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.

  

Prime Factors: Part 3

Prime Numbers:

The Amazing Mr. Five.

Of the prime numbers under 10, only 5 stands alone.  He has no other primes in his family.  As such, recognizing multiples are easy. Just look for a 5 at the end.

This is true, because any number that is divisible by 10 -- that is has a 0 at the end -- divisible by both 2 and 5.  Further determination is made after removing the 0.

For instance:

1234560 is divisible by 2 and 5 (that is, by 10). That removes 5 from the running, at least temporarily.  Here is the progression:

1234560/2*5 = 123456
123456/2    = 61728
61728/2     = 30864
30864/2     = 15432
15432/2     = 7716
7716/2      = 3858
3858/2      = 1929
1929/3      = 543
543/3       = 271

My guess is that 271 is prime. It is one over 270, which has obvious factors.  Further investigation bears me out.

The "Five and Dime" Store.

Being that 5*2 is 10, and inversely 10/2 is 5, multiplying and dividing by 5 is easy. You only need to multiply by 10  and divide by 2 to get any number multiple of 5.

123*5 = 123*10/2 = 1230/2 = 615

Using the decimal point, it works the other way:

(123/2)(10) =61.5*10 = 615

Finally, it is easy to spot numbers divisible by 25 (5^2) and 125 (5^3). This is because 100 and 1000 are multiples of 10.

For 25, there are 3 multiples: 25, 50 and 75.  For example:

123,450 is divisible by 5 and 25. 50 is 2*5*5. So 2, 5, 10 and 25 are factored out.

123,450/2*5 = 12,345
12,345/5    = 6,171
6,171/3     = 2057, an odd number.  Is it prime? Stay tuned.

Another, example:

987,625 is divisible by 125 (5*25=625).  This would also place a 5 on the end of the next factor, raising the power by another 5.

8*125 = 2^3*5^2 = 10^2 = 1000

This leaves 987,000.  Removing the zeros, we then look for the factors, if any, of 987.  3 works, though 9 doesn't.

987/3 = 329.

So, dividing an odd by an odd, we've ended up with an odd number.  Further calculations are needed, but our odd number is within reach.  20*20 is 400, setting a limit.

So, what do I know?

I know that 5 and 10 are closely related by the prime number 2.  And this makes multiplying and dividing by 5 very easy. There are no other prime numbers related to 5, making the possibility of a random number being prime at less than 25%.

Even so, mathematically, there are an infinite number of prime numbers.  This is because "infinity" can not be divided. There are an infinite number of fractions.

Very odd, so to speak, that when not all odd numbers are prime, there can be an infinite number of primes anyway.

I know my brain hurts contemplating that.

To restate the FACTS:

1. The natural number 5 is prime.

2. 5 = 10/2

3. 5*2 = 10

4. Therefore, all numbers ending in 5 or 0 are multiples of 5.

Prime Factors: Part 2

Prime Factors:

The 3 family

We have taken it as a maxim that the sum of the digits will let us know if a number is divisible by 3.  But when we put it with its family, we see a clear pattern, while at the same time "proving" the maxim.

03 => 0+(3*1) = 3.

That is quite obvious, but using the formula on bigger numbers, removing the digit to the far right and adding it to  a tenth of the digits to its left, we can see WHY our maxim is true:

2853 => 285-9 = 276 => 27-18 = 9 (3*3)

2+8+5+3 = 18 (3*6); if in doubt, keep going: 18 => 1+8 = 9

Amazingly, the multipliers progress by 3, and in the third and fourth columns below, the family relationship is abundantly clear!


03 => 0+(3*1)   = 0+03 = 03
13 => 1+(3*4)   = 1+12 = 13
23 => 2+(3*7)   = 2+21 = 23
33 => 3+(3*10) = 3+30 = 33*
43 => 4+(3*13) = 4+39 = 43
53 => 5+(3*16) = 5+48 = 53
63 => 6+(3*19) = 6+57 = 63*
73 => 7+(3+22) = 7+66 = 73
83 => 8+(3*25) = 8+75 = 83
93 => 9+(3*28) = 9+84 = 93*

Building this table is easy, especially when you remember to add 3 to the multiplier, but then again, this is the "3 family."

Column 1, start with 0, add one per row.
Column 2, 3 all the way down
Column 3, just like #1 followed by "+"
Column 4, just like 2, but proceed with "("
Column 5, start with 1, add 3 per row
Column 6, same as #3
Column 7, start with 03, add 9 per row. Close with ")"
Column 7a repeats the 3 and goes on to 8
Column 7b starts with {1}3 and goes down to {0}4
Column 8, start with 03 and add 10 per row.

Wow.  I'm not sure if that is "easy" or not.  The second half presents us with a problem of large numbers as factors.  However, notice that one of them is 13 (in the family).

A second, easier, factor presents itself on down the line.  Since 30 is obviously in the family, and when added to 13 gives us 43, we have an alternative to multiplying by 13.

43 => 4-(3*30) = 4-90 = -86  (-2*43).

The other factors for primes -- 53, 73 and 83 -- use factorable numbers: 16, 22, and 25 (two squares and 2*11).  The 3 family is a math friendly family all around.

So, what do I know?

I know that fabulous patterns emerge if one looks closely.  These patterns are demonstratable using the properties of math, and can be proven.  However, it is much easier just to experiment with the results.


Here is the "chart" with only the primes:

03 => 0+(3*1)  = 0+03 = 03 (special case 1*)
13 => 1+(3*4)  = 1+12 = 13
23 => 2+(3*7)  = 2+21 = 23

43 => 4+(3*13) = 4+39 = 43 (special case 2*)
53 => 5+(3*16) = 5+48 = 53 (16 = 4*4)

73 => 7+(3+22) = 7+66 = 73 (22 = 2*11)
83 => 8+(3*25) = 8+75 = 83 (25 = 5*5)

*1 Adding the digits and reducing to a recognizable multiple of 3 works for this.  The square of 3 is 9, leaving a bonus indicator for divisibility by 9.  If the number is even, then both 2 and 3 are prime factors.

*2 Beyond 39 (3*13), so an alternative using 30 (3*10) presents itself:

43 => 4-(3*30) = 4-90 = -86 (-2*43)

Happy calculating!

Prime Factors

Though multiplication and its inverse, division, are performed easily with all whole numbers, the principle of equivalency leaves an easier solution to working with many of them.  The practice of "finding the factors" need not end with big numbers. In taking on large numbers, it is perfectly alright to break them up into the numbers from which they came by way of multiplication.

Even Numbers.

One half of all natural numbers are "even," this is to say they can be divided by the number 2.  Two is the first "prime" factor.  Almost everyone remembers the cheer: "Two, four, six, eight, who do we appreciate?"  Well, those are the first four "even" numbers, which hypothetically go on forever.  Along with the the beginning whole number "0," these provide the clue that a number has at least three factors: 1, 2 and the number in question.

Whether you multiply an even or an odd number by an even number, the answer will be even.

For example:  1234 is an even number, and therefor is not a prime number.  Its factors are 1, 2 and 617.  Is 617 a prime number?  Well, it isn't even.  Two is the only "even" prime number, so let us move on to other "prime suspects"

Odd Numbers

The other half of all natural numbers are odd.  This does not mean they are prime, but it makes task of finding big prime numbers a little easier.  I am not one to pursue such a task.  Suffice it to say that an odd number needs to be approached with care.  It can have hidden factors just waiting to be discovered.

Taking the factor of 1234 -- 617 -- the first thing is to find that number's square root.  This is best done with a calculator, but I recognize this as close to 25x25, that is 625.  This sets the limit.  A prime factor would have to be under 25, but not by much.  23x23 has a product 529.

Dropping back to the basics, then, we start with 3.  By theorem, the sum of the digits of any number must be divisible by 3 if the number is divisible by 3. 6+1+7 = 14.  14 is not divisible by 3.

The next prime number is 5. "Counting by fives" is easy, and it reveals to numbers, one even and one odd.  Every even number ending in 0 is divisible by 2 and 5.  This is two for one!  So, 617 is not divisible by 5 either.

The last odd number under 10 is the number 7.  There is no easy way to tell if a number is divisible by 7.  In the case of 617, we see a seven, but the first two numbers return a remainder of 5, leaving 57 (not divisible by seven.  Knowing the multiples of 7 up to at least 9x7 is advisable.

Note that multiplying an odd number by an odd number will get an odd number:

1x1 = 1 5x5 = 25 9x9 = 81
3x3 = 9 5x7 = 35
3x5 =15 5x9 = 45
3x7 =21 7x7 = 49
3x9 =27 7x9 = 63

With the factors 11, 13, 17, 19, and 23, only 11x17 even remotely comes close. However though 11x17 ends in 7, it is far from 617. What about 11x27?  That gets closer, but is far short as well (270 + 27 = 297).

So, 617 is indeed prime.

The most important products to know are those of the prime numbers 2, 3, 5 and 7.  Note, standing by itself is the number 49!  The "new" answer to the universal question!

2|  4

3|  6  9

5| 10 15 25

7| 14 21 35 49

      2  3  5  7 

Finding prime factors:

Starting with 2, what are the prime factors of 7,984,356 (a totally random seven digit number!)

Immediately 2 "works."  Trying 4, we get 1,996,089. Not even, so on to 3.  These digits are 1+9+9+6+0+8+9.  Added this gives us 42; reducing further to 6

So, with factors 2x2x3, we can divide by 12 to get 666,563  Is ths as far as we can go?  Not divisible by 5, so we try 7, 11, 13 and 17.  Seventeen works, yielding 39,139.

This leaves factors of 1, 2, 3, 4, 6, 12, 17, and 39,139.

Using a handy calculator, I know that the square of that large number is just under 198.  197 is a prime number having as its square 38,809. 197 times 199, the next prime number, equals 39,203. This means there are no more prime factors of our chosen number.

The prime factors of 1,996,089 are 2,3,17 and 39,139.


So, what do I know?

Odd x odd = Odd number
Odd x even = Even number
Odd + even = Odd number
Odd + odd = Even number

A prime number is a natural number that has exactly two natural divisors: 1 and itself.

2, 3, 5 and 7 are prime numbers under 10.

Even number 2 is the powerhouse of the primes, affecting ALL even numbers.

Odd number 3 can be seen to be a factor if the sum of the digits add up to a number divisible by 3.

Odd number 5 shouts out from the end of one of its products.  If the even number 0 is there, the 2 and 5 are instantly known.

Seven times seven is forty-nine (7x7=49). Every digit between 1 and 9 shows up as the final digit of multiples of 7.  So, don't look for an easy out here.

All prime numbers larger than 2 are odd numbers. About one in four numbers is prime (in the first 200 natural numbers, at least).

The factors of any number start with one and end with the square root of that number.