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When positive integer m is divided by positive integer x, the reminder
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02 Jun 2015, 05:25
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58% (01:39) correct 42% (02:01) wrong based on 221 sessions
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When positive integer m is divided by positive integer x, the reminder
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02 Jun 2015, 08:13
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 *q 1 + 7 So from theorem 7< x ; x > 7 n = y *q 2 + 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
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Re: When positive integer m is divided by positive integer x, the reminder
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03 Jun 2015, 00:37
Bunuel wrote: 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 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.



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Re: When positive integer m is divided by positive integer x, the reminder
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03 Jun 2015, 05:35
Bunuel wrote: 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 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 abouti.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 7i.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 11i.e. Min Value of y = 12 i.e. Minimum Value of x+y = 8+12 = 20Hence III only can be True Answer: Option
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Re: When positive integer m is divided by positive integer x, the reminder
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04 Jun 2015, 07:50
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.



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Re: When positive integer m is divided by positive integer x, the reminder
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04 Jun 2015, 18:15
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
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Re: When positive integer m is divided by positive integer x, the reminder
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08 Jun 2015, 05:22



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Re: When positive integer m is divided by positive integer x, the reminder
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30 Dec 2017, 04:34
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Re: When positive integer m is divided by positive integer x, the reminder &nbs
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