# Factors of a number

1. Apr 8, 2010

### jeedoubts

1. The problem statement, all variables and given/known data
N has total 105 factors including 1 and N. then find :
a) the total no of odd factors between 1 and N.
b) if the total number of divisors of N which are multiple of 36 are 45.then the total no of odd factors between 1 and N.
c)the number of ways in which N can be resolved into 2 factors which are relatively prime to each other is equal to 4,then the total no of odd factors between 1 and N.
d) if the total number of divisors of N which are multiple of 216 are 48,then the total no of odd factors between 1 and N.

3. The attempt at a solution
total number of divisors of a number a^n1*b^n2*c^n3 is equal to (n1+1)(n2+1)(n3+1)

Last edited: Apr 9, 2010
2. Apr 9, 2010

### tiny-tim

Hi jeedoubts!

(try using the X2 and X2 tags just above the Reply box )
Well, 105 = 3*5*7, so how does that help you with a) ?

3. Apr 9, 2010

### jeedoubts

we can assume N to be a2b4c6
and if check if either of a or b or c is even or not so in all 4 answers are possible.... Ithink in that way plz tell if i'm correct...

4. Apr 9, 2010

### tiny-tim

a b and c must be primes, so only one (or zero) of them can be even …

but there doesn't seem to be enough information to answer a)

5. Apr 9, 2010

### jeedoubts

what about parts b,c and d??

6. Apr 9, 2010

### tiny-tim

Can you check the question?

Are you sure it doesn't start with 135 (rather than 105) ?

7. Apr 9, 2010

### jeedoubts

it is 105 i checked it.