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 April 2nd, 2014, 03:16 PM #11 Senior Member     Joined: May 2013 From: EspaĆ±a Posts: 151 Thanks: 4 Hello. Let "n"=number of consecutive addends, such that: $1^5=0+1=1 \ / \ n=2 \rightarrow{} \frac{n}{2}=1^2$ $3^5=5+ \cdots +22=243 \ / \ n=18 \rightarrow{} \frac{n}{2}=3^2$ $5^5=38+ \cdots +87=3125 \ / \ n=50 \rightarrow{} \frac{n}{2}=5^2$ $7^5=123+ \cdots +220=16807 \ / \ n=98 \rightarrow{} \frac{n}{2}=7^2$ $9^5=284+ \cdots +445=59049 \ / \ n=162 \rightarrow{} \frac{n}{2}=9^2$ ... In addition, the difference of the first Addend of a succession, with the end of the previous one, is a number squared: $5-1=2^2$ $38-22=4^2$ $123-87=6^2$ $284-445=8^2$ ... Regards. Thanks from johnr
 April 2nd, 2014, 03:48 PM #12 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Great stuff. Just heard from my friend. I'll be doing this mid-May or so!
 April 2nd, 2014, 04:37 PM #13 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Just glancing over the material here a bit more closely and I can see that Soroban's first demo is particularly germane, as it basically is an alternative view of the fact that the difference between the squares of the n+1st and nth triangular numbers is the cube of n+1.
 April 2nd, 2014, 08:52 PM #14 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 As a "break", give them this li'l "match moving" puzzle: II = VI : move 1 match only to end up with both sides equal I = VT : right side is now SQRT(1) Thanks from johnr
 April 3rd, 2014, 05:02 AM #15 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Maybe you can mine something from this: Thanks from johnr

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