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Find the last two digits of the given number

$1) 21^{99} $

$2) 41^{9999} $

$3) 31^{2019} $

$4) 91^{2018!} $
in Numerical Ability by Boss (47.4k points) | 76 views
0
For 1 im getting 81
0
For 2, getting 61.

please tell if these answers are correct or not
0
$1)81$

$2)61$

$3)71$

$4)01$

i have problem in $4)$?
0
yes getting 71 and 01 for second last and last what you are not getting in 4th??
0
can you explain the concept of $4)$?
0
I have a problem on $2018!$
+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]@@@@@00C0 * 90* 1 @@@@@@00 +  @@@@@@00C1 * 90*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.

0
Yeah i got it

Thanks Brother
0
Can you share how u did??
+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
thanks!
0

@Utkarsh Joshi

My concept is right??

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

+1
what u said is applicable for the specific set of problems. for digits ending with 5, exponent greater than 2 will always have 25 at the end straightforward right?
0
$5^{1}=05$

and $5^{2}=25$

       $5^{3}=..25$

     ---------------------------

     ----------------------------

$5^{N} =........25$ where $N$ belong to the Natural number.

this is right?

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