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The integers x and y are both positive, the remainder when x [#permalink]
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05 Jan 2013, 14:17
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The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+y EXCEPT A. 125 B. 101 C. 77 D. 51 E. 41 I got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually. I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks
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Last edited by Bunuel on 07 Jan 2013, 04:29, edited 2 times in total.
Edited the question.



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Re: The integers x and y are both positive, the remainder when x [#permalink]
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05 Jan 2013, 15:24
Your question "Each of the following is a possible value of ??? EXCEPT" seems to be incomplete. Which value is required to be determined?
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Re: The integers x and y are both positive, the remainder when x [#permalink]
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06 Jan 2013, 00:59
PraPon wrote: Your question "Each of the following is a possible value of ??? EXCEPT" seems to be incomplete. Which value is required to be determined? You are right. Edited. 2X+Y.
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Re: The integers x and y are both positive, the remainder when x [#permalink]
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06 Jan 2013, 01:50
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roygush wrote: The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2X+Y EXCEPT:
A)125 B)101 C)77 D)51 E)41
I got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.
I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks There are two statements in the question: 1) The remainder when x is divided by 12 is 7. This can be written as \(x=12I + 7\), where I is an integer. 2) The remainder when y is divided by 12 is 3. This can be written as \(y=12J + 3\), where J is an integer. One thing to note here is that the maximum remainder that these two equations can generate is 11. So \(2x\) will not be equal to \(12I+14\) but will be equal to \(12I+2\). Therefore \(2x + y = 12(I+J) + 5\) or any multiple of 12 + 5. 51 doesn't follows such pattern. Hence is the answer.
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Re: The integers x and y are both positive, the remainder when x [#permalink]
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07 Jan 2013, 04:34
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roygush wrote: The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+y EXCEPT
A. 125 B. 101 C. 77 D. 51 E. 41
I got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.
I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks The remainder when x is divided by 12 is 7: \(x=12q+7\) > \(2x=24q+14\). The remainder when y is divided by 12 is 3: \(y=12p+3\). \(2x+y=(24q+14)+(12p+3)=24q+12p+12+5=12(2q+p+1)+5=(multiple \ of \ 12)+5\). Only D is not a multiple of 12 plus 5. Answer: D.
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Re: The integers x and y are both positive, the remainder when x [#permalink]
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07 Jan 2013, 07:04
Bunuel wrote: roygush wrote: The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+y EXCEPT
A. 125 B. 101 C. 77 D. 51 E. 41
I got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.
I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks The remainder when x is divided by 12 is 7: \(x=12q+7\) > \(2x=24q+14\). The remainder when y is divided by 12 is 3: \(y=12p+3\). \(2x+y=(24q+14)+(12p+3)=24q+12p+12+5=12(2q+p+1)+5=(multiple \ of \ 12)+5\). Only D is not a multiple of 12 plus 5. Answer: D. bunuel, can we apply this method on any remainders problem? write the two equations and then work from there?
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Re: The integers x and y are both positive, the remainder when x [#permalink]
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07 Jan 2013, 09:44
roygush wrote: Bunuel wrote: roygush wrote: The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+y EXCEPT
A. 125 B. 101 C. 77 D. 51 E. 41
I got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.
I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks The remainder when x is divided by 12 is 7: \(x=12q+7\) > \(2x=24q+14\). The remainder when y is divided by 12 is 3: \(y=12p+3\). \(2x+y=(24q+14)+(12p+3)=24q+12p+12+5=12(2q+p+1)+5=(multiple \ of \ 12)+5\). Only D is not a multiple of 12 plus 5. Answer: D. bunuel, can we apply this method on any remainders problem? write the two equations and then work from there? I wouldn't say any, but \(dividend=divisor*quotient+remainder\) formula is indeed very handy to deal with questions about remainders. For more check GMAT Math Book chapter on remainders: remainders144665.htmlHope it helps.
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Re: The integers x and y are both positive, the remainder when x [#permalink]
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25 Jun 2015, 06:30
x= 12q+7 => 2x = 24q+14 y = 12p+3
2x+y = 24q+14+12p+3 => 12(p+2q)+17
Any answer choice from which 17 is subtracted and it doesn't turn out to be a multiple of 12 is the answer. => 5117 = 34 (not a multiple of 12).
(D) is the answer.



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Re: The integers x and y are both positive, the remainder when x [#permalink]
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30 Jul 2015, 09:04
Another way to do it is the following, which took me around 4 minutes (perhaps someone quick in calculations could do it faster, dunno...). So:
x=12q+7 = 19, 31, 43, 55, 67 etc. Doudling it: 38, 62, 126, 110, 134 etc. y=12z+3 = 15, 27, 39, 51, 63 etc.
At this point, we can already see that D is the answer, as D=51, which is a possible value of y alone. So, it is not possible that if we add x to 51 it will remain 51. So, ANS D
Do you think it is doable like this in under 2 minutes, even though it took me 4? I am asking as I am terribly slow in calculations. Performing calculations is a serious problem of mine in gmat...



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Re: The integers x and y are both positive, the remainder when x [#permalink]
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30 Jul 2015, 09:17
roygush wrote: The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+y EXCEPT
A. 125 B. 101 C. 77 D. 51 E. 41
I got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.
I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks Hi, x/12 = 7 (remainder) and y/12 = 3 (remainder). Hence the least value of x has to be 7 and least value of y is 3. Therefore, 2x+y = 17. Hence, look for an answer choice that is a multiple of 17, which is 51. Ans  D



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Re: The integers x and y are both positive, the remainder when x [#permalink]
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30 Jul 2015, 09:20
pacifist85 wrote: Another way to do it is the following, which took me around 4 minutes (perhaps someone quick in calculations could do it faster, dunno...). So:
x=12q+7 = 19, 31, 43, 55, 67 etc. Doudling it: 38, 62, 126, 110, 134 etc. y=12z+3 = 15, 27, 39, 51, 63 etc.
At this point, we can already see that D is the answer, as D=51, which is a possible value of y alone. So, it is not possible that if we add x to 51 it will remain 51. So, ANS D
Do you think it is doable like this in under 2 minutes, even though it took me 4? I am asking as I am terribly slow in calculations. Performing calculations is a serious problem of mine in gmat... Hi, x/12 = 7 (remainder) and y/12 = 3 (remainder). Hence the least value of x has to be 7 and least value of y is 3. Therefore, 2x+y = 17. Hence, look for an answer choice that is a multiple of 17, which is 51. Ans  D The above method, if used, can be solved within 2 mins. I took 1.5 mins to do so.



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Re: The integers x and y are both positive, the remainder when x [#permalink]
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30 Jul 2015, 15:27
Hi All, I'm a big fan of TESTing VALUES on this question along with a bit of 'brute force' math (pacifist85's approach showcases this tactic nicely). There is one detail worth noting and one aspect I would add to it: since this is an EXCEPT question, once you find the exception, you can stop working. The detail in pacifist85s math is that Q and Z could both be 0, so X COULD = 7 (and by extension, 2X COULD = 14) and Y COULD = 3. I would start with answer E because it's smallest (so it would have the least number of possible sums that could equal it). 41 = 38 + 3. It's possible, so it's NOT what we're looking for. 51 though…using the possible values of 2X… 2X = 14; Y would have to be 37 (which is NOT possible). 2X = 38; Y would have to be 13 (which is NOT possible). 2X = 62; this is already TOO BIG. Thus, 51 is the option that is NOT possible… Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: The integers x and y are both positive, the remainder when x [#permalink]
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15 Jul 2016, 19:36
roygush wrote: The integers x and y are both positive, the remainder when x is divided by 12 is 7, and the remainder when y is divided by 12 is 3. Each of the following is a possible value of 2x+y EXCEPT
A. 125 B. 101 C. 77 D. 51 E. 41
I got this right but it took me 2 minutes. I wonder if there is a quick way to solve this question without plugging numbers to X and Y. I wrote down like 6 numbers with remainder of 7 and 3 and found each of the numbers from the answer choices manually.
I also tried the other approach X=12Q+7, Y=12R+3 and calculated 2X+Y but couldn't understand what to do next. thanks Just for others who may be interested, the first time I did this problem it took me almost 5 minutes, but I got it right. However, upon review, I figured out that since you understand the concept, all you have to do is divide the options by 12, and the right answer will give you a different remainder from the others. Hence, all you need to do is divide at least 3 options by 12, until one gives you a different remainder than the rest. Goodluck!




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