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The integers x and y are both positive, the remainder when x [#permalink]
05 Jan 2013, 13:17
1
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Difficulty:
55% (hard)
Question Stats:
68% (03:15) correct
32% (02:06) wrong based on 194 sessions
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
Re: The integers x and y are both positive, the remainder when x [#permalink]
05 Jan 2013, 14:24
Your question "Each of the following is a possible value of ??? EXCEPT" seems to be incomplete. Which value is required to be determined? _________________
Re: The integers x and y are both positive, the remainder when x [#permalink]
06 Jan 2013, 00:50
Expert's post
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. _________________
Re: The integers x and y are both positive, the remainder when x [#permalink]
07 Jan 2013, 03:34
3
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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.
Re: The integers x and y are both positive, the remainder when x [#permalink]
07 Jan 2013, 06:04
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? _________________
Re: The integers x and y are both positive, the remainder when x [#permalink]
07 Jan 2013, 08:44
Expert's post
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.
Re: The integers x and y are both positive, the remainder when x [#permalink]
09 Sep 2014, 01:31
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Re: The integers x and y are both positive, the remainder when x [#permalink]
30 Jul 2015, 08:04
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...
Re: The integers x and y are both positive, the remainder when x [#permalink]
30 Jul 2015, 08:17
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.
Re: The integers x and y are both positive, the remainder when x [#permalink]
30 Jul 2015, 08:20
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.
Re: The integers x and y are both positive, the remainder when x [#permalink]
30 Jul 2015, 14:27
1
This post received KUDOS
Expert's post
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.
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