# Congruences

• Oct 10th 2007, 06:09 PM
edwardteo.368
Congruences
This is the problem which i'm currently stuck on, and I've been told to use principles based on mod, congruence equations and Fermat's theorum:

Show that 27 * 23^n + 17 * 10^2n is divisible by 11 for all positive integers n
• Oct 10th 2007, 07:51 PM
Soroban
Hello, edwardteo.368!

Quote:

Show that $\displaystyle 27\!\cdot\!23^n + 17\!\cdot\!10^{2n}$ is divisible by 11 for all positive integers $\displaystyle n$

Let: $\displaystyle N \:=\:27\!\cdot\!23^n + 17\!\cdot\!10^{2n}$

We have: .$\displaystyle N \;\equiv\;5\!\cdot\!1^n + 6\!\cdot\!(\text{-}1)^{2n}\;\pmod{11}$

. . . . . . . . . .$\displaystyle \equiv\;5\!\cdot\!1 + 6\!\cdot\!\left[(\text{-}1)^2\right]^n\;\pmod{11}$

. . . . . . . . . .$\displaystyle \equiv\;5 + 6\cdot1^n\;\pmod{11}$

. . . . . . . . . .$\displaystyle \equiv\;5 + 6\;\pmod{11}$

. . . . . . . . . .$\displaystyle \equiv \;11\;\pmod{11}$

. . . . . . . . . .$\displaystyle \equiv\;0\;\pmod{11}$

Therefore, $\displaystyle N$ is a multiple of 11.

• Oct 10th 2007, 07:59 PM
edwardteo.368
thnx for the help