What floor is it?

Algebra Level 3

$\left\lfloor \left| \frac{2x-3}{4} \right| \right\rfloor = 5$

What is the largest negative integer $x$ that satisfies the equation above?

-8 -9 -10 -12

4 solutions

Joshua Chin
Mar 17, 2016

Disambiguation: Note that for two negative integers $m$ and $n$ , if $|m|<|n|$ , $m$ is the larger negative integer than $n$ .

So we know that $5\le |\frac { 2x-3 }{ 4 } |<6$ .

Since $x<0$ , then we conclude that $-6<\frac { 2x-3 }{ 4 } \le -5$ . Now the possible values for $\frac { 2x-3 }{ 4 }$ are -5, -5.25, -5.5 and -5.75 for $2x-3$ to be integer, this means the possible values for $2x-3$ are -20, -21, -22 and -23 for $x$ to be integer.

But for $x$ to be an integer, $2|(2x-3+3)$ . So $2x-3$ cannot take on values of -20 or -22, can only take up values of -21 or -23.

Now we recall the definition of larger negative integer as stated above. So for the largest negative integer, we want $|2x-3|$ to be minimum. So for largest negative possible integer value, $|2x-3|=21$ and thus $x=-9$

Moderator note:

Instead of claiming that

Now the possible values for $\frac { 2x-3 }{ 4 }$ are -5, -5.25, -5.5 and -5.75 for $2x-3$ to be integer, ...

it is better to show that $-24 < 2x - 3 \leq -20$ , and since $2x-3$ is an integer, hence $2x-3 \in \{ -23, -22, -21, -20 \}$ .

The definition of "largest negative integer" should probably be clarified in the problem rather than the solution if possible (presumably just by adding the disambiguation note above).

Dietrich Geisler - 4 years, 10 months ago

after all solving i got wrong answer marking it -10. my bad! but i feel the framing is quite right as fundamentally largest -ve integer is -9

shakes the mathematical basics

Aditya Gahlawat - 4 years, 9 months ago

I don't understand why x=-10 is not a solution. If we compute x=-10 in the equation will give us 5. Here is the solution http://www.wolframalpha.com/input/?i=floor(%7C(2x-3)%2F4%7C)%3D5

Mateo Matijasevick - 5 years, 2 months ago

Hi Mateo, $x=-10$ is a solution, you are right. However we want to find the largest negative solution, and $x=-9$ is a solution as well. Please read my solution on largest negative integer.

Joshua Chin - 5 years, 2 months ago

Thank you. I passed over that.

Mateo Matijasevick - 5 years, 2 months ago

Where did I make my mistake in the following

-6 < (2x-3)/4 < -5 -24 < (2 x - 3) < -20

-21 < 2 x < -17

-11.5 < x < -8.5 which led me to the wrong answer of -8?

Kermit Rose - 2 years, 4 months ago

You didn't make any mistake on your process to get -11.5 < x < -8.5

Look at your -8. It is located outside the range. The answer must be -9

Muhamad Fachri Wijaya - 2 years, 2 months ago

Kelvin Hong
Jul 12, 2018

From the original question, we can deduce that $5\leq\bigg|\frac{2x-3}4\bigg|<6$ $20\leq|2x-3|<24$ $20\leq3-2x<24\text{ (Since x is negative.)}$ $-24<2x-3\leq-20$ $-21<2x\leq-17$ $-10.5<x\leq-8.5$ the larger negative value of $x$ is $\boxed{-9}$ .

Egor Eremeev
Jun 5, 2018

Thats strange to give a choice between 4 numbers, i can just test them! Just like i did...

Anant Dixit
Sep 23, 2017

We are searching for a negative solution of $x$ , so in this particular case, we can rewrite the expression in the problem as, $\left\lfloor - \left(\frac{2x-3}{4}\right) \right\rfloor = 5 \Rightarrow \left\lfloor \frac{3-2x}{4} \right\rfloor =5 \Rightarrow 5 \le \frac{3-2x}{4} \le 6$ With a bit of algebra, we can show $-10.5 \le x \le -8.5$ , which gives the largest possible negative value of $x$ to be $-9$ .

What floor is it?

4 solutions

Moderator note:

shakes the mathematical basics

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