How to get last two digits of 2^222 without using calculator?

Question

As per question above, how to get last two digits of 2^222 without using calculator?

I need last two digits not last digit and show me the steps to get it.

Answer 1

04 is the answer.

2^1 = 2; last digit 2
2^2 = 4; last digit 4
2^3 = 8; last digit 8
2^4 = 16; last digit 6

last digit 2 (the cycle 2-4-8-6 is repeating)
etc.
2^x will end in 6 if x divisible by 4
2^x will end in 2 if x-1 divisible by 4
2^x will end in 4 if x-2 divisible by 4
2^x will end in 8 if x-3 divisible by 4

likewise, every value of 2^x where x-2 divisible by 4 cycles:
(multiply each by 16 to get the next)
04 (2^2, 2^22, 2^42, ...)
64 (or -36)(2^6, 2^26, 2^46,...)
24 (or -76)(2^10, 2^30, 2^50, ...)
84 (or -16)(2^14, 2^34, ...)
44 (or -56)(2^18, 2^38,...)

so 2^222 ends in 04.

Answer 2

use mod 100

we know

2^13 = 8192 mod 100 or -8 mod 100 (as easy to compute with 8)

so 2^221 = (2^13)^17 = (-8)^17 = - 2^51 mod 100

multiply by 2

2^222 = -2^52 mod 100
= - (2^13)^4 mod 100
= -(-8)^4 mod 100
= -2^12 mod 100
= -4096 mod 100
= -96 mod 100
= 04

These are last 2 digits that is 04

Answer 3

Maybe this might help.
2^222 (take logs).
Log 2^222
222 Log 2
222*0∙301 029 995
66∙828 659 04 (take Anti-log).
6∙739 986 66... x 10^66
It looks like the last two digits are 66.

Answer 4

04:

2¹º = 1024 ≡ -1 (mod 25)
2²²² = 4 * 2²²º = 4 * (2¹º)²² ≡ 4 (mod 25)

so two digits of 2^222 is: 04, 29, 54, 79
On the other hand, 2²²² = 4 * 2²²º , it follow two digits of 2^222 is multiple of 4.
Thus it is 04

Answer 5

by using mod 10
only the last one digit can be found.

Answer 6

if:
2^1= ...000010
2^2= ...000100
2^3= ...001000
...

for 2^222 or 2^n where n>1
the last two digits would be 00
at least written in binary

Answer 7

only the last one digit can be found.