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

Question

If n=t^3 , when n and t are positive integers and 8, 9, 10 are each factors of n, which of the following must be a factor of n?

A. 16
B. 175
C. 225
D. 275
E. 625

Bunuel · Accepted Answer

n=t^3=integer^3  means that n is a perfect cube. A perfect cube is an integer that is the cube of an integer. For example  27=3^3 , is a perfect cube.

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.

A. 16=2^ 4 : not necessarily a factor of n.
B. 175=5^2* 7 : not necessarily a factor of n.
C. 225=3^2*5^2: since 3^3=27 and 5^3=125 are the factors of n, then 225 is also a factor of n.
D. 275=5^2* 11 : not necessarily a factor of n.
E. 625=5^ 4 : not necessarily a factor of n.

Answer: C.

P.S. Please post PS questions in PS forum.

Zarrolou · Answer

8=2^3 9=3^2 10=5*2 So n is formed by 2,3,5 . Because both t and n are positive integers we can write n=(2*3*5)^3 Now to the question: which of the following must be a factor of n? All the answers (except the correct one) are not possible combinations of (2*3*5)^3 so they are not factors Only 225=3^2*5^2 is in n , so C is the answer. Let me know if it's clear PS:welcome to the club!

alex90 · Answer

Ty

nice explanation :-D

Bunuel · Answer

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

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

dave785 · Answer

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?

Zarrolou · Answer

Of course, you're right.

Bunuel · Answer

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.

dave785 · Answer

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

mau5 · Answer

n = 8p,n = 9q and n = 10r.Thus, n^3 = t^6 = 2^4*3^2*5*k

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

225 is definately a factor of n.

C.

burgeoning · Answer

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?

ENGRTOMBA2018 · Answer

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  N  = Integer. This can only happen when  N = A^{multiple of 3}

Hope this helps.

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

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

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