Find $x$

 Posts: 53
 Joined: Tue Dec 08, 2015 4:25 pm
 Location: Bashaboo , Dhaka
Find $x$
Find $x$ such that $x^2  3x + 7 \equiv 2 (mod 11)$
"(To Ptolemy I) There is no 'royal road' to geometry."  Euclid

 Posts: 53
 Joined: Tue Dec 08, 2015 4:25 pm
 Location: Bashaboo , Dhaka
Re: Find $x$
Check my solution:
This equation can be rewritten :
$x(x3) \equiv 5 \equiv 6 (mod 11)$
But if we check the factors of $6$,we will find
that there is no solution of $x$.
This equation can be rewritten :
$x(x3) \equiv 5 \equiv 6 (mod 11)$
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 nonnegative 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 $x(x3) mod 11$ 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 $x(x3) mod 11$ 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

 Posts: 53
 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 :
$x(x3) \equiv 5 \equiv 6 (mod 11)$
If we check the factors of $6$,we will find
that the solution is $x \equiv 7$.
Correct solution:
This equation can be rewritten :
$x(x3) \equiv 5 \equiv 6 (mod 11)$
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(x3) = 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)