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Given the following module operation:

$$8x + 51\pmod{10}$$

Details:

  • $x \equiv y \pmod{m}$ is by definition equivalent to $m|(x−y)$.
  • $x \equiv y \pmod{m}$ , $x,y\in \mathbb{Z}, m \in \mathbb{Z^+}$ iff $a \pmod {m} = b \pmod{m}$.

Question 1: Can we split it as follows based on module details above:

$$8x + 51\pmod{10} = 8$$

$$51\pmod{10} = 1$$

Then,

$$8x + 1\pmod{10} = 8$$

Question 2: Is it possible to subtract numbers from the module operation $1$ from both sides:

$$8x \pmod{10} = 7$$

Question 3: Suppose we have the quantity $8x \pmod{10} = 7$, then what is the meaning that the inverse of $8 \pmod{10}$ is $7 \pmod{10}$? Given the inverse, how we can multiply both sides by it as follows as this was not covered in basic module operations in

$$7\cdot 8x \pmod{10} = 7 \cdot 7 \pmod{10}$$

We can conclude that

$$56x \pmod{10} = 7 $$

Avv
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    Where does the "$=8$" come from? – Théophile Mar 18 '21 at 14:55
  • @Théophile. Thank you for asking. It's given that the remainder for the whole quantity is 8. – Avv Mar 18 '21 at 15:04
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    In normal algebra, $r$ is the inverse of $s \iff r\times s = 1.$ In modular arithmetic, $r$ is the inverse of $s \pmod{n} \iff (r \times s) \equiv 1\pmod{n}.$ Since $8$ and $10$ are not relatively prime, $8$ has no inverse $\pmod{10}$. If you remove the common factor of $2$, and ask what is the inverse of $4 \pmod{5}$, the answer would be $4$ itself, since $4 \times 4\equiv 1\pmod{5}.$ – user2661923 Mar 18 '21 at 15:07
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    If $8x + 51 \equiv y \pmod{10}$ and $x$ is an integer then $y$ is going to have to be an odd integer, and cannot be $8$ – Henry Mar 18 '21 at 15:10
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    Please carefully review the basic properties of congruences described in the linked dupe (and its links). Your recent questions seem to imply that you are attempting to move forward without first having mastered the basics, which will be no more successful than learning to run before you have learned to walk. – Bill Dubuque Mar 18 '21 at 16:08
  • @user2661923. Thank you. – Avv Mar 18 '21 at 21:02
  • @Henry. Thank you. – Avv Mar 18 '21 at 21:02

1 Answers1

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Indeed,

$8x+51\mod 10 = [(8x\mod 10) + (51\mod 10)] \mod 10 = [(8x\mod 10)+1]\mod 10 = 8x+1\mod 10$.

That is, the modulo operation is compatible with addition (and multiplication).

The additional $\mod 10$ is necessary, e.g.

$18+16\mod 10 = [(18\mod 10) + (16\mod 10)] \mod 10 = 8+6 \mod 10 = 14\mod 10 = 4.$

Wuestenfux
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  • Thank you. I got why we need another $\bmod $ outside. I am still a little bit confused about something I found in the same answer, $7x + 8 \mod 11= 2$, so $7x \bmod 11= -6$, but I saw the solution I found adds extra $\bmod $ after $7x \bmod 11= -6 \bmod 11$? Is this the same as your explanation or different one? – Avv Mar 20 '21 at 23:00