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

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When positive integer x is divided by positive integer y, the result [#permalink]

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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
[Reveal] Spoiler:
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 two-place decimal of .bc. So this statement must be true; y cannot equal 3.
[Reveal] Spoiler: OA

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 [#permalink]

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New post 16 Sep 2016, 19:37
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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 \(\frac{x}{y} = a.bc and also \frac{(x-1)}{y} = a\)
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 [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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 \(\frac{x}{y} = a.bc and also \frac{(x-1)}{y} = a\)
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 [#permalink]

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New post 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 [#permalink]

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New post 19 Sep 2016, 12:03
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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 pre-test 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 left-hand 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 [#permalink]

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New post 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
[Reveal] Spoiler:
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 two-place 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 [#permalink]

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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 [#permalink]

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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
Re: When positive integer x is divided by positive integer y, the result   [#permalink] 08 Apr 2018, 07:18
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