Discuss this mathematical assumption

IN General N = 2^p * 3^q * 5^r * 7^s.....so on Total no of factors are = (p+1) * (q+1) * (r+1) * (s+1)...... Does anyone know about this theorem...??? If yes then explain it briefly..???

#Algebra

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered 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"

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

Let $n=p_{1}^{a_{1}}p_{2}^{a_{2}}.....p_{k}^{a_{k}}$

Now, suppose you have $k$ boxes labelled $1,2,...,k$ .Suppose the $ith$ box contains $a_{i}$ identical objects of $p_{i}$ type.

Now,you want to form all possible combinations(of any length) with those objects.

Now, consider the $ith$ box.You can include $1$ $p_{i}$ or $2$ $p_{i}'s$ or...or $a_{i}$ $p_{i}'s$ or no $p_{i}$ . So there are $a_{i}+1$ ways of choosing the objects of $p_{i}$ type, $i=1,2....,k$

Hence, in total there will be $(a_{1}+1)(a_{2}+1)....(a_{k}+1)$ combinations.

Now, see that this counting is similar to counting the number of divisors.

Souryajit Roy - 6 years, 10 months ago

understood buddy...!!! Now i gotcha it completely..!! thanks @Souryajit Roy

Rahul Jain - 6 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}`	$\boxed{123}$