20070319, 22:22  #1 
May 2004
New York City
2^{3}·23^{2} Posts 
Easy Arithmetic
Find three consecutive odd cubes whose sum is a four (decimal) digit
number with identical digits. 
20070320, 09:46  #2 
"Lucan"
Dec 2006
England
1100101001010_{2} Posts 
I found four candidates, (the four digit numbers being
1197,2403,4257,6903). What am I missing? Last fiddled with by davieddy on 20070320 at 09:46 
20070320, 13:14  #3 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
4267_{10} Posts 

20070320, 14:29  #4 
May 2003
60B_{16} Posts 
Maybe by "consecutive" he meant distinct.

20070320, 14:37  #5 
Sep 2006
Brussels, Belgium
3^{3}×61 Posts 
I did find the same candidates, but changing the problem to 3 consecutive squares I get 5555 = 41^2+43^2+45^2

20070320, 16:21  #6 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
That's quite a coincidence.
Must be what he meant. Well spotted! Last fiddled with by davieddy on 20070320 at 16:52 
20070320, 17:47  #7 
May 2004
New York City
2^{3}·23^{2} 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 20070320 at 17:48 
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