Calculate the last digits of Graham's Number Graham's number ends in a 7. It is a massive number, in theory requiring more information to store than the size of the universe itself. However it is possible to calculate the last few digits of Graham's number.

The last few digits are:

02425950695064738395657479136519351798334535362521
43003540126026771622672160419810652263169355188780
38814483140652526168785095552646051071172000997092
91249544378887496062882911725063001303622934916080
25459461494578871427832350829242102091825896753560
43086993801689249889268099510169055919951195027887
17830837018340236474548882222161573228010132974509
27344594504343300901096928025352751833289884461508
94042482650181938515625357963996189939679054966380
03222348723967018485186439059104575627262464195387


Your program may not contain these (or similar numbers), but must calculate them. It must calculate 200 digits or more.

Output to stdout. Running time of a maximum of 2 minutes on decent hardware. Shortest program wins.
dc - 21 chars

[3z202&gt;xO200^|]dsxxrp


This takes about a minute on my computer, and would take lots longer for values larger than 200. It doesn't output leading zeroes.

Here's a slightly longer but faster version (26 chars):

[3rAz^|dz205&gt;x]dsxxAz6-^%p
3[3rAz^|dz202&gt;x]dsxxAz3-^%p # or an extra character for a few less iterations