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data sufficiency

by Guest Fri Jun 08, 2007 3:19 pm

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.

What do you think the answer is? I think answer is "E", but as per book it is B.

I don't understand why? This example is taken from Kaplan 800 book.
dbernst
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by dbernst Sat Jun 09, 2007 1:34 am

Tricky Tricky Tricky.

As is often my tactic on difficult DS problems, I first approached this by plugging in numbers.

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

If x = 10 and 2y = 6, r = 4. Thus, x = 10, y = 3 and r of x/y = 1
If x = 12 and 2y = 8, r = 4. Thus, x = 12, y = 4. and r of x/y = 0

Since I found two different remainders for x/y, I eliminated AD on my AD/BCE grid.

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

If y = 5, the first integer that gives a remainder of 4 when divided by y is 9. Thus, x = 4 (since 5+4 = 9) and the remainder of x/y = 4. Keeping y = 5, the next integer that gives a remainder of 4 when divided by y is 14. Thus, x = 9. Since this simply made our original x increase by y (we added one more value of y), the remainder of x/y will remain unchanged (r still = 4). This pattern holds for all y = 5.

If y = 6, the first integer that gives a remainder of 4 when divided by y is 10. Thus, x = 4 and the remainder of x/y = 4. Keeping y = 6, the next integer that gives a remainder of 4 when divided by y is 16. Thus, x = 10. Since this simply made our original x increase by y, the remainder of x/y will remain unchanged (r still = 4). This pattern holds for all y = 6.

If y = 7, the first integer that gives a remainder of 4 when divided by y is 11. Thus, x = 4 and the remainder of x/y = 4. Keeping y = 7, the next integer that gives a remainder of 4 when divided by y is 18. Thus, x = 11. Since this simply made our original x increase by y, the remainder of x/y will remain unchanged (r still = 4). This pattern holds for all y = 7.

In essence, for the remainder the fraction from statement (2) to be 4, the difference between the numerator and the denominator (or a multiple of the demonimator) must be 4. Since, in equation (2), x is the number that accounts for this difference, the first value for x = 4 (with y>4). Thus the remaider of the first value for x/y will always be 4. To keep the remainder of 4 in (2), the numerator will always increase by multiples y. Thus, the remainder of x/y will always = 4. The correct answer is B. (There must be an algebraic method for this problem, but it is too late right now for me to see it. I or another instructor will reply with this during daylight hours!)
-dan

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.

What do you think the answer is? I think answer is "E", but as per book it is B.

I don't understand why? This example is taken from Kaplan 800 book.
Kishore
 
 

(x+y)/y

by Kishore Sun Oct 14, 2007 1:35 pm

Remainder when (x+y) is divided by y is always the same as the remainder when x is divided by y. To generaize

Remainder when (x+n.y^m) is divided by y is always the same as when x is divided by y

Since n.y^m will always contribute 0 towards the remainder.
Guest
 
 

by Guest Mon Oct 15, 2007 11:48 am

another take:

2) (x+y)/y= x/y+ y/y = x/y +1

So x/y+ 1 = a quotient (let's say K) + 4
x/y = k + 3

So the remainder is 3.
Pathik
 
 

by Pathik Sat May 03, 2008 4:28 pm

How can this be solved without picking the numbers,
Please comment on my logic.

St1 x = (2y)k + 4 - where k is any int
x = y*2k + 4
x = y*m + 4 - where m is another int

so x/y will give remainder of 4. After reading Dan's response I agree that my solution is incorrect. But I am not sure what is wrong in my approach

Thanks
Pathik
rfernandez
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by rfernandez Thu May 15, 2008 7:22 pm

How can this be solved without picking the numbers,
Please comment on my logic.

St1 x = (2y)k + 4 - where k is any int
x = y*2k + 4
x = y*m + 4 - where m is another int

so x/y will give remainder of 4. After reading Dan's response I agree that my solution is incorrect. But I am not sure what is wrong in my approach

Thanks
Pathik


Your solution makes an assumption that may or may not be true. Specifically, it mandates that the x/y result in a quotient (m) that is exactly twice the quotient of x/(2y) (whose quotient is k). Sometimes it happens to be true that m=2k, but not always.

For example, set x=14 and y=5.
x/2y = 14/10 = 1 r 4. So k=1.
x/y = 14/5 = 2 r 4. So m=2.
Here's a case where both have remainders of 4 AND m=2k.

However, if m does not equal 2k, it breaks down.
Set x=10 and and y=3.
x/2y = 10/6 = 1 r 4. So k=1.
x/y = 10/3 = 3 r 1. So m=3.
Here's a case where the remainder of x/y is not 4. And it's not 4 specifically because m does not equal 2k.

Rey
Guest
 
 

by Guest Fri May 16, 2008 7:49 am

This can be solved by picking numbers:

1) Consider the first case. Let x=4 and y=4
x/2y = 4/8 . Hence the remainder is 4
x/y = 4/4. The remainder is zero

Now let x=10 and y=3
x/2y = 10/6 . Hence the remainder is 4
x/y = 10/3 . The remainder is 1

The remainder is different for both hence option (A) is incorrect

2) Now for the second case. Let x=4 and y=8
x+y/y = 12/8 . Hence the remainder is 4
x/y = 4/8. The remainder is 4

Now let x=4 and y=7
x+y/y = 11/7. Hence the remainder is 4
x/y = 4/7. The remainder is 4

Whatever numbers you plug in ull get the same remainder for x/y in the second case.

Therefore, the answer is (B)

I hope this solves your problem
StaceyKoprince
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by StaceyKoprince Thu May 29, 2008 12:31 am

Thanks for your solution, guest! Actually, the last poster before you and Rey asked how to do it WITHOUT picking numbers :) but I'm sure this will be useful for others!
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
cutlass
 
 

by cutlass Thu May 29, 2008 12:06 pm

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

=> x = 2yn + 4

n = (x-4)/2y

All this tells us is x is even.

w/o knowing x & y, we cannot determine the remainder of x/y

2) (x+y)/y

= quotient(x/y) + remainder(x/y) + 1
= n + remainder(x/y)

As you can see, the remainder of x/y is the same as that of (x+y)/y
rfernandez
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by rfernandez Fri May 30, 2008 4:45 am

Yep, another good solution. Nice work.

Rey
sanj
 
 

by sanj Fri May 30, 2008 5:58 am

cutlass Wrote:1) When x is divided by 2y, the remainder is 4.

=> x = 2yn + 4

n = (x-4)/2y

All this tells us is x is even.

w/o knowing x & y, we cannot determine the remainder of x/y

2) (x+y)/y

= quotient(x/y) + remainder(x/y) + 1
= n + remainder(x/y)

As you can see, the remainder of x/y is the same as that of (x+y)/y


but my dear it should be quotient(x/y)*y + remainder(x/y) + 1
sanj
 
 

by sanj Fri May 30, 2008 6:09 am

it can better be understood
1) x=2ym +4
x/y = 2m +4/y cant say insuff
2) (x+y)/y = x/y + y/y
so r[ (x+y)/y] = r(x/y) +r(y/y)
and r(y/y)=0
therefor r[ (x+y)/y] = r(x/y)
hence B
rfernandez
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by rfernandez Fri Jun 06, 2008 3:43 am

Well, looks like we've explored every nook and cranny of this one. Lots of ways to get there. Nice work all.

Rey