Show (23^(37)+15^(16)) is congurent to 5 (mod 18).

Not sure how to do this need some help please!!

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- Dec 13th 2009, 11:03 AMsteph3824Need help with this congruency problem
Show (23^(37)+15^(16)) is congurent to 5 (mod 18).

Not sure how to do this need some help please!! - Dec 13th 2009, 12:46 PMtonio

I did it two ways and I get something different. The following is done modulo 18:

$\displaystyle 23^{37}+15^{16}=5^{37}+5^{16}\cdot 3^{16}=5^{16}\left(5^{21}+3^{16}\right)=\left(5^3\ right)^5\cdot 5\left(\left(5^3\right)^5+\left(3^4\right)^4\right )$ $\displaystyle =(-1)^5\cdot 5\left((-1)^5+9\right)=(-5)\left((-1)+9\right)=(-5)\cdot 8=-40=-4=14\!\!\!\!\pmod{18}$

Of course, I could be wrong...

Tonio - Dec 13th 2009, 03:03 PMaidan