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 2007-03-19, 22:22 #1 davar55     May 2004 New York City 10000011001102 Posts Easy Arithmetic Find three consecutive odd cubes whose sum is a four (decimal) digit number with identical digits.
 2007-03-20, 09:46 #2 davieddy     "Lucan" Dec 2006 England 6,451 Posts I found four candidates, (the four digit numbers being 1197,2403,4257,6903). What am I missing? Last fiddled with by davieddy on 2007-03-20 at 09:46
2007-03-20, 13:14   #3
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"Tim Sorbera"
Aug 2006
San Antonio, TX USA

102538 Posts

Quote:
 Originally Posted by davieddy I found four candidates, (the four digit numbers being 1197,2403,4257,6903). What am I missing?
I found the same candidates. Are we doing something wrong or did davar55 word the problem wrong?

 2007-03-20, 14:29 #4 Zeta-Flux     May 2003 2·761 Posts Maybe by "consecutive" he meant distinct.
 2007-03-20, 14:37 #5 S485122     Sep 2006 Brussels, Belgium 62E16 Posts I did find the same candidates, but changing the problem to 3 consecutive squares I get 5555 = 41^2+43^2+45^2
 2007-03-20, 16:21 #6 davieddy     "Lucan" Dec 2006 England 6,451 Posts That's quite a coincidence. Must be what he meant. Well spotted! Last fiddled with by davieddy on 2007-03-20 at 16:52
 2007-03-20, 17:47 #7 davar55     May 2004 New York City 106616 Posts Mea culpa. I did mean three consecutive odd squares. There's something to be said for repeating a preview before posting. And on such a simply stated problem. Last fiddled with by davar55 on 2007-03-20 at 17:48

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