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
"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.