M31-27

Question

When a positive integer  m  is divided by a positive integer  x , the remainder obtained is 7. Similarly, when a positive integer  n  is divided by a positive integer  y , the remainder obtained is 11. Which of the following could be a value of  x + y ?
 
I. 18
 
II. 19
 
III. 20

A. I only
B. II only
C. III only
D. II and III only
E. None

Bunuel · Accepted Answer

Official Solution:

When a positive integer  m  is divided by a positive integer  x , the remainder obtained is 7. Similarly, when a positive integer  n  is divided by a positive integer  y , the remainder obtained is 11. Which of the following could be a value of  x + y ?
 
I. 18
 
II. 19
 
III. 20

A. I only
B. II only
C. III only
D. II and III only
E. None

The remainder obtained from division is ALWAYS smaller than the divisor itself. Hence, we can conclude that  x  must be greater than 7, and  y  must be greater than 11. As a result, the minimum values for  x  and  y  are 8 and 12 respectively, leading to a minimum value of  x + y  equal to 20.

Answer: C

nelliegu · Answer

Hi! If x>7 and y>11, isn't it possible to say that x+y>18? Then, the answer could also be 19.
I understand that if we consider each equation separately, the answer has to be 20, but I wanted to check if we're allowed to combine the equations to get "19".

Bunuel · Answer

The point is, we are told that x and y are positive  integers . For positive integers x and y, where x > 7 and y > 11, it's not possible that x + y = 19.

nelliegu · Answer

I had missed that, thanks.

gmatzpractice · Answer

m = xa+7 and x>7
n = yb + 11 and y>11
Eliminate I. and II. 
Answer C or E

Ex. for III. 
m=15; a=1; x=8
n=23; b=1; y=12
Answer C

theeni03 · Answer

I think this is a  high-quality  question and I agree with explanation.

AkshdeepS · Answer

I think this is a  high-quality  question and I agree with explanation.

carma19 · Answer

Reminder always < Divisor .
Therefore, 7<X, 11<Y.
Least value of X = 8
Least value of Y = 12
So, lest value of X + Y = 8 + 12 = 20.
Hence,  C

raaz8522 · Answer

Bunuel ,

But I thought, diving any positive integers between 1-6 by 7 will lead to a remainder of 7. can you please tell me what I am missing?

Best Regards,

Bunuel · Answer

No. When divisor is more than dividend then the reminder equals to the dividend. For example, 1 divided by 7 gives the remainder of 1. 2 divided by 7 gives the remainder of 2...

Check the links below for more:

6. Remainders 
     Theory   
 REMAINDERS 
 Remainders: Tips and hints 
 Divisibility and Remainders on the GMAT 
 Remainders on the GMAT 
 A Remainders Shortcut for the GMAT 
 All About Negative Remainders on the GMAT 
 A Tricky Question on Negative Remainders 
 Compilation of tips and tricks to deal with remainders 
 Cyclicity and the remainders on the GMAT 
 GMAT Question Patterns: Remainders 
     Questions   
 Units digits, exponents, remainders problems  
 DS Questions 
 PS Questions

For other subjects:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

joe123x · Answer

what is the proof from the equation that the remainder is always less than the divisor?

eq: m/x = q + 7/x

Bunuel · Answer

The remainder should always be less than the divisor because that is how we define the remainder in the context of division:

If  x  and  y  are  positive integers , there exist unique integers  q  and  r , called the  quotient  and  remainder , respectively, such that  y =divisor*quotient+remainder= xq + r  and  0\leq{r}<x .

When we divide one number (the dividend) by another number (the divisor), we are essentially asking how many times the divisor can fit into the dividend. The quotient is the whole number answer to this question, and the remainder is the amount that is left over after we've taken out as many whole divisors as we can.

For example, if we divide 13 by 3, we get a quotient of 4 and a remainder of 1. This means that 3 times 4 can fit into 13, with 1 left over. The remainder is the "leftover" amount, which must be less than the divisor (in this case, less than 3) because if it were equal to or greater than the divisor, then we could fit another whole divisor into the dividend.

Therefore, the remainder must always be less than the divisor in order for the result of the division to make sense and be consistent with the way we define the operation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

bumpbot · Answer

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M31-27

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