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For any positive number x, the function [x] denotes the greatest integ

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For any positive number x, the function [x] denotes the greatest integ  [#permalink]

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New post Updated on: 26 Sep 2018, 08:35
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A
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C
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E

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Question Stats:

33% (02:35) correct 67% (02:31) wrong based on 40 sessions

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For any positive number x, the function [x] denotes the greatest integer less than or equal to x. For example, [1] = 1, [1.367] = 1 and [1.999] = 1.
If k is a positive integer such that \(k^2\) is divisible by 45 and 80, what is the units digit of \([\frac{k^3}{4000}]\)?

A. 0
B. 1
C. 7
D. 4
E. Can not be determined.

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Originally posted by GmatDaddy on 26 Sep 2018, 07:42.
Last edited by GmatDaddy on 26 Sep 2018, 08:35, edited 1 time in total.
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Re: For any positive number x, the function [x] denotes the greatest integ  [#permalink]

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New post 26 Sep 2018, 08:22
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2
GmatDaddy wrote:
For any positive number x, the function [x] denotes the greatest integer less than or equal to x. For example, [1] = 1, [1.367] = 1 and [1.999] = 1.
If k is a positive integer such that \(k^2\) is divisible by 45 and 80, what is the units digit of \([\frac{k^3}{4000}]\)?

A. 0
B. 1
C. 27
D. 54
E. Can not be determined.


Please correct your choices..
We are looking for units digit and choice C and D give you 2-digit integers.

Anyways..
k^2 divisible by 45 or 3^2*5, so k will surely be a multiple of 3*5
k^2 is also divisible by 80 or 2^4*5 or 2^3*2*5 so k is surely a multiple of 2*2*5
Therefore k will surely be a multiple of LCM of 3*5 and 2*2*5 or 2*2*3*5 or 60
Therefore k^3 = (60a)^3=216000a^3
Therefore k^3/4000=216000a^3/4000=54a^3
Now units digit will depend on a..
Say a is 1 Ans is 4
a is 7, Ans is 4*3 or 2
So cannot be determined

E
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Re: For any positive number x, the function [x] denotes the greatest integ  [#permalink]

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New post 29 Sep 2018, 17:55
GmatDaddy wrote:
For any positive number x, the function [x] denotes the greatest integer less than or equal to x. For example, [1] = 1, [1.367] = 1 and [1.999] = 1.
If k is a positive integer such that \(k^2\) is divisible by 45 and 80, what is the units digit of \([\frac{k^3}{4000}]\)?

A. 0
B. 1
C. 7
D. 4
E. Can not be determined.



If a number is divisible by 45 and 80, then it’s divisible by the least common multiple of 45 and 80, which is 720. The prime factorization of 720 is 9 x 8 x 10 = 3^2 x 2^3 x 2 x 5 = 2^4 x 3^2 x 5. Therefore, k^2 is divisible by 2^4 x 3^2 x 5 and k must be divisible by 2^2 x 3 x 5 = 60. In other words, k is a multiple of 60.

If k = 60, then k^3/4000 = 60^3/4000 = (60 x 60 x 60)/4000 = (6 x 6 x 6)/4 = 54. So the units digit of [k^3/4000] is 4.

If k = 600, then k^3/4000 = 600^3/4000 = (600 x 600 x 600)/4000 = (60 x 60 x 60)/4 = 54,000. So the units digit of [k^3/4000] is 0.

We see that the units digit of [k^3/4000] is not unique. So it can’t be determined.

Answer: E
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Re: For any positive number x, the function [x] denotes the greatest integ   [#permalink] 29 Sep 2018, 17:55
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