Question

What is the probability that a randomly chosen positive integer not exceeding 1000, is a power of another integer. For example, 64=82 or 27=33 are powers but 55 or 24 are not powers.

Answer 1

We are to compute the number of integers less than 1000 which are power of other integers.

• Perfect Squares: 1², 2² , .....
  Now the square root of 1000 is greater than 31 but less than 32. Therefore 31² would be the largest perfect square less than 1000
  Therefore we have 31 perfect squares.
• Perfect cubes: 1³, 2³, .... 10³ that is 10 such perfect cubes. But some of these cubes are perfect squares as well which we have already counted. So we cannot double count them here.
  
  The cubes of perfect squares are perfect squares as well. Out of the first 10 numbers the perfect squares are 1, 4, and 9. Therefore we need to remove cubes of these 3 numbers.
  So we have 7 numbers which are non perfect square cubes.
• Perfect 4^th power: 1⁴ , 2⁴, 3⁴, 4⁴, 5⁴ that is 5 such numbers but all these are perfect squares as well, and therefore are already counted in the perfect squares case. Therefore we wont count them here.
• Perfect 5th power: 1⁵, 2⁵, 3⁵ that is 3 such numbers but we count here only 2 as 1⁵ is already counted before. Therefore 2 such numbers.
• Perfect 6th powers already accounted in perfect squares / cubes.
• Perfect 7th power, there is only number to be taken that is 2⁷
  Therefore 1 more number here.
• Perfect 8th power already taken in perfect squares
• Perfect 9th power already taken in perfect cubes.
• No more higher powers to be considered as 2¹⁰ > 1000

Therefore total such numbers = 31 + 7 + 2 + 1 = 41

Therefore 41 is the required number of numbers here. Therefore the probability here is computed as:
= 41/1000

= 0.041

Therefore 0.041 is the required probability here.