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 563 views answer comment Share Follow See all 14 Comments See all 14 14 Comments reply Show 11 previous comments 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.