Solving System of Equations with inequalities and modulo

TF98

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Sep 17, 2013
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I know how to solve system of equations but I'm not sure what to do when modulo and inequalities are introduced. Here's an example:

x+8y+80z=704
x+y+z=90
x mod 8=0
0 ≤ x,y,z ≤ 90

I know that the answer is x=64,y=20,z=6 but I can't figure out how I would get there. I would give my attempts but I'm pretty clueless.

Could I be guided in the right direction?
 
I know how to solve system of equations but I'm not sure what to do when modulo and inequalities are introduced. Here's an example:

x+8y+80z=704
x+y+z=90
x mod 8=0
0 ≤ x,y,z ≤ 90

I know that the answer is x=64,y=20,z=6 but I can't figure out how I would get there. I would give my attempts but I'm pretty clueless.

Could I be guided in the right direction?
You have two regular equations and three unknowns, plus you have other information about your unknowns. So the question is whether that extra information is sufficient to make up for the missing regular equation. You actually have a great deal of extra information. Let's eliminate y or z temporarily because we have more information about x than either y or z.

\(\displaystyle x + y + z = 90 \implies z = 90 - x - y \implies x + 8y + 80(90 - x - y) = 704 \implies x - 80x + 8y - 80y + 7200 = 704 \implies\)

\(\displaystyle 79x + 72y = 7196.\)

There are an infinite number of solutions to the equation above, but we have other information. In particular x = 0 mod 8, which means

that x is an integer that is divisible by 8, and \(\displaystyle 0 \le x,\ y,\ z \le 90\).

\(\displaystyle x = 8w \implies w \in \mathbb I,\ 0 \le w \le 12,\ and\ 632w + 72y = 7196.\)

There may be anywhere from zero to thirteen correct answers, but you are not dealing with infinity any longer.

\(\displaystyle w = 12 \implies 632 * 12 + 72y = 7196 \implies 72y = - 388 \implies y < 0.\)

But y is not negative so w is not 12.

Carry on
 
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