## Find $x$

For students of class 6-8 (age 12 to 14)
Absur Khan Siam
Posts: 61
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

Absur Khan Siam
Posts: 61
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 $x$.
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

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### 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 $x \equiv 7(mod 11)$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

Absur Khan Siam
Posts: 61
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 $x \equiv 7$.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

thczarif
Posts: 8
Joined: Mon Sep 25, 2017 11:27 pm

### Re: Find $x$

x(x-3) = -5 = 6 (mod 11)
by testing the factors of 6 we get that x = 7 (mod 11)