Let N = 121212... up to 300 digits. What is the remainder when N is divided by 999?
12
121
216
666
Let me write the given 300 digit number as follows:
- 1212 ... 121212 = 1212 ... 121*1000 + 212
= 1212 ... 2121*(999 + 1) + 212
= 1212 ... 2121*999 + 1212 ... 2121 + 212
= 999*(Something) + 1212 ... 2333
Now the first part i.e. 999*(Something) is divisible by 999. Thus the remainder is same as the remainder when 1212 ... 333 is divided by 999.
Now note that 1212 ... 333 has 297 digits. We have to find the remainder by repeating the same method as above. But don't worry. It's not much cumbersome as it looks like!
Let's do the 2nd step.
- 1212 ... 1212333 = 1212 ... 1212*1000 + 333
= 1212 ... 1212*(999 + 1) + 333
= 1212 ... 1212*999 + 1212 ... 1212 + 333
= 999*(Something) + 1212 ... 1545
The number 1212 ... 1545 has 294 digits.
Now it is clear that in each step the number will be reduced by 3 digits. Thus there will be a total of 100 such steps. Also note that in each alternate step 121 and 212 will be removed and will be added to the remaining part. Thus in every 2 step 6 digit will be removed and (121+ 212) = 333 will be added.
Therefore after the 100 steps all the digits will be removed and 50 times 333 will remain. The remainder is nothing but the remainder when this reduced number is divided by 999.
The reduced number = 50*333 = (48 + 2)*333 = 48*333 + 2*333 = 16*999 + 666
Thus the remainder is 666.
The correct answer is D.
