MATH SOLVE

2 months ago

Q:
# The sequence 2, 3, 5, 6, 7, 10, 11, $\ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes nor perfect fifth powers (in the form of $x^{5}$, where $x$ is an integer). what is the $1000^{\mathrm{th}}$ term of the sequence?

Accepted Solution

A:

Answer: Β 1041Step-by-step explanation:Among the numbers 1β1000, there are 31 squares, so the sequence will extend to at least 1031. In those added numbers, there is another square (1024), so the sequence must extend to at least 1032.There are 7 more numbers that are cubes, but not squares, so the sequence must extend to at least 1039.And there are 2 additional numbers that are 5th powers that are not squares or cubes. Compensating for the removal of these numbers extends the end of the sequence to 1041.There are no numbers in this range that are both cubes and 5th powers and that have not already been accounted for. (The only 15th power is 1.)Hence, the 1000th number in the sequence is 1041._____This result is verified by a computer program that listed the numbers.