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When positive integer x is divided by positive integer y, the result
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Updated on: 20 Sep 2016, 01:09
Question Stats:
36% (02:41) correct 64% (02:40) wrong based on 180 sessions
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When positive integer x is divided by positive integer y, the result is the number a.bc, where a, b, and c each represent unique digits. If the remainder when xis divided by y is 1, which of the following must be true? I. b < 6 II. y < x III. y does not equal 3 A. II only B. I and II only C. II and III only D. I and III only E. I, II and III I do not understand the OA, this is the OA EDITED THE OA All three statements must be true, and can be determined using a thorough understanding of quotient and remainder in division operations. For statement I, b<6, consider the role that b plays in the decimal a.bc. The decimal places .bc are calculated by taking the remainder of 1 and dividing by the divisor of y. And in that decimal, b will tell you whether you'd round up to the nearest integer or down to the nearest integer. So if b is 6 or greater, that means that the calculation of 1 divided by y is greater than 1/2. But it can't be, because the only thing smaller than 2 for y to be is 1, in which case there would be no remainder. So the greatest that b can be is 5, in the event that y=2 For statement II, again consider that remainder of 1. If the divisor is ever greater than the dividend (when dealing with positive values)  which, in fraction form, means that the denominator is greater than the numerator  then the dividend/numerator is equal to the remainder. (Consider such a case: when 3 is divided by 4, 3 doesn't go in to 4 so all of 3 is "left over" as the remainder.) So if y were less than x, then all of x would be have to be the remainder. But that would make x=1, which cannot work because both x and y have to be positive integers, and 1 is the smallest positive integer. So there is no combination here that would allow for x=1 and y>x Therefore y must be less than x, proving this statement true. For statement 3, look again at how the decimal points are created: by taking the remainder of 1 and dividing by y. If y were 3, then you would have a repeating decimal and not the twoplace decimal of .bc. So this statement must be true; y cannot equal 3.
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Originally posted by kzivrev on 16 Sep 2016, 15:35.
Last edited by abhimahna on 20 Sep 2016, 01:09, edited 2 times in total.
Edited the OA.




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Re: When positive integer x is divided by positive integer y, the result
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07 Feb 2017, 08:55
kzivrev wrote: When positive integer x is divided by positive integer y, the result is the number a.bc, where a, b, and c each represent unique digits. If the remainder when xis divided by y is 1, which of the following must be true?
I. b < 6 II. y < x III. y does not equal 3
A. II only B. I and II only C. II and III only D. I and III only E. I, II and III
I do not understand the OA, this is the OA
.[/spoiler] Slightly different approach. \(\frac{1}{y} = \frac{bc}{100}\) \(y = \frac{100}{bc}\) \(bc\) should be a two digit factor of 100, where b and c are distinct "non zero" digits. The only factor that satisfies above criteria is 25. So our b=2 and c=5. Back to options: I. Yes. II. Not always > 1/4 = 0.25 III. Yes, because y should be a factor of \(100 = 2^2*5^2\), which corresponds to the fact that we have terminating decimal (\(y = 2^a*5^b\)). Hence D.




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Re: When positive integer x is divided by positive integer y, the result
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16 Sep 2016, 19:37
kzivrev wrote: When positive integer x is divided by positive integer y, the result is the number a.bc, where a,b, and c each represent unique digits. If the remainder when xis divided by y is 1,which of the following must be true? I. b<6 II. y<x III. Y does not equal 3
A. II only B. I and II only C. II and III only D. I and III onl E. I , II and III
Given \(\frac{x}{y} = a.bc and also \frac{(x1)}{y} = a\) this means \(y * .bc = 1\) Now Consider I) b<6, look at second equation \(y * .bc = 1\) if b >5 then the result will be greater than 1. Consider II) \(y<x\) Since \(x\) and \(y\) are positive and \(\frac{x}{y}\) gives a quotient and remainder, \(x\) has to be greater than \(y\). Consider III) \(Y\) does not equal 3 if \(Y\) is equal to 3, then the bc will not be a terminating decimal. Hence I, II and III all satisfies the statement. Answer is E



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Re: When positive integer x is divided by positive integer y, the result
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16 Sep 2016, 23:58
I have a doubt with regard to Statement 2 : What if x/y equals 1/4. The remainder is 1 and 1/4 equals 0.25 where a=0, b=2 and c=5. This also satisfies the requirement that a, b, and c represent unique digits.



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Re: When positive integer x is divided by positive integer y, the result
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Updated on: 17 Sep 2016, 12:57
aayushagrawal wrote: I have a doubt with regard to Statement 2 : What if x/y equals 1/4. The remainder is 1 and 1/4 equals 0.25 where a=0, b=2 and c=5. This also satisfies the requirement that a, b, and c represent unique digits. I'm currently confused about this as well as it seems a valid answer given all the variables are unique and the remainder is 1. Pretty sure there is an error here. Went with D. Edit: To be fair to the poster above, the answer they give in Veritas is E.
Originally posted by msandman on 17 Sep 2016, 00:18.
Last edited by msandman on 17 Sep 2016, 12:57, edited 1 time in total.



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Re: When positive integer x is divided by positive integer y, the result
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17 Sep 2016, 04:06
Senthil1981 wrote: kzivrev wrote: When positive integer x is divided by positive integer y, the result is the number a.bc, where a,b, and c each represent unique digits. If the remainder when xis divided by y is 1,which of the following must be true? I. b<6 II. y<x III. Y does not equal 3
A. II only B. I and II only C. II and III only D. I and III onl E. I , II and III
Given \(\frac{x}{y} = a.bc and also \frac{(x1)}{y} = a\) this means \(y * .bc = 1\) Now Consider I) b<6, look at second equation \(y * .bc = 1\) if b >5 then the result will be greater than 1. Consider II) \(y<x\) Since \(x\) and \(y\) are positive and \(\frac{x}{y}\) gives a quotient and remainder, \(x\) has to be greater than \(y\).
Consider III) \(Y\) does not equal 3 if \(Y\) is equal to 3, then the bc will not be a terminating decimal. Hence I, II and III all satisfies the statement. Answer is E Dude, don't you think you have missed something in the highlighted statement above. We could have x = 1 and y =4, result 0.25, which satisfies all the conditions given in the question. I am with D and not E.
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Re: When positive integer x is divided by positive integer y, the result
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17 Sep 2016, 22:20
Took the Veritas test today and selected D for this question. Pretty sure the answer is D and not E.



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Re: When positive integer x is divided by positive integer y, the result
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19 Sep 2016, 12:03
Hey everyone, As the author of the question here, I should chime in  I apologize! I put this one into an experimental slot of the practice tests before the weekend (it's Monday now) because I've seen a lot of students struggling lately with abstract Quotient/Remainder problems, and amidst all the time I spent making sure that variables, fractions, etc. were all formatted properly...I missed the fact that it should have said "nonzero" in "a, b,c and c each represent unique, NONZERO digits" (so that II would be kind of a freebie with the other two statements being trickier). So I apologize  of course, as it's written earlier in this thread, the answer is D and not E. To anyone who saw this on a practice test, much like GMAC we use experimental positions to pretest practice questions so this one wouldn't have counted toward your score. And the data is very helpful  with about 100 responses over the weekend I could quickly tell when I got to my desk this morning that something wasn't right, since 700+ scorers were getting it right less often than the lefthand side of the curve was. Sorry for the confusion  great analysis above!
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Re: When positive integer x is divided by positive integer y, the result
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27 Sep 2016, 08:47
kzivrev wrote: When positive integer x is divided by positive integer y, the result is the number a.bc, where a, b, and c each represent unique digits. If the remainder when xis divided by y is 1, which of the following must be true? I. b < 6 II. y < x III. y does not equal 3 A. II only B. I and II only C. II and III only D. I and III only E. I, II and III I do not understand the OA, this is the OA EDITED THE OA All three statements must be true, and can be determined using a thorough understanding of quotient and remainder in division operations. For statement I, b<6, consider the role that b plays in the decimal a.bc. The decimal places .bc are calculated by taking the remainder of 1 and dividing by the divisor of y. And in that decimal, b will tell you whether you'd round up to the nearest integer or down to the nearest integer. So if b is 6 or greater, that means that the calculation of 1 divided by y is greater than 1/2. But it can't be, because the only thing smaller than 2 for y to be is 1, in which case there would be no remainder. So the greatest that b can be is 5, in the event that y=2 For statement II, again consider that remainder of 1. If the divisor is ever greater than the dividend (when dealing with positive values)  which, in fraction form, means that the denominator is greater than the numerator  then the dividend/numerator is equal to the remainder. (Consider such a case: when 3 is divided by 4, 3 doesn't go in to 4 so all of 3 is "left over" as the remainder.) So if y were less than x, then all of x would be have to be the remainder. But that would make x=1, which cannot work because both x and y have to be positive integers, and 1 is the smallest positive integer. So there is no combination here that would allow for x=1 and y>x Therefore y must be less than x, proving this statement true. For statement 3, look again at how the decimal points are created: by taking the remainder of 1 and dividing by y. If y were 3, then you would have a repeating decimal and not the twoplace decimal of .bc. So this statement must be true; y cannot equal 3. Is this official explanation? I think the question is not air tight, and quite confusing... I use elimination for this question. Since II is obviously not "must be true" (with x=1 and y=4), all answers with II must be wrong, hence D is the only correct option. Fortunately there were not answers like "I only", or "III only", or else I would have got it wrong!!



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Re: When positive integer x is divided by positive integer y, the result
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08 Apr 2018, 07:18
Given xy=a.bcandalso(x−1)y=axy=a.bcandalso(x−1)y=a this means y∗.bc=1y∗.bc=1 Now Consider I) b<6, look at second equation y∗.bc=1y∗.bc=1 if b >5 then the result will be greater than 1. Consider II) y<xy<x Since xx and yy are positive and xyxy gives a quotient and remainder, xx has to be greater than yy. Consider III) YY does not equal 3 if YY is equal to 3, then the bc will not be a terminating decimal.
Hence I, II and III all satisfies the statement. Answer is E



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Re: When positive integer x is divided by positive integer y, the result
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21 May 2019, 11:55
1.The maximum value of bc=25 (remainder=1 and y=4) hence b is always less than 6 2. x=ay+1 when a=0, then y>x 3. As the x/y is terminating decimal, y can never be equal to 3
Only 1 and 3 statements must be true




Re: When positive integer x is divided by positive integer y, the result
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