Simon Singh, several of whose books I have reviewed before, had a mathematical contest where he was seeking numbers that fit this pattern:

A^(B+C)=BCA

For example, 5^(1+2)=125.

The contest closed yesterday, but it is still something people might want to play with. I found three additional solutions, listed below in AES (with the key ‘answers’). I also proved that there are no solutions other than the one above and those below.

##### Encrypted: decrypt with this AES tool

ZZZZZ WQKVD XWLVG REKEQ QONRX WDIFC HXCDB LVDSV UJJXT NDNWT DDRSJ RAQUG XPIUW NOFUD NFEGM GCQIF MFQPK FTCWR BWNFG SVWXS VHVOW CURXQ MXEKC QPETN FSOKB DPSFA KVHUW XTBQP HFRAD TGTUG JBLWN SOILW KNPQR FACRC COHIQ WUQCK TGLXD FWLEC FOQVE GFRUJ FHGJJ VTOOH BCLGE PPWTV BGDPA ECQEJ EKTPA BFTOT EWQXG RTUML WBFBN VGLWT LKXHP NMJHR JVQLW JWVFN EEUTJ FUDHS IWULC RFKID NRAUI TGDBJ VIHII LVOHE SLORI YYYYY

##### End encrypted message

{ 8 comments… read them below or add one }

A proof – how wonderful! You must share it! (if this comment box is not too narrow)

It is the dullest sort of proof: a proof by exhaustion. There aren’t so many three-digit numbers, after all.

Excel file

Someone can port it to Google Docs, if they like. Probably, the necessary formulas will still work.

Google docs

I am a bit embarrassed to put up such a crude proof.

If a reader can produce one that is more elegant (subjectively judged by me), I will send them a copy of Simon Singh’s

Fermat’s Last Theorem.Aha..Well, I wondered if there was some elegant way, but for puzzles like this, usually not.

Partly, that is because the string ‘BCA’ is mathematically meaningless.

If it was B*C*A, some kind of nice proof would be more feasible.

Well, it’s not meaningless really. You’re just looking for all solutions to the equation:

A^(B+C)=100.B + 10.C + A

subject to the constraints

0< A < 9

0 < B < 9

0 < C < 9

but it is a bit contrived, and doesn't generalize very naturally for A9.

Quite right. It hadn’t occurred to me to treat the BCA string that way.