Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 4, 2011"

Latest revision as of 10:50, 4 July 2011

$3^x-3 \equiv 0 \mod 16 \Rightarrow 3^x \equiv 3 \mod 16$ .

So $3^x$ cycles through 3,9,11,1 mod 16, $x = 1$ doesn't work so try $x = 5$ which works and we're done.

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 ==Solution==
-<math>3^x-3 \equiv 0 \mod 16 \Rightarrow 3^x \equiv 3 \mod 16</math>
+<math>3^x-3 \equiv 0 \mod 16 \Rightarrow 3^x \equiv 3 \mod 16</math>.
-So <math>3^x</math> cycles through 3,9,11,1 mod 16, <math>x = 1</math> doesn't work so try <math>x = 5</math> which works and we're done
-{{potd_solution}}
+So <math>3^x</math> cycles through 3,9,11,1 mod 16, <math>x = 1</math> doesn't work so try <math>x = 5</math> which works and we're done.