Prime Factors: Part 4

Prime Factors:

The Royal Sevens

It is said by those who study such things that the number seven is the number of completion.  This is, of course, based on the creation of the world being finished in six days and God resting on the seventh day.  I accept this, for it was codified along with the fourth commandment on tablets of stone.

It is for this reason that I have decided to call the related primes ending in the number 7 the "Royal Family" of primes.  Seven is a combination of three and the first square number, that is, the number 4.

However, the important feature is the relationship with the number 5.  It just so happens that 7 is a "mirror image" of 3. Where 3 is two less that 5, 7 is two more.  This translates into a chart very much like that of the threes.

However, in this case, subtraction back to the 5 will get the answer we need. I will start with 17 and backtrack back to 7.

What relationship does 1 have to 7 to get a something divisible by 17.  It so happens that 5 times 7 is 35 and 2 times 17 is 34.  That looks good, since they are just 1 apart.

The bigger number is the 35, so the operation will have to be subtraction.  It will look odd, but the multiples in our present format will all be negative integers.  But, then I know that -1 is just as good a factor as +1.

Here is the chart (with seven as "07" inserted)

07 => 0-(7*2) = 0-14  =  -14
17 => 1-(7*5) = 1-35  =  -34
27 => 2-(7*8) = 2-56  =  -54    
37 => 3-(7*11)= 3-77  =  -74
47 => 4-(7*14)= 4-98  =  -94
57 => 5-(7*17)= 5-119 = -114
67 => 6-(7*20)= 6-140 = -134
77 => 7-(7*23)= 7-161 = -154
87 => 8-(7*26)= 8-186 = -174
97 => 9-(7*29)= 9-203 = -194

Just like with the threes in the mirror, the multipliers increase by 3.  In this case, they start with 2 instead of 1, so the "mirror" is a bit warped.  The columns are not as neat as the numbers on the left don't match those on the right.  We actually have to do some math to see what is happening here.

When you put a candidate for division in the left column the format makes more sense.  However, with the multipliers getting into the twenties, we need to step back and concentrate on getting our math right.

Let's pick 37 as our prime.  I will use a very simple multiple of 37 to illustrate the formula:

111 => 11-(1*11) = 11-11 = 0. As a matter of fact, 37 times 3 is 111.

To get a more obvious answer, let us use the square of 37.

1369 => 136-(9*11) = 136-99 = 37.  Yep, 37 is a factor of 37!

One more thing, since 7 is 5+2, if you need to multiply something by 7, you only need to know the twos table and addition.

789 x 7 = ???
789 x 10 = 7890
7890/2 = 3945
789*2  = 1578

                          3945                         

+1578

    5523 

Or, 7 = 10-3.  With 3 = 2+1. Good old twos tables and subtract. Or, of course, just learn the seven's table. Oh, yeah, you can use a calculator!


So, what do I know?

Because of the special relationship between 5 and 2, the tables for the 3 and 7 families look much alike.  Practically mirror images, one finds a "positive" difference by adding, while the other uses the "negative" difference by subtracting.  Each progresses by a factor of 3 between decades.

And so, the chart with only the primes:

07 => 0-(7*2) = 0 -14 =  -14
17 => 1-(7*5) = 1 -35 =  -34
37 => 3-(7*11)= 3 -77 =  -74
47 => 4-(7*14)= 4 -98 =  -94
67 => 6-(7*20)= 6-140 = -134
87 => 8-(7*26)= 8-186 = -174

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.