A, B, C, D are all distinct digits. ABCD is a concatenated number

$ABCD$ = $A^3$ + $(A+C+D)^3 $ = $D^3+ (A + D)^3$

Figure out this Number and why is it Famous?

Also provide detailed reasoning to show how you figured it out.

  • $\begingroup$ Why downvotes? It is puzzling? $\endgroup$ – Uvc May 22 '19 at 8:44
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    $\begingroup$ The only reason I can think of for a downvote is that you ask these at a fast rate and it might be stale in some eyes $\endgroup$ – Adam May 22 '19 at 8:58
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    $\begingroup$ All your puzzles are pretty much the same, honestly. There's little variety. We want to see different styles from you. $\endgroup$ – greenturtle3141 May 22 '19 at 12:16
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    $\begingroup$ In addition to the lack of variety and rapid rate of questions, some of the downvotes may be because you often require a level of detailed analysis in the answers that isn't explicitly stated in the question - and wasn't initially asked for here. The no-computers tag isn't enough to make it clear to readers that you want a very high level of proof/explanation/logical analysis. I see you've accepted that in your comment below. $\endgroup$ – MichaelMaggs May 22 '19 at 14:07
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    $\begingroup$ Got it..already included in response to Garett..this is my first month..am gauging the interest of community, cater to their interests..my aim is to make math even complex one enjoyable and easy to digest through puzzles. $\endgroup$ – Uvc May 22 '19 at 14:19

The answer is

Which is the famous Taxicab number. It is the smallest integer that can be expressed as a sum of two positive integer cubes in 2 different ways.

  • $\begingroup$ No computers..answer needs logical reasoning and how it is arrived yet $\endgroup$ – Uvc May 22 '19 at 8:02

I think

ABCD is 1729, the sum of the cubes of 1 and 1+2+9=12 and also of the cubes of 9 and 1+9=10. It's famous because of the story about Hardy, Ramanujan and the taxi. I think it's rather unfair to call it a "cube" since it isn't itself a cube. No computers required precisely because the story is so famous and it was the first thing I thought of on looking at the puzzle.


here's a proof that it's the only solution. From the second equation we know that $1000A\leq 9^3+(A+9)^3$, which requires $A=1$ or $A=2$; if $A=2$ then our number has to be $9^3+11^3=2060$, which plainly doesn't work out. So $A=1$ and we have $1BCD=(C+D+1)^3+1=(D+1)^3+D^3$. Now $D$ has to be at least 8 to make that last thing have four digits, so we just have to check 8 (which doesn't work) and 9 (which does).

I have to say, though, that

precisely because this is so well known and the question didn't ask for a uniqueness proof or anything of the sort, you should accept Kruga's answer which delivered everything you asked for and I'm sure wasn't arrived at by computerized calculation.

  • $\begingroup$ Thx for the valuable input..it was my fault that I didn’t explicitly state that logical reasoning to be included..I will do few edits as per your comments so that future readers won’t be misled..thx very much. $\endgroup$ – Uvc May 22 '19 at 13:34

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