Jump to content

Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.

Of course, you can also enjoy our collection of amazing optical illusions and cool math games.

If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.

Thanks and enjoy the Den :-)
Guest Message by DevFuse

- - - - -

Adding really simple fractions

Best Answer Rob_G, 02 May 2014 - 04:44 PM

Spoiler for Well...
Go to the full post

  • Please log in to reply
1 reply to this topic

#1 bonanova



  • Moderator
  • PipPipPipPip
  • 6160 posts
  • Gender:Male
  • Location:New York

Posted 02 May 2014 - 11:34 AM

Johnny was asked to write the sum of the two fractions p/n and q/n where p, q and n are single-digit numbers.

He remembered that the numerators were to be added, but he thought maybe the denominators should be added as well.

Also, he decided that putting digits next to each other could suggest addition.

So his first try was to say that


p/n + q/n = pq/nn. (where pq and nn are two-digit numbers.)


Then he remembered the result should not depend on the order of addition,

but he saw that p/n + q/n = qp/nn was just as bad. So he decided to use them both.

His final representation of the sum was


p/n + q/n = (pq + qp)/nn.


The puzzle asks: Let p, q and n be chosen at random from {1, 2, 3, 4, 5, 6, 7, 8, 9}.

What is the expected error in Johnny's formula?

  • -1

Vidi vici veni.

#2 Rob_G


    Junior Member

  • Members
  • PipPip
  • 29 posts

Posted 02 May 2014 - 04:44 PM   Best Answer

Spoiler for Well...

  • 0

0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users