Waste less time on Facebook — follow Brilliant.
×

Number Theory

Is there a theorem that describes how many prime factors a number has? And does this this theorem have a solution on how to find them?

Note by Gabriel Kong
4 years, 6 months ago

No vote yet
2 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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

Sort by:

Top Newest

You can review Divisors of an Integer.

Calvin Lin Staff - 4 years, 6 months ago

Log in to reply

If a number \(N= p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{m}^{a_{m}} \), for primes \(p_{1},p_{2}\cdots p_{m}\) and natural numbers \(a_{1},a_{2}\cdots a_{m}\), then the number of factors that \(N\) has is denoted by \(d(N)\) and is given by the formulae: \(d(N)= (a_{1}+1) (a_{2}+1)\cdots (a_{m}+1)\).

And the proof is combinatoric. If \(S= p_{1}^{b_{1}}p_{2}^{b_{2}}\cdots p_{m}^{b_{m}}\) is a factor of \(N\) , where \(b_{1},b_{2}\cdots b_{m}\) are natural, then each \(b_{i} \leq a_{i}\) for all indices \(i\). So ecah \(b_{i}\) can be chosen in \((a_{i}+1)\) ways, so the total number of such \(S\) is \(d(n)\).

Shourya Pandey - 4 years, 6 months ago

Log in to reply

Thanks a lot! :D

Gabriel Kong - 4 years, 6 months ago

Log in to reply

What shourya pointed out was the number FACTORS of a number, not the number of PRIME factors.

Harrison Lian - 4 years, 6 months ago

Log in to reply

http://en.wikipedia.org/wiki/Divisor_function and http://mathworld.wolfram.com/Divisor.html

Lab Bhattacharjee - 4 years, 6 months ago

Log in to reply

wow! thanks.

Gabriel Kong - 4 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...