Ever worked on a large RSA problem? Here's one. It's a map to a priceless treasure. Really, I'm serious. Are you familiar with the RSA algorithm? Here are the two defining equations:

where C represents the ciphertext, P the plaintext, e and d the encryption and decryption exponents and n=pq the famous product of two primes. Well . . . here they are:

p = 67884660780899331266801378212733177571605492557891402279008830206100794305

724070115579183320113470899991239995017050831404466512248702094416187569628422

476957060834133015534669884673117608338759578775786955583153609078080317617492

041806334748209312135561009971331291

q = 65461813941200756890315580538544875968074002380447666538181878811706061599

199367765151734651031787699688330461650305673195742584479935137160110717763356

529697993251722360679278934892400601660485306787941403739827195667604072622572

609491247145476756847079434969988081

e = 44437204767003001868097045149648140876362324619394009079299043878517911969

228029732730441048069596707002372038585932832511086692388041228106144451067355

258749630096039425691840629982443130557557725704663899500194320195625005077447

381783105928165291218563682919444284530493075120031858191903986090555039105326

270661230450906287706993321646575220562750603137420024681537964020165547893902

163506054454854159497996062198113701277367291951155008551328931235354903142754

60024358644614729448187815663062174156210748429317814010864962218803

d = 24099484129657996682015695949166121225745723939763117626974031741517408918

225337787473322246843618324248340637237139890065304545599889836388365066647027

506621330916876452324799318829262123443837387546142646727332270289707290786110

033448166593418827753824680406632063622991565188554831392078940583055077596534

503437543636951589263391680220230874258651936870383122909544084899566052947848

794308396857882357053687631947989375255093394970364106222595266557392360187112

8935391724091063036390921591299547462093473942277540901972180155867

And here is the encrypted (ciphertext) treasure map:

584565320456079056472406794697025571766693520913877507241662721424514015287011

603416154188536836341015926767446223936487086729380578563273429545303354882999

335872814693052640028188293170276069519193534066874924002823402299557574111508

994858974296221386784608393260544702126406796897098959631948264896983143307205

839708632666773260086979553742525654915079519938492366869933223381055342706622

425031839180665648719174085014618369417651996764817708423914215175891883694441

939466211544062220527814839790244648360067898947829169131086041

Now after deciphering, the plaintext will be in base-127 (the ASCII code). So if I had encrypted the message "zeta" and you had run it through the RSA decryption algorithm, you'd get:

251546584

which is base-127 for "zeta". That is, the ASCII code for the letters "zeta" as a number in base 127:

All you need to do then is correctly pass the ciphertest through the RSA algorithm, split out the ASCII codes, and read the message. Mathematica has all the commands to do that if you know how. Once you do that, the treasure is yours. For true! And if you think it takes a lot of work, after successfully cutting and pasting the numbers in Mathematica, it only takes 3 very,very short commands to do this.

Note: guys, I'm pretty sure everything I said above is precisely correct but it's been a while since I worked on this so any comments/corrections are certainly welcomed.

Just want to challenge those interested in RSA.

I was however able to successfully cut and past the numbers directly from SOS, into Mathematica, and retrieve the plaintext.