Prime and Composite Number Theorem (Divisibility Rules)

How can we check if a number is prime or composite?

Let 'x' be the number you check:

x² = x * x

Say √(x) = y 

√(x) = y

y = a * b 

Now if we were to assign a value to x, like x = 123, then: 

√(123) ~ between √(121) and √(144) closer to √(121) because

123 - 121 = 2

144 - 123 = 21

2 < 21

midpoint

144-121= 23

23/2 = 11.5

and so we know that √(123) ~ is close to 11, a little higher somewhere between 11.1 or 11.09.

y = a * b 

Let's say approximate for y is 11.1

So either a or b have to be less than 11.1

1 * 11.1

2 * 5.55

...

We just have to check the numbers less than the whole number 11 to see if 123 is divisible. 

123 / 2 = WE KNOW THIS WON'T WORK BECAUSE LAST DIGIT ISN'T EVEN.

123 / 3 = WE KNOW THIS WILL WORK BECAUSE 1+2+3= 6, AND 6 IS A MULTIPLE OF 3.

WE CAN STOP HERE.

123 / 4 = WE KNOW THIS WON'T WORK BECAUSE DIDN'T SATISFY "MULTIPLE OF 2"

this is literally just 123 / (2*2) if it helps.

123 / 5 = WE KNOW THIS WON'T WORK BECAUSE LAST DIGIT ISN'T 0 NOR 5.  

123 / 6 = SINCE 123 WAS DIVISIBLE BY 3 BUT NOT 2, IT WILL NOT BE DIVISIBLE BY 6.

this is literally just 123 / (2*3) if it helps.

123 / 7 = WILL NOT WORK; CHECK DIVISIBILITY RULE BY 7 BELOW.

123 / 8 = WILL NOT WORK, NOT DIVISIBILE BY BOTH FACTORS OF 8: 2 NOR 4.

this is literally just 123 / (2*4) if it helps.

123 / 9 = WE KNOW THIS WON'T WORK BECAUSE 1+2+3 = 6, AND 6 IS NOT A FACTOR OF 9.

123 / 10 = WE KNOW THIS WON'T WORK BECAUSE LAST DIGIT ISN'T 0.

123 / 11 = WE KNOW THIS WOULDN'T WORK BECAUSE VALUE IS ONLY 2 AWAY FROM NEAREST MULTIPLE OF 11 WHICH IS 121.

Now we just have to check the first one that works. In this case, check 3.

123/3 = 41

Therefore, 123 is NOT a prime, but a composite number. 

This trick will help you find out till what number you should bother checking (dividing it by) to see if x is prime or composite.. :)

Extras:

EXPLORING THE MULTIPLE OF SEVEN RULE (really cool watch!):

https://youtu.be/UDQjn_-pDSs