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
asked Dec 21 '18 at 3:03
147pm147pm
335212
$endgroup$
|
show 2 more comments
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
asked Dec 21 '18 at 3:03
147pm147pm
335212
$endgroup$
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
|
show 2 more comments
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
asked Dec 21 '18 at 3:03
147pm147pm
335212
$endgroup$
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
asked Dec 21 '18 at 3:03
147pm147pm
335212
asked Dec 21 '18 at 3:03
147pm147pm
335212
asked Dec 21 '18 at 3:03
147pm147pm
335212
asked Dec 21 '18 at 3:03
147pm147pm
335212
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
|
show 2 more comments
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
|
show 2 more comments
1 Answer
1
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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?
answered Dec 21 '18 at 9:43
William ElliotWilliam Elliot
9,1662820
$endgroup$
$begingroup$
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
add a comment |
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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?
answered Dec 21 '18 at 9:43
William ElliotWilliam Elliot
9,1662820
$endgroup$
$begingroup$
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
add a comment |
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?
answered Dec 21 '18 at 9:43
William ElliotWilliam Elliot
9,1662820
$endgroup$
$begingroup$
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
add a comment |
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?
answered Dec 21 '18 at 9:43
William ElliotWilliam Elliot
9,1662820
$endgroup$
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?
answered Dec 21 '18 at 9:43
William ElliotWilliam Elliot
9,1662820
answered Dec 21 '18 at 9:43
William ElliotWilliam Elliot
9,1662820
answered Dec 21 '18 at 9:43
William ElliotWilliam Elliot
9,1662820
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
add a comment |
$begingroup$
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
add a comment |
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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