Mixing
Being $2011$ prime, calculate 2009! divided by 2011
$begingroup$
I'm working in the following excercise:
Being $2011$ prime, calculate $2009!$ divided by $2011$
By Wilson's theorem I have:
$$2010! equiv -1 mod 2011$$
$$2009! * 2010 equiv -1 * 2010 mod 2011$$
$$2009! equiv -2010 mod 2011 $$
$$2009! equiv 1 mod 2011$$
Checking the answer is correct but I'm not sure about step two, is that correct? any help will be really appreciated.
number-theory modular-arithmetic
asked Jan 30 at 22:17
mrazmraz
450110
$endgroup$
add a comment |
$begingroup$
I'm working in the following excercise:
Being $2011$ prime, calculate $2009!$ divided by $2011$
By Wilson's theorem I have:
$$2010! equiv -1 mod 2011$$
$$2009! * 2010 equiv -1 * 2010 mod 2011$$
$$2009! equiv -2010 mod 2011 $$
$$2009! equiv 1 mod 2011$$
Checking the answer is correct but I'm not sure about step two, is that correct? any help will be really appreciated.
number-theory modular-arithmetic
asked Jan 30 at 22:17
mrazmraz
450110
$endgroup$
1
$begingroup$
It's perfectly correct.
$endgroup$
– Bernard
Jan 30 at 22:21
1
$begingroup$
No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
$endgroup$
– fleablood
Jan 30 at 22:23
1
$begingroup$
It should be noted that 2011 is prime
$endgroup$
– J. W. Tanner
Jan 30 at 22:41
1
$begingroup$
Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
$endgroup$
– fleablood
Jan 30 at 22:49
add a comment |
$begingroup$
I'm working in the following excercise:
Being $2011$ prime, calculate $2009!$ divided by $2011$
By Wilson's theorem I have:
$$2010! equiv -1 mod 2011$$
$$2009! * 2010 equiv -1 * 2010 mod 2011$$
$$2009! equiv -2010 mod 2011 $$
$$2009! equiv 1 mod 2011$$
Checking the answer is correct but I'm not sure about step two, is that correct? any help will be really appreciated.
number-theory modular-arithmetic
asked Jan 30 at 22:17
mrazmraz
450110
$endgroup$
I'm working in the following excercise:
Being $2011$ prime, calculate $2009!$ divided by $2011$
By Wilson's theorem I have:
$$2010! equiv -1 mod 2011$$
$$2009! * 2010 equiv -1 * 2010 mod 2011$$
$$2009! equiv -2010 mod 2011 $$
$$2009! equiv 1 mod 2011$$
Checking the answer is correct but I'm not sure about step two, is that correct? any help will be really appreciated.
number-theory modular-arithmetic
number-theory modular-arithmetic
asked Jan 30 at 22:17
mrazmraz
450110
asked Jan 30 at 22:17
mrazmraz
450110
edited Jan 30 at 22:42
mraz
asked Jan 30 at 22:17
mrazmraz
450110
asked Jan 30 at 22:17
mrazmraz
450110
asked Jan 30 at 22:17
mrazmraz
450110
450110
1
$begingroup$
It's perfectly correct.
$endgroup$
– Bernard
Jan 30 at 22:21
1
$begingroup$
No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
$endgroup$
– fleablood
Jan 30 at 22:23
1
$begingroup$
It should be noted that 2011 is prime
$endgroup$
– J. W. Tanner
Jan 30 at 22:41
1
$begingroup$
Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
$endgroup$
– fleablood
Jan 30 at 22:49
add a comment |
1
$begingroup$
It's perfectly correct.
$endgroup$
– Bernard
Jan 30 at 22:21
1
$begingroup$
No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
$endgroup$
– fleablood
Jan 30 at 22:23
1
$begingroup$
It should be noted that 2011 is prime
$endgroup$
– J. W. Tanner
Jan 30 at 22:41
1
$begingroup$
Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
$endgroup$
– fleablood
Jan 30 at 22:49
1
1
$begingroup$
It's perfectly correct.
$endgroup$
– Bernard
Jan 30 at 22:21
$begingroup$
It's perfectly correct.
$endgroup$
– Bernard
Jan 30 at 22:21
1
1
$begingroup$
No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
$endgroup$
– fleablood
Jan 30 at 22:23
$begingroup$
No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
$endgroup$
– fleablood
Jan 30 at 22:23
1
1
$begingroup$
It should be noted that 2011 is prime
$endgroup$
– J. W. Tanner
Jan 30 at 22:41
$begingroup$
It should be noted that 2011 is prime
$endgroup$
– J. W. Tanner
Jan 30 at 22:41
1
1
$begingroup$
Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
$endgroup$
– fleablood
Jan 30 at 22:49
$begingroup$
Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
$endgroup$
– fleablood
Jan 30 at 22:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.
That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.
$$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$
$$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)
$$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$
(This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)
....
Instead do:
$2010! equiv -1 pmod {2011}$
$2009!*2010 = 2010! equiv -1 pmod {2011}$.
Now notice $2010 equiv -1 pmod {2011}$ so
$2009!*(-1) equiv - 1pmod {2011}$ and
$2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so
$2009! equiv 1 pmod {2011}$.
answered Jan 30 at 22:29
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
$$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
2009!equiv 1 pmod {2011}$$
You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.
answered Jan 30 at 22:22
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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active
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active
oldest
votes
$begingroup$
No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.
That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.
$$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$
$$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)
$$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$
(This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)
....
Instead do:
$2010! equiv -1 pmod {2011}$
$2009!*2010 = 2010! equiv -1 pmod {2011}$.
Now notice $2010 equiv -1 pmod {2011}$ so
$2009!*(-1) equiv - 1pmod {2011}$ and
$2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so
$2009! equiv 1 pmod {2011}$.
answered Jan 30 at 22:29
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.
That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.
$$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$
$$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)
$$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$
(This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)
....
Instead do:
$2010! equiv -1 pmod {2011}$
$2009!*2010 = 2010! equiv -1 pmod {2011}$.
Now notice $2010 equiv -1 pmod {2011}$ so
$2009!*(-1) equiv - 1pmod {2011}$ and
$2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so
$2009! equiv 1 pmod {2011}$.
answered Jan 30 at 22:29
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.
That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.
$$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$
$$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)
$$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$
(This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)
....
Instead do:
$2010! equiv -1 pmod {2011}$
$2009!*2010 = 2010! equiv -1 pmod {2011}$.
Now notice $2010 equiv -1 pmod {2011}$ so
$2009!*(-1) equiv - 1pmod {2011}$ and
$2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so
$2009! equiv 1 pmod {2011}$.
answered Jan 30 at 22:29
fleabloodfleablood
73.8k22891
$endgroup$
No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.
That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.
$$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$
$$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)
$$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$
(This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)
....
Instead do:
$2010! equiv -1 pmod {2011}$
$2009!*2010 = 2010! equiv -1 pmod {2011}$.
Now notice $2010 equiv -1 pmod {2011}$ so
$2009!*(-1) equiv - 1pmod {2011}$ and
$2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so
$2009! equiv 1 pmod {2011}$.
answered Jan 30 at 22:29
fleabloodfleablood
73.8k22891
edited Jan 30 at 22:45
answered Jan 30 at 22:29
fleabloodfleablood
73.8k22891
answered Jan 30 at 22:29
fleabloodfleablood
73.8k22891
answered Jan 30 at 22:29
fleabloodfleablood
73.8k22891
73.8k22891
add a comment |
add a comment |
$begingroup$
The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
$$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
2009!equiv 1 pmod {2011}$$
You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.
answered Jan 30 at 22:22
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
$$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
2009!equiv 1 pmod {2011}$$
You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.
answered Jan 30 at 22:22
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
$$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
2009!equiv 1 pmod {2011}$$
You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.
answered Jan 30 at 22:22
Ross MillikanRoss Millikan
301k24200375
$endgroup$
The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
$$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
2009!equiv 1 pmod {2011}$$
You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.
answered Jan 30 at 22:22
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 22:22
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 22:22
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 22:22
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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1
$begingroup$
It's perfectly correct.
$endgroup$
– Bernard
Jan 30 at 22:21
1
$begingroup$
No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
$endgroup$
– fleablood
Jan 30 at 22:23
1
$begingroup$
It should be noted that 2011 is prime
$endgroup$
– J. W. Tanner
Jan 30 at 22:41
1
$begingroup$
Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
$endgroup$
– fleablood
Jan 30 at 22:49