Easy as 1, 2, 3?

We will use the following facts (the simple consequences of the definitions of the ceiling, floor and fractional part functions):

Fact 1:

$\lfloor x\rfloor + \lbrace x \rbrace = x$

Fact 2:

$\lceil x \rceil = \lfloor x\rfloor \iff x \in \mathbb {Z}$

$\lceil x \rceil = \lfloor x\rfloor + 1 \iff x \notin \mathbb {Z}$

Now consider the following 2 cases:

$\lceil x \rceil + 2 \lfloor x\rfloor + 3 \lbrace x \rbrace = 6$

$\text {Case I: } x \in \mathbb {Z}$

$3 \lfloor x\rfloor + 3 \lbrace x \rbrace = 6$

$3x = 6$

$x = 2$

$\text {Case II: } x \notin \mathbb {Z}$

$3 \lfloor x\rfloor + 3 \lbrace x \rbrace + 1= 6$

$3x = 5$

$x = \frac {5}{3}$

$\text {Since we have exactly two solutions for x, our answer should be } \boxed {2}$

Since $\left \lceil x \right \rceil$ and $\left \lfloor x \right \rfloor$ are integers,

$3 \times \left \{ x \right \}$ is an integer.

So, we have three choices for $\left \{ x \right \}$ :

• $0$
• $\frac{1}{3}$
• $\frac{2}{3}$

Lets consider each case, and see if it has a solution:

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1) $\left \{ x \right \} = 0$

$\implies 3x=6$

$x=2$

Yes, a solution exists.

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2) $\left \{ x \right \} = \frac{1}{3}$

Let $x_0 = \left \lfloor x \right \rfloor$ (must be an integer)

$x_0 + 1 + 2x_0 + 1 = 6$

$\implies 3x_0 = 4$

No, no solution here.

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3) $\left \{ x \right \} = \frac{2}{3}$

$x_0 + 1 + 2x_0 +2 = 6$

$\implies 3x_0 = 3$

$x_0 = 1$

Yes, a solution exists.

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Therefore, $\boxed2$ solutions exist, $x = 1 \frac{2}{3}$ and $x = 2$ .

Since the RHS is an integer, the LHS must be an integer. Since $\lceil x \rceil$ and $\lfloor x \rfloor$ are integers, $3\{x\}$ must be an integer. There are two cases when $3\{x\}$ is an integer $\{x\} = 0$ and $\{x\} = \dfrac k3$ , where $k$ is either 1 or 2.

When $\{x\} = 0$ , then $\lceil x \rceil = \lfloor x \rfloor = x$ and $\lceil x \rceil +2 \lfloor x \rfloor + 3\{x\} = x + 2x + 0 = 3x = 6$ , $\implies x = 2$ .

When $\{x\} = \frac k3$ , then

$\begin{aligned} {\color{#3D99F6}\lceil x \rceil} + 2 \lfloor x \rfloor + 3\{x\} & = 6 \\ {\color{#3D99F6} \lfloor x \rfloor + 1} + 2 \lfloor x \rfloor + 3 \times \frac k3 & = 6 \\ 3\lfloor x \rfloor + 1 + k & = 6 \\ \lfloor x \rfloor & = \frac {5-k}3 \end{aligned}$

Since the LHS is an integer, RHS must also be an integer. For the RHS to be an integer, $k = 2$ , $\implies \lfloor x \rfloor = 1$ and $x = 1\frac 23$ .

Therefore, there are $\boxed{2}$ values of $x$ satisfying the equation.

First we start by looking for integer solutions. If $x$ is an integer, then $\{ x \} = 0$ . So $\lceil x \rceil + 2 \times \lfloor x \rfloor + 3 \times \{ x \} = 6 \ \implies x + 2x = 6 \ \implies x = 2$ .

Noting that $6$ is an integer, as well as $\lceil x \rceil$ , and $\lfloor x \rfloor$ , we see that $3 \times \{ x \}$ must also be an integer and thus, $x$ must be a rational number of the form $\dfrac{a}{3}$ . Since these must add up to 6, there are no negative solutions for $x$ , hence $x$ is of the form $\dfrac{a}{3}, a \in \mathbb{Z}^+$ .

$\left \lceil \dfrac{a}{3} \right \rceil + 2 \times \left \lfloor \dfrac{a}{3} \right \rfloor + 3 \times \left \{ \dfrac{a}{3} \right \} = 6 \space \ \left( \text{for } \ {a, n \in \mathbb{Z}^+ \implies \dfrac{a}{3}} = \left \{ n\dfrac{1}{3},n\dfrac{2}{3} \right \} \ \forall \ a \ | \ \dfrac{a}{3} \notin \mathbb{Z}^+ \right)$ .

$\text{Checking } \ a = 3n+1, \ n \in \mathbb{Z}^+ \ \text{ gives } \ \left \lfloor \dfrac{a}{3} \right \rfloor + 2 \times \left \lfloor \dfrac{a}{3} \right \rfloor = 5$ .

$5$ is an odd number (i.e., $5 \in \{2n+1\}, \ n \in \mathbb{Z}$ ). Therefore $\left \lceil \dfrac{a}{3} \right \rceil$ must also be odd since $2 \times \left \lfloor \dfrac{a}{3} \right \rfloor \in \{2n\}, \ n \in \mathbb{Z}$ .

Since $\left \lceil \dfrac{a}{3} \right \rceil$ is odd, then $\left \lfloor \dfrac{a}{3} \right \rfloor$ is even since $\left \lfloor \dfrac{a}{3} \right \rfloor = \left \lceil \dfrac{a}{3} \right \rceil - 1$ .

Iterating across odd positive integers, $\left \lceil \dfrac{a}{3} \right \rceil$ , we see that no $a$ satisfies $\left \lceil \dfrac{a}{3} \right \rceil + 2 \times \left \lfloor \dfrac{a}{3} \right \rfloor = 5$ .

$\text{Checking } \ a = 3n+2, \ n \in \mathbb{Z}^+ \ \text{ gives } \ \left \lceil \dfrac{a}{3} \right \rceil + 2 \times \left \lfloor \dfrac{a}{3} \right \rfloor + 3 \times \left \{ \dfrac{a}{3} \right \} = 6$ .

Iterating across all even positive integers, $\left \lceil \dfrac{a}{3} \right \rceil$ , we see that $a = 5$ satisfies the equation and thus, $x = \dfrac{5}{3}$ is a solution to $\lceil x \rceil + 2 \times \lfloor x \rfloor + 3 \times \{ x \} = 6$ .

Hence, we have $x = \left \{ \dfrac{5}{3}, 2\right \}$ .

Therefore, there are $\boxed{2}$ distinct values which satisfy the equation given.