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

Question

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

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 two-place decimal of .bc. So this statement must be true; y cannot equal 3.

vitaliyGMAT · Accepted Answer

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.

Senthil1981 · Answer

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

aayushagrawal · Answer

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.

msandman · Answer

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.

abhimahna · Answer

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.

nisharma · Answer

Took the Veritas test today and selected D for this question. Pretty sure the answer is D and not E.

VeritasPrepBrian · Answer

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!

mhngo · Answer

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!!

SidKan · Answer

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

nick1816 · Answer

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

IanStewart · Answer

This is a nice solution -- much cleaner than the one in the spoiler tag in the OP -- but it's not quite right. In the original version of the problem, we were not told that the digits were nonzero. So the number "bc" needs to be a factor of 100, but if our digits can be zero, then "bc" can be "04", for example; it's not the case that "25" is the only possible value of "bc". Here we might have x = 26 and y = 25, say, and x/y = 1.04. In the question's original wording, the only digit we can be sure is nonzero is c, because if c were zero, the decimal would be "a.b", and not "a.bc".

I gather the question meant to declare a, b and c to all be nonzero, and if that's true, then the solution to item I becomes correct, but then the solution for item II becomes incorrect, since 1/4 = 0.25 is no longer possible (the digit a is zero in that case).

bumpbot · Answer

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

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