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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? source: Optimus Prep
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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/gmatquant ... emainders/In short, if we divide N by d, with a quotient Q and remainder R, then N = d*Q + RHere, 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/gmatquant ... emainders/In short, if we divide N by d, with a quotient Q and remainder R, then N = d*Q + RHere, 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/gmatquant ... emainders/In short, if we divide N by d, with a quotient Q and remainder R, then N = d*Q + RHere, 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 a2b=1332=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 a2b is positive number, say a=22, b=7 => 2b=14 Hence a2b = 8 Hence remainder when divided by 9 =8
say a=67, b=25 => 2b=50 Hence a2b = 17 Hence remainder when divided by 9 is again 8
say a=49, b=7 => 2b=14 Hence a2b = 35 Hence remainder when divided by 9 is again 8
Answer is 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 problemwork only with remainders. a2b=42*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 => a2b = 10 or a = 22; b = 25 => a2b = 28 Division by 9 yields 1. I feel that answer should be 1. Nevertheless, going by equations above : a2b = 9(x2y)  10 (a2b)/9 = (x2y)  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/gmatquant ... emainders/In short, if we divide N by d, with a quotient Q and remainder R, then N = d*Q + RHere, 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 a2b=1332=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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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 => a2b = 10 or a = 22; b = 25 => a2b = 28 Division by 9 yields 1. I feel that answer should be 1. Nevertheless, going by equations above : a2b = 9(x2y)  10 (a2b)/9 = (x2y)  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 a2b, when you do (492*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 a2b, when you do (492*16)/2 ?? Tx, Nelson Sorry, it was a typo. Have corrected it now
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