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The integers x and y are both positive, the remainder when x [#permalink]
05 Jan 2013, 13:17

00:00

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Difficulty:

55% (hard)

Question Stats:

63% (03:50) correct
37% (01:44) wrong based on 78 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

1

This post received KUDOS

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

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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