MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

# blaat

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on 5 Apr 2013

on 5 Apr 2013

on 5 Apr 2013

on 5 Apr 2013

on 5 Apr 2013

on 5 Apr 2013

blaat submitted Solution 227404 to Problem 325. 2 b | ~ 2 b

on 5 Apr 2013

on 5 Apr 2013

on 30 Mar 2013

on 30 Mar 2013

on 30 Mar 2013

on 30 Mar 2013

on 30 Mar 2013

on 30 Mar 2013

on 30 Mar 2013

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on 30 Mar 2013

on 30 Mar 2013

on 30 Mar 2013

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on 30 Mar 2013

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on 14 Mar 2013

on 21 Jan 2013

on 13 Oct 2012

on 13 Oct 2012

on 13 Oct 2012

on 13 Oct 2012

on 13 Oct 2012

on 13 Oct 2012

on 5 Oct 2012

on 5 Oct 2012

on 5 Oct 2012

blaat submitted Solution 44094 to Problem 176. Nearest

on 17 Feb 2012

on 17 Feb 2012

on 17 Feb 2012

blaat submitted Solution 43935 to Problem 300. Remove NaN ?

on 17 Feb 2012

on 16 Feb 2012

on 15 Feb 2012

on 15 Feb 2012

blaat 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

on 14 Feb 2012

on 14 Feb 2012

on 14 Feb 2012

on 14 Feb 2012