Question

\(mn = 2010020020010002 = 2^1×7^5×11^5×13^5\) How many possible (m,n) ordered pairs are there?

Answer 1

I don't know how counting works, but it's definitely to do something with separating the list of factors into two different sets.

Answer 2

I think it is just a Combinatorics Question.

Answer 3

@ganeshie8

Answer 4

Yes it is.

For example if you have to do \(mn = 20\) then factorize \(2 \times 2 \times 5\). Find the number of ways to separate the set \(\{2,2,5\}\) into two individual unique sets.

Answer 5

\(mn = 2010020020010002 = 2^1×7^5×11^5×13^5\)


number of factors = \((1+1)(5+1)(5+1)(5+1)\)

Answer 6

you can find the possible ordered pairs ^

Answer 7

How did you calculate that?
And is, number of factors = possible ordered pairs?

Answer 8

{2,2} and {5}
{2,5} and {2}

also include
{2,2,5}, {1}
{1}, {2,2,5}

Answer 9

@ParthKohli That was for 20
How do i do it for: 2010020020010002 ?

Answer 10

generalize a method by looking at smaller numbers. I did a similar problem but that was long ago.

with that, I leave you. you're in safe hands though (ganesh!)

Answer 11

Oh wait a sec you're right !
total number of ordered pairs = total number of factors

Answer 12

Ganesh,
Why is it (432)^2.
And how did you get 432 ??

Answer 13

(1+1)(5+1)(5+1)(5+1) = 432

Answer 14

\((1+1)(5+1)(5+1)(5+1)\) should be the final answer

Answer 15

Ok, how did you get (1+1)(5+1)(5+1)(5+1) ? T.T

Answer 16

\(mn = 2010020020010002 = 2^1×7^5×11^5×13^5\)

Answer 17

Yes.

Answer 18

\(m\) can hav 2^0 or 2^1 in its prime factorization : (1+1) ways

Answer 19

\(m\) can hav 7^0 or 7^1 or 7^2 or 7^3 or 7^4 or 7^5 in its prime factorization : (5+1) ways

Answer 20

\(\cdots\)

Answer 21

What do you actually mean by that?

Answer 22

(5+1) ways ?

Answer 23

wat if
mn= (17^3*11)(...) ??

Answer 24

yes the prime power for 7, can occur in (5+1) = 6 ways in m

Answer 25

Okay. So why is

N(m,n) = 432 ?

Answer 26

In general :

if \(N = p_1^{a_1}p_2^{a_2}\cdots p_ra^r\) is the prime factorization of \(N\)
then total number of factors = \((a_1+1)(a_2+1)\cdots (a_r+1)\)

Answer 27

right !
forgot about that

Answer 28

choosing \(m\), uniquely determines the other factor\(n\),
so total number of ordered pairs equals total number of factors

Answer 29

How do I actually understand that formula?
Any link?

And I understand how No. of ordered pairs = No. of factors! :D

Answer 30

let me try explaining u quick before giving the link

Answer 31

\(N = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r} \)

Answer 32

First observation : \(p_1, ~p_2, .... p_r\) are the factors of \(N\) as they divide \(N\) evenly

Answer 33

Yes.

Answer 34

Next, think at finding how many ways the `power` of a specific prime factor can occur in a factor.

Answer 35

in our present problem, 7^5 is there in factorization of N
so, 7^0 or 7^1 or 7^2 or 7^2 or 7^3 or 7^4 or 7^5 are also factors of \(N\)

Answer 36

For 7^5

1 * mn
7 * x
49 * y
343 * z
7^4 * (some number)
7^5 * (2*13^5*11^5)

For a specific prime factor.

Answer 37

Okay. Yes. Then?

Answer 38

yes a power of 7 can occur in 6 ways

Answer 39

next look at other primes

Answer 40

p^r can occur in (r+1) ways

Answer 41

multiply them all to get all the prime power combinations

Answer 42

its like this :-
u have 20 number , first u wanna chose one # =m or 2 # =m or 3 num =m
or 4#=5 ...... or 20#=m
:3

Answer 43

Yes. All the prime factors can occur in (r+1) ways.
What about their combinations now?

Answer 44

thats the fundamental principal of counting

Answer 45

if 7^5 can be chosen in 6 ways
andd 11^5 can be chosen in 6 ways
then you can choose both of them in 6*6 ways

Answer 46

Oh I think I get it now!
"This is when the concept clicks". *__*

Thank you very much @ganeshie8 !

Answer 47

\(a_1 + 1\) ways for \(p_1^{a_1}\)
\(a_2 + 1\) ways for \(p_2^{a_2}\)
\(\cdots \)
\(a_r + 1\) ways for \(p_r^{a_r}\)


so total combinations =
\((a_1+1)(a_2+1)\cdots (a_r+1)\)

Answer 48

@ AkashdeepDeb
i couldn't correct early it should
be (1+1)(5+1)(5+1)(5+1)/2 = 216

Answer 49

For e.g 20 =2^2*5^1
So the total number of factors are (2+1)(1+1)=6 (1,2,4,5,10,20)
but the number of ways in which 20 can be expressed is 6/2 = 3
hence we have (1,20) ,(4,5) and (2,10)

Answer 50

I think ganeshie explained it well, but here is what I thought:
they want the number of ordered pairs which means (for example) (1,2) is different from (2,1), and all we have to count are the number of unique m's we can form

m can either use or not use the factor of 2: use either 1 or 2. That is 2 choices
m can use from 0 up to 5 sevens, so multiply by 6 (number of choices)
same for 11 and 13
we get 2*6*6*6= 432

Answer 51

here i don't think so to consider (1,2) and (2,1) to be different pairs...

Answer 52

The question states:
How many possible (m,n) ordered pairs are there?

*ordered* pair means the order of m and n matter, i.e. (m,n) is different from (n,m)