6.1. In a row, the numbers are written: $100^{100}, 101^{101}, 102^{102}, \ldots, 876^{876}$ (i.e., the numbers of the form $n^{n}$ for natural numbers n from 100 to 876.) How many of the written numbers are perfect cubes? (A perfect cube is the cube of an integer.)
# Answer: 262

Solution. Consider a number of the form $m^{k}$, where $m$ and $k$ are natural numbers. If $k$ is divisible by 3, then $m^{k}$ is a perfect cube. Otherwise, $m^{k}$ is a perfect cube if and only if $m$ is a perfect cube. Thus, the answer to our problem is the total number of numbers that are divisible by 3 and the perfect cubes that are not divisible by 3 in the set of numbers $\{100,101,102, \ldots, 876\}$. The number of needed numbers divisible by 3 is $(876-99) / 3=259$. The number of needed perfect cubes is exactly $3: 5^{3}, 7^{3}, 8^{3}$. Therefore, the total number of required numbers is $259+3=262$.