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2017-02-25, 23:20   #3
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C716 Posts

Quote:
 Originally Posted by MattcAnderson Hi Mersenneforum, In order to tackle this first number theory problem, (okay it is problem 84) I will try a similar problem. Find all x such that 5*x+6 is congruent to 0 mod 7 expression 1 From the reading, I notice that the greatest common divisor of 5 and 6 is 1. So the techniques presented here should apply. We make an augmented T table X 5*X 5*X mod 7 ______________________ 0 0 0 1 5 5 2 10 3 3 15 1 4 20 6 5 25 4 6 30 2 From expression 1, conclude that 5*x is congruent to 1 mod 7. My education has a gap here. I use my Maple computer tool to conclude that x is congruent to 2*x+1 mod 7. Here is my Maple code – a:=-(5*x+6) mod 7 And, again, Maple returns a= 2*x+1 mod 7. Maybe someone can help me fill in the gaps. Regards, Matt
$5x+6 \equiv 0 \pmod 7\\
5x\equiv -6 \pmod 7 \\
5\equiv -2 \pmod 7 \\
\therefore \\
5x\equiv -2x \pmod 7 \\
-6 \equiv -2x\pmod 7 \\
3 \equiv x \pmod 7 \\
$

edit:

the solution like this for 84 is similar to this:

$\text {deleted previous ramble after checking with PARI/GP} \\
12x+24 \equiv 0 \pmod {33} \\
3(4x+8)\equiv 0 \pmod {3(11)}\\
(4x+8)\equiv 0 \pmod {11}\\
4(x+2)\equiv 0 \pmod {11}\\
\text { by 0 product rule one of these has to be 0 and 4 is not 0} \\
x+2 \equiv 0 \pmod {11}\\
x\equiv -2 \pmod {11}\\
x \equiv 9 \pmod {11}\\
$

Last fiddled with by science_man_88 on 2017-02-25 at 23:41