Find the remainder of the division of the number 1000! by $10^{250}$.

#
Let's call the main odd numbers those that are not divisible by 5. According to the formula from problem 60553, \(1000!\) is divisible by \(2^{994} \cdot 5^{249} = 2^{745} \cdot 10^{249}\). Thus, this number ends with 249 zeros, and we need to find the digit preceding them. By representing each number from 1 to 1000 in the form \(2^m 5^n a\), where \(a\) is a main number, we get that this digit coincides with the last digit of the number \(2^{745} P\), where \(P\) is the product of the resulting main numbers. Note that

\(2^{745} = 2 \cdot 16^{186}\) ends in a 2, so it is sufficient to find the last digit of the number \(2P\).

For the pair \((m, n)\), where \(2^m 5^n \leq 1000\), in \(1000!\) there are factors of the form \(2^m 5^n a\), where \(a\) are main numbers not exceeding \(\left[1000 \cdot 2^{-m} 5^{-n}\right]\). Their product we denote as \(P_{m, n}\). Let \(P\) be the product of all such numbers \(P_{m, n}\).

We will compute these products.

The factors of \(P_{0,0}\) are all main numbers from 1 to 1000. Among them, there are 100 numbers ending in 1, 3, 5, 7.

Since \(3 \cdot 7 \equiv 9 \cdot 9 \equiv 1 \pmod{10}\)

\(P_{0,0} \equiv 1 \pmod{10}\)

The factors of \(P_{1,0}\) are all main numbers from 1 to 499. Similarly, \(P_{1,0} \equiv 1 \pmod{10}\).

With \(P_{2,0}\) (main numbers less than 250), the situation changes: for the number 253, there is no matching 7 (257 > 250).

Therefore, \(P_{2,0} \equiv 3 \pmod{10}\).

Similarly, \(P_{3,0} \equiv 3 \pmod{10}\) (for 123, there is no matching 7).

\(P_{4,0} \equiv 1 \pmod{10}\) (main numbers from 1 to 61).

\(P_{5,0} \equiv 9 \pmod{10}\) (main numbers from 1 to 31, the number of 9s is odd).

\(P_{6,0} \equiv 3 \cdot 9 \pmod{10}\) (main numbers \(-1, 3, 7, 9, 11, 13\)).

\(P_{7,0} \equiv 1 \pmod{10}\) (main numbers \(-1, 3, 7\)).

\(P_{8,0} = 3 \pmod{10}\) (main numbers \(-1, 3\)).

\(P_{9,0} = 9 \pmod{10}\) (the only main number is \(-1\)).

\(P_{0,1} \equiv 1 \pmod{10}\) (main numbers less than 200).

\(P_{1,1} \equiv 1 \pmod{10}\) (main numbers less than 100).

\(P_{1,2} \equiv 9 \pmod{10}\) (main numbers from 1 to 49, the number of 9s is odd).

\(P_{1,3} \equiv 3 \pmod{10}\) (main numbers from 1 to 25).

\(P_{1,4} \equiv 9 \pmod{10}\) (main numbers \(-1, 3, 7, 9, 11\)).

\(P_{1,5} = 3 \pmod{10}\) (main numbers \(-1, 3\)).

\(P_{1,6} = 3\) (main numbers \(-1, 3\)).

\(P_{1,7} = 1\) (the only main number is \(-1\)).

\(P_{0,2} \equiv 1 \pmod{10}\) (main numbers less than 40).

\(P_{1,2} \equiv 1 \pmod{10}\) (main numbers less than 20).

\(P_{2,2} \equiv 9 \pmod{10}\) (main numbers \(-1, 3, 7, 9\)).

\(P_{3,2} = 3 \pmod{10}\) (main numbers \(-1, 3\)).

\(P_{3,2} = 1\) (the only main number is \(-1\)).

\(P_{0,3} \equiv 1 \pmod{10}\) (main numbers \(-1, 3, 7\)).

\(P_{1,3} = 3 \pmod{10}\) (main numbers \(-1, 3\)).

\(P_{2,3} = P_{3,3}\) (the only main number is \(-1\)).

Thus, \(P \equiv 3^8 \cdot 9^6 = 9^{10} \equiv 1 \pmod{10}\), so the number \(2P\) ends in a 2.

## Answer

\(2 \cdot 10^{249}\)