Archive for the ‘Math’ Category

The Inevitable Knot in Your Long Extension Cord

Friday, July 28th, 2017

This knot in my extension cord occurred inadvertently and thus I assumed the highest with the highest likelihood it could be undone with a single pass through a cable end through a loop or loops with the loops arranged so only a single pass is required.  You would think this because once the cord is knotted it is hard to get another knotting function to occur on an existing knot.  The most likely scenario being a knot occuring due to an incorrect uncoiling.  Carefully looking at it I found this was true and verified it by doing just that.  The fact it looks like a pretzel dazzles a bit.  By the way the left side is the short end. To the right is the 25 foot long extension.


Initial solution

As with the above photo the left side exit strand is the short strand to be passed through in whatever method we can find.  

Stretch out the knot to get the second state shown in the second photo. Pass the left hand strand through the loop on the left.  You are left with a twisted loop that is no longer knotted. 




Schematic illustration of the simplified model for knot formation. Because of its stiffness, the string tends to coil in the box, as seen in Fig. 1, causing a number of parallel string segments to lie parallel adjacent the end segment. As discussed in the text, we model knots as forming due to a random series of braid moves of the end segment among the adjacent segments (diagrams at bottom). The overall connectivity of the segments is indicated by the dashed line.

Rectangularizing a Circle to derive the area

Saturday, March 11th, 2017

536 Puzzles: #309: A Maypole Puzzle

Saturday, February 11th, 2017

During a gale a maypole was broken in such a manner that it struck the level ground at a distance of twenty feet from the base of the pole, where it entered the earth. It was repaired, and broken by the wind a second time at a point five feet lower down, and struck the ground at a distance of thirty feet from the base. What was the original height of the pole? In neither case did the broken part become actually detached.

From this you can set up the following equations:

(h-x)^2 - x^2 = 30^2

(h-(x+5))^2 - (x+5)^2 = 20^2


h^2 - 2hx  = 30^2

h^2 - 2h(x+5) = 20^2

Subtracting the second from the first yields:

2h*5 = 30^2 - 20^2

h =50  

Solve System of Equations X+YZ=6 Y+XZ=6 Z+XY=6

Tuesday, February 7th, 2017

Research LInks

Find the solution for the system of equations:

        X+YZ=6            Y+XZ=6             Z+XY=6  

Multiply each equation by the term missing to make each second term a triple

 X^2+XYZ=6X             Y^2+XYZ=6Y             Z^2+XYZ=6Z  

So now subtract the second equation from the first: 

 X^2 -Y^2=6(X-Y)    

(X+Y)(X -Y)=6(X-Y)    


Since the equations are symmetric we should have

(X+Y)=6        (X+Z)=6           (Y+Z)=6 

Subtracting the 3rd from the 1st:

X - Z =0   or 

 X = Z     and again since the equations are symmetric  X = Y = Z  


The very first equation at the top  X+YZ=6   becomes

 X^2+X - 6=0  

 X^2+X - 6=0  

 (X+3)(X-2)=0   giving the solutions

 X = 2,-3 

Euclidean Algorithm for Greatest Common Denominator

Tuesday, January 31st, 2017

Research Links

A 24-by-60 rectangle is covered with ten 12-by-12 square tiles, where 12 is the GCD of 24 and 60. More generally, an a-by-b rectangle can be covered with square tiles of side-length c only if c is a common divisor of a and b. So the task is to find the largest square that tiles the rectangle completely.  See below:

Subtraction-based animation of the Euclidean algorithm. The initial rectangle has dimensions a = 1071 and b = 462. Squares of size 462×462 are placed within it leaving a 462×147 rectangle. This rectangle is tiled with 147×147 squares until a 21×147 rectangle is left, which in turn is tiled with 21×21 squares, leaving no uncovered area. The smallest square size, 21, is the GCD of 1071 and 462.

So the example compiled down to arithmetic steps would be as follows:

1071 = 2*462 + 147

462 = 3*147 + 21 

147 = 7*21 + 0   Thus 21 is the GCD of 1071 & 462

Langley’s Adventitious Angles

Tuesday, January 31st, 2017

Research Links

Farmer buys livestock puzzle

Sunday, January 29th, 2017

A farmer has 100 dollars to buy 100 animals.  A cow costs 10, a pig costs 5 and a chicken costs 50 cents.  How many of each does he buy?  He must use all his money and he must buy at least 1 of each type.



note Z must be even because otherwise you will get  ddd.5<> 100 in the first equation. Eliminate z by multiplying first equation by 2 and subtracting the second equation




19x+9y=100   Here a solution is only found because the solution is specified to be integers


Brute force Integer solution

Find a solution for the above and plugging into:  x+y+z=100   now plug in number until you get an even z. Try x=1:






and is even and thus the problem is done


Modulus Solution


(19x+9y)mod 9 = 100 mod 9

(19x) mod 9 + (9y) mod 9 = 1 

(19x) mod 9  = 1 

(19 mod 9)  *( x mod 9)  = 1 

 ( x )mod 9 = 1 

Two obvious solutions are 

 x=1    and  x=10    

The first solution is viable.  The second is not because there is no money left over to buy any other type of animal.

Polynomial Root Solver with Root Map in Complex Plane and Graph of Function

Friday, January 27th, 2017

Research Links

Eulers Theorem

Friday, January 27th, 2017

Research Links


Eulers theorem is very similar to Fermat's little theorem.  It is not restricted to prime number p.  It is restricted to relatively prime a & p.  It states:

a^ {psi(p)} mod p = 1  

Where psi(p)    is the Euler totient function which is the number of integers less than and coprime to p.  If you followed the proof for Fermat's little theorem then you can understand this generalization rapidly.  As before when the integer a is coprime to p you get the jumble of all the integers 1,2,3…p-1.  This was guaranteed by p being prime in Fermat's little theorem.  When you relent on that condition then you have some integers a that are not coprime to p and they will not give you a full contingent of integers. See the spread sheet clips below


2 is coprime to 15 and thus all values 0 through 14 are cycled through.  3 is not coprime to 15 and thus the gearing does not cycle through all values. 

In the spread sheet example clipped above the totient {psi(15)} = 8 

The following spread sheet snip contains the value of the modulus 1 to p-1 where p=15 in this case and the modulus of a*remainder.  As before with Fermat's little theorem the modulus values come in different order but contain identical values.


m_1 * m_2 ...m_{psi(p)}  = am_1 * am_2 ...am_{psi(p)}  

m_1 * m_2 ...m_{psi(p)}  = a^{psi(p)}*[ m_1 * m_2 ...m_{psi(p)}]  mod p

a^{psi(p)} mod p  = 1

Flash Mind Reader

Friday, January 27th, 2017

Click on the image to go there.

The flash mind reader has you compute a number in your head.  You note the symbol next to the number you compute.  You then click the crystal ball and it returns the symbol you were thinking of.  This is a classic "force card" trick. Let's compute the number:

When you pick 10a+b    you pick   a & b

Then they have you calculate 

10a+b -(a+b) = 9a   Thus your result will be limited to 1,9,18….99   They got lazy and did not go to 100 ah ha!  They could have as easily just told you to take the tens digit and multiply by 9.  But then you would have seen immediately that your end result is limited to only 10 values.

So when the page comes up these values all have the same symbol next to them. The rest of the numbers will never be arrived at so they can obscure their subterfuge by allowing the symbols associated with those numbers to be random.  When you click on the crystal ball it pops up the symbol next to the 9a force numbers and the illusion is complete.