Let's analyze the assertion:

Assertion:

The cube root of 64,000 is 40, and 64,000 is a perfect cube. The reasoning given is that "no cube can end with exactly two zeros."

Verification:

Step 1: Finding the Cube Root

Let's calculate the cube root of 64,000: $64000 = 40^3$ Calculating $40^3$: $40^3 = 40 \times 40 \times 40 = 64,000$ So, the cube root of 64,000 is indeed 40. Hence, 64,000 is a perfect cube.

Step 2: Validating the Reasoning

The statement claims that no cube can end with exactly two zeros.

Let's check why:

1. For a number to end in two zeros, it must be divisible by $100 = 10^2$.
2. A perfect cube's prime factors must occur in multiples of 3.

Since $100 = 2^2 \times 5^2$, for a perfect cube to end with two zeros, it would require an additional factor of $2 \times 5 = 10$ to make it $10^3 = 1000$. Hence, any perfect cube ending in zeros must end in at least three zeros, not just two.

Conclusion:

The reasoning is correct. A perfect cube cannot end in exactly two zeros. Therefore, the assertion that 64,000 is a perfect cube and has its cube root as 40 is correct, but 64,000 doesn't contradict the claim since it ends in three zeros.

Would you like more details or have any further questions?

Relative Questions:

1. How can we determine whether a number is a perfect cube?
2. Why does a cube require factors in groups of three?
3. Are there any other restrictions on the ending digits of perfect cubes?
4. How do cube roots differ from square roots in terms of their properties?
5. What are other examples of perfect cubes?

Tip:

When determining whether a number is a perfect cube, factorizing into prime numbers can help identify whether all exponents are multiples of three.