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Which of the following is a perfect cube?

vaibhav1221

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09 Sep 2019, 21:32

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Which of the following is a perfect cube?

(A) 39316
(B) 27006
(C) 46658
(D) 32768
(E) 64012

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chetan2u

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09 Sep 2019, 22:14

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vaibhav1221

Which of the following is a perfect cube?

(A) 39316
(B) 27006
(C) 46658
(D) 32768
(E) 64012

Whenever a question is asked on squareroots, there is a bigger chance of it testing the units digit concept, however the cuberoot generally can have any units digit

So, let us check the units digit.
As all numbers are even, let us see the possible units digit of even numbers..
(A) numbers ending in 0 - 0*0*0=0
(B) numbers ending in 2 - 2*2*2=8
(A) numbers ending in 4 -4*4*4=4
(A) numbers ending in 6 - 6*6*6=6
(A) numbers ending in 8 - 8*8*8=2
As we can see, all digits are possible..

Now, the next step would be to test the RANGE of answers, or proximity to any known cube.
A reliable GMAT source would always give the choices in some order and not the way it is given..
The smallest number in choices is 27006~\(27000=30^3\).
The largest number in choices is 64012~\(64000=40^3\).

Now, we are left with
(A) 39316...ending with 6 means ONLY possibility of 36.. check for \(36^3=46656\)..NO
(B) 27006... This can be discarded based on (A) or proximity to \(30^3\)
(C) 46658...ending with 8 means ONLY possibility of 32.. check for \(32^3=32768\)..NO
(D) 32768...Yes as \(32^3=32768\)
(E) 64012...This can be discarded due to proximity to \(40^3\)

D

vaibhav1221

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09 Sep 2019, 22:27

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chetan2u

vaibhav1221

Which of the following is a perfect cube?

(A) 39316
(B) 27006
(C) 46658
(D) 32768
(E) 64012

Hi chetan2u,

This is a self made question based on a concept of squares by veritas, extended to cubes.

As all the numbers in the options are even, the numbers are bound to be divisible by 2. With that given, the basic requirement for any number out of the five options to be a perfect cube is to have at least 3 factors of 2. The perfect cube should at least be divisible by 2^3 = 8. Only option D is divisible.

chetan2u

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09 Sep 2019, 22:36

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vaibhav1221

chetan2u

vaibhav1221

Which of the following is a perfect cube?

(A) 39316
(B) 27006
(C) 46658
(D) 32768
(E) 64012

The concept of squares extended to cube is not likely to be seen in actuals, but yes nothing wrong in knowing the concept.
Even divisibility by 8 can be checked from just the last 3 digits and only 768 is divisible by 8, but again in GMAT, generally divisibility by 2, 3, 9 and sometimes 11 is tested.

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10 Sep 2019, 01:54

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2, 6 and 8 can be the unit digit of a perfect cube. So, we have to find different approach to solve this question.

Unit digit of all the options is even; Hence, the Option that is a perfect cube must be divisible by 2^3=8.
If a number is divisible by 8, it's last 3 digits are also divisible by 8.

A. 316=320-4 is not divisible by 8. Eliminate A
B. 006 is not divisible by 8. Eliminate B
C. 658=640+18 is not divisible by 8. Eliminate C
D. 768=720+48 is divisible by 8. Keep D
E. 012 is not divisible by 8. Eliminate E

We don't have to calculate the exact cube root of 32768, as we eliminated all other options

D

vaibhav1221

Which of the following is a perfect cube?

(A) 39316
(B) 27006
(C) 46658
(D) 32768
(E) 64012

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10 Sep 2019, 02:59

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vaibhav1221

Which of the following is a perfect cube?

(A) 39316
(B) 27006
(C) 46658
(D) 32768
(E) 64012

Interesting question. Since all the given answer choices are even, the correct answer should contain 2^3=8 as its factor for it to be a perfect cube.
A) is not evenly divisible by 8, incorrect
B) is not evenly divisible by 8, incorrect
C) is not evenly divisible by 8, incorrect
D) evenly divisible by 8, CORRECT
E) is not evenly divisible by 8, incorrect

CrackverbalGMAT

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11 Sep 2019, 20:00

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For questions like these, where you have to find out whether a number is a perfect square or a perfect cube, always work with identifying squares/cubes of numbers which end with ZERO.
For example, \(10^2\) = 100, \(20^2\) = 200 and so on. Similarly, \(10^3\) = 1000, \(20^3\) = 8000, \(30^3\) = 27000 and so on.

Also, the unit digit of a perfect cube corresponds uniquely to the unit digit of its cube root.
A perfect cube ending with 8 will have its cube root ending with 2.
A perfect cube ending with 6 will have its cube root ending with 6.
A perfect cube ending with 2 will have its cube root ending with 8.

Therefore, in the options, we can say that if 39316 is a perfect cube, its cube root will have the unit digit as 6; similar conclusion can be drawn about 27006. 46658/32768, if they are perfect cubes, will have their cube root ending with 2. 64012, if it’s a perfect cube will be ending with 8.

If a number ends with 0, finding out its cube is the easiest thing to do. That is,

If x = k0, then \(x^3\) = (\(k^3\))000. For example,

\(10^3\) = 1000, \(20^3\) = 8000, \(30^3\) = 27000, \(40^3\) = 64000 and so on.
From the above examples, one thing is very clear – 27006 and 64012 are not perfect cubes. Answer options B and E can be eliminated straight away.

This also means that the cube root of the other three numbers lies between 30 and 40. But why? Because all these numbers are between 27000 and 64000.

Now, \(35^3\) will be a number above 42000. This is because \(35^2\) = 1225. Even if we ignore the 25, 1200 * 35 will give us 42000.
The only number that ends with 6, between 30 and 40, is 36. 36 is more than 35. Therefore, \(36^3\) cannot be 39316. So, 39316 cannot be a perfect cube.

The only number that ends with 2, between 30 and 40 is 32. 32 is less than 35. Therefore, \(32^3\) cannot be 46658 (since 46658 is more than \(35^3\); clearly, \(32^3\) cannot be greater than \(35^3\), right?).

Therefore, 32768 IS the perfect cube and it IS the cube of 32. You may observe here that 32768 is also closer to 27000, so clearly between 46658 and 32768, it’s only 32768 which can be the cube of 32.

As mentioned before, working with perfect cubes is not about memorizing a set of numbers. It’s about knowing certain cubes which are easy to know, and then using these numbers with sound logic to eliminate numbers which cannot be perfect cubes.
Hope this helps!