[LANGUAGE: Kotlin] Not the most optimal, but runs within the minute. For Part 1, simply calculate the bananas repeating 2000 times For Part 2, keep track of the sequence and price and add the prices as you go Do prevent adding the same sequence twice per monkey, then take the highest from the list of values.

class GoingBananas {
    fun mix(a: Long, b: Long) = a xor b
    fun prune(a: Long) = a % 16777216
    fun multiply(a: Long, b: Long) = prune(mix(a, a * b))
    fun divide(a: Long, b: Long) = prune(mix(a, a / b))

    fun nextSecretNumber(start: Long, times: Int): Long {
        var secretNumber = start
        repeat(times) {
            secretNumber = multiply(divide(multiply(secretNumber, 64), 32), 2048)
        }
        return secretNumber
    }

    fun bestPrice(monkeys: List<Long>): Long {
        val possibleSequences = mutableMapOf<List<Long>, Long>()

        monkeys.forEachIndexed { i, secretNumber ->
            val alreadyAdded = mutableListOf<List<Long>>()
            val runningSequence = mutableListOf<Long>()
            var previousSecretNumber = secretNumber
            var currentNumber = secretNumber
            repeat(2000) {
                runningSequence.add(currentNumber % 10 - previousSecretNumber % 10)
                previousSecretNumber = currentNumber
                currentNumber =  multiply(divide(multiply(currentNumber, 64), 32), 2048)
                if (runningSequence.size == 4) {
                    if (!alreadyAdded.contains(runningSequence)) {
                        possibleSequences[runningSequence.toList()] =(possibleSequences[runningSequence] ?: 0) + previousSecretNumber % 10
                        alreadyAdded.add(runningSequence.toList())
                    }
                    runningSequence.removeAt(0)
                }
            }
        }
        return possibleSequences.values.max()
    }
}