The Math Factor Podcast

HP. Happy Root 10 Day!

March 16, 2012 · answers, math puzzles, numbers, The Mathcast · Permalink

«« HO. Crazies on the Plane· · · HQ. Newton v Leibnitz »»

For procrastinators only, we celebrate √10 day! And we pose a new puzzle:

The number of 1′s in this quiz is ____

The number of 2′s in this quiz is ____

The number of 3′s in this quiz is ____

The number of 4′s in this quiz is ____

There are actually two solutions to this one, but more generally, what happens with more lines in the quiz?

Finally, here’s the link to the special issue of Nature with essays on the great Alan Turing.

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4 Comments »

1. martin said,

   March 17, 2012 at 3:42 pm

   I wrote a python program to simulate 5 (not 100) ppl getting on the plane.

   http://pastebin.com/6rhxv0yA
   the last guy  gets his seat about 0.499045  of the time.

    

2. Shawn said,

   March 19, 2012 at 1:25 am

   Show Spoiler ▼

   Quiz A The number of 1?s in this quiz is 2
   The number of 2?s in this quiz is 3
   The number of 3?s in this quiz is 2
   The number of 4?s in this quiz is 1 Quiz B
   The number of 1?s in this quiz is 3
   The number of 2?s in this quiz is 1
   The number of 3?s in this quiz is 3
   The number of 4?s in this quiz is 1

3. gpeace said,

   March 22, 2012 at 8:39 pm

   I found Show Spoiler ▼

   two solutions for 4 lines, one for 5 lines and none for 6 lines.
   In general, the sum of the numbers in the second column equals twice the number of lines, since each number in each of the two columns must be counted once.  So for 4 lines, a + b + c + d = 8 (where a is the number of 1′s, b is the number of 2′s, etc.).
   Also, we can multiply the number of 1′s in the second column (a-1, since there is one in the first column) by 1, add it to the number of 2′s in the second column (b-1) multiplied by 2, and so on.  This must also equal 8.
   1*(a-1)+2*(b-1)+3*(c-1)+4*(d-1)=8
   a + 2b + 3c + 4d = 18
   Since I only have two equations for four variables, I had to use some trial and error to get two solutions for four lines, which are:
   (a,b,c,d) = (2,3,2,1) and (a,b,c,d) = (3,1,3,1)
   For five lines, a+b+c+d+e=10 and a+2b+3c+4d+5e=25 and I found one solution:
   (a,b,c,d,e) = (3,2,3,1,1)
   I am interested to know if there are more precise general principles than my two equations that would not make it unpleasant to look for solutions beyond 6 lines.

4. Brett Barbaro said,

   July 18, 2012 at 11:33 am

   Hey! Episode HP is broken – I can only download 2.3MB worth. Any chance of getting it fixed? I can’t stand to miss even a second of your amazing show! :)

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