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Algebra Level 3

x x x x = 88 x \lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor =88 If x = a b x = \frac ab is a solution to the equation above, where a a and b b are coprime positive integers, what is a + b a+b ? Notation: \lfloor \cdot \rfloor denotes the floor function .


The answer is 29.

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1 solution

Chris Lewis
Nov 30, 2020

First note that x x can't be an integer. Now note that x x x \lfloor x \cdot \lfloor x \cdot \lfloor x \rfloor \rfloor \rfloor is an integer, say n n , so x = 88 n x=\frac{88}{n}

It's easy to see that f ( x ) = x x x x f(x)=x \cdot \lfloor x \cdot \lfloor x \cdot \lfloor x \rfloor \rfloor \rfloor is an increasing function*, and that f ( 3 ) = 81 f(3)=81 and f ( 4 ) = 256 f(4)=256 , so 3 < x < 4 3<x<4 and 22 < n 29 22<n\le 29 .

This only leaves eight possible n n to check; just one value, n = 28 n=28 , works, so x = 88 28 = 22 7 x=\frac{88}{28}=\frac{22}{7} and the answer is 29 \boxed{29} .

Incidentally, f ( π ) = 87.96 f(\pi)=87.96\ldots .


  • As Mark Hennings points out, f f is only increasing for positive x x ; but there are no solutions for negative x x (see the comments below).

x = 3.15 also satisfy this equation.

In fact there are infinitely many solutions, check my report.

Pi Han Goh - 6 months, 2 weeks ago

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I'm getting f ( 3.15 ) = 88.2 f(3.15)=88.2 ...

3.15 3.15 3.15 3.15 = 3.15 3.15 3.15 3 = 3.15 3.15 9.45 = 3.15 3.15 9 = 3.15 28.35 = 3.15 28 = 88.2 \begin{aligned} 3.15 \cdot \lfloor 3.15 \cdot \lfloor 3.15 \cdot \lfloor 3.15 \rfloor \rfloor \rfloor &= 3.15 \cdot \lfloor 3.15 \cdot \lfloor 3.15 \cdot 3 \rfloor \rfloor \\ &= 3.15 \cdot \lfloor 3.15 \cdot \lfloor 9.45 \rfloor \rfloor \\&= 3.15 \cdot \lfloor 3.15 \cdot 9 \rfloor \\ &= 3.15 \cdot \lfloor 28.35 \rfloor \\&= 3.15 \cdot 28 \\&= 88.2 \end{aligned}

Chris Lewis - 6 months, 2 weeks ago

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Arggghhhhhh, I added another floor outside of the expression.. Silly me. I'm deleting my report.

Pi Han Goh - 6 months, 2 weeks ago

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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?

Chris Lewis - 6 months, 2 weeks ago

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@Chris Lewis Here you go

Pi Han Goh - 6 months, 1 week ago

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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.

Chris Lewis - 6 months, 1 week ago

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@Chris Lewis Try the graph...

Mark Hennings - 6 months, 1 week ago

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@Mark Hennings The number of solutions to f ( x ) = k f(x)=k for positive integer k k looks interesting. It can be 0 0 , 1 1 or 2 2 ; but the sequence of k k values with two solutions looks very strange; I think it begins 1 , 31 , 40 , 41 , 112 , 118 , 1,31,40,41,112,118,\ldots (though I'm just checking these with GeoGebra and Wolfram|Alpha at the moment)

Chris Lewis - 6 months, 1 week ago

The function f f is only increasing for x 0 x \ge 0 . It is defined for all real x x . However, f ( x ) < 72 f(x) < 72 for all 3 < x < 0 -3 < x< 0 , f ( 3 ) = 81 f(-3) = 81 and f ( x ) > 111 f(x) > 111 for x < 3 x < -3 , so there are no negative solutions of f ( x ) = 88 f(x)=88 .

Mark Hennings - 6 months, 2 weeks ago

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Thanks Mark, I missed that negative values of x x were allowed. I've updated the solution.

Chris Lewis - 6 months, 1 week ago

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