Question number 23. For how many numbers n from the set 1, 2, 3, all the way up to 99 and 100, so the first 100 consecutive integers, is n to the power n, the square of a natural number. And what we can do, first of all, is demand that n is even. If n is even, then n divided by 2 is a natural number. And also, if we take n to the power n over 2, is a natural number. So we write n to the power n is equal to n to the power n over 2, and then that number we square. The exponents multiply to n, and so we have something that's mathematically equivalent. There are 50 even numbers in the set, and so we have at least 50 numbers here as candidates. But are there others that are valid? For example, odds. And it's easy to see that when n is equal to 1, everything works. 1 to the first power is 1, and that's also equal to 1 squared. Are there others? So of the odd numbers, can we, for example, use a number like 9? When n is equal to 9, 9 to the ninth power, we can rewrite as 3 squared to the ninth power. And even though this is not the square of a natural number, we can switch the order of taking exponents. We can raise 3 to the power 9 first, and then square that, so that in the end, the exponents multiply to 18, and so this works. So what we need, it looks like, are odd numbers that are perfect squares. So let's look at all the perfect squares that we have. We can have 1 squared, that's 1. 2 squared doesn't work. 3 squared works. 3 squared is 9. 4 squared doesn't work. 5 squared works, 25. 6 squared doesn't work. 7 squared, 49 will be fine. 8 squared will not work. And then 9 squared, 81 will be fine. And then we're past our range. So we take these five numbers here, these five, together with the 50 numbers, and those are all the numbers in the range from 1 to 100, for which that number raised to its own exponent is equal to the square of a natural number. So the answer here is B, 55.

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