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# help

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If $a, b, c$ are positive integers less than 13 such that

$2ab+bc+ca \equiv 0\pmod{13}$
$ab+2bc+ca \equiv 6abc\pmod{13}$
$ab+bc+2ca \equiv 8abc\pmod {13}$

then determine the remainder when $a + b + c$ is divided by $13$.

Sep 27, 2020

#1
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a + b + c leaves a remainder of 8 when divided by 13.

Sep 27, 2020
#2
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Sorry, but it says that is incorrect.

Thank you for trying to help out though!

Also, if someone wants to post a solution to this, it'd be best if they provided an explanation, because I'd like to learn from it. Thanks!

Guest Sep 27, 2020
#3
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Look at many, many solutions here:

https://www.wolframalpha.com/input/?i=%282ab%2Bbc%2Bca%29+mod+13+%3D0%2C++%28ab%2B2bc%2Bca%29+mod+13%3D+6abc%2C++%28ab%2Bbc%2B2ca%29++mod+13+%3D+8abc%2C+solve+for+a%2Cb%2Cc

Sep 27, 2020
#4
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Unfortunately, this also does not work as your input had equal signs instead of modular congruence signs. I don't believe you can input modular congruence signs into Wolfram Alpha.

Thanks for trying to help me though!

Guest Sep 28, 2020
#5
+111124
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If a, b, and c are all 0 then these three equations are all true.

And if this is true then  (a+b+c) mod13 = 0

So if there is only one solution then that solution must be 0

However, just two of them are 0 then the third one can equal to many different numbers.

So

I do not believe this has one specific answer.

Sep 28, 2020
#6
+1

I know I must be annoying for saying all of the answers are wrong, but unfortunately there's a problem with this one too. The problem states that $a, b, c$ must be positive integers less than 13, so $0$ doesn't fit this requirement. However, thank you very much for trying to solve this for me!

Guest Sep 28, 2020
#7
+111124
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Yep, good point!

Melody  Sep 29, 2020
#8
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Sep 29, 2020
#9
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a = 3, b = 6, c = 9.

Unique.

Will post a solution, of sorts, if anyone is interested.

Guest Sep 30, 2020
#10
+111124
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Thanks guest,

I have some interest.

Did you use code to do the grunt work and just test all the possibilities

Or did you actually come by the answer in a logical, mathematical (no trial and error)  way ?

Melody  Sep 30, 2020
#11
+1

Thanks Melody, it's nice to know that someone is actually interested.

Often that doesn't appear to be the case.

The equations can be written in the form

$$\displaystyle 2X+Y+Z\equiv 0 \dots\dots(1)\\ X+2Y+Z\equiv R\dots\dots(2)\\ X+Y+2Z\equiv S\dots\dots(3)$$

where each one is mod 13.

$$\displaystyle (1) - 2(3) : -Y-3Z \equiv -2S\dots\dots(4)\\ (2) - (3) :Y - Z \equiv R-S\dots\dots(5)\\ (4)+(5) : -4Z\equiv R-3S\dots\dots(6).$$

Returning to the original unknowns,

$$\displaystyle -4ca \equiv 6abc-24abc \equiv -18abc,$$

from which

$$\displaystyle9b=2+13k\quad \text{ where k is an integer}.$$

The smallest value of k for which b will be an integer is k = 4, leading to 9b = 54, b = 6.

The next suitable value for k puts b out of the permitted range.

The remaining unknowns (a = 3 and c = 9) can then be found by back substitution, (6) into (5) and then into (2) or (3).

Leave that to you.

Guest Sep 30, 2020
#12
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THANK YOU THANK YOU THANK YOU!

I really like your explanation! Thanks a lot!

Guest Sep 30, 2020
#13
+111124
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Brilliant, thanks very much!

I never would have thought to do those substitutions!

I continued with matrices, and again, I would not have thought to use matrices on a modulo question if you had not inspired me.

Melody  Sep 30, 2020