Congruences

**edwardteo.368** · October 10th 2007, 06:09 PM

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

**Soroban** · October 10th 2007, 07:51 PM

Hello, edwardteo.368!

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

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

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

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

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

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

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

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

Therefore, $N$ is a multiple of 11.

**edwardteo.368** · October 10th 2007, 07:59 PM

thnx for the help

Thread: Congruences

LinkBack

Thread Tools

Display

Congruences

Similar Math Help Forum Discussions

Congruences

Congruences

Congruences

More congruences

Congruences

Search Tags