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

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This post was BOOKMARKED

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Difficulty:

45% (medium)

Question Stats:

70% (03:20) correct
30% (01:57) wrong based on 134 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

2

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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