Defining Variables
Let's define some variables to represent the unknowns:
- Let $p$ be the current price of one candy box (in dollars).
- Let $n$ be the number of candy boxes you can buy for $\$ 180$ at the current price.

So, at the current price:

$$
\(\begin{gathered}
n \times p=180 \\
n=\frac{180}{p}
\end{gathered}\)
$$


After the price increase:
- New price per candy box: $\(p+1\)$
- Number of candy boxes you can buy now: $\(n-6\)$
- So, $\((n-6) \times(p+1)=180\)$

Setting Up the Equation
We have two expressions for $n$ :
1. $\(n=\frac{180}{p}\)$
2. From the increased price scenario: $\(n-6=\frac{180}{p+1}\)$

Substitute the first equation into the second:

$$
\(\frac{180}{p}-6=\frac{180}{p+1}\)
$$


Now, let's solve for $p$.

Solving the Equation
Starting with:

$$
\(\frac{180}{p}-6=\frac{180}{p+1}\)
$$


Let's get rid of the denominators by multiplying both sides by $p(p+1)$ :

$$
\(p(p+1)\left(\frac{180}{p}-6\right)=p(p+1)\left(\frac{180}{p+1}\right)\)
$$


Simplify both sides:
Left side:

$$
\(\begin{aligned}
p(p+1) \times \frac{180}{p}- & p(p+1) \times 6=(p+1) \times 180-6 p(p+1) \\
& =180(p+1)-6 p(p+1)
\end{aligned}\)
$$


Right side:

$$
\(\begin{gathered}
p(p+1) \times \frac{180}{p+1}=p \times 180 \\
=180 p
\end{gathered}\)
$$


So, the equation becomes:

$$
\(180(p+1)-6 p(p+1)=180 p\)
$$


Expand the left side:

$$
\(180 p+180-6 p^2-6 p=180 p\)
$$


Combine like terms:

$$
\(\begin{gathered}
\left(-6 p^2\right)+(180 p-6 p)+180=180 p \\
-6 p^2+174 p+180=180 p
\end{gathered}\)
$$


Bring all terms to one side to set the equation to zero:

$$
\(\begin{gathered}
-6 p^2+174 p+180-180 p=0 \\
-6 p^2-6 p+180=0
\end{gathered}\)
$$


We can simplify this by dividing every term by -6 :

$$
\(p^2+p-30=0\)
$$


Solving the Quadratic Equation
Now, we have:

$$
\(p^2+p-30=0\)
$$


This is a quadratic equation in the form $\(a p^2+b p+c=0\)$, where:
- $a=1$
- $b=1$
- $c=-30$

We can solve for $p$ using the quadratic formula:

$$
\(\begin{gathered}
p=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
p=\frac{-1 \pm \sqrt{1^2-4(1)(-30)}}{2(1)} \\
p=\frac{-1 \pm \sqrt{1+120}}{2} \\
p=\frac{-1 \pm \sqrt{121}}{2} \\
p=\frac{-1 \pm 11}{2}
\end{gathered}\)
$$

This gives two solutions:
1. $\(p=\frac{-1+11}{2}=\frac{10}{2}=5\)$
2. $\(p=\frac{-1-11}{2}=\frac{-12}{2}=-6\)$

Since the price can't be negative, we discard $p=-6$.
So, the current price is $p=5$ dollars.
Verifying the Answer
Let's check if this makes sense with the given information.
Current price: \$5
Number of boxes for \$180:

$$
\(n=\frac{180}{5}=36 \text { boxes }\)
$$


After price increase: \$6
Number of boxes for \$180:

$$
\(\frac{180}{6}=30 \text { boxes }\)
$$


Difference in number of boxes:

$$
\(36-30=6\)
$$


This matches the given information that 6 fewer boxes can be bought.

Checking the Options
The current price is \$5, which corresponds to option (B).
Eliminating Other Options
Just to be thorough, let's quickly see why the other options don't work.

Eliminating Other Options
Just to be thorough, let's quickly see why the other options don't work.

Option A: \$4
- Current: $\(\frac{180}{4}=45\)$ boxes
- Increased price: $\(\$ 5, \frac{180}{5}=36\)$ boxes
- Difference: 45-36 $\(=9 \neq 6\)$ (doesn't match)

Option C: \$6
- Current: $\(\frac{180}{6}=30\)$ boxes
- Increased price: $\(\$ 7, \frac{180}{7} \approx 25.71\)$ (not integer, and difference isn't 6)

Option D: \$30
- Current: $\(\frac{180}{30}=6\)$ boxes
- Increased price: $\(\$ 31, \frac{180}{31} \approx 5.81\)$ (not integer, and difference isn't 6)

Option E: \$40
- Current: $\(\frac{180}{40}=4.5\)$ boxes (not integer, already invalid)

Only option B (\$5) satisfies all conditions perfectly.
Final Answer

After carefully working through the problem and verifying each step, the current price of each candy box is:
$\(B\)$