modulo

**franios** · November 26th 2012, 01:26 PM

find the remainder when 11¹²¹ is divided by 13

**coolge** · November 26th 2012, 03:16 PM

use 11 = -2 mod 13 and 2^5 = -1 mod 13.

**Soroban** · November 26th 2012, 05:11 PM

Hello, franios!

$\text{Find the remainder when }11^{121}\text{ is divided by 13.}$

Note that: . $11 \:\equiv\: \text{-}2\text{ (mod 13)}$

Also that: . $11^6 \:=\:(\text{-}2)^6\text{ (mod 13)} \:=\:64 \quad\Rightarrow\quad 11^6 \:\equiv\: \text{-}1\text{ (mod 13)}$

Hence: . $11^{121} \:=\:11^{6\cdot20+1} \;=\;11^{6\cdot20}\cdot 11^1 \;=\;(11^6)^{20}\cdot 11$

Therefore: . $11^{121} \;\equiv\;(\text{-}1)^{20}\cdot 11\text{ (mod 13)} \;\equiv\;11\text{ (mod 13)}$

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