binomial
although i was able to get the ans by congruence modulo... any idea how to do this by binomial theorem? i tried by writing them as 19+4 and 19-4 but did not get anything useful
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@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Consider division by 19, there's no remainder by binomial. Now consider 3 and write 23 as (24-1)
how do i approach this using binomial tho
i did the exact same thing while using congruence
tried by this way for binomial
forgot to multiply 4^2 in third and fourth line second term mb
After this, if you wanna do this way, write out 19 everywhere to 18+1, then you get a double summation kinda thingy
Solve that to get some k.2^m kinda stuff
After that write it out as k.(3-1)^m
After which you get the remainder.
So yea...using 19 and 3 is better
yeahh my method is really messy 19 and 3 se sahi se ho ja rha hai
thankss
+solved @SirLancelotDuLac
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