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If x and y are non-zero integers, what is the remainder when

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If x and y are non-zero integers, what is the remainder when [#permalink] New post 13 Dec 2004, 01:46
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If x and y are non-zero integers, what is the remainder when x is divided by y?

1. x when divided by 2y gives a remainder of 4

2. (x+y) when divided by y gives a remainder of 4


Can anyone show how to solve such problems algebraically since plugging numbers can be time consuming sometimes.
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 [#permalink] New post 13 Dec 2004, 05:14
B?

1. x when divided by 2y gives a remainder of 4
x = (2y)*k + 4
Seems insufficient.

2. (x+y) when divided by y gives a remainder of 4
x+y = ky + 4
x = (k-1)y + 4
Sufficient. Remainder will be 4.
Moreover,
(x+y)/y (remaider 4) can be simplified to x/y + y/y. Since y/y never has a remainder, regardless of the value of y, the remainder when x is divided by y will always be the same as the remainder when (x+y)y.
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 [#permalink] New post 13 Dec 2004, 06:07
D.

Same explanation as above...

Quote:
1. x when divided by 2y gives a remainder of 4
x = (2y)*k + 4
Seems insufficient.

2. (x+y) when divided by y gives a remainder of 4
x+y = ky + 4
x = (k-1)y + 4
Sufficient. Remainder will be 4.


s1: x=(2y)k+4 => x = y(2k) + 4
Remainder 4.
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 [#permalink] New post 13 Dec 2004, 06:28
No, consider the following example:
If X=10, Y=3.
10/(2*3) leaves remainder 4. Great.
But 10/3 leaves remainder 1.
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 [#permalink] New post 13 Dec 2004, 07:26
Dookie wrote:
No, consider the following example:
If X=10, Y=3.
10/(2*3) leaves remainder 4. Great.
But 10/3 leaves remainder 1.


I really tried to do this algebraically, and then I broke down and plugged in. I think it's a great question, and I (personally) found it difficult to wrap my head around it without some numbers.

I tried a couple options for number 1. 24/10 and 24/5 gave the same remainder. Then I searched my head for a number that's divisible by something but not twice that something. I came up with 60/8 and 60/4.

I know this is a little sloppy, but when I laid out the algebra, I just couldn't see it.

For 2, I tried just one example and then it all made sense. Remainders are set up kind of like multiples. If I divide a number by 5 and get some remainder, then 5 more than that number will give me the same remainder. That's what number 2 is saying, and it was clear right away that it's sufficient.

B
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 [#permalink] New post 13 Dec 2004, 08:19
Dookie wrote:
No, consider the following example:
If X=10, Y=3.
10/(2*3) leaves remainder 4. Great.
But 10/3 leaves remainder 1.


I agree Dookie, my bad :oops:

I guess plugging in, is the best way to go about it...
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 [#permalink] New post 13 Dec 2004, 18:41
This might be problem that might be easily solved by substitution.

Let me try and confirm using Algebra though

x = yk1 +C1

Statement 1)
x = 2yk2 + 4

equating the original equation and this as LHS is x

yk1 + C1 = 2yk2+4
C1 = y(2k2-k1) +4

If k1 = 2k2, C1 = 4

But if k1 <>2K2 (which can happen if y can go more times when dividing x than 2y can when dividing x, for example.)

C 1 <>4 in this case

Does not work.

2) x+y = (y)k2 + 4
using the original equations and this as LHS is x

(yk1 + C1) +y = (y)k2 + 4
C1 = y(k2-1-k1) +4

We know k2 = k1 +1, as y will go one more times on x+y than it would just on x.

It is always 4

B it is.

Like I said doing this using algebra in 2 minute is tough, so substitution might be better for this one.
  [#permalink] 13 Dec 2004, 18:41
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