1Remainder for 211 is 2.
Remainder for 233 is 2.
Remainder for 255 is 2.
So that should follow as
Remainder for 219911991 is 2...
7 Answers
1
11abhishek the logic used by you is not suffice .In maths you dont go by objective approach.(although it helps me in my phase tests)
btw the philosophy, the question is a simple application of the fermat theorem
1Never heard of the Fermat Theorem......
@ jeetopper : Can u post the solution ?
21@ abhisek
is not 2^1/1 leaves remainder zero??
ur observation only holds if the power of 2 is prime..
and actually this is fermat theorem. which says a ^p = a mod p where p is prime..
A = b mod c means A-b can be divided by c or A leaves remainder b ,when divided by C( if and only if C> b).
example : 64 = 1 mod 3 also 64 = 4 mod 3,(but 1 is the remainder not 4)
Now coming to the question, in order to apply fermat theorem we have to pray that 1991 is prime.
as 1991 is not prime ,i think fermat theorem can not be applied here///
there is another theorem of fermat which says (A+B)^ P =( A^p + B^p) mod p,again p is prime.....
1Yes i agree , the 1st statement (remainder of 21/1 = 2) is incorrect....and i got the Fermat's Theorem....
1HEY GIVE ME THE ANSWER.....I WANT TO CHECK WHETHER IT IS CORRECT OR NOT...
