$$\\1+4^{34}+7^{35}\\
4^n=4,16,64,256,.... $ when n is even it ends in 6$\\
7^n=7,49,343,2401, 16807, $ So the last digits are 7,9,3,1,7,...$\\
35=4*8+3$ so the last digit of $ 7^{35}$ is 3 (3 is the third number in the sequence)$\\
$Edited: Geno is right, these three digits need to be added$\\\\
$ The last digit will be hte last digit of $ 6+3+1 $ Which is zero $$$
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