Register    Login    Search    Rss Feeds

 Page 1 of 1 [ 5 posts ] 



 
Author Message
 Post subject: Question 36; Page 39; Number properties
 Post Posted: Sun Feb 15, 2009 1:16 pm 
Offline
Course Students


Posts: 4
Q; If x and y are positive integers and x/y has a remainder of 5, what is 5, what is the smallest possible value of xy?

The answer explanation says that x should have a smallest value of 5 and I somehow fail to understand the reasoning. Can someone please explain as to why x has the smallest value of 5?
if we x/y with remainder 5, as in the question and replace with 5/6 (x=5 and y=6, as in explanation), the remainder would not be 5.


Top 
 Post subject: Re: Question 36; Page 39; Number properties
 Post Posted: Sat Feb 21, 2009 3:55 pm 
Offline


Posts: 2
ashishparmar01 wrote:
Q; If x and y are positive integers and x/y has a remainder of 5, what is 5, what is the smallest possible value of xy?

The answer explanation says that x should have a smallest value of 5 and I somehow fail to understand the reasoning. Can someone please explain as to why x has the smallest value of 5?
if we x/y with remainder 5, as in the question and replace with 5/6 (x=5 and y=6, as in explanation), the remainder would not be 5.


I have the same question, can somebody answer this?


Top 
 Post subject: Re: Question 36; Page 39; Number properties
 Post Posted: Sun Feb 22, 2009 4:31 pm 
Offline
ManhattanGMAT Staff


Posts: 901
Location: St. Louis, MO
We want to minimize xy, so we'll minimize both x and y. (That's a bit of a simplification, but it works here.)

If x/y has a remainder of 5, we know that:
(1) x has to be 5 more than a multiple of y. Mathematically, x = ny + 5, where n = an integer.
(2) y has to be an integer greater than 5.

Why rule (2)? Consider some examples.
29/6 leaves a remainder of 5 because 29 = 4*6 + 5
17/6 leaves a remainder of 5 because 17 = 3*6 + 5
25/5 leaves a remainder of 5 because 25 = 4*5 + 5. NO! That's wrong.

The remainder is what is left over after the divisor (denominator) has gone into the dividend (numerator) as many times as possible. If the divisor is 5, and the "remainder" is 5 or more, that implies that the divisor could go into the dividend at least once more. Thus, the remainder must ALWAYS be less than the divisor. For a divisor of 5, possible remainders are 0, 1, 2, 3, and 4 (not 5!). For a divisor of 6, possible remainders are 0, 1, 2, 3, 4, and 5. The minimum possible divisor in this problem is 6.

So, we know that:
(2) minimum y is 6.
(1) minimum x is 5 more than a multiple of y = 6. Thus, x = 5, 11, 16, 21, 26. The minimum x must be 5.

Minimum xy = (5)(6) = 30.

_________________
Emily Sledge
Instructor
ManhattanGMAT


Top 
 Post subject: Re: Question 36; Page 39; Number properties
 Post Posted: Mon Feb 23, 2009 5:49 pm 
Offline
Course Students


Posts: 4
Thanks for your response Emily. Appreciate the same.


Top 
 Post subject: Re: Question 36; Page 39; Number properties
 Post Posted: Wed Mar 11, 2009 1:49 pm 
Offline
ManhattanGMAT Staff


Posts: 480
Location: Durham, NC
: )


Top 
Display posts from previous:  Sort by  
 
 Page 1 of 1 [ 5 posts ] 





Who is online

Users browsing this forum: No registered users and 0 guests

 
 

 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to: