kop wrote:If K to the power of 3 (k cube) is divisible by 240, what is the least possible value of integer K?

a 12
b 30
c 60
d 90
e 120

Please explain?

240 = (2) (2) (2) (2) (3) (5), which is (2) (2) (3) (3) (5) (5) short of being a minimum cube for this kind of K.

Let's see what minimum K comes out to be when the supply is made:

K^3 = (2) (2) (2) (2) (3) (5) (2) (2) (3) (3) (5) (5)

Hence the least possible value of integer K = [spoiler](2) (2) (3) (5) = 60

Choose C[/spoiler]

Alternative Approach

We know that 240 = (2) (2) (2) (2) (3) (5)

May be we plug in the answers, start testing from C as usual, see it's

(C) 60 = (2) (2) (3) (5)

If cubed 60^3 = (2) (2) (2) (2) (3) (5) (2) (2) (2) (2) (3) (5) (2) (2) (2) (2) (3) (5) , which contains everything that is required by 240 = (2) (2) (2) (2) (3) (5). Hence, we can eliminate the bigger choices, as we're looking for the least value only. We should plug in a smaller choice to make sure if (C) is correct. Let's plug in

(B) 30 = (2) (3) (5)

If cubed 30^3 = (2) (3) (5) (2) (3) (5) (2) (3) (5), which is (2) short of what is required by 240 = (2) (2) (2) (2) (3) (5). Hence, [spoiler](C)[/spoiler] remains the correct answer.

Hi Sanju09

I did not understand the below part. could you please explain me in detail as how did you get to 2 2 3 3 5 5 short of being a minimum cube.????


which is (2) (2) (3) (3) (5) (5) short of being a minimum cube for this kind of K.

Let's see what minimum K comes out to be when the supply is made:

K^3 = (2) (2) (2) (2) (3) (5) (2) (2) (3) (3) (5) (5)

kop wrote:Hi Sanju09

I did not understand the below part. could you please explain me in detail as how did you get to 2 2 3 3 5 5 short of being a minimum cube.????


which is (2) (2) (3) (3) (5) (5) short of being a minimum cube for this kind of K.

Let's see what minimum K comes out to be when the supply is made:

K^3 = (2) (2) (2) (2) (3) (5) (2) (2) (3) (3) (5) (5)

The prime factor split of a perfect cube contains each prime appearing 3, 6, 9, 12...number of times. The four 2s in 240 are two 2s short to be six 2s; the single 3 in 240 is two 3s short to be three 3s; and the the single 5 in 240 is two 5s short to be three 5s.

kop wrote:If K to the power of 3 (k cube) is divisible by 240, what is the least possible value of integer K?

a 12
b 30
c 60
d 90
e 120

Please explain?

We can PLUG IN THE ANSWERS, which represent the value of k
Since the question stem asks for the LEAST possible value of k, start with the SMALLEST answer choice.
When the correct value is plugged in, k³/240 = integer.

A: 12³/240 = (12*12*12)/(12*20) = 144/20 = 72/10 = 7.2
Eliminate A.

B: 30³/240 = (30*30*30)/(30*8) = 900/8 = 450/4 = 225/2.
Eliminate B.

C: 60³/240 = (60*60*60)/(60*4) =3600/4 = 900.
Success!

The correct answer is C.

Got it. Thanks a lot Sanju09.
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