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If x and y are positive integers, what is the remainder when [#permalink]
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04 Aug 2010, 14:17
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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: Need solution ! [#permalink]
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04 Aug 2010, 18:53
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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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Re: Need solution ! [#permalink]
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04 Aug 2010, 19:48
isnt st 2 enugh? (X+y)=y(p)+4 so, x= y(p1)+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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Re: Need solution ! [#permalink]
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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 > \(x+y=yp+4\) > \(x=y(p1)+4\) (\(x\) is 4 more than multiple of \(y\))> this statement directly tells us that \(x\) divided by \(y\) yields remainder of 4. Sufficient. Answer: B. Hope it's clear.
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Re: Need solution ! [#permalink]
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04 Aug 2010, 23:14
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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Re: Need solution ! [#permalink]
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24 Feb 2011, 11:30
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 > \(x+y=yp+4\) > \(x=y(p1)+4\) (\(x\) is 4 more than multiple of \(y\))> this statement directly tells us that \(x\) divided by \(y\) yields remainder of 4. Sufficient. Answer: B. 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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Re: Need solution ! [#permalink]
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24 Feb 2011, 11:40
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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 > \(x+y=yp+4\) > \(x=y(p1)+4\) (\(x\) is 4 more than multiple of \(y\))> this statement directly tells us that \(x\) divided by \(y\) yields remainder of 4. Sufficient. Answer: B. 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(p1)+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 (p1), which for integer values of p can give us ANY value: any even as well as any odd. So basically x=y(p1)+4 is the same as x=yp+4. No need for double checking. Hope it's clear.
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Re: Need solution ! [#permalink]
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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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Re: Need solution ! [#permalink]
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Bunuel wrote: That's a good question. A. x=y(2k)+4, k any integer >=0. B. x=y(p1)+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 (p1), which for integer values of p can give us ANY value: any even as well as any odd. So basically x=y(p1)+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 4statement 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 4statement 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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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 ... yapplied/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. Answer (B)
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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, 08:23
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, 08:58
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, 18:53
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, 01:32
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 numberplugging: 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 < yAnalyzing St. 1When x is divided by 2y, the remainder is 4That 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 < ySimilarly, 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. 2When x + y is divided by y, the remainder is 4That 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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