Find minimum integer $n$ and $m$ such that $36^n = 16^m, n in mathbb{Z}, m in mathbb{Z}$?
$begingroup$
How would I go about finding the minimum $n$ and $m$ such that $36^n = 16^m, n in mathbb{Z}, m in mathbb{Z}$?
The practical reason for this is that I would like to find the minimum number of hex characters such that I can evenly convert it to base 36.
least-common-multiple
asked Dec 23 '18 at 17:51
RoxyRoxy
1061
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add a comment |
$begingroup$
How would I go about finding the minimum $n$ and $m$ such that $36^n = 16^m, n in mathbb{Z}, m in mathbb{Z}$?
The practical reason for this is that I would like to find the minimum number of hex characters such that I can evenly convert it to base 36.
least-common-multiple
asked Dec 23 '18 at 17:51
RoxyRoxy
1061
$endgroup$
2
$begingroup$
You might post another question explaining better, what you are looking after. Well for your question, the answer is 0.
$endgroup$
– Love Invariants
Dec 23 '18 at 18:07
add a comment |
$begingroup$
How would I go about finding the minimum $n$ and $m$ such that $36^n = 16^m, n in mathbb{Z}, m in mathbb{Z}$?
The practical reason for this is that I would like to find the minimum number of hex characters such that I can evenly convert it to base 36.
least-common-multiple
asked Dec 23 '18 at 17:51
RoxyRoxy
1061
$endgroup$
How would I go about finding the minimum $n$ and $m$ such that $36^n = 16^m, n in mathbb{Z}, m in mathbb{Z}$?
The practical reason for this is that I would like to find the minimum number of hex characters such that I can evenly convert it to base 36.
least-common-multiple
least-common-multiple
asked Dec 23 '18 at 17:51
RoxyRoxy
1061
asked Dec 23 '18 at 17:51
RoxyRoxy
1061
asked Dec 23 '18 at 17:51
RoxyRoxy
1061
asked Dec 23 '18 at 17:51
RoxyRoxy
1061
asked Dec 23 '18 at 17:51
RoxyRoxy
1061
1061
2
$begingroup$
You might post another question explaining better, what you are looking after. Well for your question, the answer is 0.
$endgroup$
– Love Invariants
Dec 23 '18 at 18:07
add a comment |
2
$begingroup$
You might post another question explaining better, what you are looking after. Well for your question, the answer is 0.
$endgroup$
– Love Invariants
Dec 23 '18 at 18:07
2
2
$begingroup$
You might post another question explaining better, what you are looking after. Well for your question, the answer is 0.
$endgroup$
– Love Invariants
Dec 23 '18 at 18:07
$begingroup$
You might post another question explaining better, what you are looking after. Well for your question, the answer is 0.
$endgroup$
– Love Invariants
Dec 23 '18 at 18:07
add a comment |
2 Answers
2
active
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$begingroup$
If $n,m , geq 1$ then $3 | 36^n = 2^{4m}$.
answered Dec 23 '18 at 17:56
Lucas HenriqueLucas Henrique
1,026414
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add a comment |
$begingroup$
This is equivalent to finding integer solutions for $3^{2n}=2^{4m-2n}$
Obviously, the only integer solutions would be $n=m=0$
answered Dec 23 '18 at 18:01
Sik Feng CheongSik Feng Cheong
1579
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add a comment |
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2 Answers
2
active
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2 Answers
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active
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$begingroup$
If $n,m , geq 1$ then $3 | 36^n = 2^{4m}$.
answered Dec 23 '18 at 17:56
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
add a comment |
$begingroup$
If $n,m , geq 1$ then $3 | 36^n = 2^{4m}$.
answered Dec 23 '18 at 17:56
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
add a comment |
$begingroup$
If $n,m , geq 1$ then $3 | 36^n = 2^{4m}$.
answered Dec 23 '18 at 17:56
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
If $n,m , geq 1$ then $3 | 36^n = 2^{4m}$.
answered Dec 23 '18 at 17:56
Lucas HenriqueLucas Henrique
1,026414
answered Dec 23 '18 at 17:56
Lucas HenriqueLucas Henrique
1,026414
answered Dec 23 '18 at 17:56
Lucas HenriqueLucas Henrique
1,026414
answered Dec 23 '18 at 17:56
Lucas HenriqueLucas Henrique
1,026414
1,026414
add a comment |
add a comment |
$begingroup$
This is equivalent to finding integer solutions for $3^{2n}=2^{4m-2n}$
Obviously, the only integer solutions would be $n=m=0$
answered Dec 23 '18 at 18:01
Sik Feng CheongSik Feng Cheong
1579
$endgroup$
add a comment |
$begingroup$
This is equivalent to finding integer solutions for $3^{2n}=2^{4m-2n}$
Obviously, the only integer solutions would be $n=m=0$
answered Dec 23 '18 at 18:01
Sik Feng CheongSik Feng Cheong
1579
$endgroup$
add a comment |
$begingroup$
This is equivalent to finding integer solutions for $3^{2n}=2^{4m-2n}$
Obviously, the only integer solutions would be $n=m=0$
answered Dec 23 '18 at 18:01
Sik Feng CheongSik Feng Cheong
1579
$endgroup$
This is equivalent to finding integer solutions for $3^{2n}=2^{4m-2n}$
Obviously, the only integer solutions would be $n=m=0$
answered Dec 23 '18 at 18:01
Sik Feng CheongSik Feng Cheong
1579
answered Dec 23 '18 at 18:01
Sik Feng CheongSik Feng Cheong
1579
answered Dec 23 '18 at 18:01
Sik Feng CheongSik Feng Cheong
1579
answered Dec 23 '18 at 18:01
Sik Feng CheongSik Feng Cheong
1579
1579
add a comment |
add a comment |
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You might post another question explaining better, what you are looking after. Well for your question, the answer is 0.
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– Love Invariants
Dec 23 '18 at 18:07