Divisibility Check in different Moduli
So I'm working on modular arithmetic and encountered the following problem:
Assume that $(a cdot b) mod m = c$ and that I know $c$. Can I still check for divisibility of $c$ by $a$? A quick example tells me that checking whether $c mod a = 0$ does not work:
$((29*200) mod 512) mod 29 = 23$
But is this check impossible if $m < a cdot b$ or is there a way to find out whether $a | c$? Maybe if $m$ and $n$ are primes?
Thanks!
elementary-number-theory modular-arithmetic
add a comment |
So I'm working on modular arithmetic and encountered the following problem:
Assume that $(a cdot b) mod m = c$ and that I know $c$. Can I still check for divisibility of $c$ by $a$? A quick example tells me that checking whether $c mod a = 0$ does not work:
$((29*200) mod 512) mod 29 = 23$
But is this check impossible if $m < a cdot b$ or is there a way to find out whether $a | c$? Maybe if $m$ and $n$ are primes?
Thanks!
elementary-number-theory modular-arithmetic
$acdot b pmod m$ or $pmod n$?
– Maged Saeed
Nov 28 '18 at 20:02
The question is not at all clear. Please give a concrete example.
– Bill Dubuque
Nov 28 '18 at 20:05
I edited the question, I hope it's clearer now.
– Aares
Nov 28 '18 at 20:13
You seek to solve $ axequiv cpmod{m} $ for which see this answer.
– Bill Dubuque
Dec 1 '18 at 16:30
add a comment |
So I'm working on modular arithmetic and encountered the following problem:
Assume that $(a cdot b) mod m = c$ and that I know $c$. Can I still check for divisibility of $c$ by $a$? A quick example tells me that checking whether $c mod a = 0$ does not work:
$((29*200) mod 512) mod 29 = 23$
But is this check impossible if $m < a cdot b$ or is there a way to find out whether $a | c$? Maybe if $m$ and $n$ are primes?
Thanks!
elementary-number-theory modular-arithmetic
So I'm working on modular arithmetic and encountered the following problem:
Assume that $(a cdot b) mod m = c$ and that I know $c$. Can I still check for divisibility of $c$ by $a$? A quick example tells me that checking whether $c mod a = 0$ does not work:
$((29*200) mod 512) mod 29 = 23$
But is this check impossible if $m < a cdot b$ or is there a way to find out whether $a | c$? Maybe if $m$ and $n$ are primes?
Thanks!
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
edited Nov 28 '18 at 21:19
asked Nov 28 '18 at 19:58
Aares
62
62
$acdot b pmod m$ or $pmod n$?
– Maged Saeed
Nov 28 '18 at 20:02
The question is not at all clear. Please give a concrete example.
– Bill Dubuque
Nov 28 '18 at 20:05
I edited the question, I hope it's clearer now.
– Aares
Nov 28 '18 at 20:13
You seek to solve $ axequiv cpmod{m} $ for which see this answer.
– Bill Dubuque
Dec 1 '18 at 16:30
add a comment |
$acdot b pmod m$ or $pmod n$?
– Maged Saeed
Nov 28 '18 at 20:02
The question is not at all clear. Please give a concrete example.
– Bill Dubuque
Nov 28 '18 at 20:05
I edited the question, I hope it's clearer now.
– Aares
Nov 28 '18 at 20:13
You seek to solve $ axequiv cpmod{m} $ for which see this answer.
– Bill Dubuque
Dec 1 '18 at 16:30
$acdot b pmod m$ or $pmod n$?
– Maged Saeed
Nov 28 '18 at 20:02
$acdot b pmod m$ or $pmod n$?
– Maged Saeed
Nov 28 '18 at 20:02
The question is not at all clear. Please give a concrete example.
– Bill Dubuque
Nov 28 '18 at 20:05
The question is not at all clear. Please give a concrete example.
– Bill Dubuque
Nov 28 '18 at 20:05
I edited the question, I hope it's clearer now.
– Aares
Nov 28 '18 at 20:13
I edited the question, I hope it's clearer now.
– Aares
Nov 28 '18 at 20:13
You seek to solve $ axequiv cpmod{m} $ for which see this answer.
– Bill Dubuque
Dec 1 '18 at 16:30
You seek to solve $ axequiv cpmod{m} $ for which see this answer.
– Bill Dubuque
Dec 1 '18 at 16:30
add a comment |
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$acdot b pmod m$ or $pmod n$?
– Maged Saeed
Nov 28 '18 at 20:02
The question is not at all clear. Please give a concrete example.
– Bill Dubuque
Nov 28 '18 at 20:05
I edited the question, I hope it's clearer now.
– Aares
Nov 28 '18 at 20:13
You seek to solve $ axequiv cpmod{m} $ for which see this answer.
– Bill Dubuque
Dec 1 '18 at 16:30