Remainders and divisors [duplicate]
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This question already has an answer here:
Basic modular arithmetic question
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Two integers $a,b$ have the same remainder when divided by $x$. What is $x$ in terms of $a$ and $b$?
Is there a way to find the answer without using modular arithmetic?
elementary-number-theory
asked Dec 7 '18 at 20:46
James PalmerJames Palmer
34
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marked as duplicate by Bill Dubuque elementary-number-theory
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Dec 7 '18 at 20:54
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This question already has an answer here:
Basic modular arithmetic question
1 answer
Two integers $a,b$ have the same remainder when divided by $x$. What is $x$ in terms of $a$ and $b$?
Is there a way to find the answer without using modular arithmetic?
elementary-number-theory
asked Dec 7 '18 at 20:46
James PalmerJames Palmer
34
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marked as duplicate by Bill Dubuque elementary-number-theory
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Dec 7 '18 at 20:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Basic modular arithmetic question
1 answer
Two integers $a,b$ have the same remainder when divided by $x$. What is $x$ in terms of $a$ and $b$?
Is there a way to find the answer without using modular arithmetic?
elementary-number-theory
asked Dec 7 '18 at 20:46
James PalmerJames Palmer
34
$endgroup$
This question already has an answer here:
Basic modular arithmetic question
1 answer
Two integers $a,b$ have the same remainder when divided by $x$. What is $x$ in terms of $a$ and $b$?
Is there a way to find the answer without using modular arithmetic?
This question already has an answer here:
Basic modular arithmetic question
1 answer
elementary-number-theory
elementary-number-theory
asked Dec 7 '18 at 20:46
James PalmerJames Palmer
34
asked Dec 7 '18 at 20:46
James PalmerJames Palmer
34
asked Dec 7 '18 at 20:46
James PalmerJames Palmer
34
asked Dec 7 '18 at 20:46
James PalmerJames Palmer
34
asked Dec 7 '18 at 20:46
James PalmerJames Palmer
34
34
marked as duplicate by Bill Dubuque elementary-number-theory
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Dec 7 '18 at 20:54
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1 Answer
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If all you know is that $a=k_1x+r$ and $b=k_2x+r$, where $a,b$, and $r$ are known, then you have three variables and two equations, which will not have a unique solution. One of the only things that I think you can really say is that $x|(a-b)$, but of course that is just the modular arithmetic you wanted to avoid.
answered Dec 7 '18 at 20:49
RandomMathGuyRandomMathGuy
1872
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That makes sense, thanks!
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– James Palmer
Dec 7 '18 at 21:02
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1 Answer
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1 Answer
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oldest
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active
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active
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$begingroup$
If all you know is that $a=k_1x+r$ and $b=k_2x+r$, where $a,b$, and $r$ are known, then you have three variables and two equations, which will not have a unique solution. One of the only things that I think you can really say is that $x|(a-b)$, but of course that is just the modular arithmetic you wanted to avoid.
answered Dec 7 '18 at 20:49
RandomMathGuyRandomMathGuy
1872
$endgroup$
$begingroup$
That makes sense, thanks!
$endgroup$
– James Palmer
Dec 7 '18 at 21:02
add a comment |
$begingroup$
If all you know is that $a=k_1x+r$ and $b=k_2x+r$, where $a,b$, and $r$ are known, then you have three variables and two equations, which will not have a unique solution. One of the only things that I think you can really say is that $x|(a-b)$, but of course that is just the modular arithmetic you wanted to avoid.
answered Dec 7 '18 at 20:49
RandomMathGuyRandomMathGuy
1872
$endgroup$
$begingroup$
That makes sense, thanks!
$endgroup$
– James Palmer
Dec 7 '18 at 21:02
add a comment |
$begingroup$
If all you know is that $a=k_1x+r$ and $b=k_2x+r$, where $a,b$, and $r$ are known, then you have three variables and two equations, which will not have a unique solution. One of the only things that I think you can really say is that $x|(a-b)$, but of course that is just the modular arithmetic you wanted to avoid.
answered Dec 7 '18 at 20:49
RandomMathGuyRandomMathGuy
1872
$endgroup$
If all you know is that $a=k_1x+r$ and $b=k_2x+r$, where $a,b$, and $r$ are known, then you have three variables and two equations, which will not have a unique solution. One of the only things that I think you can really say is that $x|(a-b)$, but of course that is just the modular arithmetic you wanted to avoid.
answered Dec 7 '18 at 20:49
RandomMathGuyRandomMathGuy
1872
answered Dec 7 '18 at 20:49
RandomMathGuyRandomMathGuy
1872
answered Dec 7 '18 at 20:49
RandomMathGuyRandomMathGuy
1872
answered Dec 7 '18 at 20:49
RandomMathGuyRandomMathGuy
1872
1872
$begingroup$
That makes sense, thanks!
$endgroup$
– James Palmer
Dec 7 '18 at 21:02
add a comment |
$begingroup$
That makes sense, thanks!
$endgroup$
– James Palmer
Dec 7 '18 at 21:02
$begingroup$
That makes sense, thanks!
$endgroup$
– James Palmer
Dec 7 '18 at 21:02
$begingroup$
That makes sense, thanks!
$endgroup$
– James Palmer
Dec 7 '18 at 21:02
add a comment |
