Question

Jim makes the following conjecture. Other than 1, there are no numbers less than 100 that are both perfect squares and perfect cubes. What is a counter example that proves his conjecture false

Answer 1

64  is a counter example that proves Jim's conjecture as false.

Explanation

The number should be less than 100 and need to be both perfect square and perfect cubes.

$64 = 8*8 = (8)^2$

So, 64 is a perfect squared number.

Now, $64= 4*4*4 =(4)^3$

So, 64 is also a perfect cubed number.