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by ceilidh.erickson » Fri Oct 04, 2013 11:05 am
If 5 is divided by 6, the result is 0 remainder 5. The remainder must be smaller than Y, but there's no rule that says that X has to be larger than Y.

If you divide X by a larger integer Y, the result will be 0 remainder X.
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by Brent@GMATPrepNow » Fri Oct 04, 2013 11:08 am
Premalatha Anand wrote:If X and Y are positive integers and x/y has a remainder of 5, what is the smallest possible value of XY?


Ans is 30
Remainder cannot be larger than Y, so Y should be at least 6.

But my que is how can the smallest possible value of X can be 5??

Please explain!
There a two important rules regarding remainders

#1) If N divided by D, leaves remainder R, then 0 < R < D

#2) If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.

For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

IMPORTANT: Notice that 1 is the smallest possible value of k. A lot of people miss this. However, 1 divided by 5 equals 0 with remainder 1. In other words, 5 divides into 1 zero times with remainder 1.

For this question, we're told that x divided by y has a remainder of 5. So, the possible values of x are: 5, 5 + y, 5 + 2y, 5 + 3y, 5 + 4y, etc.

So, the SMALLEST value of x is 5.
For the division (x divided by y) leaves a remainder of 5, y must be greater than 5.
So, the SMALLEST value of y is 6

So, the SMALLEST value of xy = (5)(6) = 30

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by Premalatha Anand » Fri Oct 04, 2013 12:07 pm
Thanks Erickson. Didn't know the rule.

Thanks brent for the detailed explanation!.
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by vinay1983 » Fri Oct 04, 2013 7:12 pm
Brent@GMATPrepNow wrote:
Premalatha Anand wrote:If X and Y are positive integers and x/y has a remainder of 5, what is the smallest possible value of XY?


Ans is 30
Remainder cannot be larger than Y, so Y should be at least 6.

But my que is how can the smallest possible value of X can be 5??

Please explain!
There a two important rules regarding remainders

#1) If N divided by D, leaves remainder R, then 0 < R < D

#2) If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.

For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

IMPORTANT: Notice that 1 is the smallest possible value of k. A lot of people miss this. However, 1 divided by 5 equals 0 with remainder 1. In other words, 5 divides into 1 zero times with remainder 1.

For this question, we're told that x divided by y has a remainder of 5. So, the possible values of x are: 5, 5 + y, 5 + 2y, 5 + 3y, 5 + 4y, etc.

So, the SMALLEST value of x is 5.
For the division (x divided by y) leaves a remainder of 5, y must be greater than 5.
So, the SMALLEST value of y is 6

So, the SMALLEST value of xy = (5)(6) = 30

Cheers,
Brent
Brent, does this mean that "R" is always less than D"
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by Brent@GMATPrepNow » Fri Oct 04, 2013 7:14 pm
vinay1983 wrote: Brent, does this mean that "R" is always less than D"
Yes.

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by [email protected] » Sat Oct 05, 2013 1:22 pm
Hi Premalatha Anand,

I think that all of the other posts in this string have provided the proper insight to answer your initial question, so I'm just going to take a moment to reiterate an important point.

Since "remainder" questions/concepts will usually show up 1-2 times on Test Day, it's worth remembering that remainders are "left over" pieces and the smallest possible numerator in these situations is the one that leads to 0 r <number>.

For example: 3/5 = 0 r 3

Keep this rule in mind any time a "remainder" question appears. It will usually make the question easier to solve.

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by Premalatha Anand » Sun Oct 06, 2013 8:16 am
Rich,

Easy to remember! Wont forget hereafter. Thanks!
Regards
Premalatha
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