Find all $n$ such that $n/d(n) = p$, a prime, where $d(n)$ is the number of positive divisors of $n$
$begingroup$
Let $d(n)$ denote the number of positive divisors of $n$. Find all $n$ such that $n/d(n) = p$, a prime.
I tried this, but only I could get two solutions.
I proceeded like this -
Suppose
$$n = p^r cdot p_1^{r_1} cdot cdots cdot p_k^{r_k}$$
where the $p_i$s are distinct primes.
Given
$$n=pcdot d(n)=p(r+1)(r_1+1)cdots (r_k+1)$$
$$p^r cdots p_1^{r_1}cdot cdots cdot p_k^{r_k}=(r+1)(r_1+1)cdots(r_k+1)$$
If $k=0$, then $p^(r-1)=r+1$. Hence $p=2$ and $r=3$, or $p=3$ and $r=2$. I.e., $n=8$, or $n=9$.
But, when I am assuming $k>0$, I am finding no clue.
For this, I need help.
Thanks in advance!
number-theory prime-numbers divisibility prime-factorization
edited Jan 20 at 14:38
Blue
48.5k870154
$endgroup$
|
show 15 more comments
$begingroup$
Let $d(n)$ denote the number of positive divisors of $n$. Find all $n$ such that $n/d(n) = p$, a prime.
I tried this, but only I could get two solutions.
I proceeded like this -
Suppose
$$n = p^r cdot p_1^{r_1} cdot cdots cdot p_k^{r_k}$$
where the $p_i$s are distinct primes.
Given
$$n=pcdot d(n)=p(r+1)(r_1+1)cdots (r_k+1)$$
$$p^r cdots p_1^{r_1}cdot cdots cdot p_k^{r_k}=(r+1)(r_1+1)cdots(r_k+1)$$
If $k=0$, then $p^(r-1)=r+1$. Hence $p=2$ and $r=3$, or $p=3$ and $r=2$. I.e., $n=8$, or $n=9$.
But, when I am assuming $k>0$, I am finding no clue.
For this, I need help.
Thanks in advance!
number-theory prime-numbers divisibility prime-factorization
edited Jan 20 at 14:38
Blue
48.5k870154
$endgroup$
$begingroup$
Why do you believe the list is sensible? For $n=84$ we get $7$. For $n=88$ we get $11$. For $n=156$ you get $13$. I stopped looking after that,
$endgroup$
– lulu
Jan 20 at 14:33
$begingroup$
then maybe you will get a general form, something like 4k, or 9k+ 1 numbers.... (i am giving an example)
$endgroup$
– user636268
Jan 20 at 14:37
$begingroup$
Seems that $9$ is the only odd such number (I have no proof yet), but there are plenty of even numbers satisfying the condition giving no obvious pattern except that $18$ seems to be the only even example which is not divisible by $4$.
$endgroup$
– Peter
Jan 20 at 14:55
$begingroup$
But this is a standard question from one of the most famous math institues of our country, Indian Statistical Institute.. Obviously there is something, which i am unable to make out
$endgroup$
– user636268
Jan 20 at 14:57
1
$begingroup$
I don't see where you have made a conjecture about numbers of the form $4k,8k$. Note: all numbers of the form $8p$ where $p$ is prime work (trivially).
$endgroup$
– lulu
Jan 20 at 15:41
|
show 15 more comments
$begingroup$
Let $d(n)$ denote the number of positive divisors of $n$. Find all $n$ such that $n/d(n) = p$, a prime.
I tried this, but only I could get two solutions.
I proceeded like this -
Suppose
$$n = p^r cdot p_1^{r_1} cdot cdots cdot p_k^{r_k}$$
where the $p_i$s are distinct primes.
Given
$$n=pcdot d(n)=p(r+1)(r_1+1)cdots (r_k+1)$$
$$p^r cdots p_1^{r_1}cdot cdots cdot p_k^{r_k}=(r+1)(r_1+1)cdots(r_k+1)$$
If $k=0$, then $p^(r-1)=r+1$. Hence $p=2$ and $r=3$, or $p=3$ and $r=2$. I.e., $n=8$, or $n=9$.
But, when I am assuming $k>0$, I am finding no clue.
For this, I need help.
Thanks in advance!
number-theory prime-numbers divisibility prime-factorization
edited Jan 20 at 14:38
Blue
48.5k870154
$endgroup$
Let $d(n)$ denote the number of positive divisors of $n$. Find all $n$ such that $n/d(n) = p$, a prime.
I tried this, but only I could get two solutions.
I proceeded like this -
Suppose
$$n = p^r cdot p_1^{r_1} cdot cdots cdot p_k^{r_k}$$
where the $p_i$s are distinct primes.
Given
$$n=pcdot d(n)=p(r+1)(r_1+1)cdots (r_k+1)$$
$$p^r cdots p_1^{r_1}cdot cdots cdot p_k^{r_k}=(r+1)(r_1+1)cdots(r_k+1)$$
If $k=0$, then $p^(r-1)=r+1$. Hence $p=2$ and $r=3$, or $p=3$ and $r=2$. I.e., $n=8$, or $n=9$.
But, when I am assuming $k>0$, I am finding no clue.
For this, I need help.
Thanks in advance!
number-theory prime-numbers divisibility prime-factorization
number-theory prime-numbers divisibility prime-factorization
edited Jan 20 at 14:38
Blue
48.5k870154
edited Jan 20 at 14:38
Blue
48.5k870154
edited Jan 20 at 14:38
Blue
48.5k870154
edited Jan 20 at 14:38
Blue
48.5k870154
edited Jan 20 at 14:38
Blue
48.5k870154
48.5k870154
asked Jan 20 at 14:28
user636268
$begingroup$
Why do you believe the list is sensible? For $n=84$ we get $7$. For $n=88$ we get $11$. For $n=156$ you get $13$. I stopped looking after that,
$endgroup$
– lulu
Jan 20 at 14:33
$begingroup$
then maybe you will get a general form, something like 4k, or 9k+ 1 numbers.... (i am giving an example)
$endgroup$
– user636268
Jan 20 at 14:37
$begingroup$
Seems that $9$ is the only odd such number (I have no proof yet), but there are plenty of even numbers satisfying the condition giving no obvious pattern except that $18$ seems to be the only even example which is not divisible by $4$.
$endgroup$
– Peter
Jan 20 at 14:55
$begingroup$
But this is a standard question from one of the most famous math institues of our country, Indian Statistical Institute.. Obviously there is something, which i am unable to make out
$endgroup$
– user636268
Jan 20 at 14:57
1
$begingroup$
I don't see where you have made a conjecture about numbers of the form $4k,8k$. Note: all numbers of the form $8p$ where $p$ is prime work (trivially).
$endgroup$
– lulu
Jan 20 at 15:41
|
show 15 more comments
$begingroup$
Why do you believe the list is sensible? For $n=84$ we get $7$. For $n=88$ we get $11$. For $n=156$ you get $13$. I stopped looking after that,
$endgroup$
– lulu
Jan 20 at 14:33
$begingroup$
then maybe you will get a general form, something like 4k, or 9k+ 1 numbers.... (i am giving an example)
$endgroup$
– user636268
Jan 20 at 14:37
$begingroup$
Seems that $9$ is the only odd such number (I have no proof yet), but there are plenty of even numbers satisfying the condition giving no obvious pattern except that $18$ seems to be the only even example which is not divisible by $4$.
$endgroup$
– Peter
Jan 20 at 14:55
$begingroup$
But this is a standard question from one of the most famous math institues of our country, Indian Statistical Institute.. Obviously there is something, which i am unable to make out
$endgroup$
– user636268
Jan 20 at 14:57
1
$begingroup$
I don't see where you have made a conjecture about numbers of the form $4k,8k$. Note: all numbers of the form $8p$ where $p$ is prime work (trivially).
$endgroup$
– lulu
Jan 20 at 15:41
$begingroup$
Why do you believe the list is sensible? For $n=84$ we get $7$. For $n=88$ we get $11$. For $n=156$ you get $13$. I stopped looking after that,
$endgroup$
– lulu
Jan 20 at 14:33
$begingroup$
Why do you believe the list is sensible? For $n=84$ we get $7$. For $n=88$ we get $11$. For $n=156$ you get $13$. I stopped looking after that,
$endgroup$
– lulu
Jan 20 at 14:33
$begingroup$
then maybe you will get a general form, something like 4k, or 9k+ 1 numbers.... (i am giving an example)
$endgroup$
– user636268
Jan 20 at 14:37
$begingroup$
then maybe you will get a general form, something like 4k, or 9k+ 1 numbers.... (i am giving an example)
$endgroup$
– user636268
Jan 20 at 14:37
$begingroup$
Seems that $9$ is the only odd such number (I have no proof yet), but there are plenty of even numbers satisfying the condition giving no obvious pattern except that $18$ seems to be the only even example which is not divisible by $4$.
$endgroup$
– Peter
Jan 20 at 14:55
$begingroup$
Seems that $9$ is the only odd such number (I have no proof yet), but there are plenty of even numbers satisfying the condition giving no obvious pattern except that $18$ seems to be the only even example which is not divisible by $4$.
$endgroup$
– Peter
Jan 20 at 14:55
$begingroup$
But this is a standard question from one of the most famous math institues of our country, Indian Statistical Institute.. Obviously there is something, which i am unable to make out
$endgroup$
– user636268
Jan 20 at 14:57
$begingroup$
But this is a standard question from one of the most famous math institues of our country, Indian Statistical Institute.. Obviously there is something, which i am unable to make out
$endgroup$
– user636268
Jan 20 at 14:57
1
1
$begingroup$
I don't see where you have made a conjecture about numbers of the form $4k,8k$. Note: all numbers of the form $8p$ where $p$ is prime work (trivially).
$endgroup$
– lulu
Jan 20 at 15:41
$begingroup$
I don't see where you have made a conjecture about numbers of the form $4k,8k$. Note: all numbers of the form $8p$ where $p$ is prime work (trivially).
$endgroup$
– lulu
Jan 20 at 15:41
|
show 15 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Observation 1. It is easy to prove (e.g. by induction) that:
$$p^n geq n+1, forall pgeq2, forall ngeq 1$$
Thus
$$p^rcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
p(r+1)cdot color{blue}{(r_1+1)...(r_k+1)}geq
p^r cdot color{blue}{(r_1+1)...(r_k+1)}$$
or
$$p(r+1)geq p^r iff r+1geq p^{r-1}$$
These are the only $(r,p)$ combinations possible $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$ and finally $(1,p), forall p>3, p$ - prime.
Observation 2. Let's check the following case $(1,p), forall p>3, p$ - prime (i.e. $r=1$).
$$pcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
pcdot 2cdot color{blue}{(r_1+1)...(r_k+1)} iff \
p_1^{r_1}cdot ... cdot p_k^{r_k}=
2cdot (r_1+1)...(r_k+1) iff ...$$
or one of the $p_i=2$, for simplicity let's say $p_1=2$.
$$... iff 2^{r_1-1}cdot color{blue}{p_2^{r_2} ... cdot p_k^{r_k}}=
(r_1+1)color{blue}{(r_2+1)...(r_k+1)}geq
2^{r_1-1}cdot color{blue}{(r_2+1)...(r_k+1)}$$
or again
$$(r_1+1)geq 2^{r_1-1}$$
or $r_1 in {1,2,3}$.
This reduces the problem to the following cases $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$.
answered Jan 20 at 16:28
rtybasertybase
11k21533
$endgroup$
$begingroup$
Wrong solution. Check 8,18
$endgroup$
– user636268
Jan 20 at 16:29
$begingroup$
Which solution? This is an observation which narrows down the search space.
$endgroup$
– rtybase
Jan 20 at 16:30
1
$begingroup$
Sorry i didnt read that
$endgroup$
– user636268
Jan 20 at 16:31
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Observation 1. It is easy to prove (e.g. by induction) that:
$$p^n geq n+1, forall pgeq2, forall ngeq 1$$
Thus
$$p^rcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
p(r+1)cdot color{blue}{(r_1+1)...(r_k+1)}geq
p^r cdot color{blue}{(r_1+1)...(r_k+1)}$$
or
$$p(r+1)geq p^r iff r+1geq p^{r-1}$$
These are the only $(r,p)$ combinations possible $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$ and finally $(1,p), forall p>3, p$ - prime.
Observation 2. Let's check the following case $(1,p), forall p>3, p$ - prime (i.e. $r=1$).
$$pcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
pcdot 2cdot color{blue}{(r_1+1)...(r_k+1)} iff \
p_1^{r_1}cdot ... cdot p_k^{r_k}=
2cdot (r_1+1)...(r_k+1) iff ...$$
or one of the $p_i=2$, for simplicity let's say $p_1=2$.
$$... iff 2^{r_1-1}cdot color{blue}{p_2^{r_2} ... cdot p_k^{r_k}}=
(r_1+1)color{blue}{(r_2+1)...(r_k+1)}geq
2^{r_1-1}cdot color{blue}{(r_2+1)...(r_k+1)}$$
or again
$$(r_1+1)geq 2^{r_1-1}$$
or $r_1 in {1,2,3}$.
This reduces the problem to the following cases $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$.
answered Jan 20 at 16:28
rtybasertybase
11k21533
$endgroup$
$begingroup$
Wrong solution. Check 8,18
$endgroup$
– user636268
Jan 20 at 16:29
$begingroup$
Which solution? This is an observation which narrows down the search space.
$endgroup$
– rtybase
Jan 20 at 16:30
1
$begingroup$
Sorry i didnt read that
$endgroup$
– user636268
Jan 20 at 16:31
add a comment |
$begingroup$
Observation 1. It is easy to prove (e.g. by induction) that:
$$p^n geq n+1, forall pgeq2, forall ngeq 1$$
Thus
$$p^rcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
p(r+1)cdot color{blue}{(r_1+1)...(r_k+1)}geq
p^r cdot color{blue}{(r_1+1)...(r_k+1)}$$
or
$$p(r+1)geq p^r iff r+1geq p^{r-1}$$
These are the only $(r,p)$ combinations possible $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$ and finally $(1,p), forall p>3, p$ - prime.
Observation 2. Let's check the following case $(1,p), forall p>3, p$ - prime (i.e. $r=1$).
$$pcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
pcdot 2cdot color{blue}{(r_1+1)...(r_k+1)} iff \
p_1^{r_1}cdot ... cdot p_k^{r_k}=
2cdot (r_1+1)...(r_k+1) iff ...$$
or one of the $p_i=2$, for simplicity let's say $p_1=2$.
$$... iff 2^{r_1-1}cdot color{blue}{p_2^{r_2} ... cdot p_k^{r_k}}=
(r_1+1)color{blue}{(r_2+1)...(r_k+1)}geq
2^{r_1-1}cdot color{blue}{(r_2+1)...(r_k+1)}$$
or again
$$(r_1+1)geq 2^{r_1-1}$$
or $r_1 in {1,2,3}$.
This reduces the problem to the following cases $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$.
answered Jan 20 at 16:28
rtybasertybase
11k21533
$endgroup$
$begingroup$
Wrong solution. Check 8,18
$endgroup$
– user636268
Jan 20 at 16:29
$begingroup$
Which solution? This is an observation which narrows down the search space.
$endgroup$
– rtybase
Jan 20 at 16:30
1
$begingroup$
Sorry i didnt read that
$endgroup$
– user636268
Jan 20 at 16:31
add a comment |
$begingroup$
Observation 1. It is easy to prove (e.g. by induction) that:
$$p^n geq n+1, forall pgeq2, forall ngeq 1$$
Thus
$$p^rcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
p(r+1)cdot color{blue}{(r_1+1)...(r_k+1)}geq
p^r cdot color{blue}{(r_1+1)...(r_k+1)}$$
or
$$p(r+1)geq p^r iff r+1geq p^{r-1}$$
These are the only $(r,p)$ combinations possible $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$ and finally $(1,p), forall p>3, p$ - prime.
Observation 2. Let's check the following case $(1,p), forall p>3, p$ - prime (i.e. $r=1$).
$$pcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
pcdot 2cdot color{blue}{(r_1+1)...(r_k+1)} iff \
p_1^{r_1}cdot ... cdot p_k^{r_k}=
2cdot (r_1+1)...(r_k+1) iff ...$$
or one of the $p_i=2$, for simplicity let's say $p_1=2$.
$$... iff 2^{r_1-1}cdot color{blue}{p_2^{r_2} ... cdot p_k^{r_k}}=
(r_1+1)color{blue}{(r_2+1)...(r_k+1)}geq
2^{r_1-1}cdot color{blue}{(r_2+1)...(r_k+1)}$$
or again
$$(r_1+1)geq 2^{r_1-1}$$
or $r_1 in {1,2,3}$.
This reduces the problem to the following cases $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$.
answered Jan 20 at 16:28
rtybasertybase
11k21533
$endgroup$
Observation 1. It is easy to prove (e.g. by induction) that:
$$p^n geq n+1, forall pgeq2, forall ngeq 1$$
Thus
$$p^rcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
p(r+1)cdot color{blue}{(r_1+1)...(r_k+1)}geq
p^r cdot color{blue}{(r_1+1)...(r_k+1)}$$
or
$$p(r+1)geq p^r iff r+1geq p^{r-1}$$
These are the only $(r,p)$ combinations possible $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$ and finally $(1,p), forall p>3, p$ - prime.
Observation 2. Let's check the following case $(1,p), forall p>3, p$ - prime (i.e. $r=1$).
$$pcdot color{blue}{p_1^{r_1}cdot ... cdot p_k^{r_k}}=
pcdot 2cdot color{blue}{(r_1+1)...(r_k+1)} iff \
p_1^{r_1}cdot ... cdot p_k^{r_k}=
2cdot (r_1+1)...(r_k+1) iff ...$$
or one of the $p_i=2$, for simplicity let's say $p_1=2$.
$$... iff 2^{r_1-1}cdot color{blue}{p_2^{r_2} ... cdot p_k^{r_k}}=
(r_1+1)color{blue}{(r_2+1)...(r_k+1)}geq
2^{r_1-1}cdot color{blue}{(r_2+1)...(r_k+1)}$$
or again
$$(r_1+1)geq 2^{r_1-1}$$
or $r_1 in {1,2,3}$.
This reduces the problem to the following cases $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$.
answered Jan 20 at 16:28
rtybasertybase
11k21533
edited Jan 20 at 17:18
answered Jan 20 at 16:28
rtybasertybase
11k21533
answered Jan 20 at 16:28
rtybasertybase
11k21533
answered Jan 20 at 16:28
rtybasertybase
11k21533
11k21533
$begingroup$
Wrong solution. Check 8,18
$endgroup$
– user636268
Jan 20 at 16:29
$begingroup$
Which solution? This is an observation which narrows down the search space.
$endgroup$
– rtybase
Jan 20 at 16:30
1
$begingroup$
Sorry i didnt read that
$endgroup$
– user636268
Jan 20 at 16:31
add a comment |
$begingroup$
Wrong solution. Check 8,18
$endgroup$
– user636268
Jan 20 at 16:29
$begingroup$
Which solution? This is an observation which narrows down the search space.
$endgroup$
– rtybase
Jan 20 at 16:30
1
$begingroup$
Sorry i didnt read that
$endgroup$
– user636268
Jan 20 at 16:31
$begingroup$
Wrong solution. Check 8,18
$endgroup$
– user636268
Jan 20 at 16:29
$begingroup$
Wrong solution. Check 8,18
$endgroup$
– user636268
Jan 20 at 16:29
$begingroup$
Which solution? This is an observation which narrows down the search space.
$endgroup$
– rtybase
Jan 20 at 16:30
$begingroup$
Which solution? This is an observation which narrows down the search space.
$endgroup$
– rtybase
Jan 20 at 16:30
1
1
$begingroup$
Sorry i didnt read that
$endgroup$
– user636268
Jan 20 at 16:31
$begingroup$
Sorry i didnt read that
$endgroup$
– user636268
Jan 20 at 16:31
add a comment |
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$begingroup$
Why do you believe the list is sensible? For $n=84$ we get $7$. For $n=88$ we get $11$. For $n=156$ you get $13$. I stopped looking after that,
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– lulu
Jan 20 at 14:33
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then maybe you will get a general form, something like 4k, or 9k+ 1 numbers.... (i am giving an example)
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– user636268
Jan 20 at 14:37
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Seems that $9$ is the only odd such number (I have no proof yet), but there are plenty of even numbers satisfying the condition giving no obvious pattern except that $18$ seems to be the only even example which is not divisible by $4$.
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– Peter
Jan 20 at 14:55
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But this is a standard question from one of the most famous math institues of our country, Indian Statistical Institute.. Obviously there is something, which i am unable to make out
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– user636268
Jan 20 at 14:57
1
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I don't see where you have made a conjecture about numbers of the form $4k,8k$. Note: all numbers of the form $8p$ where $p$ is prime work (trivially).
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– lulu
Jan 20 at 15:41