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lys-0071 [83]
3 years ago
14

Jim makes the following conjecture. Other than 1, there are no numbers less than 100 that are both perfect squares and perfect c

ubes. What is a counter example that proves his conjecture false
Mathematics
1 answer:
attashe74 [19]3 years ago
5 0

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

<u><em>Explanation</em></u>

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.

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