what will be the sum of 202020202 4 considering last 5 digits

last five digits is remainder dividing 10^5
(202020202)^4 mod10^5
=>(20202)^4 mod10^5
=>(20202)^2(20202)^2 mod10^5
=>20804*20804 mod 10^5 since ABmodN=(AmodN*BmodN)modN
=>432806416 mod 10^5
06416 is the remainder and they are the last five digits
sum is 17
8 years agoHelpfull: Yes(7) No(0)
the ans. is = 17
8 years agoHelpfull: Yes(4) No(1)
considering last 5 digits and multiplying again with the last 5 digits of answer finally we get 62416 in last 5 digs as result which sums to 19 hence ans
8 years agoHelpfull: Yes(2) No(1)
ans is 60960
8 years agoHelpfull: Yes(1) No(1)
2^4 (101010101)^4
Now Take last 5 digits 10101

10101^2= take last 5 digits of the result =30201 again 30201*10101=60301 last 5 digit
60301*10101=00401 last 5

Now 004011*2^4=06416 last 5 (ans)
8 years agoHelpfull: Yes(1) No(2)
17 is the right ans
2^4*(101010101)^4 then after solving this eq. we get last 5 digit will be 06416.
after that the sum of last digit will be 17.
7 years agoHelpfull: Yes(0) No(0)
By the calculator ==1.6656325684300937290963763216064e+33
So answer is :1+6+0+6+4=17
according to trick---
202020202^4 = 2^4(101010101)^4
choose last five digits for 101010101=>10101^4
10101*10101=30201(take only last five digit)
30201*30201= 00401(take only last five digit)
so 16*00401=6416
->6+4+1+6=17
7 years agoHelpfull: Yes(0) No(1)

what will be the sum of (202020202)^4 considering last 5 digits