Question

Pick a number greater than 20 and less than 150 that is not a prime number. Give all of its prime factors. How does this help you find all the possible factors of the number? (first right answer gets medal) :))

Answer 1

nice exercise :) prime numbers are like the building blocks of all the numbers, they can be used to generate ANY number..

pick your fav number between 20 and 150

Answer 2

i think so... @cheyennestaggs whats your fav number in the given range ? :)

Answer 3

35 @ganeshie8

Answer 4

good :) what are the prime factors of 35 ?

Answer 5

i think 5 and 7

Answer 6

yes !

Answer 7

\[\large 35 = 5^1 \times 7^1\]

Answer 8

since 35 is uniquely made up of these factors, you must have one of these prime numbers in all the factors of 35

Answer 9

you have one 5 and one 7 in the prime factorization of 35

this means you can have : 5^0 or 5^1
and 7^0 or 7^1

in a factor of 5

Answer 10

so total possible factors will be below :

```
5^0 7^0 = 1
5^0 7^1 = 7
5^1 7^0 = 5
5^1 7^1 = 35
```

Answer 11

So there are exactly 4 factors of 35 -
let me know if something is not clear

Answer 12

and indeed this is true
http://www.wolframalpha.com/input/?i=factors+of+35

Answer 13

I got it. Thank you! :))

Answer 14

we can generalize this, and it may be exciting to know that there is an interesting number theoretic function which says the exact same thing : \(\large \tau \)

Answer 15

its called the `divisor function`
wana see how to find the number of divisors a number has ? :)

Answer 16

yes :)

Answer 17

lets make it a theorem :
say \(\large N\) is a number with below prime factorization
\[\large N = p_1^{e_1}p_2^{e_2}p_3^{e_3}\cdots p_r^{e_r}\]

then, number of factors of \(N\) can be given as below :
\[\large \text{# of factors of N} = \tau = (e_1+1)(e_2+1)(e_3+1)\cdots(e_r+1)\]

Answer 18

the proof is trivial -
\(p_1\) can have \(0, 1, 2\cdots e_1\) powers : \(e_1 + 1\)
\(p_2\) can have \(0, 1, 2\cdots e_2\) powers : \(e_2 + 1\)
\(p_3\) can have \(0, 1, 2\cdots e_3\) powers : \(e_3 + 1\)
\(\cdots \)
\(p_r\) can have \(0, 1, 2\cdots e_r\) powers : \(e_r + 1\)

Answer 19

So total possible combinations will be :
\((e_1+1)(e_2+1)(e_3+1)\cdots(e_r+1)\)

Answer 20

see if that makes more or less sense...

Answer 21

For example : how many factors does the number 36 have ?

Answer 22

well we start by writing out its prime factorization :
\[\large 36 = 4\times 9 = 2^2 3^2\]

so, number of factors = \(\large (2+1)(2+1) = 9\)

Answer 23

see if that makes more or less sense..