Quote:
Originally Posted by sharulgoel please give a detailed solution..i dont know the correct answer will find it out
but if u got the answer then plz share the solution too.
thanks |
I thin the ans is 666.
Two appraoches can be used
1) 1212121212...3000 digits = N
reprsenting in form of coprinmes
N=37*k1 = 27* k2 + 18
Solving this for k2=24, we get N=666.
2) 1212121212... 300 digits can be written as
=12121212(297 digit)*1000 + 212 mod 999
=12121212(297 digit)*(999+1) + 212 mod 999
=12121212(297 digit)*1 + 212 mod 999
=12121212(294 digit)*1000 + 121 + 212 mod 999
=12121212(294 digit)*(999+1) + 121 + 212 mod 999
= ......
( 121 + 212 ) * 50 mod 999 = 666.