Is 13 divisible by this incredibly huge number?

**BlackBlaze** · January 25th 2010, 07:50 PM

Does 13 divide 6^13 + 33^49?

I can't seem to think of a way to make this number more manageable.
(2^13 * 3^13) + (3^49 * 11^49)
3^13 can be dropped off for our purposes.

2^13 + 3^36 * 11^49

...And then what. Hrm.

**Drexel28** · January 25th 2010, 08:00 PM

Originally Posted by BlackBlaze

Does 13 divide 6^13 + 33^49?

I can't seem to think of a way to make this number more manageable.
(2^13 * 3^13) + (3^49 * 11^49)
3^13 can be dropped off for our purposes.

2^13 + 3^36 * 11^49

...And then what. Hrm.

$13$ is prime and so $\phi(13)=12$ . Also, $33\equiv 7\text{ mod }13$ so that $6^{13}+33^{49}\equiv 6^{13}+7^{49}\text{ mod }13$ . Since $(13,7)=(13,6)=1$ we have that $6^{13}+7^{49}=6\cdot 6^{\phi(13)}+7\cdot 7^{4\phi(13)}\equiv 6+7=13\equiv 0\text{ mod }13$ . So, yes it is divisible.

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