Algebraic treatment of finding a dividend given quotient, divisor, and remainder












0












$begingroup$


Given a divisor of $19$ and a remainder of $11$ and a quotient of $37$ where we want to calculate the dividend, I intuitively guess that the formula is



$$frac{x-11}{19} = 37$$



giving $714$. Was I asleep in class when they talked about a formal way to handle this issue? Can someone give a more abstract theoretical explanation? It seems mod should be in this somewhere.






algebra-precalculus







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Why would you need modular arithmetic to do this? By definition, $x=19(37)+11$.
    $endgroup$
    – John Douma
    Dec 21 '18 at 3:06










  • $begingroup$
    It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive.
    $endgroup$
    – 147pm
    Dec 21 '18 at 3:27










  • $begingroup$
    "It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive" Not for something this basic, there isn't.
    $endgroup$
    – fleablood
    Dec 21 '18 at 3:58






  • 1




    $begingroup$
    For every integer $N $ and positive integer $d $ there are two unique integers $q $ and $r $ so that $0le r <d $ and $N=qd+r $. And that is as deep and abstract as it gets. We can come up the algebraic terms (the integers is a unique factorization domain) but that's the end all and be all.
    $endgroup$
    – fleablood
    Dec 21 '18 at 4:04










  • $begingroup$
    @fleablood: Yes, that's the number theory-esque approach I was looking for. Thanks. I come from the programming world where I instinctively want to see such things in algorithm-friendly ways.
    $endgroup$
    – 147pm
    Dec 21 '18 at 5:25
















0












$begingroup$


Given a divisor of $19$ and a remainder of $11$ and a quotient of $37$ where we want to calculate the dividend, I intuitively guess that the formula is



$$frac{x-11}{19} = 37$$



giving $714$. Was I asleep in class when they talked about a formal way to handle this issue? Can someone give a more abstract theoretical explanation? It seems mod should be in this somewhere.






algebra-precalculus







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Why would you need modular arithmetic to do this? By definition, $x=19(37)+11$.
    $endgroup$
    – John Douma
    Dec 21 '18 at 3:06










  • $begingroup$
    It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive.
    $endgroup$
    – 147pm
    Dec 21 '18 at 3:27










  • $begingroup$
    "It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive" Not for something this basic, there isn't.
    $endgroup$
    – fleablood
    Dec 21 '18 at 3:58






  • 1




    $begingroup$
    For every integer $N $ and positive integer $d $ there are two unique integers $q $ and $r $ so that $0le r <d $ and $N=qd+r $. And that is as deep and abstract as it gets. We can come up the algebraic terms (the integers is a unique factorization domain) but that's the end all and be all.
    $endgroup$
    – fleablood
    Dec 21 '18 at 4:04










  • $begingroup$
    @fleablood: Yes, that's the number theory-esque approach I was looking for. Thanks. I come from the programming world where I instinctively want to see such things in algorithm-friendly ways.
    $endgroup$
    – 147pm
    Dec 21 '18 at 5:25














0












0








0





$begingroup$


Given a divisor of $19$ and a remainder of $11$ and a quotient of $37$ where we want to calculate the dividend, I intuitively guess that the formula is



$$frac{x-11}{19} = 37$$



giving $714$. Was I asleep in class when they talked about a formal way to handle this issue? Can someone give a more abstract theoretical explanation? It seems mod should be in this somewhere.






algebra-precalculus







share|cite|improve this question









$endgroup$




Given a divisor of $19$ and a remainder of $11$ and a quotient of $37$ where we want to calculate the dividend, I intuitively guess that the formula is



$$frac{x-11}{19} = 37$$



giving $714$. Was I asleep in class when they talked about a formal way to handle this issue? Can someone give a more abstract theoretical explanation? It seems mod should be in this somewhere.






algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 3:03









147pm147pm

335212




335212








  • 4




    $begingroup$
    Why would you need modular arithmetic to do this? By definition, $x=19(37)+11$.
    $endgroup$
    – John Douma
    Dec 21 '18 at 3:06










  • $begingroup$
    It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive.
    $endgroup$
    – 147pm
    Dec 21 '18 at 3:27










  • $begingroup$
    "It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive" Not for something this basic, there isn't.
    $endgroup$
    – fleablood
    Dec 21 '18 at 3:58






  • 1




    $begingroup$
    For every integer $N $ and positive integer $d $ there are two unique integers $q $ and $r $ so that $0le r <d $ and $N=qd+r $. And that is as deep and abstract as it gets. We can come up the algebraic terms (the integers is a unique factorization domain) but that's the end all and be all.
    $endgroup$
    – fleablood
    Dec 21 '18 at 4:04










  • $begingroup$
    @fleablood: Yes, that's the number theory-esque approach I was looking for. Thanks. I come from the programming world where I instinctively want to see such things in algorithm-friendly ways.
    $endgroup$
    – 147pm
    Dec 21 '18 at 5:25














  • 4




    $begingroup$
    Why would you need modular arithmetic to do this? By definition, $x=19(37)+11$.
    $endgroup$
    – John Douma
    Dec 21 '18 at 3:06










  • $begingroup$
    It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive.
    $endgroup$
    – 147pm
    Dec 21 '18 at 3:27










  • $begingroup$
    "It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive" Not for something this basic, there isn't.
    $endgroup$
    – fleablood
    Dec 21 '18 at 3:58






  • 1




    $begingroup$
    For every integer $N $ and positive integer $d $ there are two unique integers $q $ and $r $ so that $0le r <d $ and $N=qd+r $. And that is as deep and abstract as it gets. We can come up the algebraic terms (the integers is a unique factorization domain) but that's the end all and be all.
    $endgroup$
    – fleablood
    Dec 21 '18 at 4:04










  • $begingroup$
    @fleablood: Yes, that's the number theory-esque approach I was looking for. Thanks. I come from the programming world where I instinctively want to see such things in algorithm-friendly ways.
    $endgroup$
    – 147pm
    Dec 21 '18 at 5:25








4




4




$begingroup$
Why would you need modular arithmetic to do this? By definition, $x=19(37)+11$.
$endgroup$
– John Douma
Dec 21 '18 at 3:06




$begingroup$
Why would you need modular arithmetic to do this? By definition, $x=19(37)+11$.
$endgroup$
– John Douma
Dec 21 '18 at 3:06












$begingroup$
It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive.
$endgroup$
– 147pm
Dec 21 '18 at 3:27




$begingroup$
It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive.
$endgroup$
– 147pm
Dec 21 '18 at 3:27












$begingroup$
"It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive" Not for something this basic, there isn't.
$endgroup$
– fleablood
Dec 21 '18 at 3:58




$begingroup$
"It just seems like there might be a deeper mathematical truth to this than just what seems to be intuitive" Not for something this basic, there isn't.
$endgroup$
– fleablood
Dec 21 '18 at 3:58




1




1




$begingroup$
For every integer $N $ and positive integer $d $ there are two unique integers $q $ and $r $ so that $0le r <d $ and $N=qd+r $. And that is as deep and abstract as it gets. We can come up the algebraic terms (the integers is a unique factorization domain) but that's the end all and be all.
$endgroup$
– fleablood
Dec 21 '18 at 4:04




$begingroup$
For every integer $N $ and positive integer $d $ there are two unique integers $q $ and $r $ so that $0le r <d $ and $N=qd+r $. And that is as deep and abstract as it gets. We can come up the algebraic terms (the integers is a unique factorization domain) but that's the end all and be all.
$endgroup$
– fleablood
Dec 21 '18 at 4:04












$begingroup$
@fleablood: Yes, that's the number theory-esque approach I was looking for. Thanks. I come from the programming world where I instinctively want to see such things in algorithm-friendly ways.
$endgroup$
– 147pm
Dec 21 '18 at 5:25




$begingroup$
@fleablood: Yes, that's the number theory-esque approach I was looking for. Thanks. I come from the programming world where I instinctively want to see such things in algorithm-friendly ways.
$endgroup$
– 147pm
Dec 21 '18 at 5:25










1 Answer
1






active

oldest

votes


















0












$begingroup$

Guessing is not math. A number x is divided by

19 with a result of 37 and a remainder of 11.



Thus x/19 = 37 + 11/19. So use simple algebra

to solve for x. Do you know how to do that?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    As I stated above, I was looking for the Euclidean division treatment. I took a number theory course and this problem probably reminded me of it.
    $endgroup$
    – 147pm
    Dec 21 '18 at 19:22










  • $begingroup$
    Number theory not needed. x = 11 (mod 19) is the hard way to solve the problem.
    $endgroup$
    – William Elliot
    Dec 21 '18 at 23:00












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Guessing is not math. A number x is divided by

19 with a result of 37 and a remainder of 11.



Thus x/19 = 37 + 11/19. So use simple algebra

to solve for x. Do you know how to do that?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    As I stated above, I was looking for the Euclidean division treatment. I took a number theory course and this problem probably reminded me of it.
    $endgroup$
    – 147pm
    Dec 21 '18 at 19:22










  • $begingroup$
    Number theory not needed. x = 11 (mod 19) is the hard way to solve the problem.
    $endgroup$
    – William Elliot
    Dec 21 '18 at 23:00
















0












$begingroup$

Guessing is not math. A number x is divided by

19 with a result of 37 and a remainder of 11.



Thus x/19 = 37 + 11/19. So use simple algebra

to solve for x. Do you know how to do that?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    As I stated above, I was looking for the Euclidean division treatment. I took a number theory course and this problem probably reminded me of it.
    $endgroup$
    – 147pm
    Dec 21 '18 at 19:22










  • $begingroup$
    Number theory not needed. x = 11 (mod 19) is the hard way to solve the problem.
    $endgroup$
    – William Elliot
    Dec 21 '18 at 23:00














0












0








0





$begingroup$

Guessing is not math. A number x is divided by

19 with a result of 37 and a remainder of 11.



Thus x/19 = 37 + 11/19. So use simple algebra

to solve for x. Do you know how to do that?






share|cite|improve this answer









$endgroup$



Guessing is not math. A number x is divided by

19 with a result of 37 and a remainder of 11.



Thus x/19 = 37 + 11/19. So use simple algebra

to solve for x. Do you know how to do that?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 21 '18 at 9:43









William ElliotWilliam Elliot

9,1662820




9,1662820












  • $begingroup$
    As I stated above, I was looking for the Euclidean division treatment. I took a number theory course and this problem probably reminded me of it.
    $endgroup$
    – 147pm
    Dec 21 '18 at 19:22










  • $begingroup$
    Number theory not needed. x = 11 (mod 19) is the hard way to solve the problem.
    $endgroup$
    – William Elliot
    Dec 21 '18 at 23:00


















  • $begingroup$
    As I stated above, I was looking for the Euclidean division treatment. I took a number theory course and this problem probably reminded me of it.
    $endgroup$
    – 147pm
    Dec 21 '18 at 19:22










  • $begingroup$
    Number theory not needed. x = 11 (mod 19) is the hard way to solve the problem.
    $endgroup$
    – William Elliot
    Dec 21 '18 at 23:00
















$begingroup$
As I stated above, I was looking for the Euclidean division treatment. I took a number theory course and this problem probably reminded me of it.
$endgroup$
– 147pm
Dec 21 '18 at 19:22




$begingroup$
As I stated above, I was looking for the Euclidean division treatment. I took a number theory course and this problem probably reminded me of it.
$endgroup$
– 147pm
Dec 21 '18 at 19:22












$begingroup$
Number theory not needed. x = 11 (mod 19) is the hard way to solve the problem.
$endgroup$
– William Elliot
Dec 21 '18 at 23:00




$begingroup$
Number theory not needed. x = 11 (mod 19) is the hard way to solve the problem.
$endgroup$
– William Elliot
Dec 21 '18 at 23:00


















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