Mathematics question urgent

the numbers 1 to 2011 are written side by side as follows 1234 ............... 20102011 . if the number is divided by 9, then what is the remainder?

A) 0
B) 1
C ) 2
D ) 3

There is a theorem that says that the remainder of a number that is divided by 9 is the same as the remainder when the sum of its digits is divided by 9.

So the remainder of N = 1234...20102011 when divided by 9 is the same as the remainder when

(1) + (2) + (3) + ... + (2 + 0 + 1 + 0) + (2 + 0 + 1 + 1) is divided by 9.

We group this sum into groups of 9

( (1) + (2) + ... + (9) ) +
( (1+ 0) + (1 + 1) + (1 + 2) + ... + (1 + 8) ) +
( (1 + 9) + (2 + 0) + (2 + 1) + ... + (2 + 7) ) +
.
.
.
( (1 + 9 + 9 + 9) + (2 + 0 + 0 + 0) + ... + (2 + 0 + 0 + 7) )

and the last group only consists of

((2 + 0 + 0 + 8) + (2 + 0 + 0 + 9) + (2 + 0 + 1 + 0) + (2 + 0 + 1 + 1) )

Inside of each group except the last, the remainder when the sum is divided by 9 will be 0.

The last sum is 10 + 11 + 12 + 13 = 46; which has a remainder of 1 when divided by 9.

So the answer is B.

As we run from 1, 2, 3, ..., 2011, the numbers cycle through 1, 2, 3, ..., 8, 0 modulo 9. We also know that the numbers belong to the same modulo class as their digit sums. So, if we concatenate the digits of an entire cycle, the digit sum will be:

1 + 2 + ... + 8 + 0 = 0 (mod 9)

Therefore, an entire cycle will add to give a multiple of 9. So, where does 2011 fall in this cycle? Well:

2011 = 2 + 0 + 1 + 1 = 4 (mod 9)

So, the last four numbers, 2008, 2009, 2010, and 2011 form a partial cycle of 1, 2, 3, 4 (mod 9). Since the other numbers' digits sum to a multiple of 9, this large number will be congruent to:

1 + 2 + 3 + 4 = 1 (mod 9)

Therefore, the answer is B).

1 is the answer...
to check whether a number is divisible by 9,we will add the component numbers in it and see if the result is divisible by it...
eg.lets look at the no 738...the sum of its component 7+3+8=18,which is divisible by 9 and hence 738 is divisible by 9..
in this case,the sum of the component number is 1+2+3+4+....2010+2011 which is equal to 2023066(it can be obtained by the formula tha sum to n term from one is n*(n+1)/2..here it is 2011*2012/2)
now we have to check whether 2023066 is divisible by 9..2+0+2+3+0+6+6=19...18 is divisible by 9..so we get 1 as the remainder if we divide 19 by 9..hence we get 1 as the remainder if we divide 2023066.hence we also get 1 as the remainder if we divide the given figure with 9..

The answer is B: 1

Adding up all the numbers from 1 to 2011 we get 2023066
Then we add up all the digits together and we get 19
dividing 19 by 9, we get a remainder of 1.