Find # Odd Factors of a Number: 1 to N

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The discussion revolves around finding the number of odd factors of a number N, given that N has 105 total factors. Participants explore various scenarios, including the number of divisors that are multiples of 36 and 216, and the conditions under which N can be expressed as a product of relatively prime factors. The conversation highlights the formula for calculating the total number of divisors based on prime factorization and raises questions about the sufficiency of the provided information for solving the problems. Ultimately, there is uncertainty regarding the calculations and the interpretation of the problem statement. The thread emphasizes the complexity of determining odd factors without additional context or clarity.
jeedoubts
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1. Homework Statement
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)
 
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Hi jeedoubts! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
jeedoubts said:
N has total 105 factors including 1 and N. then find :
a) the total no of odd factors between 1 and N.
…
total number of divisors of a number a^n1*b^n2*c^n3 is equal to (n1+1)(n2+1)(n3+1)

Well, 105 = 3*5*7, so how does that help you with a) ? :smile:
 
tiny-tim said:
Hi jeedoubts! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)


Well, 105 = 3*5*7, so how does that help you with a) ? :smile:


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 please tell if I'm correct...
 
jeedoubts said:
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 please tell if I'm correct...

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) :confused:
 
tiny-tim said:
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) :confused:


what about parts b,c and d??:confused::confused:
 
Can you check the question?

Are you sure it doesn't start with 135 (rather than 105) ?
 
tiny-tim said:
Can you check the question?

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

it is 105 i checked it.
 

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