3^98 dividedo by 5 , the remainder ?Please send method/ trick & video also in hindi

3^98.By solving to the log..Log(3^98)=98log3=98(0.4771)=Anti(46.7558)=5.699×10^46=56990×10^42=56990×10^42÷5=11398×10^4=remainder=0

We can write 3^49 as (3^2)^49 and it can be further written as (10-1)^49 and after binomial expansion every term is divisible by 5 except the last term which is -1, if we add and subtract 4 then after that remainder is 4