Easy Arithmetic
Find three consecutive odd cubes whose sum is a four (decimal) digit
number with identical digits. 
I found four candidates, (the four digit numbers being
1197,2403,4257,6903). What am I missing? 
[quote=davieddy;101480]I found four candidates, (the four digit numbers being
1197,2403,4257,6903). What am I missing?[/quote] I found the same candidates. Are we doing something wrong or did davar55 word the problem wrong? 
Maybe by "consecutive" he meant distinct.

I did find the same candidates, but changing the problem to 3 consecutive squares I get 5555 = 41^2+43^2+45^2

That's quite a coincidence.
Must be what he meant. Well spotted! 
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. :blush: 
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