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The integers x and y are both positive, the remainder when x is divide Updated on: 11 Oct 2017, 04:34
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.
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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

Show Spoiler
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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The integers x and y are both positive, the remainder when x is divide 07 Jan 2013, 04:34
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roygush
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
Show more

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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The integers x and y are both positive, the remainder when x is divide 25 Jun 2015, 06:30
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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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The integers x and y are both positive, the remainder when x is divide 19 Apr 2018, 16:15
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roygush
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
Show more

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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The integers x and y are both positive, the remainder when x is divide 06 Jan 2013, 01:50
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roygush
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
Show more

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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The integers x and y are both positive, the remainder when x is divide 07 Jan 2013, 07:04
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roygush
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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bunuel, can we apply this method on any remainders problem?
write the two equations and then work from there?
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The integers x and y are both positive, the remainder when x is divide 07 Jan 2013, 09:44
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roygush
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?
Show more

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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The integers x and y are both positive, the remainder when x is divide 30 Jul 2015, 09:04
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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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The integers x and y are both positive, the remainder when x is divide 30 Jul 2015, 09:17
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roygush
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
Show more


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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The integers x and y are both positive, the remainder when x is divide 30 Jul 2015, 09:20
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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...
Show more

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

Final Answer:
Show Spoiler
D


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The integers x and y are both positive, the remainder when x is divide 15 Jul 2016, 19:36
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roygush
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
Show more

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 28 Jun 2018, 08:08
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roygush
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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Please move to PS
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roygush
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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Done. Thank you.
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The integers x and y are both positive, the remainder when x is divide 21 May 2020, 04:06
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roygush
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

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

2x + y = 2(12m + 7) + 12n + 3 = 24m + 14 + 12n + 3 = 24m + 12n + 17 = 12(2m + n + 1) + 5

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.

Answer: D
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The integers x and y are both positive, the remainder when x is divide 12 Jun 2020, 17:24
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roygush
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,

2x + y = 2(12m + 7) + 12n + 3 = 24m + 14 + 12n + 3 = 24m + 12n + 17 = 12(2m + n + 1) + 5

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.

Answer: D
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ScottTargetTestPrep

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
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The integers x and y are both positive, the remainder when x is divide 29 Jun 2020, 08:57
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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


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

2x + y = 2(12m + 7) + 12n + 3 = 24m + 14 + 12n + 3 = 24m + 12n + 17 = 12(2m + n + 1) + 5

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.

Answer: D

ScottTargetTestPrep

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
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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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The integers x and y are both positive, the remainder when x is divide 10 Sep 2024, 20:16
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Theory: Dividend = Divisor*Quotient + Remainder

The remainder when x is divided by 12 is 7
x -> Dividend
12 -> Divisor
q -> Quotient (Assume)
7 -> Remainders
=> x = 12*q + 7
=> x = 12q + 7 ...(1)

The remainder when y is divided by 12 is 3
y -> Dividend
12 -> Divisor
t -> Quotient (Assume)
3 -> Remainders
=> y = 12*t + 3
=> y = 12t + 3 ...(2)

=> 2x+y = 2*(12q + 7) + 12t + 3 [ From (1) and (2) ]
= 24q + 14 + 12t + 3 = 24q + 12t + 12 + 5 = 12*(2q+t+1) + 5

=> 2x+y = a Multiple of 12 + 5
=> 2x+y when divided by 12 should give 5 as remainder.

A. \(\frac{125}{12}\) remainder is 5 => possible
B. \(\frac{101}{12}\) remainder is 5 => possible
C. \(\frac{77}{12}\) remainder is 5 => possible
D. \(\frac{51}{12}\) remainder is 3 => NOT possible
E. \(\frac{41}{12}\) remainder is 5 => possible

So, Answer will be D
Hope it helps!

Watch following video to MASTER Remainders

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The integers x and y are both positive, the remainder when x is divide 18 Jan 2026, 15:42
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x=12Q+7 and y=12z+3

We want to know what the possible values are for 2x+y

I substituted the x and y into the equation and got 24Q+12z+17=

I quickly listed the multiples of 12 and 24, only 5 of each. So 120 for 24 and 60 for 12.
I used the answers and subtracted 17 from each.

51-17 = 34 was the only one that could not be produced from my multiples list.
roygush
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

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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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