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(a) $10X + 12Y − 8Z ≡ 14 \pmod 8$

(b) $3X − 15Y + 24Z ≡ 13 \pmod {81}$

So I see this is a system of equations and I take the equations and did:

$(b)-3(a)$ which yields:

$33X +21Y ≡ -29 \pmod {81-8}$ ?

I'm unsure of if I can directly compute $\mod{(81-8)}$ however. Ideally I would want both mods to be the same number I believe.

gt6989b
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Bugcatcher123
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  • You can't subtract different congruencies like that. – Alan Oct 09 '17 at 23:31
  • Yea I was thinking as much. But how do I go about "changing" the congruencies to solve this system? – Bugcatcher123 Oct 09 '17 at 23:33
  • No idea, that's beyond the number theory I've had. The first thing you can do is reduce the first one though, mod 8, 10 is congruent to 2, 12 to 4, -8 to 0, 14 to 6, which gets you to $2x+4y$ is congruent to 6 mod 8 – Alan Oct 09 '17 at 23:35

1 Answers1

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$(b)$ does not have a solution in integers, since $3X - 15Y + 24Z$ is always a multiple of $3$, so the remainder when it is divided by $81$(which is a multiple of three) must also be a multiple of $3$. However, that number is given to be $13$, which is not a multiple of three.

On the other hand, $(a)$ has a solution in integers when considered in isolation : just take $X = Y=Z = 1$, for example. Naturally, both $(a)$ and $(b)$ can't be satisfied simultaneously, as $b$ can never be satisfied.

On a general note, if you are given two congruences mod different numbers, then you cannot add these congruences.

Sarvesh Ravichandran Iyer
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