Getting It Right

Geometry: Part 2

The Right Angle

In order to make sure your patio is squared, whether it is going to be a rectangle or a square, you will
need the cord three times the length of the distance to the distance to the first peg.

You really don't even have to measure the distance to the first peg to assure a square corner, but having a tape measure will make this easier.  Let's say your patio is to extend 12 feet (or, if you'd rather go metric, 4 meters).  This means you will need 36 feet (or twelve meters) of cord to easily square up your patio.  The cord will be a little bit longer to leave room for securing it to the pegs.

As pictured in figure #1, the tripling of the length is done by walking back and forth between the pegs with the spool of cord.

Once you have the full length (#2), walk that back to the first peg (#3) and then repeat (#4).  In the end, the four strands will be 9 feet (3 meters) long.  If you began against a wall, just lay the 9 foot piece against the wall and put the third peg down.

Fasten the 9-foot length to the third peg and the end of full cord to the first peg.  Taking the full cord, walk to the second peg, moving it to where it provides a taut line between all the pegs.

This works because in every triangle with a right angle the side opposite that angle has a length that is the square root of the sum of the squares of the other two sides.  It so happens that this ratio is found first in whole numbers with 3, 4, and 5.  And so, any multiples of these numbers produces similar triangles.  If you are going to be doing a lot of building, you could make your own "square" tool using a yardstick or even a twelve inch ruler.  Or, of course, just buy one when you get a chance!

So, what do I know?

A right triangle can be constructed based on the formula

a² + b² = c²

That is, when side c is opposite the right angle, the squares of sides a and b add up to the square of side c.

This manifests itself the ratio of 3:4:5.  That is to say, 9 + 16 = 25.  In whole numbers this only works with multiples of these three numbers.  The ratio using 1:2 would render an irrational number: the square root of 5.  Likewise 2:3 would need the square root of 13!  It only works with these three adjacent whole numbers!

Prime Factors: Part 6

Prime Factors:

The Chart

Finally, Down to the basic formulas.  In this chart, or more specifically, diagram, one can find all the parts of the formulas used to determine whether a large number is divisible by a particular prime number.

Usually, a large number is reduced first by dividing by 2, 3, 5 and 7.  Then, an estimate of the square root of the remaining number gives the upper limit for searching for prime factors.  The chart works for the double digit primes as well as for "03" and "07."

The "rule" of the "sum of the digits" for the 3 can be derived from the X3 quadrant, assuming the format of "03."  Likewise, plugging "07" into the X7 quadrant gives the "trick" or "rule" for finding out if a number is divisible by that prime number.

To review:

For divisibility

 by 2: Look for the even number on the end.

 by 3: The sum of the digits will be divisible by 3.

 by 5: Look for the 5 (or 0, but see 2) at the end of the number.

 by 7: Using the formula for the X7 primes: For "07"10x+n = x-(n*(0+2))

          That is, for divisibility by 7, for any number "10x+n" take twice n from x until you get to a multiple of 7.

Double digit primes below 100 are on the chart.  Larger primes can be added remembering to put them in the right "family."

101 is in the X1 family, Therefore, plugging it into the formula: 10-(1*10) => K*X1.  Of course, with the zero in the middle, numbers up to 10,000 are easy to spot as multiples of this prime!

8989 => 898-90 = 808. Of course, that choice was influenced by my building the chart.  I put up a reduplicative whole number!

Really big odd numbers always give us a challenge.  Once divisibility by 3 and 5 are eliminated, the chart come into play.  I am going to blindly punch in 7 digits, making sure that the end number is not even or a 5.

7351613.  7+3+5+1+6+1+3 = 26.  So, not divisible by 3.

Let us try 7.

7351613 => 735161-(3*2)=> 735155=> 735155-(5*2)=> 735145=> 7351-(5*2)=> 734-2=> 732. 

732 is not divisible by 7.

What about 11?

7351613 => 735161-3=> 73515-8=> 73507=>7350-7=> 7343=> 431=> 42.  Nope.

So, 13?

7351613 => 735161+(3*4) = 735173 => 73517+(3*4) = 73539 => 7353+(9*4) = 7381 => 738+4 =

742 => 74+(2*4) = 74+8 = 82.  13 does not work. 

Which of the two attempts got closer?  42 is 2 away from 44, while 82 is 4 away from 78.  Perhaps the X1 family offers a better chance.

What of 31, which requires a multiplier of 3, two away from 1.

7351613 => 735161-3*6=> 7350-9*3=> 732-3*3=> 72-1*3 = 69.  69 = 62+7.  Further away!

So, let us first estimate a square root by dividing the seven digits into groups of two starting from the right.

7 35 16 13.  The square root will be a number between 2000 and 3000.  2,500*2,500 = 6,250,000.

So, my guess is that the limit of possibilities lay somewhere around 2750.  Ending in a 3, the number is not an exact square.  A lot of work to go, but I think I've written enough.

So, what do I know?

The search for primes continues past 97.  But, placing all possible primes in the four "families" makes the job a bit easier.  Once the one gets past division by 7, at least without a calculator, having derived a family formula helps to discover larger primes by elimination.

I have not memorized the above chart, but I did use simple math to derive the formulae.  The exercise of my brain has staved off mental deficiencies for at least a few months!

I don't know if finding the largest prime number yet is worth the effort, but perhaps using this chart will help someone in the search.

I know that the chart will help an inquiring mind find proof that 7,351,613 is not prime.  I challenge a reader, whether a friend or an random reader, to send me the prime factor.  You DO have a clue.