Smallest Positive Integer N for which tow(n)=6 is True

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Homework Help Overview

The discussion revolves around finding the smallest positive integer \( n \) such that \( \tau(n) = 6 \), where \( \tau(n) \) represents the number of divisors of \( n \). Participants explore the relationship between the prime factorization of \( n \) and the divisor function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the form of \( n \) based on the equation \( \tau(n) = (1 + a_1)(1 + a_2) \cdots (1 + a_k) \) and consider examples of integers that satisfy this condition. There is an emphasis on generating examples and making conjectures based on those examples.

Discussion Status

Several participants have proposed potential forms for \( n \), such as \( p^5 \) and \( p^2p \). There is an ongoing exploration of whether these forms are exhaustive and how to prove the smallest \( n \) that meets the criteria. Some participants express uncertainty about the next steps in the reasoning process.

Contextual Notes

Participants are working within the constraints of the problem, focusing on the divisor function and its implications for the structure of \( n \). There is a recognition that proving the conjectured smallest \( n \) is a necessary next step in the discussion.

kathrynag
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Find the form of all n in N satifying tow(n)=6. (sorry don't know how to write this Tex). What is the smallest positive integer n for which this is true?

tow(n)=6 so 6=(1+a1)(1+a2)----(1+ak) where n=p1^a1p2^a2---pk^ak
 
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[tex]\tau[/tex](n)=6
We have [tex]\tau[/tex](n)=(1+[tex]\alpha_{1}[/tex])(1+[tex]\alpha_{2}[/tex])...(1+[tex]\alpha_{k}[/tex]) for k=[tex]p^{\alpha_{1}}p^{\alpha_{2}}...p^{\alpha_{k}}[/tex]
So 6=(1+[tex]\alpha_{1}[/tex])(1+[tex]\alpha_{2}[/tex])...(1+[tex]\alpha_{k}[/tex])
I just don't knw about the next step...
 
There is no standard next step. You have to think about it. Can you find any examples of integers n such that tau(n)=6? You should be able to find a lot. Use those examples to make a guess. Now try and find an argument to prove your guess is correct. This is really a basic creative math question. There's no cookbook method.
 
Well 2^5 works
(5+1)=6
3^1*7^2
(1+1)(2+1)=2*3=6

So n=p^5 or n=p^2p^1
 
n=2^2*3=4*3=12 smallest possible n?
 
kathrynag said:
Well 2^5 works
(5+1)=6
3^1*7^2
(1+1)(2+1)=2*3=6

So n=p^5 or n=p^2p^1

Now that's a good start. Yes, 32=2^5 works. And 3^1*7^2=147 works. Any tentative conclusions so far?
 
I think it's something of the form n=p^5 or n=p^2*p
I think those are the only forms that work
 
kathrynag said:
I think it's something of the form n=p^5 or n=p^2*p
I think those are the only forms that work

I would write that as p^5 or p1^2*p2. There are two different primes there. And sure, you can only factor 6 as 6*1 or 2*3. Now keep going.
 
kathrynag said:
n=2^2*3=4*3=12 smallest possible n?

Sure it is. Now can you prove it? I checked your page once and you want to be a mathematician, right? Mathematicians prove things. This an easy one. Try an omit the '?', by giving me an argument why it's true.
 
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