Oh my! 100th powers!

Let N be the number of positive integers m for which

31002100(modm)3^{100}\equiv2^{100} \pmod m

Then N can be expressed as 2k2^k

Find the value of kk.

Details For those who don't know Mod notation :

ab(modc)a\equiv b \pmod c if and only if "aa" and "bb" give the SAME remainder when divided by cc.

And this information is enough to find the trick which makes the problem easy.


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