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Re: 0 < y < x. x and y are odd integers. [#permalink]
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Very simple.
Remember: Remainder is always an integer and is always less than the divisor. Thus.

Quantity A: Product of two odds is always odd. Thus odd/2 will always yield remainder of 1. Hence Quantity A = 1.

Quantity B: since x > y, this means that Quantity B will always be larger than 1.

Hence Quantity B is greater.
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Re: 0 < y < x. x and y are odd integers. [#permalink]
hi, i am having some problem with this question,
what if, y = 2, x = 3
Then, QTY A: 6/2 = 3
QTY B: 3/2

what am I missing?
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Re: 0 < y < x. x and y are odd integers. [#permalink]
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sooshii wrote:
hi, i am having some problem with this question,
what if, y = 2, x = 3
Then, QTY A: 6/2 = 3
QTY B: 3/2

what am I missing?


x and y are odd integers. both

Y CANNOT be 2 which is an even number.

I hope now is clear
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Re: 0 < y < x. x and y are odd integers. [#permalink]
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I think it is a mistake here or something wrong with my approach. Please correct me.

0 < y < x

Also y and x are ODD.

First: The remainder when xy divided by 2
It always has a remainder 1.


Now: x/y

0 < y < x

y = 3 and x = 5 ===> the remainder is 2
BUT!
y = 3 and x = 9 ===> the remainder is 0.

So I think the answer should be D not B.
Am I missing some provided limitation?
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Re: 0 < y < x. x and y are odd integers. [#permalink]
Expert Reply
y = 3 and x = 5 ===> the remainder is 2

x*y=3*5=15/2= remainder 1

BUT!
y = 3 and x = 9 ===> the remainder is 0.

3*9=27/2 = remainder 1

More on remainder here https://gre.myprepclub.com/forum/gre-qu ... tml#p51989
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Re: 0 < y < x. x and y are odd integers. [#permalink]
In your two examples, you divided it by 2. The divisor is y, and y must be an odd number.


Carcass wrote:
y = 3 and x = 5 ===> the remainder is 2

x*y=3*5=15/2= remainder 1

BUT!
y = 3 and x = 9 ===> the remainder is 0.

3*9=27/2 = remainder 1

More on remainder here https://gre.myprepclub.com/forum/gre-qu ... tml#p51989
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0 < y < x. x and y are odd integers. [#permalink]
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Expert Reply
QA says

The remainder when xy divided by 2

means that the two numbers x and y multiplied together we have a result and then the same result is divided by two

Aside the fact that who wrote the question missed xy is divided by two
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Re: 0 < y < x. x and y are odd integers. [#permalink]
why isn't the answer "D"

A is always 1
but B could be 0 (e.g. 9/3) or 2 (5/3) or even more like 4 (9/5); so A sometimes more than B and sometimes less than B. Wouldn't that mean D is the answer?

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0 < y < x. x and y are odd integers. [#permalink]
Expert Reply
I do not know why this question turned into tricky when instead it is very simple as question.
For example the explanation provided above, aside mine tells us

Quote:
Remember: Remainder is always an integer and is always less than the divisor. Thus.

Quantity A: Product of two odds is always odd. Thus odd/2 will always yield remainder of 1. Hence Quantity A = 1.

Quantity B: since x > y, this means that Quantity B will always be larger than 1.

Hence Quantity B is greater.


This is conceptually, using just theory.

Or you can pick numbers

The result will be the same

You also miss the point of the question: the two quantities ask you

QA the remainder

QB the dividend, the result of the division. In your case: 9/3=3 and the remainder is 0. However in QB what you are looking for is the result of the division , the dividend which is 3

In fact. in QB x/y is simply a division. we do not care of the remainder. It is just a division.

All the mistakes above are made because people carrying the information we want is QA towards QB, and this is WRONG
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0 < y < x. x and y are odd integers. [#permalink]
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