53. Find a three-digit number, every integer power of which ends in the three digits that make up the original number.
53. Let \( N \) be the desired number; then, in particular, \( N^2 - N \) ends with three zeros, i.e., it is divisible by 1000. Since \( N^2 - N = N(N-1) \), and \( N \) and \( N-1 \) are coprime numbers, this is only possible if one of these numbers is divisible by 8, and the other by 125 (neither of these numbers can be divisible by 1000 on its own, as \( N \) is a three-digit number).

If \( N \) is a three-digit number divisible by 125, then \( N-1 \), as it is easy to verify, will be divisible by 8 only if \( N = 625, N-1 = 624 \). It is also easy to verify that if \( N-1 \) is a three-digit number divisible by 125, then \( N \) will be divisible by 8 only if \( N-1 = 375, N = 376 \).

Notice now that since \( N^{k-1} - 1 \) for any integer \( k \geqslant 2 \) is divisible by \( N-1 \), then \( N^k - N = N(N^{k-1} - 1) \) for any integer \( k \) will be divisible by \( N(N-1) = N^2 - N \); therefore, if \( N^2 - N \) ends with three zeros, then \( N^k - N \) for any integer \( k \geqslant 2 \) will also end with three zeros, i.e., \( N^k \) will end with the same three digits as \( N \). From this, it follows that the numbers 625 and 376 (and only they) satisfy the conditions of the problem.