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GmatDaddy
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For any positive number x, the function [x] denotes the greatest integ Updated on: 26 Sep 2018, 08:35
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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41% (02:41) correct 59% (02:48) wrong based on 140 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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E
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For any positive number x, the function [x] denotes the greatest integ 26 Sep 2018, 08:22
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GmatDaddy
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.
Show more

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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For any positive number x, the function [x] denotes the greatest integ 29 Sep 2018, 17:55
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GmatDaddy
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.
Show more


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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For any positive number x, the function [x] denotes the greatest integ 23 May 2021, 02:58
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GmatDaddy
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
Show more

Hi Scott,

Can I get your help to understand the sentence below:

"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."

5 is single. How do we root that?

Thank you.
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For any positive number x, the function [x] denotes the greatest integ 05 Aug 2021, 16:01
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joyousjoyce
ScottTargetTestPrep
GmatDaddy
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

Hi Scott,

Can I get your help to understand the sentence below:

"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."

5 is single. How do we root that?

Thank you.
Show more
Response:

If k^2 is divisible by 5, then since k is an integer, k^2 must be divisible by 5^2. To see why this must be the case, let’s suppose for a moment that k^2 is divisible by 5 but not by 5^2. If this was true, when you took the square root of k^2, you wouldn't obtain an integer even if every other prime factor of k^2 came in pairs. That’s how we can conclude that k must contain at least one factor of 5.

Another way of obtaining the same result is using the following rule: “the exponents of the prime factors of a perfect square must be even”. We know 5 is a factor of k^2, so there will be a 5 among the prime factors of k^2. Using the above rule, we know the exponent of 5 will be even, so the exponent of 5 is at least 2. That is why k^2 must be divisible by 5^2 and k must be divisible by 5.
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