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In short, if we divide N by d, with a quotient Q and remainder R, then N = d*Q + R

Here, we divide a by 9, get some unknown quotient (call it x) and a remainder of 4. That gives us: a = 9x + 4 Then, we divide b by 9, get some unknown quotient (call it y) and a remainder of 7. That gives us: b = 9y + 7

Then, a − 2b = (9x + 4) − 2*(9y + 7) = 9x + 4 − 18y − 14 = 9x − 18y − 10 That's what we are going to divide by 9. Well, the part (9x − 18y) is a multiple of 9, so 9 goes into that with no remainder. It's hard to tell what the remainder is if we divide -10 by 9: here's a way to think of it. Take any multiple of 9, any at all, subtract 10, and find the remainder when you divide by 9.

18 - 10 = 8 --- remainder = 8 54 = 10 = 44 --- divide by 9, and the remainder = 8

So, the remainder when (a − 2b) is divided by 9 is 8.

In short, if we divide N by d, with a quotient Q and remainder R, then N = d*Q + R

Here, we divide a by 9, get some unknown quotient (call it x) and a remainder of 4. That gives us: a = 9x + 4 Then, we divide b by 9, get some unknown quotient (call it y) and a remainder of 7. That gives us: b = 9y + 7

Then, a − 2b = (9x + 4) − 2*(9y + 7) = 9x + 4 − 18y − 14 = 9x − 18y − 10 That's what we are going to divide by 9. Well, the part (9x − 18y) is a multiple of 9, so 9 goes into that with no remainder. It's hard to tell what the remainder is if we divide -10 by 9: here's a way to think of it. Take any multiple of 9, any at all, subtract 10, and find the remainder when you divide by 9.

18 - 10 = 8 --- remainder = 8 54 = 10 = 44 --- divide by 9, and the remainder = 8

So, the remainder when (a − 2b) is divided by 9 is 8.

In short, if we divide N by d, with a quotient Q and remainder R, then N = d*Q + R

Here, we divide a by 9, get some unknown quotient (call it x) and a remainder of 4. That gives us: a = 9x + 4 Then, we divide b by 9, get some unknown quotient (call it y) and a remainder of 7. That gives us: b = 9y + 7

Then, a − 2b = (9x + 4) − 2*(9y + 7) = 9x + 4 − 18y − 14 = 9x − 18y − 10 That's what we are going to divide by 9. Well, the part (9x − 18y) is a multiple of 9, so 9 goes into that with no remainder. It's hard to tell what the remainder is if we divide -10 by 9: here's a way to think of it. Take any multiple of 9, any at all, subtract 10, and find the remainder when you divide by 9.

18 - 10 = 8 --- remainder = 8 54 = 10 = 44 --- divide by 9, and the remainder = 8

So, the remainder when (a − 2b) is divided by 9 is 8.

Does all this make sense? Mike

Hi Mike, I tried to solve this problem by taking values a=13 and b=16. Therefore a-2b=13-32=-19 -19 divided by 9 leaves a remainder of -1. I am not sure whether a negative number can be a remainder. Please help.

Re: The positive integers a and b leave remainders of 4 and 7 [#permalink]

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09 Aug 2014, 11:15

Another way might be 'substitution'? e.g a = 13 ; b = 16 => a-2b = -10 or a = 22; b = 25 => a-2b = -28 Division by 9 yields -1. I feel that answer should be |1|. Nevertheless, going by equations above :- a-2b = 9(x-2y) - 10 (a-2b)/9 = (x-2y) - 10/9 ; it all boils to 10/9. isn't it?

In short, if we divide N by d, with a quotient Q and remainder R, then N = d*Q + R

Here, we divide a by 9, get some unknown quotient (call it x) and a remainder of 4. That gives us: a = 9x + 4 Then, we divide b by 9, get some unknown quotient (call it y) and a remainder of 7. That gives us: b = 9y + 7

Then, a − 2b = (9x + 4) − 2*(9y + 7) = 9x + 4 − 18y − 14 = 9x − 18y − 10 That's what we are going to divide by 9. Well, the part (9x − 18y) is a multiple of 9, so 9 goes into that with no remainder. It's hard to tell what the remainder is if we divide -10 by 9: here's a way to think of it. Take any multiple of 9, any at all, subtract 10, and find the remainder when you divide by 9.

18 - 10 = 8 --- remainder = 8 54 = 10 = 44 --- divide by 9, and the remainder = 8

So, the remainder when (a − 2b) is divided by 9 is 8.

Does all this make sense? Mike

Hi Mike, I tried to solve this problem by taking values a=13 and b=16. Therefore a-2b=13-32=-19 -19 divided by 9 leaves a remainder of -1. I am not sure whether a negative number can be a remainder. Please help.

the remainder cannot be negative. \(0\leq remainder<divisor\)
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

Re: The positive integers a and b leave remainders of 4 and 7 [#permalink]

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09 Aug 2014, 11:34

1

This post received KUDOS

kishgau wrote:

Another way might be 'substitution'? e.g a = 13 ; b = 16 => a-2b = -10 or a = 22; b = 25 => a-2b = -28 Division by 9 yields -1. I feel that answer should be |1|. Nevertheless, going by equations above :- a-2b = 9(x-2y) - 10 (a-2b)/9 = (x-2y) - 10/9 ; it all boils to 10/9. isn't it?

to -10/9 that have remainder -1+9=8
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

Re: The positive integers a and b leave remainders of 4 and 7 [#permalink]

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09 Aug 2014, 23:16

This question led me to multiple thread where the -ve remainder is debated. Some have quoted wiki that doesn't have any reservation about a -ve remainder. However, as per GMAT prep, remainder is always >0.

There may be another way to look at this. If we draw number line, plot -10 on the line and take an intercept of |-9| , it would leave one unit distance before -10. I don't think we'd take two intercepts and then count the distance between -18 and -10.

This concept looks weird but as long as GMAT feels it right, I'd go with that.

Re: The positive integers a and b leave remainders of 4 and 7 [#permalink]

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20 Jun 2016, 11:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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