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# If x and y are positive integers, what is the remainder when

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Director
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If x and y are positive integers, what is the remainder when  [#permalink]

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04 Aug 2010, 15:17
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If x and y are positive integers, what is the remainder when x is divided by y?

(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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16 Nov 2011, 22:50
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nusmavrik wrote:
If x and y are positive integers, what is the remainder when x is divided by y?
(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4

If you understand the concept of divisibility well, you can pretty much do this orally in less than 30 secs with a little bit of visualization. Divisibility involves grouping. Check these out first since I am explaining using the concept discussed in these posts:
http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/

Stmnt 1: When x is divided by 2y, the remainder is 4

When you divide x by 2y, you make groups with 2y balls in each and you have 4 balls leftover.
Instead, if you divide x by y, you may have 4 balls leftover or you may have fewer balls if y is less than or equal to 4 i.e. say if y = 3, you could make another group of 3 balls and you will have only 1 ball leftover. So you could have different remainders. Not sufficient.

Stmnt 2: When x + y is divided by y, the remainder is 4

When you make groups of y balls each from (x+y), the y balls make 1 group and you are left with x balls. If the remainder is 4, it means when you make groups of y balls each from x balls, you have 4 balls leftover.
Since the question asks us: how many balls are leftover when you make groups of y balls from x balls, you get your answer directly as '4'.
Sufficient.

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04 Aug 2010, 19:53
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1. IMO B

the second one:
(x+y)/y= b+4,
x/y+y/y=b+4,
x/y+1=b+4
or x/y=b+3. the remainder is 3
with this one we agree, it is sufficient.
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04 Aug 2010, 20:48
isnt st 2 enugh?
(X+y)=y(p)+4
so, x= y(p-1)+4; so means remainder of x/y is 4

Please correct me if am wrng. For st1, I cant find an x/y using x=2y(m)+4
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04 Aug 2010, 22:46
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nusmavrik wrote:
If x and y are positive integers, what is the remainder when x is divided by y?
(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4

Positive integer $$x$$ divided by positive integer $$y$$ yields remainder of $$r$$ can be expressed as $$x=yq+r$$. Question is $$r=?$$

(1) When x is divided by 2y, the remainder is 4. If $$x=20$$ and $$y=8$$ (satisfies the given statemnet as 20 divided by 2*8=16 yields reminder of 4), then $$x$$ divided by $$y$$ yields $$r=4$$ (20 divided by 8 yields remainder of 4) BUT if $$x=10$$ and $$y=3$$ (satisfies the given statemnet as 10 divided by 2*3=6 yields reminder of 4), then $$x$$ divided by $$y$$ yields $$r=1$$ (10 divided by 3 yields remainder of 1). Two different answers. Not sufficient.

(2) When x + y is divided by y, the remainder is 4 --> $$x+y=yp+4$$ --> $$x=y(p-1)+4$$ ($$x$$ is 4 more than multiple of $$y$$)--> this statement directly tells us that $$x$$ divided by $$y$$ yields remainder of 4. Sufficient.

Hope it's clear.
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05 Aug 2010, 00:14
trapped !

Its the quotient that decreases (minus 1). The remainder is unaffected.

E.g (3 + 5) /5 ----> remainder = 3, quotient = 1
3/5 ------> remainder = 3, quotient = 0

onedayill wrote:
1. IMO B

the second one:
(x+y)/y= b+4,
x/y+y/y=b+4,
x/y+1=b+4
or x/y=b+3. the remainder is 3
with this one we agree, it is sufficient.
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24 Feb 2011, 12:30
Bunuel wrote:
nusmavrik wrote:
If x and y are positive integers, what is the remainder when x is divided by y?
(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4

Positive integer $$x$$ divided by positive integer $$y$$ yields remainder of $$r$$ can be expressed as $$x=yq+r$$. Question is $$r=?$$

(1) When x is divided by 2y, the remainder is 4. If $$x=20$$ and $$y=8$$ (satisfies the given statemnet as 20 divided by 2*8=16 yields reminder of 4), then $$x$$ divided by $$y$$ yields $$r=4$$ (20 divided by 8 yields remainder of 4) BUT if $$x=10$$ and $$y=3$$ (satisfies the given statemnet as 10 divided by 2*3=6 yields reminder of 4), then $$x$$ divided by $$y$$ yields $$r=1$$ (10 divided by 3 yields remainder of 1). Two different answers. Not sufficient.

(2) When x + y is divided by y, the remainder is 4 --> $$x+y=yp+4$$ --> $$x=y(p-1)+4$$ ($$x$$ is 4 more than multiple of $$y$$)--> this statement directly tells us that $$x$$ divided by $$y$$ yields remainder of 4. Sufficient.

Hope it's clear.

I'm confused. With the same proof given for the second point, could I not convince myself that 1 would suffice?
eg: x = yp + r
From (1) I could say x = 2yp + 4
(or) x = (2p)y + 4
Clearly this statement tells us that x is 4 more than a multiple of y as well. Why can I not convince myself at this step that the statement would suffice? Although, the statement does not suffice since the logic fails when you plug in (x=10,y=3) .
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24 Feb 2011, 12:40
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bugSniper wrote:
Bunuel wrote:
nusmavrik wrote:
If x and y are positive integers, what is the remainder when x is divided by y?
(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4

Positive integer $$x$$ divided by positive integer $$y$$ yields remainder of $$r$$ can be expressed as $$x=yq+r$$. Question is $$r=?$$

(1) When x is divided by 2y, the remainder is 4. If $$x=20$$ and $$y=8$$ (satisfies the given statemnet as 20 divided by 2*8=16 yields reminder of 4), then $$x$$ divided by $$y$$ yields $$r=4$$ (20 divided by 8 yields remainder of 4) BUT if $$x=10$$ and $$y=3$$ (satisfies the given statemnet as 10 divided by 2*3=6 yields reminder of 4), then $$x$$ divided by $$y$$ yields $$r=1$$ (10 divided by 3 yields remainder of 1). Two different answers. Not sufficient.

(2) When x + y is divided by y, the remainder is 4 --> $$x+y=yp+4$$ --> $$x=y(p-1)+4$$ ($$x$$ is 4 more than multiple of $$y$$)--> this statement directly tells us that $$x$$ divided by $$y$$ yields remainder of 4. Sufficient.

Hope it's clear.

I'm confused. With the same proof given for the second point, could I not convince myself that 1 would suffice?
eg: x = yp + r
From (1) I could say x = 2yp + 4
(or) x = (2p)y + 4
Clearly this statement tells us that x is 4 more than a multiple of y as well. Why can I not convince myself at this step that the statement would suffice? Although, the statement does not suffice since the logic fails when you plug in (x=10,y=3) .

That's a good question.
A. x=y(2k)+4, k any integer >=0.
B. x=y(p-1)+4, p any integer >=0.

Why A is not sufficient to determine the remainder and B is? Why did I use number plugging to show this in the first case and didn't in the second?

If we are told that x divided by y gives a remainder of 4, means x=yp+4 where p is integer >=0. We don't know x and y so p (quotient) can be any integer.

Look at equation A, the quotient is 2k, 2k is even. It can be rephrased as x divided by y will give the remainder of 4 IF quotient is even. But what about the cases when quotient is odd? We don't know that so we must check to determine this.

As for B. Quotient here is (p-1), which for integer values of p can give us ANY value: any even as well as any odd. So basically x=y(p-1)+4 is the same as x=yp+4. No need for double checking.

Hope it's clear.
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24 Feb 2011, 12:53
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B is sufficient by using the rule of remainders additive property:
(x+y)/y leaves a remainder of 4.
Means: remainder left by x/y + remainder left by y/y = 4
remainder left by x/y+0=4
remainder left by x/y=4

At least B is Sufficient.
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11 Nov 2011, 21:24
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Bunuel wrote:
That's a good question.
A. x=y(2k)+4, k any integer >=0.
B. x=y(p-1)+4, p any integer >=0.

Why A is not sufficient to determine the remainder and B is? Why did I use number plugging to show this in the first case and didn't in the second?

If we are told that x divided by y gives a remainder of 4, means x=yp+4 where p is integer >=0. We don't know x and y so p (quotient) can be any integer.

Look at equation A, the quotient is 2k, 2k is even. It can be rephrased as x divided by y will give the remainder of 4 IF quotient is even. But what about the cases when quotient is odd? We don't know that so we must check to determine this.

As for B. Quotient here is (p-1), which for integer values of p can give us ANY value: any even as well as any odd. So basically x=y(p-1)+4 is the same as x=yp+4. No need for double checking.

Hope it's clear.

Bunuel could we say the following? (Having in mind that the Remainder depends on the divisor)

(1) When x is divided by 2y, the remainder is 4

statement 1 ----> x=2*y*k+4, k integer

Therefore because the divisor has to be larger (not equal because it is stated that a reminder exists) than the remainder: 2*y>4 --> y>2 -->y>=3

If y (divisor) is smaller than 4 then the remainder changes and if it is larger than 4 the remainder is 4.

For example:
if y=3 then x=2*3*k+3+1, R=1
if y=4 then x=4*2*k+4+0, R=0
(if y=5 then x=2*5k+4, R=4)

Therefore Insufficient.

2) When x + y is divided by y, the remainder is 4

statement 2 ----> x+y=y*k+4, k integer

We are told that the remainder is 4, therefore y>=5! Which means that remainder will always be 4.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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17 Nov 2011, 09:23
Karishma...Really impressive reply..!! Very precise..and easily understandable...went through your post too...had never imagined division from this perspective...i think this is a better approach to these questions..! Thanks for sharing!
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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17 Nov 2011, 09:58
I must admit,initially,i did get trapped in option A & B both and would've answered both are sufficient,but carefully after evaluating,realized, only B suffices.
A doesn't.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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17 Dec 2011, 19:53
Yes a tricky question. Karishma, thank you for the detailed explanation. The concept of "grouping" applied to division, although new to me, is easily understandable and very simple indeed.
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Re: If x and y are positive integers, what is the remainder when  [#permalink]

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29 May 2015, 02:32
Quote:
If x and y are positive integers, what is the remainder when x is divided by y?

(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4

Here's an alternate solution to this question, without using number-plugging:

The question asks us the remainder for x/y

This means, in the expression x = ky + r . . . (1) (where k is the quotient and r is the remainder)

We have to find the value of r, where 0 =<r < y

Analyzing St. 1
When x is divided by 2y, the remainder is 4

That is, x = (m)(2y) + 4 . . . (2)

Now, upon comparing Equations (1) and (2), can we say r = 4? Let's see:

We know that remainder is always less than the divisor.

Since r is the remainder in Equation 1, we can be sure that r < y

Similarly, since 4 is the remainder in Equation 2, we can be sure that 4 < 2y. This gives us, y > 2

So, y can be 3, 4, 5, 6, 7 etc.

Now, if y = 3, then r must be less than 3. So, upon comparing (1) and (2), you'll not say that r = 4. Instead, you'll break down 4 into 3 + 1. So, if y = 3, remainder r = 1

If y = 4, then r must be less than 4. So, upon comparing (1) and (2), you'll not say that r = 4. Instead, you'll write 4 as 4 + 0. So, if y = 4, remainder r = 0

If y > 4, then 4 becomes a valid value for remainder r. So, in this case, upon comparing (1) and (2), we can say that r = 4.

Thus, we see that r can be 0, 1 or 4. Since we have not been able to determine a unique value of r, St. 1 is not sufficient.

Analyzing St. 2
When x + y is divided by y, the remainder is 4

That is, x + y = qy + 4
Or, x = (q - 1)y + 4 . . . (3)

Equation (3) conveys that when x is divided by y, the remainder is 4. So, St. 2 is sufficient to find the value of the remainder.

Please note that the way we processed St. 1 was different from the way we processed St. 2, because in St. 1 the divisor was 2y, whereas in St. 2, the divisor was y (the same divisor as in the question)

Hope this alternate solution was useful!

Best Regards

Japinder
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