wats the remainder

u still wrong dude

8² = 4 mod 10

but

8⁴=6 mod 10

dude u r forgetting (aⁿ)^m=a^mn

see from fermat's little theorem,
2^{199}≡ 2 mod(199)
so 2^{1990}≡1024 mod(199)
now as the last digit of 2^{1990} is 4,
we can write 2^{1990}≡1020+4=1024 mod(10)
as 199 and 10 are relatively prime, we can write 2^{1990}≡1024 mod(1990)

try by congruences.........bino sey toh ho hi jayega i guess

is the ans 1024 subhomo.

waise xyz , try out these :

http://www.goiit.com/jforum.htm?module=search&action=search&username=&searchIn=6,3,1,7,12&tmp
SearchIn=6,3,1,7,12&search_keywords=find+the+remainder+when
+13^99+is+divided+by+81&match_type=all&category_version=0&x=0&y=0

by the way anybody noticing any pattern?

nnishant bhaiyan ,

i don see any help with congruencies here , i mean don see a simple method or any pattern to finish off the q

extremely sorry guys you are right i should have said
if aⁿ = c ( mod b)
then a^{n X any integer} = c(mod b)
take for example
4² = 6(mod 10)
then 4⁴ = 6(mod 10)
4⁶ = 6(mod 10) etc.

see this..... http://targetiit.com/iit-jee-forum/posts/typically-olympiad-stuff-9676.html

@saumya u hav done rong....
try using eulers theorem

such tat 13⁵⁴≡1mod81....since since 13 and 81 r coprime \varphi (81)=81(1-1/3)=54.........

13³ = 10 (mod 81)
so 13⁹⁹ = 10 (mod 81)
so the remainder is 10.