Smallest value of n given its sixth divisor

[Mr Davis 97]

Homework Statement

All of the divisors of $n$ are in increasing order: $1=d_1 < d_2 < \dots < d_t = n$. We know that $d_6=15$. What is the smallest possible value of $n$?

Homework Equations

The Attempt at a Solution

Here is my reasoning. We have the chain $1 < d_2 < d_3 < d_4 < d_5 < 15 < n$, where we make $15$ the largest factor that's not $n$. Since $15~|~n$, we have that $5~|~n$ and $3~|~n$. Hence, we have to put $3$ and $5$ somewhere. The minimal sequence is then $1 < 2 < 3 < 4 < 5 < 15 < n$, so $n=2\cdot 3\cdot 4 \cdot 5 \cdot 15 =1800$

 

[fresh_42]

Why not $60$? $1,2,3,4,5,15 \,|\,60$ and $t=6$.

Edit: Doesn't work, since $6,10$ and $12$ are not listed. But this applies to your example, too.

 

[Mr Davis 97]

Why not $60$? $1,2,3,4,5,15 \,|\,60$ and $t=6$.

Edit: Doesn't work, since $6,10$ and $12$ are not listed. But this applies to your example, too.

I'm not seeing what you mean. Where does my logic go bad?

 

[fresh_42]

I'm not seeing what you mean. Where does my logic go bad?

All of the divisors of $n$ are in increasing order: $1=d_1 < d_2 < \dots < d_t = n$. We know that $d_6=15$.

So in case of $n=60$ as well as $n=1800$ we have $1=d_1<2=d_2<3=d_3<4=d_4<5=d_5<6=d_6<10=d_7<12=d_8<15=d_9\neq d_6$.
In case of $n=1800$ the list also includes $d_7=8$ and $d_8=9$ which shifts the indices even further.

 

[Mr Davis 97]

So in case of $n=60$ as well as $n=1800$ we have $1=d_1<2=d_2<3=d_3<4=d_4<5=d_5<6=d_6<10=d_7<12=d_8<15=d_9\neq d_6$.
In case of $n=1800$ the list also includes $d_7=8$ and $d_8=9$ which shifts the indices even further.

I see. Any idea on how to proceed then?

 

[fresh_42]

I tried $2,3,4$. But then we have $2\cdot 3=6$ and $2\cdot 4=8$ and $3\cdot 4=12$, too. These are already six numbers smaller than $15$. Besides that, with $15$ we have an automatic divisor $5$. So $1,3,5,15$ are set. Etc...

 

[Mr Davis 97]

What about if we have 1,3,4,5,12,15?

 

[fresh_42]

You cannot have $4$ without $2$.

 

[Mr Davis 97]

You cannot have $4$ without $2$.

So then what about 1,3,5,7,11,15?

 

[WWGD]

So then what about 1,3,5,7,11,15?

Then you need a 9: 15*3=45. I think you need to use pairwise relatively-prme numbers.

 

[fresh_42]

So then what about 1,3,5,7,11,15?

Looks good, but I'm not sure whether this is already the minimal solution.

 

[WWGD]

Looks good, but I'm not sure whether this is already the minimal solution.

Look at my post #10. I think he needs to use pairwise relatively prime.

 

[fresh_42]

Look at my post #10. I think he needs to use pairwise relatively prime.

That wasn't a valid objection, because we do not have to use a prime twice. $3\,|\,n$ and $15\,|\,n$ doesn't make $9$ a divisor. I even think that we can substitute $11$ by $9$ in his example.

 

[haruspex]

I see. Any idea on how to proceed then?

You know 3 and 5 must appear in the list.
Next question should be whether 2 can appear. If it does, what else must come before 15?