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Old 2007-03-19, 22:22   #1
davar55
 
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Default Easy Arithmetic

Find three consecutive odd cubes whose sum is a four (decimal) digit
number with identical digits.
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Old 2007-03-20, 09:46   #2
davieddy
 
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"Lucan"
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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
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Old 2007-03-20, 13:14   #3
Mini-Geek
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"Tim Sorbera"
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Quote:
Originally Posted by davieddy View Post
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?
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Old 2007-03-20, 14:29   #4
Zeta-Flux
 
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Maybe by "consecutive" he meant distinct.
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Old 2007-03-20, 14:37   #5
S485122
 
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I did find the same candidates, but changing the problem to 3 consecutive squares I get 5555 = 41^2+43^2+45^2
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Old 2007-03-20, 16:21   #6
davieddy
 
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That's quite a coincidence.
Must be what he meant. Well spotted!

Last fiddled with by davieddy on 2007-03-20 at 16:52
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Old 2007-03-20, 17:47   #7
davar55
 
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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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