If n=t^3 , when n and t are positive integers and 8, 9, 10

Ty nice explanation

The least value of n is (2*3*5)^3.

Also, 225=3^2*5^2 not 5^3.

This one doesn't quite make sense to me.

I factored it down to 2*2*2 (8), 3*3 (9) and 2*5 (10).

Those are factors of N, not T.

The 2 in (10) i cancelled out since we're already using one in (8), and then multiplied it out and got 360, which wasn't an answer choice.

Is there a mathematical rule that limits the possible factors of a perfect cube?

Of course, you're right.

The powers of the primes of a perfect cube are always multiples of 3. For example, \(1,000=2^3*5^3\): the powers of 2 and 5 are multiples of 3.

Now, since \(8=2^3\), \(9=3^2\), and \(10=2*5\) are factors of n, then 2, 3, and 5 are factors of n, therefore \(2^3=8\), \(3^3=27\) and \(5^3=125\) must be the factors of n.

Ah, that makes perfect sense. Thank you for your help!

n = 8p,n = 9q and n = 10r.Thus,\(n^3 = t^6 = 2^4*3^2*5*k\)[k is some positive integers]

Thus, for t to be a positive integer, k must atleast have the value = \(2^2*3^4*5^5\)

Thus,\(n = t^3 = 2^3*3^2*5^2*y\) [y is a positive integer]

225 is definately a factor of n.

C.

Hi Bunuel,
Thanks for the explanation..
The rule you have used - is this a general rule?
The powers of the primes of a perfect cube are always multiples of 3. For example, 1,000=23∗53
Is this true for all powers, i.e., the powers of the primes of a perfect square are always multiples of 2, for perfect (rasied to 4) they are multiples of 4 and so on?

Let me try to explain.

By definition of a cube or a square , a number is a perfect square when N = \(A^{2p}\) and a number is a perfect cube when N = \(A^{3p}\), where p is an integer.

Thus for making a perfect square you need to raise the power of primes to multiples of 2 and for making a perfect cube you need to raise the power of primes to multiples of 3.

Alternately, you can see it this way: A perfect square N when taken the square root of (ie number raised to the power of 1/2), should give you an integer. In other words, if N is a perfect square, then \(\sqrt N\) = Integer. This can only happen when \(N = A^{multiple of 2}\)

Also, a perfect cube N when taken the cube root of (ie number raised to power of 1/3), should give you an integer. In other words, if N is a perfect cube, then \(\sqrt[3] N\) = Integer. This can only happen when \(N = A^{multiple of 3}\)

Hope this helps.

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