FYI: we cannot execute "matlabpool open" on the labs. ;)
What it does: it finds the maximum number of occurrences for each prime number smaller than x in the factorisations of the numbers 1:x (e.g., for x = 10, the maximum number of occurrences for 2 is 3, since 2*2*2 = 8). If the product of all prime factors taken to the power of their maximum # of occurrences is then taken, the smallest number that is divisible by 1:10 is obtained. So for x = 10: 2 * 2 * 2 * 3 * 3 * 5 * 7 = 2520.
This code works but the server is too slow...
Remove all the consonants
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Sum of first n terms of a harmonic progression
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Make a random, non-repeating vector.
Siblings of a graphics object
Project Euler: Problem 10, Sum of Primes
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