### Instruction: The number $85^9 - 21^9 + 6^9$ is divisible by an integer between 2000 and 3000. compute that integer.### Response: This problem looks tricky, since the numbers are very large and it is not clear how to factorize them or find their remainders when divided by a number in the range 2000-3000.
However, I notice that all three numbers have a common factor of 3, since the sum of their digits is divisible by 3.
So I can rewrite the expression as $3(28^9 - 7^9 + 2^9)$.
Now, I wonder if I can use some modular arithmetic to simplify the problem further.
I recall that Fermat's little theorem states that if p is a prime number and a is not divisible by p, then $a^{p-1} \equiv 1 \pmod p$.
This means that, for example, $28^8 \equiv 1 \pmod 29$, since 29 is prime and 28 is not divisible by 29.
Similarly, $7^8 \equiv 1 \pmod 29$ and $2^8 \equiv 1 \pmod 29$.
So I can use this fact to reduce the powers of 28, 7, and 2 modulo 29.
For instance, $28^9 \equiv 28 \times 28^8 \equiv 28 \times 1 \equiv 28 \pmod 29$.
Likewise, $7^9 \equiv 7 \times 7^8 \equiv 7 \times 1 \equiv 7 \pmod 29$.
And $2^9 \equiv 2 \times 2^8 \equiv 2 \times 1 \equiv 2 \pmod 29$.
Therefore, $28^9 - 7^9 + 2^9 \equiv 28 - 7 + 2 \equiv 23 \pmod 29$.
Multiplying both sides by 3, we get $3(28^9 - 7^9 + 2^9) \equiv 3 \times 23 \equiv 69 \pmod 29$.
This means that the original expression is divisible by any divisor of 69 that is also a multiple of 29.
The only such number in the range 2000-3000 is 29 times 69, which is 2001.