52. Olya wrote fractions of the form $1 / n$ on cards, where $n-$ are all possible divisors of the number $6^{100}$ (including one and the number itself). She arranged these cards in some order. After that, she wrote the number on the first card on the board, then the sum of the numbers on the first and second cards, then the sum of the numbers on the first three cards, and so on, finally, the sum of the numbers on all the cards. Each sum Olya wrote on the board as an irreducible fraction. What is the smallest number of different denominators that could appear on the board?

(O. Ivanova)
52. Answer: two denominators. Let's represent all fractions in the form $a_{n} / 6^{100}$, then $a_{1}, a_{2}, \ldots$ are again all divisors of the number $6^{100}$, each appearing once. Let the partial sums be denoted by $S_{n} / 6^{100}$. Then the denominator of the irreducible representation of the partial sum depends on the powers of the prime factors 2 and 3 in the numerator $S_{n}$.

It is not difficult to understand that among $S_{n}$ there are both even and odd numbers, so in some $S_{n}$, the factor 2 enters with a zero power, and in others, with a non-zero power. Therefore, the answer in the problem is at least 2.

Let's provide an example of an arrangement with two different denominators. First, write down the divisor $a_{1}=1$. Then write down in any order all divisors that are multiples of 2 and 3. Up to this point, all partial sums $S_{n}$ are neither divisible by 2 nor by 3, i.e., they contain 2 and 3 in the same (zero) power, and the last of them gives a remainder of 1 when divided by both 2 and 3. It remains to write down all divisors of the form $2^{k}$ and $3^{k}$ ( $k$ from 1 to 100). First, write down the powers of two, alternating $2^{k} \equiv 1(\bmod 3)$ and $2^{k} \equiv 2(\bmod 3)$ (i.e., powers with even and odd exponents). All the resulting partial sums will still be odd and not divisible by three. Finally, write down the powers of three, alternating their remainders modulo 4 so that all remaining partial sums are not divisible by 4. In this case, they will remain not divisible by three, and the factor 2 will enter them no more than in the first power! Thus, in this example, the factor 3 enters all $S_{n}$ in the zero power, and the factor 2 in the zero or first power.