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$$15\cdot25 \pmod {11}\equiv 4\cdot3 \pmod {11}$$

How does it work?

Full example

$$3^{(11-1)} \pmod{11} = 3\cdot27\cdot27\cdot27 \pmod{11}= 3\cdot5\cdot5\cdot5 \pmod{11} = 15\cdot25 \pmod{11}= 4\cdot3 \pmod{11} =1 \pmod{11}$$

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  • $\begingroup$ Do you know what mod even means? $\endgroup$
    – kingW3
    May 21, 2017 at 13:29
  • $\begingroup$ @kingW3 What did make you feel that the OP doesn't know it? $\endgroup$
    – Ramil
    May 21, 2017 at 13:49
  • $\begingroup$ @Ramil Well the result is kinda trivial for someone that is not a beginner in modular arithmetic, in case he doesn't really know what mod means it would be greater to explain in length, while if he knows what mod is then a simple $15\cdot 25=(11+4)\cdot (22+3)=11\cdot(22+3+8)+4\cdot3\equiv 4\cdot 3\pmod{11}$ would suffice. $\endgroup$
    – kingW3
    May 21, 2017 at 13:59
  • $\begingroup$ @kingW3 Yes, probably the OP is a beginner, but this doesn't mean that he doesn't know the definition. Of course, the question is trivial, but the answer to it doesn't follow immediately from the definition. $\endgroup$
    – Ramil
    May 21, 2017 at 14:08

3 Answers 3

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This is actually really simple. Take this for example:$$a \equiv x (\text{mod } n),b \equiv y(\text{mod }n) $$ Then knowing the definition of module we can write $a$ and $b$ like this:$$a= kn + x, b =ln + y$$ We can say that:$$ab = kln^2 + kyn + lxn + xy \Rightarrow ab \equiv xy (\text{mod n})$$ Everything else from modular arithmetic can be derived this way. Try it yourself!

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$a\equiv b \pmod{11}$ means that $11$ divides $a-b$. So if we want to see that $$(a+11)(b+11)\equiv ab \pmod{11}$$ we should check that $11$ is a divisor of $$(a+11)(b+11)-ab=ab+11a+11b+11*11-ab=11a+11b+121$$ This is clearly the case (in your example we have that $a=4$ and $b=14$). You can do this again for $b$ and you see that $15\cdot 25 \equiv 3 \cdot 4 \pmod{11}$.

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$$ab \pmod n = ((a\pmod n)\times( b\pmod n)) \pmod n$$

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