Last digit

$2^{35}=(1024)^{3}\times{32}\equiv4^{3}\times{2^{5}}\equiv64\times{32}\equiv4\times{2}\equiv8\pmod{10}$

to find last digit in b^i where b is base ans i is index.

Always find remainder of index ÷ 4,

if remainder is 1 then unit digit is, unit digit of base^1

if remainder is 2 then unit digit is, unit digit of base^2

if remainder is 3 then unit digit is, unit digit of base^3

if remainder is 0 then unit digit is, unit digit of base^4

in 2^35, 35 ÷ 4, remainder is 3

so unit digit is unit digit of 2^3 = 8

let's make a list using brute force:

• $2^1 \equiv 2 \pmod {10}$
• $2^2 \equiv 4 \pmod {10}$
• $2^3 \equiv 8 \pmod {10}$

• $2^4 \equiv 6 \pmod {10}$

• $2^5 \equiv 2 \pmod {10}$

• $2^6 \equiv 4 \pmod {10}$
• $2^7 \equiv 8 \pmod {10}$
• $2^8 \equiv 6 \pmod {10}$

the pattern here is $[2,4,8,6,2,4,8,6\dots6]$

so we need to divide the exponent by 4

$35 \equiv 3 \pmod 4$ and if the remainder was $3$ then the last digit is $8$

1
2
3
4
5
6
7
8

if exponent%4==1:
    return 2
else if exponent%4==2:
    return 4
else if exponent%4==3:
   return 8
else:
    return 6

$2^1=2$

$2^2=4$

$2^3=8$

$2^4=16$

$2^5=32$

The units digits repeat in every cycle of $4$ . So we divide $35$ by $4$ . We have

$\dfrac{35}{4}=8$ remainder $3$ . So the units digit is $8$ .