Prime root

Today, I will invent something (ridiculous).

I will invent the prime root (unofficially).

You have to take a number's prime factors and add them.

And Then take that number's prime factors and add them.

Continue doing that until you get a prime number. That is (unofficially) the prime root.

Say for example we have 15. The prime factors are 3 and 5. Add them and we get 8. The prime factors are 2 but there are 3 of them, So 2+2+2=6. The prime factors of it is 2 and 3. Add them and we get 5. STOP! We got a prime number. So the prime root of 15 is 5 (unofficially).

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

How you got this step :

The prime factors are 2 but there are 3 of them, So 2+2+2=6.

Ram Mohith - 2 years, 10 months ago

Because 2^3 =8 and we can just multiply 2 by 3 to get the sum of the factors (unofficially).

Mohammad Farhat - 2 years, 10 months ago

Remember! This is just nonsense

Mohammad Farhat - 2 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$