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The integers x and y are both positive, the remainder when x is divide

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The integers x and y are both positive, the remainder when x is divide  [#permalink]

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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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Originally posted by roygush on 05 Jan 2013, 14:17.
Last edited by Bunuel on 11 Oct 2017, 04:34, edited 3 times in total.
Edited the question.
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Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

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New post 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 is divide  [#permalink]

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New post 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 is divide  [#permalink]

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New post 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 is divide  [#permalink]

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New post 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: remainders-144665.html

Hope it helps.
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Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

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New post 25 Jun 2015, 06:30
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1
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. => 51-17 = 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 is divide  [#permalink]

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New post 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 is divide  [#permalink]

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New post 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 is divide  [#permalink]

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New post 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 is divide  [#permalink]

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New post 30 Jul 2015, 15:27
1
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:

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Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

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New post 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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The integers x and y are both positive, the remainder when x is divide  [#permalink]

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New post 19 Apr 2018, 16:15
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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


One approach is to test values

IMPORTANT: When it comes to remainders, we have a nice rule that says:
If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Okay, onto the question....

The remainder when x is divided by 12 is 7
Possible values of x = 7, 19, 31, 43, 55, ...
So, possible values of 2x are: 14, 38, 62, 86, 110, ...

The remainder when y is divided by 12 is 3
Possible values of y are: 3, 15, 28, 40, 52, ...

Each of the following is a possible value of 2x + y EXCEPT

A) 125 This equals 110 + 15. ELIMINATE A
B) 101 This equals 86 + 15. ELIMINATE B
C) 77 This equals 62 + 15. ELIMINATE C
D) 51
E) 41 This equals 38 + 3. ELIMINATE E

Answer: D

Cheers,
Brent

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Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

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New post 28 Jun 2018, 08:08
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Please move to PS
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.

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Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

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New post 28 Jun 2018, 08:38
BrushMyQuant wrote:
Please move to PS
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.

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Re: The integers x and y are both positive, the remainder when x is divide  [#permalink]

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Re: The integers x and y are both positive, the remainder when x is divide   [#permalink] 02 Jul 2019, 04:20
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