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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

Respuesta :

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.

[tex]64 = 8*8 = (8)^2[/tex]

So, 64 is a perfect squared number.

Now, [tex]64= 4*4*4 =(4)^3[/tex]

So, 64 is also a perfect cubed number.

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