IBM have a monthly riddle set up called Ponder This. The riddles are mathematical  too difficult for me :)
Here's part of Novermber's (cute) riddle, followed by a Perl solution that surprised me for being so short and, to my perlverted mind, straightforward. I've only begun looking at that site, but I bet other riddles there can be solved with Perl, too.
The numbers 1 to 9999 (decimal base) were written on a paper. Then the paper was partially eaten by worms. It happened that just those parts of paper with digit "0" were eaten. Consequently the numbers 1200 and 3450 appear as 12 and 345 respectively, whilst the number 6078 appears as two separate numbers 6 and 78. What is the sum of the numbers appearing on the wormeaten paper?
perl le '$sum += $_ for map { split /0/ } 1 .. 9999; print $sum'
37359000

Re: Ponder This: the wormeaten page by BrowserUk (Pope) on Dec 16, 2004 at 23:16 UTC 
Cute. Though it would be a tad more efficient, (and w complient)if it were /0+/
"But you should never overestimate the ingenuity of the sceptics to come up with a counterargument." Myles Allen
"Think for yourself!"  Abigail
"Time is a poor substitute for thought"theorbtwo
"Efficiency is intelligent laziness." David Dunham
"Memory, processor, disk in that order on the hardware side. Algorithm, algorithm, algorithm on the code side."  tachyon
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Re: Ponder This: the wormeaten page by Anonymous Monk on Dec 17, 2004 at 10:59 UTC 
perl le '$_ = "@{[1 .. 9999]}"; s/[ 0]+/+/g; print eval'

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perl le 'print eval join'+',map{split/0+/}1..9999'
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But I suppose this is cheating: perl le 'map$\+=$_,map/[^0]+/g,1..9999;print'
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perl le '$_=join 0,1..9999;y/0/+/s;print eval'

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Re: Ponder This: the wormeaten page by zentara (Archbishop) on Dec 17, 2004 at 12:17 UTC 
perlverted :) I learned a new word today.
I'm not really a human, but I play one on earth.
flash japh
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Re: Ponder This: the wormeaten page by hv (Parson) on Dec 17, 2004 at 13:58 UTC 
This can also be solved by arithmetic quite readily.
The sum of the digits 1 .. 9 is 45.
Each digit appears 1000 times in each of the units, 10s, 100s and 1000s places.
The units always count as themselves: total 1000 * 45.
The tens are followed by a zero 1/10 of the time: total (100 + 900 * 10) * 45.
The hundreds are directly followed by a zero 1/10 of the time, and followed by a nonzero digit then a zero 1/10 of the remaining: total (100 + 90 * 10 + 810 * 100) * 45.
Similarly for the thousands: total (100 + 90 * 10 + 81 * 100 + 729 * 1000) * 45.
So the sum is (1300 + 1080 * 10 + 891 * 100 + 729 * 1000) * 45 = 830200 * 45 = 37,359,000
This approach can readily be adapted to base b, and beyond to the eat digits 0 .. x1 required for part 2 of the puzzle.
Hugo
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Re: Ponder This: the wormeaten page by Anonymous Monk on Jan 26, 2005 at 12:48 UTC 
Of course you take a lateral approach and say that the "wormeaten paper" is the missing part and therefore the sumof the zeros would of course be zero?  [reply] 

But if you do that, you have to consider that the zeros on the page may form a pattern that looks like a number. Now you have to go over the many possible dimentions of the page and check them all! Thankfully, you have a head start, since you can use this OCR code; all you have to account for now is rotation.
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