When positive integer m is divided by positive integer x, the reminder

Question

When positive integer m is divided by positive integer x, the reminder is 7 and when positive integer n is divided by positive integer y, the reminder is 11. Which of the following is a possible value for x + y?

I. 18
II. 19
III. 20

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

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:

The remainder is ALWAYS less than the divisor, thus x > 7 and y > 11. Therefore the least values of x and y are 8 and 12, respectively, making the least value of x + y equal to 20.

Answer: C.

askhere · Answer

Euclid's Division Lemma states that "Given positive integers a and b, there exists unique integers q and r satisfying
a = bq+ r where 0 ≤ r < b" (q is quotient and r is remainder)

From the question :
m = x *q1 + 7 So from theorem 7< x ; x > 7
n = y *q2 + 11 So from theorem 11< y ; y > 11

So the least value of x = 8 & least value of y = 12
Least value of x+y = 20.

Out of given choices only III is possible.

Answer C

NavenRk · Answer

Since the reminders are 7 & 11 respectively, the minimum values for x & y are 8 and 12 respectively (y>11) (x>7)
 
Value of x+y = 12+8 =20. Hence C.

GMATinsight · Answer

TIP FOR SUCH QUESTIONS:  Make Algebraic Equation in the beginning to understand how to proceed further. Then Start thinking the possible values of variables asked about

i.e. "When positive integer m is divided by positive integer x, the reminder is 7" can be understood as

m = ax +7  which means the  value of x must be greater than 7  as the remainder 7 is possible only when the Divisor is bigger than the Remainder 7

i.e. Min Value of x = 8

AND

i.e. "When positive integer n is divided by positive integer y, the reminder is 11" can be understood as

n = by +11  which means the  value of y must be greater than 11  as the remainder 11 is possible only when the Divisor is bigger than the Remainder 11

i.e. Min Value of y = 12

i.e. Minimum Value of x+y = 8+12 = 20

Hence III only can be True

Answer: Option  C

tjerkrintjema · Answer

Maybe I'm going in too fast without thinking of the alternatives.
But when m/x yields remainder 7, you can assume x > 7
Similarly when n/y yields remainder 11, you can assume y > 11

Therefore the least possible would be x=8 and y=12.
Adding those together yields 20.
Only answer satisfying this would be C.

But like I said, maybe I overlooked things.

EMPOWERgmatRichC · Answer

Hi tjerkrintjema,

You've handled this question perfectly. As you can see from the other posts, there are a variety of different ways to approach this prompt (most GMAT questions can be approached in more than one way). Beyond the immediate goal of trying to get a question correct, an additional goal should always be to take an approach that is fastest/easiest; you've done that here.

GMAT assassins aren't born, they're made,
Rich

ScottTargetTestPrep · Answer

Let’s review a simple divisibility rule, using division by 3 as an example: The possible remainders when a number is divided by 3 are 0, 1, or 2. Note that the largest possible remainder when division by 3 is performed is 1 less than 3, which is 2. In other words, if we have a remainder 2 from a division problem, we know that the minimum value of the divisor is 2 + 1 = 3.

So, in general, when division by w is performed and a remainder of r is obtained, we know that the minimum value of the divisor w is r + 1. Thus, when division by x yields a remainder of 7, then the minimum value of x is 7 + 1 = 8. Similarly, when division by y yields a remainder of 11, then the minimum value of y is 11 + 1 = 12. Thus, the smallest value of x + y is 8 + 12 = 20, so only statement III can be true.

Answer: C

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When positive integer m is divided by positive integer x, the reminder

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