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

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Manager
Joined: 01 Sep 2012
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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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Question Stats:

65% (03:12) correct 35% (01:48) wrong based on 274 sessions

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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
[Reveal] Spoiler: OA

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IMPOSSIBLE IS NOTHING

Last edited by Bunuel on 07 Jan 2013, 04:29, edited 2 times in total.
Edited the question.
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Posts: 39719
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.

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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
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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…

[Reveal] Spoiler:
D

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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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Posts: 128
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, 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.

bunuel, can we apply this method on any remainders problem?
write the two equations and then work from there?
_________________

If my answer helped, dont forget KUDOS!

IMPOSSIBLE IS NOTHING

Math Expert
Joined: 02 Sep 2009
Posts: 39719
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.

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 [#permalink]

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09 Sep 2014, 02:31
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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. => 51-17 = 34 (not a multiple of 12).

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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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Joined: 08 Nov 2014
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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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Posts: 3
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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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!
Re: The integers x and y are both positive, the remainder when x   [#permalink] 15 Jul 2016, 19:36
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