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.

×

Problem Loading...

Note Loading...

Set Loading...