4n) = 4n(n+1)+1. This expression when divided by 8 always gives reminder 1 for any value of n >=0.

Now, Statement 1 says that When (x^2 + y^2) is divided by 8, the remainder is 5 (4 + 1). Hence x^2 when divided by 8 should give remainder 4, as 1 is coming from y^2. So, x^2 = 8k + 4. x = 2 * [(2k + 1)^(1/2)].
So, its clear that x can never be divided by 4, as [(2k + 1)^(1/2)] can never be even. SUFFICIENT.

B is clearly INUFFICIENT.

Hi Sherloked,
I think you missed something here. Question is asking about x, not x^2.
x will "never" be divisible by 4.

Thank you.

Given: X, Y >0 and Y is ODD
To find : Is X divisible by 4

Solution:

Statement 1: When (x^2 + y^2) is divided by 8, the remainder is 5.

(x^2 + y^2) could take any of the following values 5, 13, 21, 29......

When (x^2 + y^2) = 13 then Y=3 and X=2 and 13 divided by 8 leaves a remainder of 5
X is not Divisible by 4

When (x^2 + y^2) = 29 then Y=5 and X = 2 and 29 divided by 8 leaves a remainder of 5
X is not divisible by 4

Statement 1 is sufficient


Statement 2: x – y = 3

When Y=3 and X = 6 then X-Y = 3
X is not divisible by 4

When Y=1 and X = 4 then X-Y=3
X is divisible by 4

Statement 2 is not sufficient

Hence answer is A

Hi AmoyV,

With Fact 1, you have an interesting theory, but you have NO proof that your theory impacts the question that was ASKED. If you do a little more work, you should be able to prove that Fact 1 is actually SUFFICIENT.

Be careful about assuming the "I don't know anything about <blank>" logic means that a Fact is insufficient. DS questions are designed to test you on specific concepts, so it's always better to have proof that you're correct, than to just assume that you are (and possibly miss out on some easy points when your thinking is incomplete).

GMAT assassins aren't born, they're made,
Rich

Given x and y are positive integers and y is odd

Statement 1:
x^2 + y^2 = 8q + 5
x^2 + (2k+1)^2 = 8q +5
x^2 + 4k^2+4k+1 = 8q + 5
x^2 + 4k^2+4k = 8q + 4
x^2 = 8q + 4 - 4k^2 - 4k
x^2 = 4 [(2q+1) - (k(k+ 1))]
x^2 = 4 [odd - even]
x^2 = 4 * odd square
x = 2 * odd integer; x is not a multiple of 4
Sufficient

Statement 2:
x - y = 3
x = y +3; x could be 4, 6, 8, 10... so x could or couldn't be a multiple of 4
Not Sufficient

Answer A

VERITAS PREP OFFICIAL SOLUTION:

As of now, we don’t know any specific properties of squares of odd and even integers. However, we do have a good (presumably!) understanding of divisibility. To recap quickly, divisibility is nothing but grouping. To take an example, if we 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 the quotient and 1 is the remainder. For more on these concepts, check out our previous posts on divisibility.

Coming back to our question,
First thing that comes to mind is that if y is odd, y = (2k + 1).

We have no information on x so let’s proceed to the two statements.

Statement 1: When (x^2 + y^2) is divided by 8, the remainder is 5.

The statement tell us something about y^2 so let’s get that.
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), k(k+1) will be even. So 4k(k+1) will be divisible by 4*2 i.e. by 8. So when y^2 is divided by 8, it will leave a remainder 1.

When y^2 is divided by 8, remainder is 1. To get a remainder of 5 when x^2 + y^2 is divided by 8, we should get a remainder of 4 when x^2 is divided by 8. So x must be even. If x were odd, the remainder when x^2 were divided by 8 would have been 1. So we know that x is divisible by 2 but we don’t know whether it is divisible by 4 yet.

x^2 = 8a + 4 (when x^2 is divided by 8, it leaves remainder 4)
x^2 = 4(2a + 1)
So x = 2*?Odd Number

Square root of an odd number will be an odd number so we can see that x is even but not divisible by 4. This statement alone is sufficient to say that x is NOT divisible by 4.

Statement 2: x – y = 3

Since y is odd, we can say that x will be even (Since Even – Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say. This statement alone is not sufficient.

Answer (A)

So could you point out the takeaway from this question?

Note that when we were analyzing y, we used no information other than that it is odd. We found out that the square of any odd number when divided by 8 will always yield a remainder of 1.

Now what can you say about the square of an even number? Say you have an even number x.
x = 2a
x^2 = 4a^2

This tells us that x^2 will be divisible by 4 i.e. we can make groups of 4 with nothing leftover. What happens when we try to make groups of 8? We join two groups of 4 each to make groups of 8. If the number of groups of 4 is even, we will have no remainder leftover. If the number of groups of 4 is odd, we will have 1 group leftover i.e. 4 leftover. So when the square of an even number is divided by 8, the remainder is either 0 or 4.

Looking at it in another way, we can say that if a is odd, x^2 will be divisible by 4 and will leave a remainder of 4 when divided by 8. If a is even, x^2 will be divisible by 16 and will leave a remainder of 0 when divided by 8.

Takeaways
– The square of any odd number when divided by 8 will always yield a remainder of 1.
– The square of any even number will be either divisible by 4 but not by 8 or it will be divisible by 16 (obvious from the fact that squares have even powers of prime factors so 2 will have a power of 2 or 4 or 6 etc). In the first case, the remainder when it is divided by 8 will be 4; in the second case the remainder will be 0.

Hi anitesh,
Yes, I solved for values of x^2, and then wrote the same, that x is not divisible by 4, since all the values of x^2, give values of x which can not be divisible for 4.

Hope I didnt confuse you

Sorry my fault..You didnot confuse me at all

Nice Question (So a bump up)

Given: y is positive odd. (1,3,5....)

Asked: x divisible by 4? (x=4/8/12/16.......)

Statement 1:
(1) When (x^2 + y^2) is divided by 8, the remainder is 5.

If y=1 then and (x^2+1) has to leave a remainder of 5 when divided by 8 then x ^2 can be 4 or 12.

Try with y=3 you will find value of x is divisible by 4.

Sufficient.

(2) x – y = 3

Remember y is odd.

y=1 then x=4 Works

y=3 then x=6 . Doesn't Work.

In sufficient.

Answer (A)

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