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# The positive integers a and b leave remainders of 4 and 7

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

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08 Aug 2014, 15:40
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The positive integers a and b leave remainders of 4 and 7, respectively, when divided by 9. What is the remainder when a − 2b is divided by 9?

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

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08 Aug 2014, 16:09
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bytatia wrote:
The positive integers a and b leave remainders of 4 and 7, respectively, when divided by 9. What is the remainder when a − 2b is divided by 9?

Dear bytatia,
I'm happy to respond.

This is a problem ripe for the Rebuilding the Dividend Formula. See:
http://magoosh.com/gmat/2012/gmat-quant ... emainders/

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

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08 Aug 2014, 16:59
mikemcgarry wrote:
bytatia wrote:
The positive integers a and b leave remainders of 4 and 7, respectively, when divided by 9. What is the remainder when a − 2b is divided by 9?

Dear bytatia,
I'm happy to respond.

This is a problem ripe for the Rebuilding the Dividend Formula. See:
http://magoosh.com/gmat/2012/gmat-quant ... emainders/

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

Perfect! Yes it does! Thank you.

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

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09 Aug 2014, 02:01
mikemcgarry wrote:
bytatia wrote:
The positive integers a and b leave remainders of 4 and 7, respectively, when divided by 9. What is the remainder when a − 2b is divided by 9?

Dear bytatia,
I'm happy to respond.

This is a problem ripe for the Rebuilding the Dividend Formula. See:
http://magoosh.com/gmat/2012/gmat-quant ... emainders/

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.

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

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09 Aug 2014, 08:04
a=9x+4
so a could be 4, 13,22, 31,40,49,58,67,76,85

b=9y + 7
so b could be 7,16,25,34,43

Selecting any random number for a and b such that a-2b is positive number,
say a=22, b=7 => 2b=14
Hence a-2b = 8 Hence remainder when divided by 9 =8

say a=67, b=25 => 2b=50
Hence a-2b = 17 Hence remainder when divided by 9 is again 8

say a=49, b=7 => 2b=14
Hence a-2b = 35 Hence remainder when divided by 9 is again 8

(corrected my initial silly calculation mistake. Brain fart !!)

Last edited by romitsn on 09 Aug 2014, 15:48, edited 1 time in total.

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

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09 Aug 2014, 10:30
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If your divisor doesn't change through the problem-work only with remainders.

a-2b=4-2*7=-10

Of course -10 could not be the remainder. Just add a multiple of 9 such that your remainder stays in interval 0...8.

-10+18=8

The correct answer is 8.
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Re: The positive integers a and b leave remainders of 4 and 7 [#permalink]

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09 Aug 2014, 10:55
smyarga -- I understand your point.

But that still does not explain the problem with my initial approach. Can you explain.

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

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09 Aug 2014, 10:59
romitsn wrote:
smyarga -- I understand your point.

But that still does not explain the problem with my initial approach. Can you explain.

You just made a mistake with calculations. For the first one the correct sequence is
4, 13, 22, 31,...
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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?

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

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09 Aug 2014, 11:33
desaichinmay22 wrote:
mikemcgarry wrote:
bytatia wrote:
The positive integers a and b leave remainders of 4 and 7, respectively, when divided by 9. What is the remainder when a − 2b is divided by 9?

Dear bytatia,
I'm happy to respond.

This is a problem ripe for the Rebuilding the Dividend Formula. See:
http://magoosh.com/gmat/2012/gmat-quant ... emainders/

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. $$0\leq remainder<divisor$$
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And yes, I like kudos:)

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

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09 Aug 2014, 11:34
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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
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Re: The positive integers a and b leave remainders of 4 and 7 [#permalink]

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09 Aug 2014, 11:48
ah ...stupid silly mistake..sorry and thanks !

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

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09 Aug 2014, 14:22
The answer is 8, 9x - 18y - 10, cancel out 9x and 18 (they aren't remainders), take a multiple of 9, 18, subract 10, and bam you get 8

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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.

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

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13 Aug 2014, 02:21
Picking up friendly numbers

a = 45 + 4 = 49

b = 9+7 = 16

$$\frac{49 - 2*16}{9} = \frac{17}{9}$$.......... gives remainder = 8
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Last edited by PareshGmat on 11 Jan 2015, 07:17, edited 1 time in total.

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

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11 Jan 2015, 00:06
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Hello Paresh GMAT,
From where did you get the divisor 2 in a-2b, when you do (49-2*16)/2 ??
Tx, Nelson

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

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11 Jan 2015, 07:18
nelson1972 wrote:
Hello Paresh GMAT,
From where did you get the divisor 2 in a-2b, when you do (49-2*16)/2 ??
Tx, Nelson

Sorry, it was a typo. Have corrected it now
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Re: The positive integers a and b leave remainders of 4 and 7 [#permalink]

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Re: The positive integers a and b leave remainders of 4 and 7   [#permalink] 20 Jun 2016, 11:25
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