# Number reduction before mod [closed]

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

How does it work?

Full example

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

• Do you know what mod even means? May 21, 2017 at 13:29
• @kingW3 What did make you feel that the OP doesn't know it? May 21, 2017 at 13:49
• @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. May 21, 2017 at 13:59
• @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. May 21, 2017 at 14:08

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!

$a\equiv b \pmod{11}$ 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 $15\cdot 25 \equiv 3 \cdot 4 \pmod{11}$.

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