What is the Remainder when Dividing a Number with 1000 Ones by 1001?

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The problem involves finding the remainder when a number consisting of 1000 ones is divided by 1001. The calculation shows that the number can be expressed as (10^999 - 1)/9, and through modular arithmetic, it is determined that 10^1000 - 1 is congruent to 990 modulo 1001. The elegant solution presented involves recognizing patterns in the digits and simplifying the problem to finding the remainder of 1111 when divided by 1001, which also results in 110. The discussion confirms that the remainder is 110, and acknowledges the neatness of the alternative solution provided. The conclusion is that both methods lead to the same remainder of 110.
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here is the problem:
1. what is the remainder if 11...1 is divided by 1001? (the dividend has 1000 1's)

this is what i did (please tell me if there is any other methods of doing it):

11...1

=10^{999} + 10^{998} + \cdots + 10^{0}

=\frac{10^{999} - 1}{9}

i noticed that
1000 = 10^{3} \equiv -1 \left(\bmod 1001\right)

10^{6} \equiv 1 \left(\bmod 1001\right)

10^{996} \equiv 1 \left(\bmod 1001\right)

10^{996}.10^{4} \equiv 10^{4} \equiv 991 \left(\bmod 1001\right)

10^{1000} - 1 \equiv 990 \left(\bmod 1001\right)

\frac{10^{999} - 1}{9} \equiv 110 \left(\bmod 1001\right)

is there any problem in my method? is there any other easier way of doing it?
 
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There is an elegant solution that looks like this:

100100
010010
001001

each of those is obviously divisble by 1001, and their sum is

111111

there are 6 1's, so we can delete all but the first 1000 mod 6 =4 1s and we only need to work out the remainder of 1111 on division by 1001, and that is obviously 110.
 
thanks a lot matt grime. your solution is really neat.
 
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