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Will take some time ?? (Posted on 2018-05-18) Difficulty: 4 of 5
Found on the web:
begin
260723233115686*1931
43744839742282591947
1181326544136*5138222
186378732807587076747
519650114814905002347

Any three of these numbers add up to a perfect square.
end

I have no means to check whether it is a true statement and revised
just 3 possible combinations by adding mod 100 the 2 last digits, i.e. 47+47+47= 41, 47+47+22=96 and 47+47+31=25.
I leave it to the solvers to check the 10 possible combinations and to explain what means were used and what was the runtime.
To make it more interesting 2 distinct digits I've replaced by an asterisk "*".

Have fun finding if the statement is true!
If so - for what values of the asterisks?

See The Solution Submitted by Ady TZIDON    
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Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): computer solution Comment 5 of 5 |
(In reply to re(2): computer solution by Jer)

The source code file was typed on a computer (actually copied from the UBASIC source file and modified) and uploaded to Google Drive, then downloaded to the phone. It did require some debugging on the phone but that was not really much more difficult than debugging on a computer, as they were syntax tweaks as the logic was already worked out..


It was motivated by speed curiosity mostly. 

  Posted by Charlie on 2018-05-20 20:44:06
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