Last 2 digits of an Expression

Last 2 digits is the remainder when Divided by 100.
This can be found usind Binomial theorem.
The method described below is quick and is
an extension of the Binomial Theorem.

For expressions ending in '5'
Last 2 digits will always be 25.
For '1'
I will first so some examples and then follow them with an
explanation.
Take 41 ^ 73
The last digit/ the right most digit will always be 1
For the last but one digit take the ten's digit of the base i.e 4 in the above
example and multiply it will the unit's digit of the power i.e 3 in the example.
So 2.
The last 2 digits will be 21.
Some more examples.
31 ^ 123 =
Right most is one.
Last but one will be 3 (ten's digit of the base) * 3 (unit's digit of the power)=9
So last 2 digits will be 91
81 ^ 1423 = 41
Last but one here will be 8*3


For Numbers ending in 3,7,9

From the power cycles,
3 ^ 4k/4k+4 and 7 ^ 4k/4k+4 , 9 ^ 2k will end in 1.
So using this we can convert them into expressions ending in 1.
233 ^ 56
The last 2 digits will depend on 33 ^ 56
This becomes (33 ^ 4 ) ^ 14
33 ^ 4 is (33 ^ 2 ) * (33 ^ 2)
I have already written about quicker methods to get squares.
33 ^ 2 will end in 89
Finally we get 21 ^ 14
This can be found using the method explained for 1.


For expressions ending in even numbers

Some pointers to remember
1024 ^ even number will always end in 76
1024 ^ odd number will always end in 24
1024 is 2 ^ 10
For most expressions ending in an even number,
we can take 2 ^ x as common and express that as 10k+
Say we have 74 ^ 288
74 is 2 * 37
so we get (2 ^ 288)*(37^288)
2 ^ 288 will be (2 ^ 10)^28 * (2 ^ 8)
We can proceed with this
and use earlier methods to get 37 ^ 288.


Agreed it's not an easy method by any means.