Easy Arithmetic

davar55 · 2007-03-19, 22:22

Find three consecutive odd cubes whose sum is a four (decimal) digit
number with identical digits.

davieddy · 2007-03-20, 09:46

I found four candidates, (the four digit numbers being
1197,2403,4257,6903).
What am I missing?

Mini-Geek · 2007-03-20, 13:14

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?

Zeta-Flux · 2007-03-20, 14:29

Maybe by "consecutive" he meant distinct.

S485122 · 2007-03-20, 14:37

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

davieddy · 2007-03-20, 16:21

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

davar55 · 2007-03-20, 17:47

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.

2007-03-19, 22:22	#1
davar55 May 2004 New York City 10210₈ 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 2×3×13×83 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, 16:21	#6
davieddy "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 2007-03-20 at 16:52

2007-03-20, 17:47	#7
davar55 May 2004 New York City 2³×23² 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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2007-03-20, 14:29	#4
Zeta-Flux May 2003 7×13×17 Posts	Maybe by "consecutive" he meant distinct.

2007-03-20, 14:37	#5
S485122 Sep 2006 Brussels, Belgium 3×7×79 Posts	I did find the same candidates, but changing the problem to 3 consecutive squares I get 5555 = 41^2+43^2+45^2