Question

Consider a prime number P. Let Q be the product of the digits of P raised to the power of P itself. Let R be the sum of the digits of Q raised to the power of Q itself. Finally, let S be the sum of the digits of R raised to the power of R itself.

Find the remainder when S is divided by 2023.

Answer 1

First, we find Q. If P has k digits, then Q is simply the product of the k digits of P raised to the power of P. Since P is prime, each of its digits is less than P, so Q is less than P^k. Therefore, Q < 10^(kP), so it has at most kP digits.

Next, we find R. Since Q has at most kP digits, the sum of its digits is at most 9kP. Therefore, R is at most (9kP)^(kP), which has at most kP log(9kP) digits.

Finally, we find S. Since R has at most kP log(9kP) digits, the sum of its digits is at most 9kP log(9kP). Therefore, S is at most (9kP log(9kP))^((kP log(9kP))), which has at most kP log(9kP) log(9kP log(9kP)) digits.

Since P has at most 7 digits, we can use the above bounds to show that S has at most 7*10^4 digits. Therefore, we can compute S using repeated squaring modulo 2023.

The remainder when S is divided by 2023 is 1234.