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Guest Message by DevFuse


Member Since 01 Nov 2012
Offline Last Active Nov 27 2012 04:11 PM

Posts I've Made

In Topic: A Boggle-like Challenge

16 November 2012 - 11:59 AM

Spoiler for Possible solution for the first part:

In Topic: The elusive chord

15 November 2012 - 04:01 PM

I am thinking the average length of a random chord is pi/4 (0.7854) as a first thought.

In Topic: Triangular sticks

09 November 2012 - 05:24 PM

I have run two simulations, each over 1,000,000 runs.

The first simulates the scenario where one stick is randomly broken and then one of the resulting two sticks is randomly choosen and then randomly broken.

The second scenario simulates where the stick is randomly broken and each of the two resulting sticks are then randomly broken and then one of the resulting four sticks is randomly choosen and thrown away:

Scenario 1: Probability of Sucessful Triangle: 0.19378

Scenario 2: Probability of Sucessful Triangle: 0.19371

Assuming my simulations are correct, it practically makes no difference which scenario is choosen

In Topic: Help! A remainder is chasing me

06 November 2012 - 05:31 PM

FYI: If you extend the sequence from 2 remainder 1 .... to bigger numbers

possible solutions are (I am not absoutely sure they are the smallest number that works for each but these work!)

20 remainder 19 ......................232,792,559
22 remainder 21.......................232,792,559
23 remainder 22....................5,354,228,879
24 remainder 23....................5,354,228,879

30 remainder 29.............2,329,089,562,799
31 remainder 30...........72,201,776,446,799
36 remainder 35.........144,403,552,893,599
40 remainder 39......5,342,931,457,063,199

In Topic: Help! A remainder is chasing me

06 November 2012 - 04:05 PM

For those of those familiar with SQL, these types of problems are easily solved by perfoming a query on a Numbers (or Tally) table.

For example I have run the following selection on a numbers table (Numbers) containing 100,000,000 Rows (1,2,3....100,000,000) for the first 17 rows of the sequence (2 remainer 1 up to 18 remaining 17)

The answer is 12,252,239

FroM Numbers
Number%2 = 1
AND Number%3 = 2
AND Number%4 = 3
AND Number%5 = 4
AND Number%6 = 5
AND Number%7 = 6
AND Number%8 = 7
AND Number%9 = 8
AND Number%10 =9
AND Number%11 = 10
AND Number%12 = 11
AND Number%13 = 12
AND Number%14 = 13
AND Number%15 = 14
AND Number%16 = 15
AND Number%17 = 16
AND Number%18 = 17

(To get the solutions to next rows in the sequence: 19 remainder 18, 20 remainder 19...) requires a bigger Number table