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December 23rd, 2013, 11:14 AM   #1
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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?

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
 problem.jpg (17.5 KB, 360 views)

 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.
 December 23rd, 2013, 05:33 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,586 Thanks: 1038 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
December 24th, 2013, 12:39 AM   #4
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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
And k = 2013/20132014, the denominator is 20132014, not 2013*2014.

 December 24th, 2013, 02:09 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,586 Thanks: 1038 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.
December 24th, 2013, 03:07 AM   #6
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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
 solution.jpg (79.8 KB, 313 views)

 December 24th, 2013, 03:26 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,586 Thanks: 1038 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...
December 24th, 2013, 03:39 AM   #8
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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.

 January 11th, 2014, 08:35 PM #9 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70 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

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