Floor Functions With Unlucky 13

Algebra Level 5

$\begin{aligned} \text{ Equation 1 }: &&\quad \lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor = 13 \\ \text{ Equation 2 }: &&\quad \lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor+ \lfloor 8x \rfloor = 13 \end{aligned}$

Do the equations above have real solutions?

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

1: No, 2: No 1: Yes, 2: No 1: Yes, 2: Yes 1: No, 2: Yes

2 solutions

Denton Young
Feb 21, 2016

For equation 1 :

Let us start bounding the range of $x$ first.

If $x \geq 2$ , then $\lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor \geq 2 +4 + 8 = 14$ , which is already larger than 13, thus $x$ has to be less than 2.

If $x < 2$ , then the maximum value of $\lfloor x \rfloor , \lfloor 2 x \rfloor , \lfloor 4x \rfloor$ is at most 1, 3 and 7 respectively and thus the maximum value of these three terms is $\lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor \leq 1 + 3 + 7 = 11$ , which is already less than 13, thus $x$ has to be larger than 2.

But the above tells us these intervals $x\geq 2$ and $x< 2$ are not solutions for $x$ . So there is no solution of $x$ .

Similarly, For equation 2 :

Let us start bounding the range of $x$ first.

If $x \geq 1$ , then $\lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor+ \lfloor 8x \rfloor \geq 1+2 +4 + 8 = 15$ , which is already larger than 13, thus $x$ has to be less than 1.

If $x < 1$ , then the maximum value of $\lfloor x \rfloor , \lfloor 2 x \rfloor , \lfloor 4x \rfloor , \lfloor 8x \rfloor$ is at most 0, 1, 3 and 7 respectively and thus the maximum value of these three terms is $\lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor \leq 0 +1+ 3 + 7 = 11$ , which is already less than 13, thus $x$ has to be larger than 1.

But the above tells us these intervals $x\geq 2$ and $x< 2$ are not solutions for $x$ . So there is no solution of $x$ .

Moderator note:

Good clear solution. How can we describe the range of the function $\lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor$ ?

Nice solution!

Akhash Raja Raam - 5 years, 3 months ago

How can we describe the range of the function $\lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor$ ?

Calvin Lin Staff - 5 years, 3 months ago

The word that comes to mind is "discontinuous". As x approaches an integer it does a sudden jump at every transition. At 2.999 the value is 18: when x hits 3 it suddenly jumps to 21, bypassing 19 and 20. At x = 9.999 the value is 67: when x hits 10 the value jumps to 70, bypassing 68 and 69. And so on.

Denton Young - 5 years, 3 months ago

More accurately, what is the range of integer values that the function $\lfloor x \rfloor + \lfloor 2 x \rfloor + \lfloor 4x \rfloor$ can take on?

E.g. Can it be 1000? 2016? Why, or why not? What is a necessary and sufficient condition?

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – the value can be any integer that is 0, 1, 3, or 4 (mod 7)

Denton Young - 5 years, 3 months ago

@Denton Young – Great! That is indeed the classification.

Can you explain why?

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – The equation "resets" mod 7 every time we hit a new integer (if x is an integer, the value is 7x.) So we only need to consider what happens as x goes from 0 to 1. At 0, the value is 0. when x hits 1/4, the value jumps to 1. Then when x hits 1/2, both the second and third terms increase, so the value goes to 3. When x hits 3/4, the value of the third term jumps by 1, so the value goes to 4.and when x hits 1, all three terms jump, putting us at 7 and "resetting" the function.

Denton Young - 5 years, 3 months ago

Yashas Ravi
Sep 24, 2019

If we subtract Equation $1$ from Equation $2$ , we get $\lfloor(8x)\rfloor=0$ . This means that $0.125≥x≥0$ . If we plug in the maximum value of $0.125$ into both equations, then the final result is $0$ instead of $13$ . Thus, there are no solutions.

Floor Functions With Unlucky 13

2 solutions

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