Explanation:
Okay, so I need to find the value of $ b $ in the direction vector $ \begin{pmatrix} -7 \\ b \end{pmatrix} $ for the given line in the Asymptote figure. Let me try to visualize the problem first.

The Asymptote code draws a coordinate system with axes from -8 to 8. Points A and B are given as (-5, 4) and (-1, 3) respectively. Then, there are points L and R, which are the intersections of the line AB with the vertical lines x = -8 and x = 8. The line LR is drawn in red, which is the line we're interested in. The question says that this line has a direction vector of the form $ \begin{pmatrix} -7 \\ b \end{pmatrix} $, and we need to find $ b $.

First, let me recall that the direction vector of a line can be found by taking two points on the line and subtracting their coordinates. Since the line LR is determined by points A and B, maybe I can find the direction vector from A to B and then relate it to the given direction vector?

Wait, but the line LR is the same as the line AB, right? Because L and R are points where the line AB intersects the vertical lines x = -8 and x = 8. So, the line LR is just the line AB extended to those vertical lines. Therefore, the direction vector of line LR should be the same as the direction vector of line AB.

But the problem says the direction vector is of the form $ \begin{pmatrix} -7 \\ b \end{pmatrix} $. So, if I can find the direction vector of AB, then I can compare it to this form and find $ b $.

Let me compute the direction vector of AB. Point A is (-5, 4) and point B is (-1, 3). The direction vector from A to B is $ \vec{AB} = B - A = (-1 - (-5), 3 - 4) = (-1 + 5, -1) = (4, -1) $.

Wait, so the direction vector is (4, -1). But the problem says the direction vector is (-7, b). Hmm, so these two vectors should be scalar multiples of each other because they are direction vectors of the same line.

So, if $ \begin{pmatrix} 4 \\ -1 \end{pmatrix} $ is a direction vector, then any scalar multiple of this vector is also a direction vector. Therefore, there exists some scalar $ k $ such that:

$ -7 = 4k $ and $ b = -1k $.

Let me solve for $ k $ first. From the x-component: $ -7 = 4k $, so $ k = -7/4 $. Then, substituting into the y-component: $ b = -1 * (-7/4) = 7/4 $.

Therefore, $ b = 7/4 $. Is that correct? Let me check.

Alternatively, maybe I need to find the direction vector of the line LR, which is the same as AB. But perhaps the problem is referring to a different direction vector? Wait, the problem says "the following line is parameterized, so that its direction vector is of the form...", so maybe the line LR is parameterized with direction vector (-7, b). But since direction vectors can be scaled, we need to find the correct scalar multiple.

Alternatively, maybe I need to find the slope of the line LR and then relate it to the direction vector. Let me try that approach as well to verify.

The slope of line AB can be calculated as $ m = (y_B - y_A)/(x_B - x_A) = (3 - 4)/(-1 - (-5)) = (-1)/(4) = -1/4 $. Therefore, the slope is -1/4.

A direction vector corresponding to this slope would have a rise over run of -1/4. So, if the direction vector is (Δx, Δy), then Δy / Δx = -1/4. Therefore, Δy = (-1/4)Δx.

But the given direction vector is (-7, b). So, if Δx = -7, then Δy = (-1/4)(-7) = 7/4. Therefore, b = 7/4.

Same result as before. So that seems consistent. Therefore, $ b = 7/4 $.

But let me make sure that the line LR is indeed the same as line AB. The Asymptote code defines points L and R as the intersections of line AB with the vertical lines x = -8 and x = 8. Therefore, line LR is the same line as AB, just extended to those vertical lines. Therefore, the direction vector of LR is the same as AB. Therefore, the direction vector is indeed (4, -1) or any scalar multiple.

But the problem says the direction vector is of the form (-7, b). So, we need to find b such that (-7, b) is a direction vector for the line. Since direction vectors can be scaled, we can find the scalar multiple between (4, -1) and (-7, b).

So, if we have two direction vectors, they must be scalar multiples. Therefore, there exists a scalar k such that:

-7 = 4k

and

b = -1k

So solving for k from the first equation: k = -7/4. Then, substituting into the second equation: b = -1*(-7/4) = 7/4.

Therefore, b is 7/4.

Alternatively, maybe I can parametrize the line LR and find its direction vector. Let me try that approach.

First, let's find the equation of line AB. Since we have two points A(-5, 4) and B(-1, 3), we can write the equation of the line.

We already found the slope m = -1/4. So, using point-slope form with point A:

$ y - 4 = -\frac{1}{4}(x + 5) $

Simplifying:

$ y = -\frac{1}{4}x - \frac{5}{4} + 4 $

$ y = -\frac{1}{4}x - \frac{5}{4} + \frac{16}{4} $

$ y = -\frac{1}{4}x + \frac{11}{4} $

So the equation of the line is $ y = -\frac{1}{4}x + \frac{11}{4} $.

Now, points L and R are the intersections of this line with x = -8 and x = 8, respectively. Let me find coordinates of L and R.

For point L, x = -8:

$ y = -\frac{1}{4}(-8) + \frac{11}{4} = 2 + \frac{11}{4} = \frac{8}{4} + \frac{11}{4} = \frac{19}{4} $

Therefore, L is (-8, 19/4).

For point R, x = 8:

$ y = -\frac{1}{4}(8) + \frac{11}{4} = -2 + \frac{11}{4} = -\frac{8}{4} + \frac{11}{4} = \frac{3}{4} $

Therefore, R is (8, 3/4).

Now, the line LR goes from (-8, 19/4) to (8, 3/4). Let me compute the direction vector from L to R.

The direction vector $ \vec{LR} = R - L = (8 - (-8), 3/4 - 19/4) = (16, -16/4) = (16, -4) $.

Wait, that's (16, -4). But the problem says the direction vector is of the form (-7, b). So, how does this relate?

Well, direction vectors can be scaled. So, (16, -4) is a direction vector. Let me see if this is a scalar multiple of (-7, b).

So, if (16, -4) = k*(-7, b), then:

16 = -7k

and

-4 = b*k

From the first equation, k = -16/7. Then, substituting into the second equation:

-4 = b*(-16/7) => b = (-4) / (-16/7) = (4) / (16/7) = 4*(7/16) = 28/16 = 7/4.

Same result! So, b is 7/4.

Therefore, regardless of the method, I get b = 7/4.

But let me check again. The direction vector from L to R is (16, -4). If we simplify this vector by dividing by 4, we get (4, -1), which matches the direction vector from A to B. So, that's consistent.

But the problem says the direction vector is of the form (-7, b). So, if we take the direction vector (16, -4) and find a scalar multiple that gives us (-7, b), we can do that by scaling.

Alternatively, since direction vectors can be any scalar multiple, we can take the direction vector (16, -4) and find a scalar multiple that has an x-component of -7.

So, let me think. Let me denote the direction vector as (Δx, Δy). We know that Δx / Δy = slope. Wait, but slope is rise over run, which is Δy / Δx.

Alternatively, since we have direction vector (16, -4), which is equivalent to (4, -1). So, if we want a direction vector with x-component -7, we can find the scalar multiple.

Let me think of direction vectors as being proportional. So, if (4, -1) is a direction vector, then any vector (4k, -k) is also a direction vector. So, if we want the x-component to be -7, then 4k = -7 => k = -7/4. Therefore, the y-component would be -k = -(-7/4) = 7/4. Therefore, the direction vector is (-7, 7/4). Therefore, b = 7/4.

Same answer again.

Alternatively, maybe I can parametrize the line LR with parameter t. Let me try that.

Let me parametrize the line LR. Let me take point L (-8, 19/4) as the starting point. The direction vector is (16, -4), so parametric equations would be:

x = -8 + 16t

y = 19/4 - 4t

But the problem says the direction vector is of the form (-7, b). So, if we can write the parametric equations with direction vector (-7, b), then we need to find t such that the direction vector is scaled appropriately.

Alternatively, maybe the problem is referring to a different parametrization. For example, if we want the direction vector to be (-7, b), then we can find the relation between the direction vector (16, -4) and (-7, b).

So, since direction vectors are scalar multiples, there exists a scalar λ such that:

-7 = 16λ

and

b = -4λ

From the first equation, λ = -7/16. Then, substituting into the second equation:

b = -4*(-7/16) = 28/16 = 7/4.

Same result again.

Therefore, regardless of the method, I get b = 7/4.

But let me check once more with coordinates. Let me compute the vector from L to R, which is (16, -4). If I want a direction vector of (-7, b), then the direction vector (-7, b) should be parallel to (16, -4). Therefore, the cross product should be zero.

In two dimensions, the cross product of vectors (a, b) and (c, d) is ad - bc. If they are parallel, then ad - bc = 0.

So, for vectors (-7, b) and (16, -4):

(-7)(-4) - (b)(16) = 0

28 - 16b = 0

16b = 28

b = 28 / 16 = 7/4.

Yes! That's another way to check. So, cross product is zero, so they are parallel. Therefore, b = 7/4.

Therefore, all methods lead to the same answer.

But just to make sure, let me check if there's any chance that the direction vector is from R to L instead of L to R. If we take vector RL instead of LR, then it would be (-16, 4). Then, if we want a direction vector of (-7, b), then:

-7 = -16λ

and

b = 4λ

From first equation, λ = (-7)/(-16) = 7/16. Then, b = 4*(7/16) = 28/16 = 7/4. Same result.

So regardless of direction, the value of b is 7/4.

Therefore, I think confidently that the answer is 7/4.

But let me check once again with the original points A and B. The direction vector from A to B is (4, -1). If we want a direction vector of (-7, b), then since (4, -1) and (-7, b) are scalar multiples, then:

(4, -1) = k*(-7, b)

So, 4 = -7k => k = -4/7

Then, -1 = b*k => -1 = b*(-4/7) => b = (-1)*(-7/4) = 7/4.

Same answer.

Therefore, all methods confirm that b is 7/4.

I think that's thorough enough. I can't see any mistakes in my reasoning. All approaches lead to the same answer. So, I think the answer is 7/4.

**Final Answer**
The value of $ b $ is \boxed{\dfrac{7}{4}}.

To determine the value of $ b $ in the direction vector $ \begin{pmatrix} -7 \\ b \end{pmatrix} $, we analyze the geometry of the line described in the Asymptote diagram.

---

### Step 1: Understand the Line

The line in question is defined by the points $ A = (-5, 4) $ and $ B = (-1, 3) $. The Asymptote code then defines points $ L $ and $ R $ as the intersections of the line $ AB $ with the vertical lines $ x = -8 $ and $ x = 8 $, respectively. The red line $ LR $ is the same as the line $ AB $, just extended to those vertical lines.

Thus, the direction vector of the line $ LR $ is the same as the direction vector of the line $ AB $.

---

### Step 2: Compute the Direction Vector of Line AB

The direction vector from point $ A $ to point $ B $ is:

$$
\vec{AB} = B - A = (-1 - (-5), 3 - 4) = (4, -1)
$$

So, $ \vec{AB} = \begin{pmatrix} 4 \\ -1 \end{pmatrix} $ is a direction vector of the line.

---

### Step 3: Relate to the Given Direction Vector

We are told that the direction vector of the line is of the form $ \begin{pmatrix} -7 \\ b \end{pmatrix} $. Since direction vectors of the same line are scalar multiples of each other, there exists a scalar $ k $ such that:

$$
\begin{pmatrix} -7 \\ b \end{pmatrix} = k \cdot \begin{pmatrix} 4 \\ -1 \end{pmatrix}
$$

This gives us the system of equations:

$$
-7 = 4k \quad \text{and} \quad b = -k
$$

Solving the first equation:

$$
k = \frac{-7}{4}
$$

Substitute into the second equation:

$$
b = -\left( \frac{-7}{4} \right) = \frac{7}{4}
$$

---

### Step 4: Verification via Parametric Form

Alternatively, we can find the direction vector of the line $ LR $ by computing the vector from point $ L $ to point $ R $, where:

- $ L = (-8, \frac{19}{4}) $
- $ R = (8, \frac{3}{4}) $

Then:

$$
\vec{LR} = R - L = (8 - (-8), \frac{3}{4} - \frac{19}{4}) = (16, -4)
$$

This vector is also a direction vector of the line. To match the given form $ \begin{pmatrix} -7 \\ b \end{pmatrix} $, we find a scalar $ \lambda $ such that:

$$
\begin{pmatrix} -7 \\ b \end{pmatrix} = \lambda \cdot \begin{pmatrix} 16 \\ -4 \end{pmatrix}
$$

Solving:

$$
-7 = 16\lambda \Rightarrow \lambda = \frac{-7}{16}
$$

$$
b = -4\lambda = -4 \cdot \left( \frac{-7}{16} \right) = \frac{28}{16} = \frac{7}{4}
$$

---

### Final Answer

All methods confirm that the value of $ b $ is:

$$
\boxed{\dfrac{7}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ b $ in the direction vector $ \begin{pmatrix} -7 \\ b \end{pmatrix} $ for the given line in the Asymptote figure. Let me try to visualize the problem first.

The Asymptote code draws a coordinate system with axes from -8 to 8. Points A and B are given as (-5, 4) and (-1, 3) respectively. Then, there are points L and R, which are the intersections of the line AB with the vertical lines x = -8 and x = 8. The line LR is drawn in red, which is the line we're interested in. The question says that this line has a direction vector of the form $ \begin{pmatrix} -7 \\ b \end{pmatrix} $, and we need to find $ b $.

First, let me recall that the direction vector of a line can be found by taking two points on the line and subtracting their coordinates. Since the line LR is determined by points A and B, maybe I can find the direction vector from A to B and then relate it to the given direction vector?

Wait, but the line LR is the same as the line AB, right? Because L and R are points where the line AB intersects the vertical lines x = -8 and x = 8. So, the line LR is just the line AB extended to those vertical lines. Therefore, the direction vector of line LR should be the same as the direction vector of line AB.

But the problem says the direction vector is of the form $ \begin{pmatrix} -7 \\ b \end{pmatrix} $. So, if I can find the direction vector of AB, then I can compare it to this form and find $ b $.

Let me compute the direction vector of AB. Point A is (-5, 4) and point B is (-1, 3). The direction vector from A to B is $ \vec{AB} = B - A = (-1 - (-5), 3 - 4) = (-1 + 5, -1) = (4, -1) $.

Wait, so the direction vector is (4, -1). But the problem says the direction vector is (-7, b). Hmm, so these two vectors should be scalar multiples of each other because they are direction vectors of the same line.

So, if $ \begin{pmatrix} 4 \\ -1 \end{pmatrix} $ is a direction vector, then any scalar multiple of this vector is also a direction vector. Therefore, there exists some scalar $ k $ such that:

$ -7 = 4k $ and $ b = -1k $.

Let me solve for $ k $ first. From the x-component: $ -7 = 4k $, so $ k = -7/4 $. Then, substituting into the y-component: $ b = -1 * (-7/4) = 7/4 $.

Therefore, $ b = 7/4 $. Is that correct? Let me check.

Alternatively, maybe I need to find the direction vector of the line LR, which is the same as AB. But perhaps the problem is referring to a different direction vector? Wait, the problem says "the following line is parameterized, so that its direction vector is of the form...", so maybe the line LR is parameterized with direction vector (-7, b). But since direction vectors can be scaled, we need to find the correct scalar multiple.

Alternatively, maybe I need to find the slope of the line LR and then relate it to the direction vector. Let me try that approach as well to verify.

The slope of line AB can be calculated as $ m = (y_B - y_A)/(x_B - x_A) = (3 - 4)/(-1 - (-5)) = (-1)/(4) = -1/4 $. Therefore, the slope is -1/4.

A direction vector corresponding to this slope would have a rise over run of -1/4. So, if the direction vector is (Δx, Δy), then Δy / Δx = -1/4. Therefore, Δy = (-1/4)Δx.

But the given direction vector is (-7, b). So, if Δx = -7, then Δy = (-1/4)(-7) = 7/4. Therefore, b = 7/4.

Same result as before. So that seems consistent. Therefore, $ b = 7/4 $.

But let me make sure that the line LR is indeed the same as line AB. The Asymptote code defines points L and R as the intersections of line AB with the vertical lines x = -8 and x = 8. Therefore, line LR is the same line as AB, just extended to those vertical lines. Therefore, the direction vector of LR is the same as AB. Therefore, the direction vector is indeed (4, -1) or any scalar multiple.

But the problem says the direction vector is of the form (-7, b). So, we need to find b such that (-7, b) is a direction vector for the line. Since direction vectors can be scaled, we can find the scalar multiple between (4, -1) and (-7, b).

So, if we have two direction vectors, they must be scalar multiples. Therefore, there exists a scalar k such that:

-7 = 4k

and

b = -1k

So solving for k from the first equation: k = -7/4. Then, substituting into the second equation: b = -1*(-7/4) = 7/4.

Therefore, b is 7/4.

Alternatively, maybe I can parametrize the line LR and find its direction vector. Let me try that approach.

First, let's find the equation of line AB. Since we have two points A(-5, 4) and B(-1, 3), we can write the equation of the line.

We already found the slope m = -1/4. So, using point-slope form with point A:

$ y - 4 = -\frac{1}{4}(x + 5) $

Simplifying:

$ y = -\frac{1}{4}x - \frac{5}{4} + 4 $

$ y = -\frac{1}{4}x - \frac{5}{4} + \frac{16}{4} $

$ y = -\frac{1}{4}x + \frac{11}{4} $

So the equation of the line is $ y = -\frac{1}{4}x + \frac{11}{4} $.

Now, points L and R are the intersections of this line with x = -8 and x = 8, respectively. Let me find coordinates of L and R.

For point L, x = -8:

$ y = -\frac{1}{4}(-8) + \frac{11}{4} = 2 + \frac{11}{4} = \frac{8}{4} + \frac{11}{4} = \frac{19}{4} $

Therefore, L is (-8, 19/4).

For point R, x = 8:

$ y = -\frac{1}{4}(8) + \frac{11}{4} = -2 + \frac{11}{4} = -\frac{8}{4} + \frac{11}{4} = \frac{3}{4} $

Therefore, R is (8, 3/4).

Now, the line LR goes from (-8, 19/4) to (8, 3/4). Let me compute the direction vector from L to R.

The direction vector $ \vec{LR} = R - L = (8 - (-8), 3/4 - 19/4) = (16, -16/4) = (16, -4) $.

Wait, that's (16, -4). But the problem says the direction vector is of the form (-7, b). So, how does this relate?

Well, direction vectors can be scaled. So, (16, -4) is a direction vector. Let me see if this is a scalar multiple of (-7, b).

So, if (16, -4) = k*(-7, b), then:

16 = -7k

and

-4 = b*k

From the first equation, k = -16/7. Then, substituting into the second equation:

-4 = b*(-16/7) => b = (-4) / (-16/7) = (4) / (16/7) = 4*(7/16) = 28/16 = 7/4.

Same result! So, b is 7/4.

Therefore, regardless of the method, I get b = 7/4.

But let me check again. The direction vector from L to R is (16, -4). If we simplify this vector by dividing by 4, we get (4, -1), which matches the direction vector from A to B. So, that's consistent.

But the problem says the direction vector is of the form (-7, b). So, if we take the direction vector (16, -4) and find a scalar multiple that gives us (-7, b), we can do that by scaling.

Alternatively, since direction vectors can be any scalar multiple, we can take the direction vector (16, -4) and find a scalar multiple that has an x-component of -7.

So, let me think. Let me denote the direction vector as (Δx, Δy). We know that Δx / Δy = slope. Wait, but slope is rise over run, which is Δy / Δx.

Alternatively, since we have direction vector (16, -4), which is equivalent to (4, -1). So, if we want a direction vector with x-component -7, we can find the scalar multiple.

Let me think of direction vectors as being proportional. So, if (4, -1) is a direction vector, then any vector (4k, -k) is also a direction vector. So, if we want the x-component to be -7, then 4k = -7 => k = -7/4. Therefore, the y-component would be -k = -(-7/4) = 7/4. Therefore, the direction vector is (-7, 7/4). Therefore, b = 7/4.

Same answer again.

Alternatively, maybe I can parametrize the line LR with parameter t. Let me try that.

Let me parametrize the line LR. Let me take point L (-8, 19/4) as the starting point. The direction vector is (16, -4), so parametric equations would be:

x = -8 + 16t

y = 19/4 - 4t

But the problem says the direction vector is of the form (-7, b). So, if we can write the parametric equations with direction vector (-7, b), then we need to find t such that the direction vector is scaled appropriately.

Alternatively, maybe the problem is referring to a different parametrization. For example, if we want the direction vector to be (-7, b), then we can find the relation between the direction vector (16, -4) and (-7, b).

So, since direction vectors are scalar multiples, there exists a scalar λ such that:

-7 = 16λ

and

b = -4λ

From the first equation, λ = -7/16. Then, substituting into the second equation:

b = -4*(-7/16) = 28/16 = 7/4.

Same result again.

Therefore, regardless of the method, I get b = 7/4.

But let me check once more with coordinates. Let me compute the vector from L to R, which is (16, -4). If I want a direction vector of (-7, b), then the direction vector (-7, b) should be parallel to (16, -4). Therefore, the cross product should be zero.

In two dimensions, the cross product of vectors (a, b) and (c, d) is ad - bc. If they are parallel, then ad - bc = 0.

So, for vectors (-7, b) and (16, -4):

(-7)(-4) - (b)(16) = 0

28 - 16b = 0

16b = 28

b = 28 / 16 = 7/4.

Yes! That's another way to check. So, cross product is zero, so they are parallel. Therefore, b = 7/4.

Therefore, all methods lead to the same answer.

But just to make sure, let me check if there's any chance that the direction vector is from R to L instead of L to R. If we take vector RL instead of LR, then it would be (-16, 4). Then, if we want a direction vector of (-7, b), then:

-7 = -16λ

and

b = 4λ

From first equation, λ = (-7)/(-16) = 7/16. Then, b = 4*(7/16) = 28/16 = 7/4. Same result.

So regardless of direction, the value of b is 7/4.

Therefore, I think confidently that the answer is 7/4.

But let me check once again with the original points A and B. The direction vector from A to B is (4, -1). If we want a direction vector of (-7, b), then since (4, -1) and (-7, b) are scalar multiples, then:

(4, -1) = k*(-7, b)

So, 4 = -7k => k = -4/7

Then, -1 = b*k => -1 = b*(-4/7) => b = (-1)*(-7/4) = 7/4.

Same answer.

Therefore, all methods confirm that b is 7/4.

I think that's thorough enough. I can't see any mistakes in my reasoning. All approaches lead to the same answer. So, I think the answer is 7/4.

**Final Answer**
The value of $ b $ is \boxed{\dfrac{7}{4}}.

To determine the value of $ b $ in the direction vector $ \begin{pmatrix} -7 \\ b \end{pmatrix} $, we analyze the geometry of the line described in the Asymptote diagram.

---

### Step 1: Understand the Line

The line in question is defined by the points $ A = (-5, 4) $ and $ B = (-1, 3) $. The Asymptote code then defines points $ L $ and $ R $ as the intersections of the line $ AB $ with the vertical lines $ x = -8 $ and $ x = 8 $, respectively. The red line $ LR $ is the same as the line $ AB $, just extended to those vertical lines.

Thus, the direction vector of the line $ LR $ is the same as the direction vector of the line $ AB $.

---

### Step 2: Compute the Direction Vector of Line AB

The direction vector from point $ A $ to point $ B $ is:

$$
\vec{AB} = B - A = (-1 - (-5), 3 - 4) = (4, -1)
$$

So, $ \vec{AB} = \begin{pmatrix} 4 \\ -1 \end{pmatrix} $ is a direction vector of the line.

---

### Step 3: Relate to the Given Direction Vector

We are told that the direction vector of the line is of the form $ \begin{pmatrix} -7 \\ b \end{pmatrix} $. Since direction vectors of the same line are scalar multiples of each other, there exists a scalar $ k $ such that:

$$
\begin{pmatrix} -7 \\ b \end{pmatrix} = k \cdot \begin{pmatrix} 4 \\ -1 \end{pmatrix}
$$

This gives us the system of equations:

$$
-7 = 4k \quad \text{and} \quad b = -k
$$

Solving the first equation:

$$
k = \frac{-7}{4}
$$

Substitute into the second equation:

$$
b = -\left( \frac{-7}{4} \right) = \frac{7}{4}
$$

---

### Step 4: Verification via Parametric Form

Alternatively, we can find the direction vector of the line $ LR $ by computing the vector from point $ L $ to point $ R $, where:

- $ L = (-8, \frac{19}{4}) $
- $ R = (8, \frac{3}{4}) $

Then:

$$
\vec{LR} = R - L = (8 - (-8), \frac{3}{4} - \frac{19}{4}) = (16, -4)
$$

This vector is also a direction vector of the line. To match the given form $ \begin{pmatrix} -7 \\ b \end{pmatrix} $, we find a scalar $ \lambda $ such that:

$$
\begin{pmatrix} -7 \\ b \end{pmatrix} = \lambda \cdot \begin{pmatrix} 16 \\ -4 \end{pmatrix}
$$

Solving:

$$
-7 = 16\lambda \Rightarrow \lambda = \frac{-7}{16}
$$

$$
b = -4\lambda = -4 \cdot \left( \frac{-7}{16} \right) = \frac{28}{16} = \frac{7}{4}
$$

---

### Final Answer

All methods confirm that the value of $ b $ is:

$$
\boxed{\dfrac{7}{4}}
$$

Solution:
The line passes through $\begin{pmatrix} -5 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 3 \end{pmatrix},$ so its direction vector is proportional to
\[\begin{pmatrix} -1 \\ 3 \end{pmatrix} - \begin{pmatrix} -5 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}.\]To get an $x$-coordinate of $-7,$ we can multiply this vector by the scalar $-\frac{7}{4}.$  This gives us
\[-\frac{7}{4} \begin{pmatrix} 4 \\ -1 \end{pmatrix} = \begin{pmatrix} -7 \\ 7/4 \end{pmatrix}.\]Therefore, $b = \boxed{\frac{7}{4}}.$


Explanation:
Okay, so I need to find the units digit of 4$, which is defined as a superfactorial. The problem says that n$ is equal to a tower of n! exponents, with n! levels. So, for n=4, that would be 4! raised to the power of 4! raised to the power of ... and so on, with a total of 4! levels. Let me write that down to make sure I understand.

First, let me compute 4! because that's going to be the number of exponents in the tower. 4! is 4×3×2×1 = 24. So, 4$ is a tower of 24 copies of 4! (which is 24). So, the expression is 24^24^24^...^24 with 24 levels. That's a massive number! Obviously, I can't compute it directly, so I need to find a way to determine just the units digit.

The units digit of a number is the same as the number modulo 10. So, I need to compute this tower modulo 10. However, since the tower is so tall, I need to find some patterns or properties that can help me simplify this.

First, let me recall that for exponents, Euler's theorem might be useful. Euler's theorem states that if a and n are coprime, then a^φ(n) ≡ 1 mod n, where φ(n) is Euler's totient function. Since we're working modulo 10, φ(10) = φ(2×5) = φ(2)×φ(5) = 1×4 = 4. So, φ(10) = 4. However, Euler's theorem requires that a and n are coprime. So, if the base of the exponent is coprime with 10, then we can use this theorem. But if not, we need to handle it differently.

But first, let me note that the tower is 24^24^24^...^24 with 24 levels. Let me denote this tower as T. So, T = 24^T', where T' is the tower of 24s with 23 levels. But since we need to compute T mod 10, maybe we can find a pattern or use properties of exponents modulo 10.

But before that, let me check if 24 and 10 are coprime. 24 and 10 share a common factor of 2, so they are not coprime. Therefore, Euler's theorem doesn't apply directly here. However, maybe I can break down the modulus into prime power components using the Chinese Remainder Theorem. Since 10 = 2 × 5, and 2 and 5 are coprime, if I can find T mod 2 and T mod 5, then I can combine them to get T mod 10.

Let me try that approach. Let's compute T mod 2 and T mod 5 separately.

First, T mod 2:

Since 24 is even, any power of 24 will also be even. Therefore, 24^anything is congruent to 0 mod 2. Therefore, T mod 2 = 0.

Now, T mod 5:

This is more complicated. Let me think. Since 24 mod 5 is 4, because 24 divided by 5 is 4 with a remainder of 4. So, T mod 5 is equivalent to 4^T' mod 5, where T' is the tower of 24s with 23 levels. So, we need to compute 4^T' mod 5.

But again, 4 and 5 are coprime, so Euler's theorem applies here. φ(5) = 4, so 4^4 ≡ 1 mod 5. Therefore, 4^k mod 5 cycles every 4. Therefore, if we can find T' mod φ(5) = 4, then we can compute 4^(T' mod 4) mod 5. However, if T' mod 4 is 0, then it would be 4^4 ≡ 1 mod 5, so we need to adjust accordingly.

But T' is the tower of 24s with 23 levels. Let me denote T' as 24^T'' where T'' is the tower of 24s with 22 levels. So, T' = 24^T''. Therefore, to compute T' mod 4, we need to compute 24^T'' mod 4.

But 24 mod 4 is 0, since 24 is divisible by 4. Therefore, 24^anything is 0 mod 4, as long as the exponent is at least 1. Since T'' is a tower of 24s with 22 levels, which is certainly at least 1, so T' mod 4 = 0.

Therefore, going back, T' mod 4 = 0, so 4^T' mod 5 = 4^(0) mod 5? Wait, no. Wait, Euler's theorem says that 4^φ(5) = 4^4 ≡ 1 mod 5. Therefore, 4^k ≡ 4^(k mod 4) mod 5, but only if we adjust for the exponent. Wait, actually, more precisely, if we have 4^k mod 5, since φ(5)=4, then 4^k ≡ 4^(k mod 4) * 4^(4 * m) ≡ 4^(k mod 4) * (1)^m ≡ 4^(k mod 4) mod 5. However, if k mod 4 is 0, then it's 4^0 = 1 mod 5. Wait, but 4^4 ≡ 1 mod 5, so 4^(4m) ≡ 1^m = 1 mod 5. Therefore, if k is a multiple of 4, then 4^k ≡ 1 mod 5. Therefore, since T' mod 4 = 0, then 4^T' ≡ 1 mod 5. Therefore, T mod 5 = 1.

Therefore, combining the two results:

T ≡ 0 mod 2

T ≡ 1 mod 5

We need to find a number x such that x ≡ 0 mod 2 and x ≡ 1 mod 5. Let's solve this system.

Numbers congruent to 1 mod 5 are 1, 6, 11, 16, 21, 26,... Now, which of these are even? 6, 16, 26,... So, the solutions are numbers congruent to 6 mod 10. Therefore, x ≡ 6 mod 10. Therefore, the units digit is 6.

Wait, but hold on. Let me check my steps again because this seems straightforward, but maybe I made a mistake in the exponents.

First, let me recap:

We have T = 24^T' where T' is a tower of 24s with 23 levels.

We need T mod 10. We split into mod 2 and mod 5.

For mod 2: 24 is 0 mod 2, so any power is 0 mod 2. So T ≡ 0 mod 2.

For mod 5: T ≡ 4^T' mod 5. Then, since φ(5)=4, we need T' mod 4. T' is 24^T'' mod 4. Since 24 ≡ 0 mod 4, so 24^anything ≥1 is 0 mod 4. Therefore, T' ≡ 0 mod 4. Therefore, 4^T' ≡ 4^0 ≡ 1 mod 5? Wait, but 4^T' mod 5. If T' ≡ 0 mod 4, then 4^T' ≡ (4^4)^k ≡ 1^k ≡ 1 mod 5. So yes, that's correct. Therefore, T ≡ 1 mod 5.

Therefore, solving x ≡ 0 mod 2 and x ≡ 1 mod 5 gives x ≡ 6 mod 10. Therefore, the units digit is 6. So, the answer is 6?

Wait, but let me check with smaller numbers to see if my approach is correct. Let me take a smaller case where I can compute manually.

For example, let's take n=1. Then 1$ is a tower of 1! = 1 level of 1! = 1. So, 1$ = 1. Units digit is 1. Let's check with our method.

Compute 1$ mod 10. 1! = 1. So, tower of 1 level: 1. So, 1 mod 10 is 1. Correct.

Another example: n=2. 2$ is a tower of 2! = 2 levels of 2! = 2. So, 2^2 = 4. Units digit 4. Let's check with our method.

Compute T mod 2 and mod 5.

T = 2^2 = 4. T mod 2 = 0, T mod 5 = 4. Then solving x ≡ 0 mod 2 and x ≡ 4 mod 5. The solution is 4 mod 10. Correct.

Another example: n=3. 3$ is a tower of 3! = 6 levels of 3! = 6. So, 6^6^6^6^6^6. Let's compute its units digit.

But maybe I can compute it step by step. However, since it's a tower of 6s, let's try to compute mod 10.

But let's try our method. First, compute T mod 2 and mod 5.

First, T mod 2: 6 is even, so any power is 0 mod 2. So T ≡ 0 mod 2.

T mod 5: 6 mod 5 = 1. So T = 1^T' mod 5. Since 1^anything is 1 mod 5. Therefore, T ≡ 1 mod 5. Therefore, solving x ≡ 0 mod 2 and x ≡ 1 mod 5 gives x ≡ 6 mod 10. Therefore, units digit 6. Let me check with actual computation.

Compute the tower 6^6^6^6^6^6. Let's compute the units digit. Since the units digit of 6^anything is always 6. Because 6^1=6, 6^2=36, 6^3=216, etc. So regardless of the exponent, the units digit is 6. Therefore, the units digit is 6. Which matches our result. So our method works here.

Another example: n=4. Wait, that's the original problem. But let me check with n=4. Wait, but according to our previous calculation, the units digit is 6. But let me see if that's correct.

But let me think again. Wait, for n=4, the tower is 24^24^...^24 with 24 levels. But when we calculated mod 5, we found that T ≡ 1 mod 5 and T ≡ 0 mod 2, leading to 6 mod 10. However, let me check if there's a mistake in the step where we calculated T' mod 4.

Wait, T' is the tower of 24s with 23 levels. So, T' = 24^T''. T'' is the tower with 22 levels. So, to compute T' mod 4, since 24 ≡ 0 mod 4, then 24^anything ≥1 is 0 mod 4. Therefore, T' ≡ 0 mod 4. Therefore, 4^T' mod 5 is 4^0 mod 5? Wait, no. Wait, 4^T' mod 5. Since φ(5)=4, and T' ≡ 0 mod 4, then 4^T' ≡ 4^(4k) ≡ (4^4)^k ≡ 1^k ≡ 1 mod 5. Therefore, yes, that's correct. Therefore, T ≡ 1 mod 5. Therefore, combining with T ≡ 0 mod 2 gives 6 mod 10. So, the units digit is 6. But let me check with another approach.

Alternatively, maybe I can compute the tower modulo 10 directly, but considering that exponents can be simplified.

But given that the tower is 24^24^...^24 with 24 levels. Let me think about the exponents. Since we need to compute this modulo 10, and since 10 = 2 * 5, and we already did the Chinese Remainder Theorem approach, which gave 6. But let me check if there's another way.

Alternatively, since 24 is congruent to 4 mod 5 and 0 mod 2. But maybe I can think about the tower modulo 10. However, since the tower is so tall, maybe we can find a pattern in the exponents.

But let me think again. Let me try to compute the tower modulo 10. However, since the tower is 24^T', and T' is a tower of 24s with 23 levels. Let me note that 24 mod 10 is 4, so the entire tower is equivalent to 4^T' mod 10. Wait, but earlier I split into mod 2 and mod 5. However, if I try to compute 4^T' mod 10 directly, maybe I can find a pattern.

But 4^k mod 10 cycles every 2: 4^1=4, 4^2=16→6, 4^3=64→4, 4^4=256→6, etc. So, the cycle is 4,6,4,6,... So, if the exponent is odd, it's 4 mod 10; if even, 6 mod 10. Therefore, if I can determine whether T' is odd or even, then I can find 4^T' mod 10.

But T' is the tower of 24s with 23 levels. Let me check if T' is even or odd. Since 24 is even, and any power of an even number is even. Therefore, T' is even. Therefore, T' is even, so 4^T' mod 10 is 6. Therefore, the units digit is 6. Which matches our previous result. So, this seems to confirm that the answer is 6.

Wait, so this approach is simpler. Let me explain again:

Since the entire tower T is 24^T', and 24 mod 10 is 4, so T ≡ 4^T' mod 10. Then, since 4^k mod 10 cycles every 2, with period 2: 4,6,4,6,... So, if T' is even, then 4^T' ≡ 6 mod 10; if T' is odd, then 4^T' ≡ 4 mod 10. Therefore, we need to check if T' is even or odd.

But T' is a tower of 24s with 23 levels. Since 24 is even, and any power of an even number is even. Therefore, regardless of the number of levels (as long as it's at least 1), the tower T' is even. Therefore, T' is even, so 4^T' mod 10 is 6. Therefore, the units digit is 6. Therefore, the answer is 6. So, this seems to confirm the previous result.

But why did we go through the Chinese Remainder Theorem earlier? Because when I first thought about mod 5, I considered that 4^T' mod 5 is 1, and mod 2 is 0, leading to 6 mod 10. But here, by directly looking at mod 10, we can see that since T' is even, the result is 6. So, both methods agree.

Therefore, the units digit is 6. However, let me check once more with another example. Let's take n=2 again. The tower is 2^2 = 4. The units digit is 4. According to the second method: T = 2^T', where T' is the tower with 1 level (since n! = 2, so 2 levels? Wait, n=2, n! = 2, so the tower has 2 levels. So, T = 2^2 = 4. Then, T' is the tower with 1 level, which is 2. Therefore, T' = 2. Then, 4^T' mod 10 is 4^2 = 16 mod 10 = 6? Wait, but the actual value is 4. Wait, this contradicts. Wait, so there's a problem here.

Wait, hold on. Wait, for n=2, the tower is 2!^2! = 2^2 = 4. So, T is 4. But according to the second method, I said that T = 4^T', where T' is the tower with n! -1 levels. Wait, but in this case, n=2, n! = 2, so T is 2^T', where T' is the tower with 1 level, which is 2. Therefore, T = 2^2 = 4. However, when I tried to compute 4^T' mod 10, I was wrong. Wait, why?

Because in the case of n=4, the tower is 24^T', where T' is the tower with 23 levels. However, in the case of n=2, the tower is 2^T', where T' is the tower with 1 level (which is 2). Therefore, T = 2^T', not 4^T'. Wait, so my mistake was in the second approach when I said that T ≡ 4^T' mod 10. That was incorrect. Because in reality, T is 24^T', but 24 mod 10 is 4, so T ≡ 4^T' mod 10. However, in the case of n=2, the tower is 2^T', where T' is 2, so T = 2^2 = 4. However, 2 mod 10 is 2, not 4. Therefore, my mistake was in the second approach when I thought that T ≡ 4^T' mod 10. That was only true for n=4, where the base is 24, which is 4 mod 10. For n=2, the base is 2, which is 2 mod 10. Therefore, my second approach was specific to n=4. Therefore, in the case of n=4, since the base is 24 ≡ 4 mod 10, then T ≡ 4^T' mod 10. However, in the case of n=2, the base is 2, so T ≡ 2^T' mod 10. Therefore, my second approach was correct for n=4, but I confused it with n=2. Therefore, in the case of n=4, since the base is 4 mod 10, and T' is even, then 4^T' mod 10 is 6. Therefore, the answer is 6. However, in the case of n=2, since the base is 2 mod 10, and T' is 2, which is even, then 2^T' mod 10 is 2^2 = 4 mod 10, which is correct. Therefore, my second approach was correct for n=4. Therefore, the answer is 6.

But let me check another example where the base is 4 mod 10. Let's take n=3. Wait, n=3, n! = 6. So, the tower is 6^6^...^6 with 6 levels. Let me compute its units digit. As before, since 6^anything has units digit 6. Therefore, the units digit is 6. Let me check with the second approach. The tower is 6^T', where T' is the tower with 5 levels. Since 6 mod 10 is 6, and 6^k mod 10 is always 6. Therefore, regardless of T', the units digit is 6. Therefore, that works. So, if the base is 6, the units digit is always 6. Similarly, if the base is 4, then depending on the exponent's parity, it's 4 or 6. Therefore, in our case, for n=4, since the base is 4 mod 10, and the exponent T' is even, then the units digit is 6. Therefore, the answer is 6.

Therefore, after checking with multiple examples and approaches, I think the answer is 6. However, let me just make sure that T' is indeed even. T' is the tower of 24s with 23 levels. Since 24 is even, and any power of an even number is even, so regardless of the number of levels (as long as it's at least 1), the tower is even. Therefore, T' is even. Therefore, 4^T' mod 10 is 6. Therefore, the units digit is 6.

But let me think again about the tower. For example, if we have a tower of 24s with 23 levels, which is 24^24^...^24. Since the first exponent is 24, which is even, so the entire tower is 24 raised to an even number. Therefore, 24^even. But 24 is even, so 24^even is divisible by 2^even, which is certainly even. Therefore, T' is even. Therefore, 4^T' mod 10 is 6. Therefore, the units digit is 6.

Therefore, I think the answer is 6. However, let me check if there's a possibility that T' is 0 mod φ(10) or something else. Wait, but in the first approach, we used Chinese Remainder Theorem and found that T ≡ 6 mod 10. In the second approach, we found the same. Therefore, both methods confirm that the units digit is 6.

But just to be thorough, let me check with another angle. Let me try to compute the tower modulo 10, considering that exponents can be simplified.

We have T = 24^T', where T' is the tower with 23 levels. Since 24 ≡ 4 mod 10, T ≡ 4^T' mod 10. As mentioned, 4^k mod 10 cycles every 2: 4, 6, 4, 6,... So, if T' is even, it's 6; if odd, 4. Since T' is even, as established, then T ≡ 6 mod 10. Therefore, units digit is 6.

Alternatively, if I wanted to compute T' mod 2, since the exponent in 4^T' mod 10 depends on T' mod 2. Wait, but T' is even, so T' mod 2 = 0. Therefore, 4^T' mod 10 = 4^0 mod 10? No, that's not correct. Wait, no. The cycle is every 2. So, if T' is even, then T' = 2k, so 4^{2k} = (4^2)^k = 16^k. 16 mod 10 is 6, so 16^k mod 10 is 6^k mod 10. But 6^k mod 10 is always 6. Therefore, 4^{even} mod 10 is 6. Therefore, regardless of the exponent, as long as it's even, it's 6. Therefore, that's another way to see it. Therefore, since T' is even, the result is 6. Therefore, the units digit is 6.

Therefore, after multiple approaches and checks, I feel confident that the units digit of 4$ is 6.

**Final Answer**
\boxed{6}

To determine the **units digit** of the superfactorial $ 4\$ $, we begin by understanding the definition:

$$
n\$ = \underbrace{n!^{{n!}^{{\cdot}^{{\cdot}^{{\cdot}^{n!}}}}}}_{n!}
$$

This means that $ 4\$ $ is a **tower of exponents** with $ 4! = 24 $ levels, each level being $ 24 $. So:

$$
4\$ = 24^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \quad \text{(with 24 levels of 24)}
$$

We are asked to find the **units digit** of this number, which is equivalent to computing:

$$
4\$ \mod 10
$$

---

### Step 1: Reduce the base modulo 10

Since the base of the tower is 24, and we are interested in the **units digit**, we reduce 24 modulo 10:

$$
24 \equiv 4 \mod 10
$$

So, the entire tower becomes:

$$
4^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \mod 10
$$

Let’s denote the tower of 24s with 23 levels as $ T' $. Then:

$$
4\$ = 4^{T'} \mod 10
$$

---

### Step 2: Analyze the exponent $ T' \mod 2 $

We now need to determine the **parity** of the exponent $ T' $, because the powers of 4 modulo 10 cycle every 2:

- $ 4^1 = 4 $
- $ 4^2 = 16 \equiv 6 \mod 10 $
- $ 4^3 = 64 \equiv 4 \mod 10 $
- $ 4^4 = 256 \equiv 6 \mod 10 $

So, the pattern is: **4, 6, 4, 6, ...**

Thus, if the exponent is **even**, the result is **6**; if **odd**, it is **4**.

So, we need to determine whether $ T' $ is even or odd.

---

### Step 3: Determine the parity of $ T' $

Recall that $ T' $ is a tower of 24s with 23 levels:

$$
T' = 24^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \quad \text{(23 levels)}
$$

Since 24 is even, and any power of an even number is even, the entire tower $ T' $ is **even**. Therefore:

$$
T' \equiv 0 \mod 2
$$

---

### Step 4: Compute $ 4^{T'} \mod 10 $

Since $ T' $ is even, we use the pattern from earlier:

$$
4^{\text{even}} \equiv 6 \mod 10
$$

Thus:

$$
4\$ \equiv 6 \mod 10
$$

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the units digit of 4$, which is defined as a superfactorial. The problem says that n$ is equal to a tower of n! exponents, with n! levels. So, for n=4, that would be 4! raised to the power of 4! raised to the power of ... and so on, with a total of 4! levels. Let me write that down to make sure I understand.

First, let me compute 4! because that's going to be the number of exponents in the tower. 4! is 4×3×2×1 = 24. So, 4$ is a tower of 24 copies of 4! (which is 24). So, the expression is 24^24^24^...^24 with 24 levels. That's a massive number! Obviously, I can't compute it directly, so I need to find a way to determine just the units digit.

The units digit of a number is the same as the number modulo 10. So, I need to compute this tower modulo 10. However, since the tower is so tall, I need to find some patterns or properties that can help me simplify this.

First, let me recall that for exponents, Euler's theorem might be useful. Euler's theorem states that if a and n are coprime, then a^φ(n) ≡ 1 mod n, where φ(n) is Euler's totient function. Since we're working modulo 10, φ(10) = φ(2×5) = φ(2)×φ(5) = 1×4 = 4. So, φ(10) = 4. However, Euler's theorem requires that a and n are coprime. So, if the base of the exponent is coprime with 10, then we can use this theorem. But if not, we need to handle it differently.

But first, let me note that the tower is 24^24^24^...^24 with 24 levels. Let me denote this tower as T. So, T = 24^T', where T' is the tower of 24s with 23 levels. But since we need to compute T mod 10, maybe we can find a pattern or use properties of exponents modulo 10.

But before that, let me check if 24 and 10 are coprime. 24 and 10 share a common factor of 2, so they are not coprime. Therefore, Euler's theorem doesn't apply directly here. However, maybe I can break down the modulus into prime power components using the Chinese Remainder Theorem. Since 10 = 2 × 5, and 2 and 5 are coprime, if I can find T mod 2 and T mod 5, then I can combine them to get T mod 10.

Let me try that approach. Let's compute T mod 2 and T mod 5 separately.

First, T mod 2:

Since 24 is even, any power of 24 will also be even. Therefore, 24^anything is congruent to 0 mod 2. Therefore, T mod 2 = 0.

Now, T mod 5:

This is more complicated. Let me think. Since 24 mod 5 is 4, because 24 divided by 5 is 4 with a remainder of 4. So, T mod 5 is equivalent to 4^T' mod 5, where T' is the tower of 24s with 23 levels. So, we need to compute 4^T' mod 5.

But again, 4 and 5 are coprime, so Euler's theorem applies here. φ(5) = 4, so 4^4 ≡ 1 mod 5. Therefore, 4^k mod 5 cycles every 4. Therefore, if we can find T' mod φ(5) = 4, then we can compute 4^(T' mod 4) mod 5. However, if T' mod 4 is 0, then it would be 4^4 ≡ 1 mod 5, so we need to adjust accordingly.

But T' is the tower of 24s with 23 levels. Let me denote T' as 24^T'' where T'' is the tower of 24s with 22 levels. So, T' = 24^T''. Therefore, to compute T' mod 4, we need to compute 24^T'' mod 4.

But 24 mod 4 is 0, since 24 is divisible by 4. Therefore, 24^anything is 0 mod 4, as long as the exponent is at least 1. Since T'' is a tower of 24s with 22 levels, which is certainly at least 1, so T' mod 4 = 0.

Therefore, going back, T' mod 4 = 0, so 4^T' mod 5 = 4^(0) mod 5? Wait, no. Wait, Euler's theorem says that 4^φ(5) = 4^4 ≡ 1 mod 5. Therefore, 4^k ≡ 4^(k mod 4) mod 5, but only if we adjust for the exponent. Wait, actually, more precisely, if we have 4^k mod 5, since φ(5)=4, then 4^k ≡ 4^(k mod 4) * 4^(4 * m) ≡ 4^(k mod 4) * (1)^m ≡ 4^(k mod 4) mod 5. However, if k mod 4 is 0, then it's 4^0 = 1 mod 5. Wait, but 4^4 ≡ 1 mod 5, so 4^(4m) ≡ 1^m = 1 mod 5. Therefore, if k is a multiple of 4, then 4^k ≡ 1 mod 5. Therefore, since T' mod 4 = 0, then 4^T' ≡ 1 mod 5. Therefore, T mod 5 = 1.

Therefore, combining the two results:

T ≡ 0 mod 2

T ≡ 1 mod 5

We need to find a number x such that x ≡ 0 mod 2 and x ≡ 1 mod 5. Let's solve this system.

Numbers congruent to 1 mod 5 are 1, 6, 11, 16, 21, 26,... Now, which of these are even? 6, 16, 26,... So, the solutions are numbers congruent to 6 mod 10. Therefore, x ≡ 6 mod 10. Therefore, the units digit is 6.

Wait, but hold on. Let me check my steps again because this seems straightforward, but maybe I made a mistake in the exponents.

First, let me recap:

We have T = 24^T' where T' is a tower of 24s with 23 levels.

We need T mod 10. We split into mod 2 and mod 5.

For mod 2: 24 is 0 mod 2, so any power is 0 mod 2. So T ≡ 0 mod 2.

For mod 5: T ≡ 4^T' mod 5. Then, since φ(5)=4, we need T' mod 4. T' is 24^T'' mod 4. Since 24 ≡ 0 mod 4, so 24^anything ≥1 is 0 mod 4. Therefore, T' ≡ 0 mod 4. Therefore, 4^T' ≡ 4^0 ≡ 1 mod 5? Wait, but 4^T' mod 5. If T' ≡ 0 mod 4, then 4^T' ≡ (4^4)^k ≡ 1^k ≡ 1 mod 5. So yes, that's correct. Therefore, T ≡ 1 mod 5.

Therefore, solving x ≡ 0 mod 2 and x ≡ 1 mod 5 gives x ≡ 6 mod 10. Therefore, the units digit is 6. So, the answer is 6?

Wait, but let me check with smaller numbers to see if my approach is correct. Let me take a smaller case where I can compute manually.

For example, let's take n=1. Then 1$ is a tower of 1! = 1 level of 1! = 1. So, 1$ = 1. Units digit is 1. Let's check with our method.

Compute 1$ mod 10. 1! = 1. So, tower of 1 level: 1. So, 1 mod 10 is 1. Correct.

Another example: n=2. 2$ is a tower of 2! = 2 levels of 2! = 2. So, 2^2 = 4. Units digit 4. Let's check with our method.

Compute T mod 2 and mod 5.

T = 2^2 = 4. T mod 2 = 0, T mod 5 = 4. Then solving x ≡ 0 mod 2 and x ≡ 4 mod 5. The solution is 4 mod 10. Correct.

Another example: n=3. 3$ is a tower of 3! = 6 levels of 3! = 6. So, 6^6^6^6^6^6. Let's compute its units digit.

But maybe I can compute it step by step. However, since it's a tower of 6s, let's try to compute mod 10.

But let's try our method. First, compute T mod 2 and mod 5.

First, T mod 2: 6 is even, so any power is 0 mod 2. So T ≡ 0 mod 2.

T mod 5: 6 mod 5 = 1. So T = 1^T' mod 5. Since 1^anything is 1 mod 5. Therefore, T ≡ 1 mod 5. Therefore, solving x ≡ 0 mod 2 and x ≡ 1 mod 5 gives x ≡ 6 mod 10. Therefore, units digit 6. Let me check with actual computation.

Compute the tower 6^6^6^6^6^6. Let's compute the units digit. Since the units digit of 6^anything is always 6. Because 6^1=6, 6^2=36, 6^3=216, etc. So regardless of the exponent, the units digit is 6. Therefore, the units digit is 6. Which matches our result. So our method works here.

Another example: n=4. Wait, that's the original problem. But let me check with n=4. Wait, but according to our previous calculation, the units digit is 6. But let me see if that's correct.

But let me think again. Wait, for n=4, the tower is 24^24^...^24 with 24 levels. But when we calculated mod 5, we found that T ≡ 1 mod 5 and T ≡ 0 mod 2, leading to 6 mod 10. However, let me check if there's a mistake in the step where we calculated T' mod 4.

Wait, T' is the tower of 24s with 23 levels. So, T' = 24^T''. T'' is the tower with 22 levels. So, to compute T' mod 4, since 24 ≡ 0 mod 4, then 24^anything ≥1 is 0 mod 4. Therefore, T' ≡ 0 mod 4. Therefore, 4^T' mod 5 is 4^0 mod 5? Wait, no. Wait, 4^T' mod 5. Since φ(5)=4, and T' ≡ 0 mod 4, then 4^T' ≡ 4^(4k) ≡ (4^4)^k ≡ 1^k ≡ 1 mod 5. Therefore, yes, that's correct. Therefore, T ≡ 1 mod 5. Therefore, combining with T ≡ 0 mod 2 gives 6 mod 10. So, the units digit is 6. But let me check with another approach.

Alternatively, maybe I can compute the tower modulo 10 directly, but considering that exponents can be simplified.

But given that the tower is 24^24^...^24 with 24 levels. Let me think about the exponents. Since we need to compute this modulo 10, and since 10 = 2 * 5, and we already did the Chinese Remainder Theorem approach, which gave 6. But let me check if there's another way.

Alternatively, since 24 is congruent to 4 mod 5 and 0 mod 2. But maybe I can think about the tower modulo 10. However, since the tower is so tall, maybe we can find a pattern in the exponents.

But let me think again. Let me try to compute the tower modulo 10. However, since the tower is 24^T', and T' is a tower of 24s with 23 levels. Let me note that 24 mod 10 is 4, so the entire tower is equivalent to 4^T' mod 10. Wait, but earlier I split into mod 2 and mod 5. However, if I try to compute 4^T' mod 10 directly, maybe I can find a pattern.

But 4^k mod 10 cycles every 2: 4^1=4, 4^2=16→6, 4^3=64→4, 4^4=256→6, etc. So, the cycle is 4,6,4,6,... So, if the exponent is odd, it's 4 mod 10; if even, 6 mod 10. Therefore, if I can determine whether T' is odd or even, then I can find 4^T' mod 10.

But T' is the tower of 24s with 23 levels. Let me check if T' is even or odd. Since 24 is even, and any power of an even number is even. Therefore, T' is even. Therefore, T' is even, so 4^T' mod 10 is 6. Therefore, the units digit is 6. Which matches our previous result. So, this seems to confirm that the answer is 6.

Wait, so this approach is simpler. Let me explain again:

Since the entire tower T is 24^T', and 24 mod 10 is 4, so T ≡ 4^T' mod 10. Then, since 4^k mod 10 cycles every 2, with period 2: 4,6,4,6,... So, if T' is even, then 4^T' ≡ 6 mod 10; if T' is odd, then 4^T' ≡ 4 mod 10. Therefore, we need to check if T' is even or odd.

But T' is a tower of 24s with 23 levels. Since 24 is even, and any power of an even number is even. Therefore, regardless of the number of levels (as long as it's at least 1), the tower T' is even. Therefore, T' is even, so 4^T' mod 10 is 6. Therefore, the units digit is 6. Therefore, the answer is 6. So, this seems to confirm the previous result.

But why did we go through the Chinese Remainder Theorem earlier? Because when I first thought about mod 5, I considered that 4^T' mod 5 is 1, and mod 2 is 0, leading to 6 mod 10. But here, by directly looking at mod 10, we can see that since T' is even, the result is 6. So, both methods agree.

Therefore, the units digit is 6. However, let me check once more with another example. Let's take n=2 again. The tower is 2^2 = 4. The units digit is 4. According to the second method: T = 2^T', where T' is the tower with 1 level (since n! = 2, so 2 levels? Wait, n=2, n! = 2, so the tower has 2 levels. So, T = 2^2 = 4. Then, T' is the tower with 1 level, which is 2. Therefore, T' = 2. Then, 4^T' mod 10 is 4^2 = 16 mod 10 = 6? Wait, but the actual value is 4. Wait, this contradicts. Wait, so there's a problem here.

Wait, hold on. Wait, for n=2, the tower is 2!^2! = 2^2 = 4. So, T is 4. But according to the second method, I said that T = 4^T', where T' is the tower with n! -1 levels. Wait, but in this case, n=2, n! = 2, so T is 2^T', where T' is the tower with 1 level, which is 2. Therefore, T = 2^2 = 4. However, when I tried to compute 4^T' mod 10, I was wrong. Wait, why?

Because in the case of n=4, the tower is 24^T', where T' is the tower with 23 levels. However, in the case of n=2, the tower is 2^T', where T' is the tower with 1 level (which is 2). Therefore, T = 2^T', not 4^T'. Wait, so my mistake was in the second approach when I said that T ≡ 4^T' mod 10. That was incorrect. Because in reality, T is 24^T', but 24 mod 10 is 4, so T ≡ 4^T' mod 10. However, in the case of n=2, the tower is 2^T', where T' is 2, so T = 2^2 = 4. However, 2 mod 10 is 2, not 4. Therefore, my mistake was in the second approach when I thought that T ≡ 4^T' mod 10. That was only true for n=4, where the base is 24, which is 4 mod 10. For n=2, the base is 2, which is 2 mod 10. Therefore, my second approach was specific to n=4. Therefore, in the case of n=4, since the base is 24 ≡ 4 mod 10, then T ≡ 4^T' mod 10. However, in the case of n=2, the base is 2, so T ≡ 2^T' mod 10. Therefore, my second approach was correct for n=4, but I confused it with n=2. Therefore, in the case of n=4, since the base is 4 mod 10, and T' is even, then 4^T' mod 10 is 6. Therefore, the answer is 6. However, in the case of n=2, since the base is 2 mod 10, and T' is 2, which is even, then 2^T' mod 10 is 2^2 = 4 mod 10, which is correct. Therefore, my second approach was correct for n=4. Therefore, the answer is 6.

But let me check another example where the base is 4 mod 10. Let's take n=3. Wait, n=3, n! = 6. So, the tower is 6^6^...^6 with 6 levels. Let me compute its units digit. As before, since 6^anything has units digit 6. Therefore, the units digit is 6. Let me check with the second approach. The tower is 6^T', where T' is the tower with 5 levels. Since 6 mod 10 is 6, and 6^k mod 10 is always 6. Therefore, regardless of T', the units digit is 6. Therefore, that works. So, if the base is 6, the units digit is always 6. Similarly, if the base is 4, then depending on the exponent's parity, it's 4 or 6. Therefore, in our case, for n=4, since the base is 4 mod 10, and the exponent T' is even, then the units digit is 6. Therefore, the answer is 6.

Therefore, after checking with multiple examples and approaches, I think the answer is 6. However, let me just make sure that T' is indeed even. T' is the tower of 24s with 23 levels. Since 24 is even, and any power of an even number is even, so regardless of the number of levels (as long as it's at least 1), the tower is even. Therefore, T' is even. Therefore, 4^T' mod 10 is 6. Therefore, the units digit is 6.

But let me think again about the tower. For example, if we have a tower of 24s with 23 levels, which is 24^24^...^24. Since the first exponent is 24, which is even, so the entire tower is 24 raised to an even number. Therefore, 24^even. But 24 is even, so 24^even is divisible by 2^even, which is certainly even. Therefore, T' is even. Therefore, 4^T' mod 10 is 6. Therefore, the units digit is 6.

Therefore, I think the answer is 6. However, let me check if there's a possibility that T' is 0 mod φ(10) or something else. Wait, but in the first approach, we used Chinese Remainder Theorem and found that T ≡ 6 mod 10. In the second approach, we found the same. Therefore, both methods confirm that the units digit is 6.

But just to be thorough, let me check with another angle. Let me try to compute the tower modulo 10, considering that exponents can be simplified.

We have T = 24^T', where T' is the tower with 23 levels. Since 24 ≡ 4 mod 10, T ≡ 4^T' mod 10. As mentioned, 4^k mod 10 cycles every 2: 4, 6, 4, 6,... So, if T' is even, it's 6; if odd, 4. Since T' is even, as established, then T ≡ 6 mod 10. Therefore, units digit is 6.

Alternatively, if I wanted to compute T' mod 2, since the exponent in 4^T' mod 10 depends on T' mod 2. Wait, but T' is even, so T' mod 2 = 0. Therefore, 4^T' mod 10 = 4^0 mod 10? No, that's not correct. Wait, no. The cycle is every 2. So, if T' is even, then T' = 2k, so 4^{2k} = (4^2)^k = 16^k. 16 mod 10 is 6, so 16^k mod 10 is 6^k mod 10. But 6^k mod 10 is always 6. Therefore, 4^{even} mod 10 is 6. Therefore, regardless of the exponent, as long as it's even, it's 6. Therefore, that's another way to see it. Therefore, since T' is even, the result is 6. Therefore, the units digit is 6.

Therefore, after multiple approaches and checks, I feel confident that the units digit of 4$ is 6.

**Final Answer**
\boxed{6}

To determine the **units digit** of the superfactorial $ 4\$ $, we begin by understanding the definition:

$$
n\$ = \underbrace{n!^{{n!}^{{\cdot}^{{\cdot}^{{\cdot}^{n!}}}}}}_{n!}
$$

This means that $ 4\$ $ is a **tower of exponents** with $ 4! = 24 $ levels, each level being $ 24 $. So:

$$
4\$ = 24^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \quad \text{(with 24 levels of 24)}
$$

We are asked to find the **units digit** of this number, which is equivalent to computing:

$$
4\$ \mod 10
$$

---

### Step 1: Reduce the base modulo 10

Since the base of the tower is 24, and we are interested in the **units digit**, we reduce 24 modulo 10:

$$
24 \equiv 4 \mod 10
$$

So, the entire tower becomes:

$$
4^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \mod 10
$$

Let’s denote the tower of 24s with 23 levels as $ T' $. Then:

$$
4\$ = 4^{T'} \mod 10
$$

---

### Step 2: Analyze the exponent $ T' \mod 2 $

We now need to determine the **parity** of the exponent $ T' $, because the powers of 4 modulo 10 cycle every 2:

- $ 4^1 = 4 $
- $ 4^2 = 16 \equiv 6 \mod 10 $
- $ 4^3 = 64 \equiv 4 \mod 10 $
- $ 4^4 = 256 \equiv 6 \mod 10 $

So, the pattern is: **4, 6, 4, 6, ...**

Thus, if the exponent is **even**, the result is **6**; if **odd**, it is **4**.

So, we need to determine whether $ T' $ is even or odd.

---

### Step 3: Determine the parity of $ T' $

Recall that $ T' $ is a tower of 24s with 23 levels:

$$
T' = 24^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \quad \text{(23 levels)}
$$

Since 24 is even, and any power of an even number is even, the entire tower $ T' $ is **even**. Therefore:

$$
T' \equiv 0 \mod 2
$$

---

### Step 4: Compute $ 4^{T'} \mod 10 $

Since $ T' $ is even, we use the pattern from earlier:

$$
4^{\text{even}} \equiv 6 \mod 10
$$

Thus:

$$
4\$ \equiv 6 \mod 10
$$

---

### Final Answer

$$
\boxed{6}
$$

Solution: Since $4! = 4\times 3 \times 2 \times 1 = 24$, we need to evaluate the units digit of $4\$ = \underbrace{24^{24^{\cdot^{\cdot^\cdot}}}}_{24}$. The exponent of the base $24$ is an even number, say $2n$. Then, $4\$ = 24^{2n} = 576^n$. The units digit of the product of two numbers with units digit $6$ is equal to $6$. Thus, the desired answer is $\boxed{6}$.


Explanation:
Okay, so I need to figure out how many two-digit prime numbers have digits that add up to 8. Let me start by recalling what two-digit primes are. Two-digit numbers range from 10 to 99, right? And primes are numbers greater than 1 that have no divisors other than 1 and themselves.

First, maybe I should list all two-digit primes and then check which ones have digits summing to 8. But that might take a while. Maybe there's a smarter way. Let me think.

Alternatively, I can think about all two-digit numbers where the sum of the digits is 8 and then check which of those are prime. That might be more efficient. Let me try that approach.

So, a two-digit number can be represented as 10*a + b, where a is the tens digit and b is the units digit. Both a and b are digits, so a is from 1 to 9 (since it's a two-digit number) and b is from 0 to 9. The sum of the digits is a + b = 8. So, I need to find all pairs (a, b) where a is between 1 and 9, b is between 0 and 9, and a + b = 8. Then, for each such number, check if it's prime.

Let me list all possible pairs (a, b):

Starting with a = 1: then b = 8 - 1 = 7. So the number is 17.

a = 2: b = 6. Number is 26.

a = 3: b = 5. Number is 35.

a = 4: b = 4. Number is 44.

a = 5: b = 3. Number is 53.

a = 6: b = 2. Number is 62.

a = 7: b = 1. Number is 71.

a = 8: b = 0. Number is 80.

a = 9: b = -1. Wait, that's not possible because b can't be negative. So a can't be 9 here. So the possible numbers are from a=1 to a=8.

So the numbers are: 17, 26, 35, 44, 53, 62, 71, 80.

Now, I need to check which of these numbers are prime.

Let me recall that a prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So let's check each number:

1. 17: Let me check if it's prime. The square root of 17 is about 4.12, so I need to check divisibility by primes up to 4. Divided by 2? No, it's odd. Divided by 3? 1+7=8, which isn't divisible by 3. Divided by 5? Doesn't end with 0 or 5. Divided by 7? 7*2=14, 17-14=3, not divisible by 7. So 17 is prime.

2. 26: Even number, ends with 6, so divisible by 2. Not prime.

3. 35: Ends with 5, so divisible by 5. Not prime.

4. 44: Even number, divisible by 2. Not prime.

5. 53: Let's check. Square root of 53 is about 7.28. Check divisibility by 2, 3, 5, 7. 53 is odd, not divisible by 2. 5+3=8, not divisible by 3. Doesn't end with 5, so not divisible by 5. Divided by 7? 7*7=49, 53-49=4, not divisible by 7. So 53 is prime.

6. 62: Even number, ends with 2. Divisible by 2. Not prime.

7. 71: Check if prime. Square root is about 8.43. Check divisibility by 2, 3, 5, 7. 71 is odd, not divisible by 2. 7+1=8, not divisible by 3. Doesn't end with 5. Divided by 7? 7*10=70, 71-70=1, not divisible by 7. Next prime is 11, but 11*6=66, 11*7=77, which is higher. So 71 is prime.

8. 80: Ends with 0, divisible by 10. Not prime.

So from the list, the primes are 17, 53, 71. That's three numbers. Wait, let me check again to make sure I didn't miss any.

Wait, let me check 17, 53, 71. Are there any others?

Wait, let me check 26: 26 is 2*13, not prime. 35 is 5*7. 44 is 4*11. 62 is 2*31. 80 is 8*10. So only 17, 53, 71. So three primes. Therefore, the answer is 3?

Wait, but let me check again. Maybe I missed some numbers? Let me check the list again.

Original numbers generated: 17, 26, 35, 44, 53, 62, 71, 80. That's 8 numbers. Checked all of them. Only 17, 53, 71 are primes. So answer is 3?

Wait, but let me think again. Maybe I made a mistake in generating the numbers? Let me check the pairs again.

a + b = 8, with a from 1-9, b from 0-9.

So:

a=1, b=7: 17

a=2, b=6: 26

a=3, b=5: 35

a=4, b=4: 44

a=5, b=3: 53

a=6, b=2: 62

a=7, b=1: 71

a=8, b=0: 80

a=9, b=-1: invalid. So that's correct. So 8 numbers. So only three primes. So answer is 3?

But wait, let me check if there are other two-digit primes with digits summing to 8. Maybe I missed some numbers?

Wait, for example, is there a prime like 107? No, that's three-digit. Wait, no. Wait, maybe numbers like 89? Wait, 8 + 9 = 17, which is not 8. Wait, 89 is a prime, but digits sum to 17. Not relevant here.

Wait, maybe I need to check if there are other numbers where digits sum to 8. Wait, for example, 17, 71, 53, 35, etc. Wait, but 35 is not prime, 53 is prime, 71 is prime, 17 is prime. So that's three numbers. So answer is 3?

But let me check again. Let me check 17, 53, 71. Are there any others?

Wait, what about 89? No, digits sum to 17. 71 is already there. 17 is there. 53 is there. What about 107? No, three-digit. How about 19? 1+9=10. Not 8. 29? 2+9=11. 37? 3+7=10. 47? 4+7=11. 59? 5+9=14. 67? 6+7=13. 79? 7+9=16. 89? 17. 97? 16. So none of those. How about numbers like 13? 1+3=4. 11? 1+1=2. 19? 10. 23? 5. 29? 11. 31? 4. 37? 10. 41? 5. 43? 7. 47? 11. 53? 8. 59? 14. 61? 7. 67? 13. 71? 8. 73? 10. 79? 16. 83? 11. 89? 17. 97? 16. So the only two-digit primes with digits summing to 8 are 17, 53, 71. So three numbers. Therefore, the answer is 3?

Wait, but let me check if there are other numbers. For example, 80 is not prime. 44 is not. 26 is not. 35 is not. So yeah, only three. So answer is 3?

But wait, let me check if 17, 53, 71 are the only ones. Wait, is there a prime like 107? No, three-digit. How about 100? No, not prime. Wait, maybe I need to check if there are other numbers where digits add to 8. Wait, for example, 80 is 8+0=8, but not prime. 71 is 7+1=8. 53 is 5+3=8. 17 is 1+7=8. What about numbers like 80, 71, 62, 53, 44, 35, 26, 17. So those are all numbers with digits adding to 8. So only three primes. So answer is 3.

But wait, let me check if there are any other two-digit primes that I might have missed. For example, 101 is three-digit. Wait, maybe I need to check if there are other numbers. Wait, for example, 19? No, sum is 10. 29? 11. 39? 12. 49? 13. 59? 14. 69? 15. 79? 16. 89? 17. 99? 18. So none of those. How about numbers like 10? 1+0=1. 11? 2. 12? 3. 13? 4. 14? 5. 15? 6. 16? 7. 17? 8. So 17 is there. Then 26, 35, 44, 53, 62, 71, 80. So yeah, that's all. So only three primes.

Wait, but let me check 17 again. 17 is prime. 53 is prime. 71 is prime. Are there any others? Let me check 17, 53, 71. What about 107? No, three-digit. 199? No. So no. So answer is 3.

But wait, I just thought of something. What about numbers where the digits are reversed? For example, 17 and 71. Both are primes. But we already considered them. So that's covered.

Alternatively, maybe I made a mistake in checking if 53 is prime? Let me check again. 53 divided by 2? No. Divided by 3? 5+3=8, not divisible by 3. Divided by 5? Doesn't end with 0 or 5. Divided by 7? 7*7=49, 7*8=56. 53-49=4, so not divisible by 7. Divided by 11? 11*4=44, 11*5=55. 53 is between them, not divisible by 11. So yes, 53 is prime.

Similarly, 71: Divided by 2? No. Divided by 3? 7+1=8, not divisible by 3. Divided by 5? No. Divided by 7? 7*10=70, 71-70=1, not divisible by 7. Divided by 11? 11*6=66, 11*7=77. 71 is between, not divisible by 11. So yes, prime.

And 17 is prime. So yeah, three numbers.

Wait, but let me check if there's another number. For example, 89? No, sum is 17. 97? 16. 79? 16. 67? 13. 59? 14. 47? 11. 37? 10. 29? 11. 19? 10. So none of those.

Wait, maybe I need to check numbers like 10? No, sum is 1. 11? 2. 12? 3. 13? 4. 14? 5. 15? 6. 16? 7. 17? 8. So 17 is there. Then 26, 35, 44, 53, 62, 71, 80. So yeah, that's all.

Therefore, the answer is 3.

But wait, hold on. Let me check if there are any other two-digit primes with digits summing to 8. Maybe I missed a number? Let me think. For example, 89? No. 98? 9+8=17. 89? 17. 79? 16. 73? 10. 71? 8. 67? 13. 61? 7. 59? 14. 53? 8. 47? 11. 43? 7. 41? 5. 37? 10. 31? 4. 29? 11. 23? 5. 19? 10. 17? 8. So yeah, only 17, 53, 71.

Therefore, the answer is 3.

But wait, let me check once more. Maybe I made a mistake in considering numbers. For example, 17, 53, 71. Are there any others? Let me check 80. Not prime. 44? Not. 26? Not. 35? Not. So yeah.

Alternatively, maybe I need to check if there are numbers with leading zeros? But no, two-digit numbers can't have leading zeros. So a is from 1-9.

Therefore, I think the answer is 3.

But wait, let me check online? Wait, no, I can't do that. But maybe I can think again.

Wait, another way: list all two-digit primes and check their digit sums.

List of two-digit primes:

11, 13, 17, 19,

23, 29,

31, 37, 39 (not prime),

41, 43, 47,

53, 59,

61, 67, 71, 73, 79,

83, 89,

97.

Wait, let me list them properly:

Starting from 11:

11, 13, 17, 19,

23, 29,

31, 37,

41, 43, 47,

53, 59,

61, 67, 71, 73, 79,

83, 89,

97.

Now, check which of these have digits summing to 8.

Let me go through each:

11: 1+1=2. No.

13: 1+3=4. No.

17: 1+7=8. Yes! That's one.

19: 1+9=10. No.

23: 2+3=5. No.

29: 2+9=11. No.

31: 3+1=4. No.

37: 3+7=10. No.

41: 4+1=5. No.

43: 4+3=7. No.

47: 4+7=11. No.

53: 5+3=8. Yes! Second one.

59: 5+9=14. No.

61: 6+1=7. No.

67: 6+7=13. No.

71: 7+1=8. Yes! Third one.

73: 7+3=10. No.

79: 7+9=16. No.

83: 8+3=11. No.

89: 8+9=17. No.

97: 9+7=16. No.

So only 17, 53, 71. So three primes. Therefore, the answer is 3.

Therefore, I think the answer is 3.

But just to make sure, let me check if there are any other primes in the list that I might have missed. For example, 101 is three-digit. 103? Three-digit. So no. So yeah, only those three.

Therefore, the number of two-digit primes with digits summing to 8 is 3.

**Final Answer**
The number of two-digit primes with digits summing to 8 is \boxed{3}.

To determine how many **two-digit prime numbers** have digits that sum to **8**, we can follow a structured approach:

---

### Step 1: Understand the Problem

We are looking for **two-digit prime numbers** such that the **sum of their digits is 8**. A two-digit number can be written as $10a + b$, where:

- $a$ is the tens digit (from 1 to 9)
- $b$ is the units digit (from 0 to 9)

We are given the condition:

$$
a + b = 8
$$

We will generate all such two-digit numbers and then check which of them are **prime**.

---

### Step 2: Generate All Two-Digit Numbers with Digit Sum 8

We solve $a + b = 8$ for $a$ from 1 to 8 (since $a = 9$ would make $b = -1$, which is invalid):

- $a = 1 \Rightarrow b = 7 \Rightarrow 17$
- $a = 2 \Rightarrow b = 6 \Rightarrow 26$
- $a = 3 \Rightarrow b = 5 \Rightarrow 35$
- $a = 4 \Rightarrow b = 4 \Rightarrow 44$
- $a = 5 \Rightarrow b = 3 \Rightarrow 53$
- $a = 6 \Rightarrow b = 2 \Rightarrow 62$
- $a = 7 \Rightarrow b = 1 \Rightarrow 71$
- $a = 8 \Rightarrow b = 0 \Rightarrow 80$

So the list of numbers is:
**17, 26, 35, 44, 53, 62, 71, 80**

---

### Step 3: Check Which of These Are Prime

Now, we check which of these numbers are **prime**:

- **17**: Prime ✅
- **26**: Not prime (divisible by 2) ❌
- **35**: Not prime (divisible by 5) ❌
- **44**: Not prime (divisible by 2) ❌
- **53**: Prime ✅
- **62**: Not prime (divisible by 2) ❌
- **71**: Prime ✅
- **80**: Not prime (divisible by 2) ❌

---

### Step 4: Count the Primes

From the list, the **prime numbers** with digit sum 8 are:

- 17
- 53
- 71

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many two-digit prime numbers have digits that add up to 8. Let me start by recalling what two-digit primes are. Two-digit numbers range from 10 to 99, right? And primes are numbers greater than 1 that have no divisors other than 1 and themselves.

First, maybe I should list all two-digit primes and then check which ones have digits summing to 8. But that might take a while. Maybe there's a smarter way. Let me think.

Alternatively, I can think about all two-digit numbers where the sum of the digits is 8 and then check which of those are prime. That might be more efficient. Let me try that approach.

So, a two-digit number can be represented as 10*a + b, where a is the tens digit and b is the units digit. Both a and b are digits, so a is from 1 to 9 (since it's a two-digit number) and b is from 0 to 9. The sum of the digits is a + b = 8. So, I need to find all pairs (a, b) where a is between 1 and 9, b is between 0 and 9, and a + b = 8. Then, for each such number, check if it's prime.

Let me list all possible pairs (a, b):

Starting with a = 1: then b = 8 - 1 = 7. So the number is 17.

a = 2: b = 6. Number is 26.

a = 3: b = 5. Number is 35.

a = 4: b = 4. Number is 44.

a = 5: b = 3. Number is 53.

a = 6: b = 2. Number is 62.

a = 7: b = 1. Number is 71.

a = 8: b = 0. Number is 80.

a = 9: b = -1. Wait, that's not possible because b can't be negative. So a can't be 9 here. So the possible numbers are from a=1 to a=8.

So the numbers are: 17, 26, 35, 44, 53, 62, 71, 80.

Now, I need to check which of these numbers are prime.

Let me recall that a prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So let's check each number:

1. 17: Let me check if it's prime. The square root of 17 is about 4.12, so I need to check divisibility by primes up to 4. Divided by 2? No, it's odd. Divided by 3? 1+7=8, which isn't divisible by 3. Divided by 5? Doesn't end with 0 or 5. Divided by 7? 7*2=14, 17-14=3, not divisible by 7. So 17 is prime.

2. 26: Even number, ends with 6, so divisible by 2. Not prime.

3. 35: Ends with 5, so divisible by 5. Not prime.

4. 44: Even number, divisible by 2. Not prime.

5. 53: Let's check. Square root of 53 is about 7.28. Check divisibility by 2, 3, 5, 7. 53 is odd, not divisible by 2. 5+3=8, not divisible by 3. Doesn't end with 5, so not divisible by 5. Divided by 7? 7*7=49, 53-49=4, not divisible by 7. So 53 is prime.

6. 62: Even number, ends with 2. Divisible by 2. Not prime.

7. 71: Check if prime. Square root is about 8.43. Check divisibility by 2, 3, 5, 7. 71 is odd, not divisible by 2. 7+1=8, not divisible by 3. Doesn't end with 5. Divided by 7? 7*10=70, 71-70=1, not divisible by 7. Next prime is 11, but 11*6=66, 11*7=77, which is higher. So 71 is prime.

8. 80: Ends with 0, divisible by 10. Not prime.

So from the list, the primes are 17, 53, 71. That's three numbers. Wait, let me check again to make sure I didn't miss any.

Wait, let me check 17, 53, 71. Are there any others?

Wait, let me check 26: 26 is 2*13, not prime. 35 is 5*7. 44 is 4*11. 62 is 2*31. 80 is 8*10. So only 17, 53, 71. So three primes. Therefore, the answer is 3?

Wait, but let me check again. Maybe I missed some numbers? Let me check the list again.

Original numbers generated: 17, 26, 35, 44, 53, 62, 71, 80. That's 8 numbers. Checked all of them. Only 17, 53, 71 are primes. So answer is 3?

Wait, but let me think again. Maybe I made a mistake in generating the numbers? Let me check the pairs again.

a + b = 8, with a from 1-9, b from 0-9.

So:

a=1, b=7: 17

a=2, b=6: 26

a=3, b=5: 35

a=4, b=4: 44

a=5, b=3: 53

a=6, b=2: 62

a=7, b=1: 71

a=8, b=0: 80

a=9, b=-1: invalid. So that's correct. So 8 numbers. So only three primes. So answer is 3?

But wait, let me check if there are other two-digit primes with digits summing to 8. Maybe I missed some numbers?

Wait, for example, is there a prime like 107? No, that's three-digit. Wait, no. Wait, maybe numbers like 89? Wait, 8 + 9 = 17, which is not 8. Wait, 89 is a prime, but digits sum to 17. Not relevant here.

Wait, maybe I need to check if there are other numbers where digits sum to 8. Wait, for example, 17, 71, 53, 35, etc. Wait, but 35 is not prime, 53 is prime, 71 is prime, 17 is prime. So that's three numbers. So answer is 3?

But let me check again. Let me check 17, 53, 71. Are there any others?

Wait, what about 89? No, digits sum to 17. 71 is already there. 17 is there. 53 is there. What about 107? No, three-digit. How about 19? 1+9=10. Not 8. 29? 2+9=11. 37? 3+7=10. 47? 4+7=11. 59? 5+9=14. 67? 6+7=13. 79? 7+9=16. 89? 17. 97? 16. So none of those. How about numbers like 13? 1+3=4. 11? 1+1=2. 19? 10. 23? 5. 29? 11. 31? 4. 37? 10. 41? 5. 43? 7. 47? 11. 53? 8. 59? 14. 61? 7. 67? 13. 71? 8. 73? 10. 79? 16. 83? 11. 89? 17. 97? 16. So the only two-digit primes with digits summing to 8 are 17, 53, 71. So three numbers. Therefore, the answer is 3?

Wait, but let me check if there are other numbers. For example, 80 is not prime. 44 is not. 26 is not. 35 is not. So yeah, only three. So answer is 3?

But wait, let me check if 17, 53, 71 are the only ones. Wait, is there a prime like 107? No, three-digit. How about 100? No, not prime. Wait, maybe I need to check if there are other numbers where digits add to 8. Wait, for example, 80 is 8+0=8, but not prime. 71 is 7+1=8. 53 is 5+3=8. 17 is 1+7=8. What about numbers like 80, 71, 62, 53, 44, 35, 26, 17. So those are all numbers with digits adding to 8. So only three primes. So answer is 3.

But wait, let me check if there are any other two-digit primes that I might have missed. For example, 101 is three-digit. Wait, maybe I need to check if there are other numbers. Wait, for example, 19? No, sum is 10. 29? 11. 39? 12. 49? 13. 59? 14. 69? 15. 79? 16. 89? 17. 99? 18. So none of those. How about numbers like 10? 1+0=1. 11? 2. 12? 3. 13? 4. 14? 5. 15? 6. 16? 7. 17? 8. So 17 is there. Then 26, 35, 44, 53, 62, 71, 80. So yeah, that's all. So only three primes.

Wait, but let me check 17 again. 17 is prime. 53 is prime. 71 is prime. Are there any others? Let me check 17, 53, 71. What about 107? No, three-digit. 199? No. So no. So answer is 3.

But wait, I just thought of something. What about numbers where the digits are reversed? For example, 17 and 71. Both are primes. But we already considered them. So that's covered.

Alternatively, maybe I made a mistake in checking if 53 is prime? Let me check again. 53 divided by 2? No. Divided by 3? 5+3=8, not divisible by 3. Divided by 5? Doesn't end with 0 or 5. Divided by 7? 7*7=49, 7*8=56. 53-49=4, so not divisible by 7. Divided by 11? 11*4=44, 11*5=55. 53 is between them, not divisible by 11. So yes, 53 is prime.

Similarly, 71: Divided by 2? No. Divided by 3? 7+1=8, not divisible by 3. Divided by 5? No. Divided by 7? 7*10=70, 71-70=1, not divisible by 7. Divided by 11? 11*6=66, 11*7=77. 71 is between, not divisible by 11. So yes, prime.

And 17 is prime. So yeah, three numbers.

Wait, but let me check if there's another number. For example, 89? No, sum is 17. 97? 16. 79? 16. 67? 13. 59? 14. 47? 11. 37? 10. 29? 11. 19? 10. So none of those.

Wait, maybe I need to check numbers like 10? No, sum is 1. 11? 2. 12? 3. 13? 4. 14? 5. 15? 6. 16? 7. 17? 8. So 17 is there. Then 26, 35, 44, 53, 62, 71, 80. So yeah, that's all.

Therefore, the answer is 3.

But wait, hold on. Let me check if there are any other two-digit primes with digits summing to 8. Maybe I missed a number? Let me think. For example, 89? No. 98? 9+8=17. 89? 17. 79? 16. 73? 10. 71? 8. 67? 13. 61? 7. 59? 14. 53? 8. 47? 11. 43? 7. 41? 5. 37? 10. 31? 4. 29? 11. 23? 5. 19? 10. 17? 8. So yeah, only 17, 53, 71.

Therefore, the answer is 3.

But wait, let me check once more. Maybe I made a mistake in considering numbers. For example, 17, 53, 71. Are there any others? Let me check 80. Not prime. 44? Not. 26? Not. 35? Not. So yeah.

Alternatively, maybe I need to check if there are numbers with leading zeros? But no, two-digit numbers can't have leading zeros. So a is from 1-9.

Therefore, I think the answer is 3.

But wait, let me check online? Wait, no, I can't do that. But maybe I can think again.

Wait, another way: list all two-digit primes and check their digit sums.

List of two-digit primes:

11, 13, 17, 19,

23, 29,

31, 37, 39 (not prime),

41, 43, 47,

53, 59,

61, 67, 71, 73, 79,

83, 89,

97.

Wait, let me list them properly:

Starting from 11:

11, 13, 17, 19,

23, 29,

31, 37,

41, 43, 47,

53, 59,

61, 67, 71, 73, 79,

83, 89,

97.

Now, check which of these have digits summing to 8.

Let me go through each:

11: 1+1=2. No.

13: 1+3=4. No.

17: 1+7=8. Yes! That's one.

19: 1+9=10. No.

23: 2+3=5. No.

29: 2+9=11. No.

31: 3+1=4. No.

37: 3+7=10. No.

41: 4+1=5. No.

43: 4+3=7. No.

47: 4+7=11. No.

53: 5+3=8. Yes! Second one.

59: 5+9=14. No.

61: 6+1=7. No.

67: 6+7=13. No.

71: 7+1=8. Yes! Third one.

73: 7+3=10. No.

79: 7+9=16. No.

83: 8+3=11. No.

89: 8+9=17. No.

97: 9+7=16. No.

So only 17, 53, 71. So three primes. Therefore, the answer is 3.

Therefore, I think the answer is 3.

But just to make sure, let me check if there are any other primes in the list that I might have missed. For example, 101 is three-digit. 103? Three-digit. So no. So yeah, only those three.

Therefore, the number of two-digit primes with digits summing to 8 is 3.

**Final Answer**
The number of two-digit primes with digits summing to 8 is \boxed{3}.

To determine how many **two-digit prime numbers** have digits that sum to **8**, we can follow a structured approach:

---

### Step 1: Understand the Problem

We are looking for **two-digit prime numbers** such that the **sum of their digits is 8**. A two-digit number can be written as $10a + b$, where:

- $a$ is the tens digit (from 1 to 9)
- $b$ is the units digit (from 0 to 9)

We are given the condition:

$$
a + b = 8
$$

We will generate all such two-digit numbers and then check which of them are **prime**.

---

### Step 2: Generate All Two-Digit Numbers with Digit Sum 8

We solve $a + b = 8$ for $a$ from 1 to 8 (since $a = 9$ would make $b = -1$, which is invalid):

- $a = 1 \Rightarrow b = 7 \Rightarrow 17$
- $a = 2 \Rightarrow b = 6 \Rightarrow 26$
- $a = 3 \Rightarrow b = 5 \Rightarrow 35$
- $a = 4 \Rightarrow b = 4 \Rightarrow 44$
- $a = 5 \Rightarrow b = 3 \Rightarrow 53$
- $a = 6 \Rightarrow b = 2 \Rightarrow 62$
- $a = 7 \Rightarrow b = 1 \Rightarrow 71$
- $a = 8 \Rightarrow b = 0 \Rightarrow 80$

So the list of numbers is:
**17, 26, 35, 44, 53, 62, 71, 80**

---

### Step 3: Check Which of These Are Prime

Now, we check which of these numbers are **prime**:

- **17**: Prime ✅
- **26**: Not prime (divisible by 2) ❌
- **35**: Not prime (divisible by 5) ❌
- **44**: Not prime (divisible by 2) ❌
- **53**: Prime ✅
- **62**: Not prime (divisible by 2) ❌
- **71**: Prime ✅
- **80**: Not prime (divisible by 2) ❌

---

### Step 4: Count the Primes

From the list, the **prime numbers** with digit sum 8 are:

- 17
- 53
- 71

---

### Final Answer

$$
\boxed{3}
$$

Solution:
First, we list all the two-digit numbers with a digits sum of 8:

17, 26, 35, 44, 53, 62, 71, 80

Clearly, 26, 44, 62, and 80 are not prime.  35 is not prime, but 17, 53, and 71 are (we can verify this by dividing each number by all the primes less than its square root (why is this?)).  Hence, the answer is $\boxed{3}$ two-digit primes.