My Math Forum  

Go Back   My Math Forum > High School Math Forum > Elementary Math

Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion


Reply
 
LinkBack Thread Tools Display Modes
December 23rd, 2013, 11:14 AM   #1
Newbie
 
Joined: Dec 2013

Posts: 5
Thanks: 0

Tricky problem I need help with

The problem is too difficult to write in text, so I just uploaded a picture of it.
Basically, the task is to get the value of this expression. I know, that the value is 12 (thanks to wolframalpha ), but how is it possible to get that?
I just figured out, that the value is the same if you change the expression like this:
2010*2009*2014-2012*2008*2013
or even like this:
10*9*14-12*8*13
What's the explanation?

Thanks in advance.

P.S. Sorry about my english, I don't live in Britain or so, just posted this on the first forum that came across me on google.
Attached Images
File Type: jpg problem.jpg (17.5 KB, 360 views)
artursm4 is offline  
 
December 23rd, 2013, 11:37 AM   #2
Newbie
 
Joined: Dec 2013

Posts: 5
Thanks: 0

Re: Tricky problem I need help with

Oh, and please tell me if I have posted this in the wrong section.
artursm4 is offline  
December 23rd, 2013, 05:33 PM   #3
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 13,282
Thanks: 931

Re: Tricky problem I need help with

Let k = 2013/(2013*2014); so k = 1/2014 : why that silly repeating term..."tricky" teacher?!

(2010 + k)(2009 + k)(2014 + k) - (2012 + k)(2008 + k)(2013 + k) = 12
Denis is offline  
December 24th, 2013, 12:39 AM   #4
Newbie
 
Joined: Dec 2013

Posts: 5
Thanks: 0

Re: Tricky problem I need help with

Quote:
Originally Posted by Denis
Let k = 2013/(2013*2014); so k = 1/2014 : why that silly repeating term..."tricky" teacher?!

(2010 + k)(2009 + k)(2014 + k) - (2012 + k)(2008 + k)(2013 + k) = 12
Thanks for your reply, but I'm afraid I didn't understand.
And k = 2013/20132014, the denominator is 20132014, not 2013*2014.
artursm4 is offline  
December 24th, 2013, 02:09 AM   #5
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 13,282
Thanks: 931

Re: Tricky problem I need help with

http://www.wolframalpha.com/input/?i=x% ... 3D1%2F2014

k can be any value; 12 is always the result.
Denis is offline  
December 24th, 2013, 03:07 AM   #6
Newbie
 
Joined: Dec 2013

Posts: 5
Thanks: 0

Re: Tricky problem I need help with

Quote:
Originally Posted by Denis
http://www.wolframalpha.com/input/?i=x%3D%282010%2Bk%29*%282009%2Bk%29*%282014%2Bk%2 9-%282012%2Bk%29*%282008%2Bk%29*%282013%2Bk%29%2Ck%3 D1%2F2014
I already dealt with the problem.
But thanks, You made me think in a bit different way.
My solution is in the picture below.
Attached Images
File Type: jpg solution.jpg (79.8 KB, 313 views)
artursm4 is offline  
December 24th, 2013, 03:26 AM   #7
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 13,282
Thanks: 931

Re: Tricky problem I need help with

OR (drop the "2000"):
(9 + k)(10 + k)(14 + k) = k^3 + 33k^2 + 356k + 1260 [1]
(8 + k)(12 + k)(13 * k) = k^3 + 33k^2 + 356k + 1248 [2]

[1] - [2] = 12

Notice that 9+10+14 = 8+12+13

Must have something to do with number theory...
Denis is offline  
December 24th, 2013, 03:39 AM   #8
Newbie
 
Joined: Dec 2013

Posts: 5
Thanks: 0

Re: Tricky problem I need help with

Quote:
Originally Posted by Denis
OR (drop the "2000"):
(9 + k)(10 + k)(14 + k) = k^3 + 33k^2 + 356k + 1260 [1]
(8 + k)(12 + k)(13 * k) = k^3 + 33k^2 + 356k + 1248 [2]

[1] - [2] = 12

Notice that 9+10+14 = 8+12+13

Must have something to do with number theory...
Definitely.
artursm4 is offline  
January 11th, 2014, 08:35 PM   #9
Senior Member
 
Joined: Jan 2014
From: The backwoods of Northern Ontario

Posts: 374
Thanks: 68

Re: Tricky problem I need help with

By the distributive property the entire number phrase is equal to:

(2010+2009+2014)/20132014 (2012+2008+2013)/20132014)

=6033/20132014 6033/20132014

=0
Timios is offline  
Reply

  My Math Forum > High School Math Forum > Elementary Math

Tags
problem, tricky



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Tricky Probability Problem pksinghal Probability and Statistics 1 February 24th, 2016 03:25 PM
a tricky probability problem davedave Probability and Statistics 1 March 17th, 2014 07:52 PM
Tricky exponent problem. FrozenNumber Algebra 3 November 17th, 2013 05:16 PM
Tricky problem and I'm just stumped cups Calculus 2 October 30th, 2012 01:13 PM
a very long tricky problem... wuzhe Elementary Math 5 July 26th, 2011 04:20 PM





Copyright © 2018 My Math Forum. All rights reserved.