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Apply Todayblaat submitted a Comment to Solution 41338

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.

on 14 Feb 2012

blaat submitted Solution 41340 to Problem 290. Make one big string out of two smaller strings

on 14 Feb 2012

blaat submitted Solution 41338 to Problem 239. Project Euler: Problem 5, Smallest multiple

on 14 Feb 2012

blaat submitted Solution 41262 to Problem 250. Project Euler: Problem 10, Sum of Primes

on 14 Feb 2012

blaat submitted Solution 41250 to Problem 240. Project Euler: Problem 6, Natural numbers, squares and sums.

on 14 Feb 2012

blaat submitted Solution 40660 to Problem 114. Check to see if a Sudoku Puzzle is Solved

on 14 Feb 2012

blaat submitted Solution 40637 to Problem 314. Find the sum of the elements in the "second" diagonal

on 14 Feb 2012

blaat submitted Solution 38671 to Problem 138. Number of 1s in the Binary Representation of a Number

on 12 Feb 2012

blaat submitted Solution 38647 to Problem 195. Program an exclusive OR operation with logical operators

on 12 Feb 2012

blaat submitted Solution 38218 to Problem 158. Is my wife right? Now with even more wrong husband

on 12 Feb 2012

blaat submitted Solution 37566 to Problem 110. Make an N-dimensional Multiplication Table

on 11 Feb 2012

blaat submitted a Comment to Solution 33563

Nice :) I was trying to think of clever ways to use convolution, but didn't manage. This should be the leading solution instead of the lookup based on the test cases..

on 11 Feb 2012