Find # Odd Factors of a Number: 1 to N

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Homework Help Overview

The problem involves determining the number of odd factors of a number N, given that N has a total of 105 factors. The discussion includes various conditions related to the divisors of N, such as those that are multiples of 36 and 216, as well as the number of ways N can be expressed as a product of relatively prime factors.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the total number of factors and the prime factorization of N. There are attempts to express N in terms of its prime factors and to analyze the implications of having certain divisors. Questions arise regarding the sufficiency of information provided to solve the problem, particularly for part a).

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions about the factorization of N and discussing the implications of the given conditions. Some express uncertainty about the information provided, particularly regarding the total number of factors, and seek clarification on the problem statement.

Contextual Notes

There is a discussion about whether the total number of factors should be 135 instead of 105, indicating potential confusion or misinterpretation of the problem statement.

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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