fix n,k belongs to N and consider p1 and p2, both primes. find how many divisors

b= p1^k p2^n has and prove your result.

conjecture and provide some justification how many divisors there are for a number that can be represented as the product of m arbitrary primes, each raised to a different power.

Question

fix n,k belongs to N and consider p1 and p2, both primes. find how many divisors

b= p1^k p2^n has and prove your result.

conjecture and provide some justification how many divisors there are for a number that can be represented as the product of m arbitrary primes, each raised to a different power.

ganeshie8 · Answer

Hey!

anonymous · Answer

what's up?

ganeshie8 · Answer

have you figured out the solution ?

anonymous · Answer

not yet

ganeshie8 · Answer

tried anything ?

ganeshie8 · Answer

Let $b = 2^1*3^2$

is it easy to list all the positive divisors of $b$ ?

anonymous · Answer

i don't know how to start

ganeshie8 · Answer

what are the divisors of $18$ ?

ganeshie8 · Answer

does 6 divide 18 ?

anonymous · Answer

1, 2, 3, 6, 9, 18

anonymous · Answer

ok how can i know the form in general please?

anonymous · Answer

i tried some other products and i got 6 divisors each time!

anonymous · Answer

oh no for 2^3 * 3=24, it has 8 divisors which are 	1, 2, 3, 4, 6, 8, 12, 24

anonymous · Answer

i cannot figure out the formula to know how many divisors?

ganeshie8 · Answer

Excellent! you're very close to a formula...

ganeshie8 · Answer

Consider $b = 2^3*3$

ganeshie8 · Answer

The key thing to observe here is that any divisor of $b$ will be of form $2^a*3^b$

where $0\le a\le 3$ and $0\le b\le 1$

ganeshie8 · Answer

for example,
a = 2, b = 1 gives

$2^2*3^1 = 12$ as one divisor

ganeshie8 · Answer

$0\le a\le 3$ 

how many choices are there for $a$ ?

anonymous · Answer

4

anonymous · Answer

2

ganeshie8 · Answer

good
$0\le b\le 1$
 how many choices are there for $b$ ?

ganeshie8 · Answer

Yes, 4*2 = ?

anonymous · Answer

8

ganeshie8 · Answer

so, there are 8 divisors of 2^3*3

ganeshie8 · Answer

you can choose "2" in "3+1" ways
you can choose "3" in "1+1" ways

overall, number of divisors is given by (3+1)(1+1)

ganeshie8 · Answer

lets do another example

ganeshie8 · Answer

Find the number of divisors of $2^2*5^2$

anonymous · Answer

0≤a≤2 3 choices
0≤b≤2 3 choices
3*3=9

anonymous · Answer

so the number of divisors is (a+1)*(b+1)

anonymous · Answer

in general  (a+1)*(b+1)...(n+1)

anonymous · Answer

cool!
can you help me please how to prove that ?

anonymous · Answer

oh i got a question
	3^2 * 5=45 has 5 divisors: 	1, 3, 5, 15, 45 	
but if i did the formula
0≤a≤2    3 choices
 0≤b≤1   2 choices
3*2=6 not 5
why is it different?

ganeshie8 · Answer

looks you have missed one divisor

ganeshie8 · Answer

9 ?

anonymous · Answer

oh yeah thanks