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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 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

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 is divide
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07 Jan 2013, 03:34

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10

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 is divide
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06 Jan 2013, 00:50

3

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
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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 is divide
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07 Jan 2013, 08:44

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.

Re: The integers x and y are both positive, the remainder when x is divide
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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 is divide
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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 is divide
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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 is divide
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30 Jul 2015, 14:27

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.

Re: The integers x and y are both positive, the remainder when x is divide
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15 Jul 2016, 18:36

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.

The integers x and y are both positive, the remainder when x is divide
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19 Apr 2018, 15:15

Top Contributor

1

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

Re: The integers x and y are both positive, the remainder when x is divide
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28 Jun 2018, 07:08

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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.

Re: The integers x and y are both positive, the remainder when x is divide
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28 Jun 2018, 07:38

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.

Re: The integers x and y are both positive, the remainder when x is divide
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21 May 2020, 03:06

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

We see that x = 12m + 7 and y = 12n + 3 where m and n are non-negative integers. Therefore,

We see that 2x + y is 5 more than a multiple of 12, and every number in the given choices is 5 more than a multiple of 12 except 51 (notice that 51 is 3 more than 48). Therefore, 51 is the correct answer.

The integers x and y are both positive, the remainder when x is divide
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12 Jun 2020, 16:24

ScottTargetTestPrep 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

We see that x = 12m + 7 and y = 12n + 3 where m and n are non-negative integers. Therefore,

We see that 2x + y is 5 more than a multiple of 12, and every number in the given choices is 5 more than a multiple of 12 except 51 (notice that 51 is 3 more than 48). Therefore, 51 is the correct answer.

Re: The integers x and y are both positive, the remainder when x is divide
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29 Jun 2020, 07:57

Andrewcoleman wrote:

ScottTargetTestPrep 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

We see that x = 12m + 7 and y = 12n + 3 where m and n are non-negative integers. Therefore,

We see that 2x + y is 5 more than a multiple of 12, and every number in the given choices is 5 more than a multiple of 12 except 51 (notice that 51 is 3 more than 48). Therefore, 51 is the correct answer.

what is the strategy and reasoning to include some of the remainder into the parenthesis that is multiplied by 12.

It feels like a good idea because you're including part of the remainder into the mix of two equations.

I'd be interested in learning the reason why

Response:

As a matter of fact, you have to include as much as you can into the parenthesis (which happens to be the quotient). Recall the definition of division: when is x divided by y, we say that k is the quotient and r is the remainder if 1) x = yk + r, and 2) r < y. The reason I included as much as I could into the parentheses is to satisfy the second requirement.

For instance, if you divide 13 by 4, you’ll get a quotient of 3 and a remainder of 1. On the other hand, 13 can also be expressed as 13 = 4*2 + 5. Why don’t we say the quotient is 2 and the remainder is 5? It is because the remainder has to be less than the divisor (i.e. the second requirement). So, if we were presented with the equation 13 = 4*2 + 5, we would have to write 5 = 4 + 1 and include one more 4 in the quotient: 13 = 4*2 + 4 + 1 = 4(2 + 1) + 1 = 4*3 + 1.

What I’ve done in the solution is very similar to the simplified example above. We know that 2x + y equals 24m + 12n + 17 and we are interested in the remainder when 2x + y is divided by 12. So, we group all multiples of 12 together with the greatest multiple of 12 that can be squeezed out of 17 (which is also 12) and write: 2x + y = 24m + 12n + 12 + 5 = 12(2m + n + 1) + 5. Now, 2x + y is expressed as “some multiple of 12 + some number less than 12” and we can conclude that the number less than 12 (which is 5) is the remainder from the division of 2x + y by 12.
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