Last digit

6 power any number ends with 6, same for 5 also. 4 power if odd then end with 4 else end with 6 . so,adding last digit of number 6+5+4+9+2+1=27

Let $l_{d}(x)$ denotre the last digit of x, that is, $l_{d}(x)=x\:mod\:10$ .

You can see that, for $n \in \mathbb{N}$ , $l_{d}(6^{n})=6$ , and $l_{d}(5^{n})=5$ .

For $x \in (\mathbb{N} \backslash 2\mathbb{N})$ , $l_{d}(4^{x})=4$ .

$3^{2^{1}}=9, 2^{1}=2$ , and $1=1$ .

Therefore, as $l_{d}(x)=(l_{d}(y)+l_{d}(z))\:mod\:10$ , where $x=y+z$ , $l_{d}(6^{5^{4^{3^{2^{1}}}}}+5^{4^{3^{2^{1}}}}+4^{3^{2^{1}}}+3^{2^{1}}+2^{1}+1)=(l_{d}(6^{5^{4^{3^{2^{1}}}}})+l_{d}(5^{4^{3^{2^{1}}}})+l_{d}(4^{3^{2^{1}}})+l_{d}(3^{2^{1}})+l_{d}(2^{1})+l_{d}(1))\:mod\:10$ .

By the above, we know this is to be equal to $(6+5+4+9+2+1)\:mod\:10=27\:mod\:10=7$ .