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+1 vote

+2

2018! will be a number having more than two zeroes in the end.Let that number be @@@@@@00.

(91)^{2018!}

= (90+1)^{2018!}

= (90+1)^{@@@@@@@00 }

We will expand this.

[email protected]@@@@@00C_{0} * 90^{0 }* 1 ^{@@@@@@00} + @@@@@@00C_{1} * 90^{1 }*1^{@@@@@@@00-1 }...... so on

= 1 + [email protected]@@@00 + [email protected]@##$$ 00 + ...... so on so only first term will contribute for last two digits. Hence last two digits will be 01.

+2

Yes

$1)21^{99}$

If a number has unit digit =1, this case applies

first, take unit digit and write in the unit place. __1

second, take the tens place of the number and multiplied to a unit digit of the power

in this example, $2*9=18$ take an only unit digit from this and write in the tens place.

we get last $2$ digits = 81

Similarly for $2)41^{9999}$

gives last $2$ digits $=61$

$3)31^{2019}$

gives last $2$ digits $=71$

$4)91^{2018!}$

$(91)^{@@@@...00}$ [10!=3,628,800,so 10! or greater number factorial gives atleast two zeros in the last $2$ digits]

first, take unit digit of the number and write in the unit place __ 1

second, take tense digit of the number and multiplied with a unit digit of the power.

$9*0=0$ write this in the tense place.

So,finally last $2$ digits $= 01$

I hope you understand!!

0

My concept is right??

I have the problem of finding the last $2$ digits of $5^{6525656}$ and $5^{0000001}$?

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