<s>[INST] Number in tens place Can you figure out the digit in the tens place of the solution to $4^{2015} \cdot 9^{2016}$ without using a calculator?
By tens place I mean, for example if you have number $2451$, the number in tens place here is $5$.
I know the answer, but I don't know how to get it, so if you got any ideas, please share. :)
 [/INST] You have
$$4^{11}=4194304\equiv 4 \pmod{100}$$
And since $2015=183\times 11+2$, and $183=16\times 11+7$, you have
$$4^{2015}=4^{11\times183+2}=4^2\times(4^{11})^{183}\equiv 4^2\times 4^{183} \pmod{100}$$
$$=4^2\times4^{11\times16+7}=4^9\times(4^{11})^{16}\equiv 4^9\times4^{16}=4^{25}=4^3\times(4^{11})^2\equiv4^5 \pmod{100}$$
$$=1024\equiv 24\pmod{100}$$
Likewise,
$$9^{10}=3486784401\equiv1\pmod{100}$$
Thus
$$9^{2016}=9^6\times (9^{10})^{201}\equiv 9^6=531441\equiv 41\pmod{100}$$
Therefore
$$4^{2015}\times9^{2016}\equiv 24\times41=984\equiv 84\pmod{100}$$
And the digit in tens place is $8$.
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