Last digit

Finding the last $n$ digits amounts to finding the remainder when divided by $10^{n}$ :

Lets consider some powers of 2 relative to 35:

$35 = 5 \times 7 \\ 2^5 = 32 \equiv 2 \pmod {10} \\ 2^7 = 2^5 \times 2^2 \equiv 2 \times 4 \equiv 8 \pmod {10}$

Now lets solve for $2^{35}$ in $\pmod {10}$

$2^5 \equiv 2 \pmod {10} \\ \therefore 2^{35} \text{ which can be written as } (2^5)^7 \equiv 2^7 \equiv 8 \pmod {10} \\ \text{ Hence the answer is } 8..$

When 2 is multiplied by itself repeatedly, the last digit of the answers goes in a pattern of 2, 4, 8, 6 and repeats itself. Since 2 is raised to 35, divide 35 by 4; the answer is 8.75; there is a remainder of 3. The 3rd number in the pattern is 8. Therefore, the last digit is 8.

$2^5=2(mod 10)$ $→2^{35}=2^5*2^2(mod 10)$ $→2^{35}= 2*2*2( mod 10)$ $→2^{35}=8 (mod 10)$

2^35=2^10 )^3+2^5 =1024^3×2^5 =4^3×32 =64×32 & 4×2 = 8

$35 \equiv 3 ( \mod 4 )$ , so $2 ^ { n } =$ repeating the pattern of last digits, 2, 4, 8, and 6, so it will be $\boxed { 8 }$ .

2 ^ 5 (mod 10) = 2 i s the minimum mod value we can get. here 35 / 5 = 7 . So ( 2 ^ 5 ) ^ 7 . And now putting the mod value of 2 ^ 5 = 2 we get, ( 2 ) ^ 7 = 128 Now 128 (mod 10) = 8 . So the last digit is 8!

as 36^0.5 = 6 we can just use 6'es and divide by 2 since it is 35. (((2^6)^6)/2)(mod10)