Find $x$
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Find $x$ such that $x^2 - 3x + 7 \equiv 2 (mod 11)$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
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- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
Re: Find $x$
Check my solution:
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
But if we check the factors of $6$,we will find
that there is no solution of $x$.
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
But if we check the factors of $6$,we will find
that there is no solution of $x$.
Last edited by Absur Khan Siam on Sun Feb 05, 2017 10:09 pm, edited 1 time in total.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Re: Find $x$
$x = 11n + 7$ for any non-negative integer n.
This can be proved very easily. $x \equiv {0,1,2,3,4,5,6,7,8,9,10}$. Now checking the values for $x(x-3) mod 11$ gives the only solution when $x \equiv 7(mod 11)$.
This can be proved very easily. $x \equiv {0,1,2,3,4,5,6,7,8,9,10}$. Now checking the values for $x(x-3) mod 11$ gives the only solution when $x \equiv 7(mod 11)$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
- Charles Caleb Colton
- Charles Caleb Colton
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- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
Re: Find $x$
I missed the point.Thanks to dhasan.
Correct solution:
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
If we check the factors of $6$,we will find
that the solution is $x \equiv 7$.
Correct solution:
This equation can be rewritten :
$x(x-3) \equiv -5 \equiv 6 (mod 11)$
If we check the factors of $6$,we will find
that the solution is $x \equiv 7$.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Re: Find $x$
x(x-3) = -5 = 6 (mod 11)
by testing the factors of 6 we get that x = 7 (mod 11)
by testing the factors of 6 we get that x = 7 (mod 11)