x ⌊ x ⌊ x ⌊ x ⌋ ⌋ ⌋ = 8 8 If x = b a is a solution to the equation above, where a and b are coprime positive integers, what is a + b ? Notation: ⌊ ⋅ ⌋ denotes the floor function .
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x = 3.15 also satisfy this equation.
In fact there are infinitely many solutions, check my report.
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I'm getting f ( 3 . 1 5 ) = 8 8 . 2 ...
3 . 1 5 ⋅ ⌊ 3 . 1 5 ⋅ ⌊ 3 . 1 5 ⋅ ⌊ 3 . 1 5 ⌋ ⌋ ⌋ = 3 . 1 5 ⋅ ⌊ 3 . 1 5 ⋅ ⌊ 3 . 1 5 ⋅ 3 ⌋ ⌋ = 3 . 1 5 ⋅ ⌊ 3 . 1 5 ⋅ ⌊ 9 . 4 5 ⌋ ⌋ = 3 . 1 5 ⋅ ⌊ 3 . 1 5 ⋅ 9 ⌋ = 3 . 1 5 ⋅ ⌊ 2 8 . 3 5 ⌋ = 3 . 1 5 ⋅ 2 8 = 8 8 . 2
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Arggghhhhhh, I added another floor outside of the expression.. Silly me. I'm deleting my report.
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@Pi Han Goh – No worries, I figured it might be that! By the way, I'm not very happy with my solution here; any ideas to avoid checking so many cases at the end?
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@Pi Han Goh – Very nice! This seems to be an interesting function - I don't feel I have a good intuition for it yet.
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@Mark Hennings – The number of solutions to f ( x ) = k for positive integer k looks interesting. It can be 0 , 1 or 2 ; but the sequence of k values with two solutions looks very strange; I think it begins 1 , 3 1 , 4 0 , 4 1 , 1 1 2 , 1 1 8 , … (though I'm just checking these with GeoGebra and Wolfram|Alpha at the moment)
The function f is only increasing for x ≥ 0 . It is defined for all real x . However, f ( x ) < 7 2 for all − 3 < x < 0 , f ( − 3 ) = 8 1 and f ( x ) > 1 1 1 for x < − 3 , so there are no negative solutions of f ( x ) = 8 8 .
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Thanks Mark, I missed that negative values of x were allowed. I've updated the solution.
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First note that x can't be an integer. Now note that ⌊ x ⋅ ⌊ x ⋅ ⌊ x ⌋ ⌋ ⌋ is an integer, say n , so x = n 8 8
It's easy to see that f ( x ) = x ⋅ ⌊ x ⋅ ⌊ x ⋅ ⌊ x ⌋ ⌋ ⌋ is an increasing function*, and that f ( 3 ) = 8 1 and f ( 4 ) = 2 5 6 , so 3 < x < 4 and 2 2 < n ≤ 2 9 .
This only leaves eight possible n to check; just one value, n = 2 8 , works, so x = 2 8 8 8 = 7 2 2 and the answer is 2 9 .
Incidentally, f ( π ) = 8 7 . 9 6 … .