1 votes 1 votes Find the last two digits of the given number $21^{99} $ $41^{9999} $ $31^{2019} $ $91^{2018!} $ Quantitative Aptitude quantitative-aptitude number-system descriptive + – Lakshman Bhaiya asked Oct 18, 2018 • edited Jan 20, 2020 by Lakshman Bhaiya Lakshman Bhaiya 622 views answer comment Share Follow See all 14 Comments See all 14 14 Comments reply Utkarsh Joshi commented Oct 18, 2018 i edited by Utkarsh Joshi Oct 18, 2018 reply Follow Share For 1 im getting 81 0 votes 0 votes Utkarsh Joshi commented Oct 18, 2018 reply Follow Share For 2, getting 61. please tell if these answers are correct or not 0 votes 0 votes Lakshman Bhaiya commented Oct 18, 2018 reply Follow Share $1)81$ $2)61$ $3)71$ $4)01$ i have problem in $4)$? 0 votes 0 votes Utkarsh Joshi commented Oct 18, 2018 reply Follow Share yes getting 71 and 01 for second last and last what you are not getting in 4th?? 0 votes 0 votes Lakshman Bhaiya commented Oct 18, 2018 reply Follow Share can you explain the concept of $4)$? 0 votes 0 votes Lakshman Bhaiya commented Oct 18, 2018 reply Follow Share I have a problem on $2018!$ 0 votes 0 votes Utkarsh Joshi commented Oct 18, 2018 reply Follow Share 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. =@@@@@@00C0 * 900 * 1 @@@@@@00 + @@@@@@00C1 * 901 *1@@@@@@@00-1 ...... so on = 1 + !!!@@@@00 + !!@@##$$ 00 + ...... so on so only first term will contribute for last two digits. Hence last two digits will be 01. 2 votes 2 votes Lakshman Bhaiya commented Oct 18, 2018 reply Follow Share Yeah i got it Thanks Brother 0 votes 0 votes Utkarsh Joshi commented Oct 18, 2018 reply Follow Share Can you share how u did?? 0 votes 0 votes Lakshman Bhaiya commented Oct 18, 2018 reply Follow Share 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!! 2 votes 2 votes Utkarsh Joshi commented Oct 18, 2018 reply Follow Share thanks! 0 votes 0 votes Lakshman Bhaiya commented Oct 18, 2018 reply Follow Share @Utkarsh Joshi My concept is right?? I have the problem of finding the last $2$ digits of $5^{6525656}$ and $5^{0000001}$? 0 votes 0 votes Utkarsh Joshi commented Oct 18, 2018 reply Follow Share 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? 1 votes 1 votes Lakshman Bhaiya commented Oct 18, 2018 reply Follow Share $5^{1}=05$ and $5^{2}=25$ $5^{3}=..25$ --------------------------- ---------------------------- $5^{N} =........25$ where $N$ belong to the Natural number. this is right? 0 votes 0 votes Please log in or register to add a comment.