http://www.mapleprimes.com/posts/43792-Solve-The-Latest-IBM-Ponder-This-Challenge en-us 2026 Maplesoft, A Division of Waterloo Maple Inc. Maplesoft Document System Wed, 10 Jun 2026 21:25:28 GMT Wed, 10 Jun 2026 21:25:28 GMT The latest comments added to the Post, Solve the Latest IBM "Ponder This" Challenge with Maple http://www.mapleprimes.com/images/mapleprimeswhite.jpg http://www.mapleprimes.com/posts/43792-Solve-The-Latest-IBM-Ponder-This-Challenge http://www.mapleprimes.com/posts/43792-Solve-The-Latest-IBM-Ponder-This-Challenge?ref=Feed:MaplePrimes:Solve the Latest IBM "Ponder This" Challenge with Maple:Comments#comment80795 Some 20 years ago, a restricted version of this challenge was posed in a science show on German public TV ("Kopf um Kopf"): arrange the digits 1,..,9 so that the number resulting from the first two digits (from the left) is divisible by 2, etc.. A friend of mine and me used a Sinclair Spectrum (8 bit home computer) and nested loops in BASIC to find the result... Just for fun, we didn't actually take part in the competition. I still know the result, so I won't write a Maple proc, but maybe anyone else might be interested. The latest comments added to the Post, Solve the Latest IBM "Ponder This" Challenge with Maple 80795 Tue, 02 Aug 2005 12:39:09 Z Thomas Richard Thomas Richard http://www.mapleprimes.com/posts/43792-Solve-The-Latest-IBM-Ponder-This-Challenge?ref=Feed:MaplePrimes:Solve the Latest IBM "Ponder This" Challenge with Maple:Comments#comment80794 You're the first one to attach a .mpl . We'll clear this up. It's not our intent to overly restrict file types. T4. The latest comments added to the Post, Solve the Latest IBM "Ponder This" Challenge with Maple 80794 Tue, 02 Aug 2005 16:15:58 Z Tom 4 Tom 4 http://www.mapleprimes.com/posts/43792-Solve-The-Latest-IBM-Ponder-This-Challenge?ref=Feed:MaplePrimes:Solve the Latest IBM "Ponder This" Challenge with Maple:Comments#comment86938 Modifying the code to solve that version (same problem but without 0) is easy (a two character change); I can see how you could remember (most of the) solution after so long. I assume Sinclair Basic didn't have recursion, handling it with loops seems tedious. The latest comments added to the Post, Solve the Latest IBM "Ponder This" Challenge with Maple 86938 Tue, 02 Aug 2005 17:45:35 Z Joe Riel Joe Riel http://www.mapleprimes.com/posts/43792-Solve-The-Latest-IBM-Ponder-This-Challenge?ref=Feed:MaplePrimes:Solve the Latest IBM "Ponder This" Challenge with Maple:Comments#comment86937 Sorry about that. I have to manually enter all extensions that are allowed, and I missed .mpl. It has been added to the list. ____ William Spaetzel Applications Developer, Maplesoft The latest comments added to the Post, Solve the Latest IBM "Ponder This" Challenge with Maple 86937 Tue, 02 Aug 2005 18:27:00 Z Will Will http://www.mapleprimes.com/posts/43792-Solve-The-Latest-IBM-Ponder-This-Challenge?ref=Feed:MaplePrimes:Solve the Latest IBM "Ponder This" Challenge with Maple:Comments#comment90113 That's right, Spectrum BASIC did not have recursion. I don't recall the details of our implementation. Even if I find the backup (on audio tape or floppy disk), it won't be readable any more. AFAIR, we used a string to store the number. Contrary to my intention, I've written a Maple worksheet, using a set of digits (the natural choice for such a brute force approach, IMHO). Quick, but ugly. I'll post it in my blog. The latest comments added to the Post, Solve the Latest IBM "Ponder This" Challenge with Maple 90113 Tue, 02 Aug 2005 21:50:02 Z Thomas Richard Thomas Richard http://www.mapleprimes.com/posts/43792-Solve-The-Latest-IBM-Ponder-This-Challenge?ref=Feed:MaplePrimes:Solve the Latest IBM "Ponder This" Challenge with Maple:Comments#comment80771 The probability that a K-digit number is divisible by K is approximately 1/K. The probability that first (K-1) digit of it is divisible by (K-1) is approximately 1/(K-1), etc... That gives a probability that a K-digit number satisfies the problem conditions, approximately 1/K! Now, there are (b-1)*b^(K-1) K-digit numbers in base b. Thus, the expected number of them, satisfying the problem conditions, is (b-1)*b^(K-1)/K! Approximating factorial by Stirling formula, we get the following asymptotics of K with expected number 1, K(b) ~ (b-1)*e that is pretty close to the calculated maximal possible values of K. The latest comments added to the Post, Solve the Latest IBM "Ponder This" Challenge with Maple 80771 Sat, 06 Aug 2005 20:30:39 Z alec alec