Explanation:
Okay, so I need to figure out how many blue marbles were in the bag originally. Let me start by understanding the problem.

We have a bag with yellow and blue marbles. The current ratio of blue to yellow marbles is 4:3. Then, if we add 5 blue marbles and remove 3 yellow marbles, the new ratio becomes 7:3. I need to find the original number of blue marbles.

Hmm, ratios can sometimes be tricky, but I think I can set up some equations here. Let me denote the number of blue marbles as B and the number of yellow marbles as Y.

The initial ratio is blue to yellow is 4:3. So that means B/Y = 4/3. Alternatively, I can write this as 3B = 4Y. Maybe that's useful later.

Then, after adding 5 blue marbles and removing 3 yellow marbles, the new ratio is 7:3. So the new number of blue marbles is B + 5, and the new number of yellow marbles is Y - 3. Therefore, the new ratio is (B + 5)/(Y - 3) = 7/3.

So now I have two equations:

1) 3B = 4Y

2) 3(B + 5) = 7(Y - 3)

Let me check that. The second equation comes from cross-multiplying the ratio (B + 5)/(Y - 3) = 7/3. So yes, multiplying both sides by 3(Y - 3) gives 3(B + 5) = 7(Y - 3).

Okay, so now I have two equations with two variables. Let me write them again:

Equation 1: 3B = 4Y

Equation 2: 3B + 15 = 7Y - 21

Wait, expanding equation 2: 3(B + 5) = 3B + 15, and 7(Y - 3) = 7Y - 21. So yes, equation 2 is 3B + 15 = 7Y - 21.

Now, since equation 1 is 3B = 4Y, maybe I can substitute 3B from equation 1 into equation 2. Let me try that.

From equation 1, 3B = 4Y. Substitute into equation 2:

4Y + 15 = 7Y - 21

Let me solve for Y. Subtract 4Y from both sides:

15 = 3Y - 21

Then add 21 to both sides:

36 = 3Y

Divide both sides by 3:

Y = 12

Okay, so there are 12 yellow marbles originally. Now, let me find B using equation 1: 3B = 4Y. So 3B = 4*12 = 48. Therefore, B = 48 / 3 = 16.

Wait, so originally there were 16 blue marbles? Let me check if this works with the second condition.

Original blue marbles: 16, original yellow marbles: 12. After adding 5 blue marbles, blue becomes 16 + 5 = 21. After removing 3 yellow marbles, yellow becomes 12 - 3 = 9. Then the ratio is 21:9. Let me simplify that. Divide numerator and denominator by 3: 7:3. Which matches the given ratio. Perfect! So that seems to check out.

But let me just go through the steps again to make sure I didn't make any mistakes.

Starting with the equations:

Original ratio: B/Y = 4/3 => 3B = 4Y. Correct.

After adding and removing marbles: (B + 5)/(Y - 3) = 7/3 => cross multiply to 3(B + 5) = 7(Y - 3). Correct.

Expanding that gives 3B + 15 = 7Y - 21. Then substituting 3B from equation 1 (which is 4Y) into equation 2: 4Y + 15 = 7Y - 21. Subtract 4Y: 15 = 3Y - 21. Add 21: 36 = 3Y => Y = 12. Then B = (4/3)*Y = (4/3)*12 = 16.

Yes, that seems correct. Then checking the final ratio: 16 + 5 = 21 blue, 12 - 3 = 9 yellow. 21/9 = 7/3. Correct.

So the original number of blue marbles is 16. Therefore, the answer should be 16.

But let me think again if there's another way to approach this problem, just to confirm. Maybe using variables differently?

Alternatively, let me think in terms of ratios. The original ratio is 4:3. Let me assume that the number of blue marbles is 4x and yellow marbles is 3x for some positive integer x. Then, after adding 5 blue marbles and removing 3 yellow marbles, the new numbers are 4x + 5 and 3x - 3. The new ratio is 7:3. Therefore:

(4x + 5)/(3x - 3) = 7/3

Cross multiplying:

3(4x + 5) = 7(3x - 3)

Which is 12x + 15 = 21x - 21

Subtract 12x both sides:

15 = 9x - 21

Add 21:

36 = 9x => x = 4

Therefore, original number of blue marbles is 4x = 4*4 = 16. Same answer.

So that's another way to think about it, using variables as multiples of the ratio. This also gives 16. So that confirms the answer again.

Therefore, I think 16 is correct.

But let me check if there's any possible mistake here. For example, could there be another solution? Let me see.

Suppose I didn't assume that the original numbers are multiples of 4 and 3. But since ratios are given, it's standard to represent them as variables multiplied by a common factor. So that approach is valid.

Alternatively, if I didn't use substitution, but solved the equations differently. Let me try solving the two equations again:

Original equations:

1) 3B = 4Y

2) 3B + 15 = 7Y - 21

From equation 1, we can express B as (4Y)/3. Then substitute into equation 2:

3*(4Y/3) + 15 = 7Y - 21

Simplify left side: 4Y + 15 = 7Y - 21

Which is same as before. So same result.

Therefore, regardless of the method, the answer is 16.

Another check: Let me think about the numbers. Original marbles: blue 16, yellow 12. Adding 5 blue makes 21, removing 3 yellow makes 9. 21:9 reduces to 7:3. Correct.

So, seems solid.

I think that's thorough enough. I don't see any mistakes in my reasoning. So the original number of blue marbles is 16.

**Final Answer**
The number of blue marbles before adding more is \boxed{16}.

To determine the original number of blue marbles in the bag, we begin by defining variables for the quantities of blue and yellow marbles:

- Let $ B $ represent the number of blue marbles.
- Let $ Y $ represent the number of yellow marbles.

---

### Step 1: Use the initial ratio

We are told that the initial ratio of blue to yellow marbles is $ 4:3 $. This gives us the equation:

$$
\frac{B}{Y} = \frac{4}{3} \quad \Rightarrow \quad 3B = 4Y \tag{1}
$$

---

### Step 2: Use the modified ratio after changes

After adding 5 blue marbles and removing 3 yellow marbles, the new ratio becomes $ 7:3 $. This gives us the second equation:

$$
\frac{B + 5}{Y - 3} = \frac{7}{3} \quad \Rightarrow \quad 3(B + 5) = 7(Y - 3) \tag{2}
$$

Expanding equation (2):

$$
3B + 15 = 7Y - 21
$$

---

### Step 3: Solve the system of equations

From equation (1), we can express $ B $ in terms of $ Y $:

$$
B = \frac{4Y}{3}
$$

Substitute this into equation (2):

$$
3\left(\frac{4Y}{3}\right) + 15 = 7Y - 21
$$

Simplify:

$$
4Y + 15 = 7Y - 21
$$

Solving for $ Y $:

$$
15 + 21 = 7Y - 4Y \quad \Rightarrow \quad 36 = 3Y \quad \Rightarrow \quad Y = 12
$$

Now substitute $ Y = 12 $ back into equation (1) to find $ B $:

$$
3B = 4Y = 4 \times 12 = 48 \quad \Rightarrow \quad B = \frac{48}{3} = 16
$$

---

### Step 4: Verify the solution

- Original marbles: $ B = 16 $, $ Y = 12 $
- After changes: $ B + 5 = 21 $, $ Y - 3 = 9 $
- New ratio: $ \frac{21}{9} = \frac{7}{3} $, which matches the given ratio.

---

### Final Answer

$$
\boxed{16}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many blue marbles were in the bag originally. Let me start by understanding the problem.

We have a bag with yellow and blue marbles. The current ratio of blue to yellow marbles is 4:3. Then, if we add 5 blue marbles and remove 3 yellow marbles, the new ratio becomes 7:3. I need to find the original number of blue marbles.

Hmm, ratios can sometimes be tricky, but I think I can set up some equations here. Let me denote the number of blue marbles as B and the number of yellow marbles as Y.

The initial ratio is blue to yellow is 4:3. So that means B/Y = 4/3. Alternatively, I can write this as 3B = 4Y. Maybe that's useful later.

Then, after adding 5 blue marbles and removing 3 yellow marbles, the new ratio is 7:3. So the new number of blue marbles is B + 5, and the new number of yellow marbles is Y - 3. Therefore, the new ratio is (B + 5)/(Y - 3) = 7/3.

So now I have two equations:

1) 3B = 4Y

2) 3(B + 5) = 7(Y - 3)

Let me check that. The second equation comes from cross-multiplying the ratio (B + 5)/(Y - 3) = 7/3. So yes, multiplying both sides by 3(Y - 3) gives 3(B + 5) = 7(Y - 3).

Okay, so now I have two equations with two variables. Let me write them again:

Equation 1: 3B = 4Y

Equation 2: 3B + 15 = 7Y - 21

Wait, expanding equation 2: 3(B + 5) = 3B + 15, and 7(Y - 3) = 7Y - 21. So yes, equation 2 is 3B + 15 = 7Y - 21.

Now, since equation 1 is 3B = 4Y, maybe I can substitute 3B from equation 1 into equation 2. Let me try that.

From equation 1, 3B = 4Y. Substitute into equation 2:

4Y + 15 = 7Y - 21

Let me solve for Y. Subtract 4Y from both sides:

15 = 3Y - 21

Then add 21 to both sides:

36 = 3Y

Divide both sides by 3:

Y = 12

Okay, so there are 12 yellow marbles originally. Now, let me find B using equation 1: 3B = 4Y. So 3B = 4*12 = 48. Therefore, B = 48 / 3 = 16.

Wait, so originally there were 16 blue marbles? Let me check if this works with the second condition.

Original blue marbles: 16, original yellow marbles: 12. After adding 5 blue marbles, blue becomes 16 + 5 = 21. After removing 3 yellow marbles, yellow becomes 12 - 3 = 9. Then the ratio is 21:9. Let me simplify that. Divide numerator and denominator by 3: 7:3. Which matches the given ratio. Perfect! So that seems to check out.

But let me just go through the steps again to make sure I didn't make any mistakes.

Starting with the equations:

Original ratio: B/Y = 4/3 => 3B = 4Y. Correct.

After adding and removing marbles: (B + 5)/(Y - 3) = 7/3 => cross multiply to 3(B + 5) = 7(Y - 3). Correct.

Expanding that gives 3B + 15 = 7Y - 21. Then substituting 3B from equation 1 (which is 4Y) into equation 2: 4Y + 15 = 7Y - 21. Subtract 4Y: 15 = 3Y - 21. Add 21: 36 = 3Y => Y = 12. Then B = (4/3)*Y = (4/3)*12 = 16.

Yes, that seems correct. Then checking the final ratio: 16 + 5 = 21 blue, 12 - 3 = 9 yellow. 21/9 = 7/3. Correct.

So the original number of blue marbles is 16. Therefore, the answer should be 16.

But let me think again if there's another way to approach this problem, just to confirm. Maybe using variables differently?

Alternatively, let me think in terms of ratios. The original ratio is 4:3. Let me assume that the number of blue marbles is 4x and yellow marbles is 3x for some positive integer x. Then, after adding 5 blue marbles and removing 3 yellow marbles, the new numbers are 4x + 5 and 3x - 3. The new ratio is 7:3. Therefore:

(4x + 5)/(3x - 3) = 7/3

Cross multiplying:

3(4x + 5) = 7(3x - 3)

Which is 12x + 15 = 21x - 21

Subtract 12x both sides:

15 = 9x - 21

Add 21:

36 = 9x => x = 4

Therefore, original number of blue marbles is 4x = 4*4 = 16. Same answer.

So that's another way to think about it, using variables as multiples of the ratio. This also gives 16. So that confirms the answer again.

Therefore, I think 16 is correct.

But let me check if there's any possible mistake here. For example, could there be another solution? Let me see.

Suppose I didn't assume that the original numbers are multiples of 4 and 3. But since ratios are given, it's standard to represent them as variables multiplied by a common factor. So that approach is valid.

Alternatively, if I didn't use substitution, but solved the equations differently. Let me try solving the two equations again:

Original equations:

1) 3B = 4Y

2) 3B + 15 = 7Y - 21

From equation 1, we can express B as (4Y)/3. Then substitute into equation 2:

3*(4Y/3) + 15 = 7Y - 21

Simplify left side: 4Y + 15 = 7Y - 21

Which is same as before. So same result.

Therefore, regardless of the method, the answer is 16.

Another check: Let me think about the numbers. Original marbles: blue 16, yellow 12. Adding 5 blue makes 21, removing 3 yellow makes 9. 21:9 reduces to 7:3. Correct.

So, seems solid.

I think that's thorough enough. I don't see any mistakes in my reasoning. So the original number of blue marbles is 16.

**Final Answer**
The number of blue marbles before adding more is \boxed{16}.

To determine the original number of blue marbles in the bag, we begin by defining variables for the quantities of blue and yellow marbles:

- Let $ B $ represent the number of blue marbles.
- Let $ Y $ represent the number of yellow marbles.

---

### Step 1: Use the initial ratio

We are told that the initial ratio of blue to yellow marbles is $ 4:3 $. This gives us the equation:

$$
\frac{B}{Y} = \frac{4}{3} \quad \Rightarrow \quad 3B = 4Y \tag{1}
$$

---

### Step 2: Use the modified ratio after changes

After adding 5 blue marbles and removing 3 yellow marbles, the new ratio becomes $ 7:3 $. This gives us the second equation:

$$
\frac{B + 5}{Y - 3} = \frac{7}{3} \quad \Rightarrow \quad 3(B + 5) = 7(Y - 3) \tag{2}
$$

Expanding equation (2):

$$
3B + 15 = 7Y - 21
$$

---

### Step 3: Solve the system of equations

From equation (1), we can express $ B $ in terms of $ Y $:

$$
B = \frac{4Y}{3}
$$

Substitute this into equation (2):

$$
3\left(\frac{4Y}{3}\right) + 15 = 7Y - 21
$$

Simplify:

$$
4Y + 15 = 7Y - 21
$$

Solving for $ Y $:

$$
15 + 21 = 7Y - 4Y \quad \Rightarrow \quad 36 = 3Y \quad \Rightarrow \quad Y = 12
$$

Now substitute $ Y = 12 $ back into equation (1) to find $ B $:

$$
3B = 4Y = 4 \times 12 = 48 \quad \Rightarrow \quad B = \frac{48}{3} = 16
$$

---

### Step 4: Verify the solution

- Original marbles: $ B = 16 $, $ Y = 12 $
- After changes: $ B + 5 = 21 $, $ Y - 3 = 9 $
- New ratio: $ \frac{21}{9} = \frac{7}{3} $, which matches the given ratio.

---

### Final Answer

$$
\boxed{16}
$$

Solution: Let $x$ be the number of blue marbles and $y$ the number of yellow marbles before I added more. We are given that the ratio of blue to yellow is 4:3, so $\dfrac{x}{y}=\dfrac{4}{3}$. Additionally, after we add blue marbles and remove yellow marbles the total number of blue marbles and yellow marbles will be x+5 and y-3 respectively. We're given that at this point the ratio will be $7:3$, so $\dfrac{x+5}{y-3}=\dfrac{7}{3}$. Cross multiplying the first equation gives $3x=4y$ and cross multiplying the second gives $3(x+5)=7(y-3)$. Solving two linear equations on two variables is routine; we get the solution $y=12$, $x=16$. Since $x$ represents the number of blue marbles before more were added, the answer to the problem is just $\boxed{16}$.


Explanation:
Okay, so I need to find the angle ABC, which is the angle at point B formed by points A, B, and C. Let me recall how to approach this. I think the angle between two lines can be found using vectors. So, maybe I need to find vectors BA and BC, then use the dot product formula to find the angle between them. Let me try to remember the exact steps.

First, let me note down the coordinates of the points again to make sure I have them right. Point A is (1, -11, 2), point B is (3, -4, 1), and point C is (-2, 1, -1). So angle ABC is at point B, so the two sides of the angle are BA and BC. Therefore, I need vectors BA and BC.

To find vector BA, I subtract coordinates of A from B? Wait, no. Wait, vector BA goes from B to A, so it's A - B. Similarly, vector BC goes from B to C, so it's C - B. Let me confirm that. Yes, if you have a vector from point X to point Y, it's Y - X. So vector BA is A - B, and vector BC is C - B. Let me compute those.

First, vector BA: A is (1, -11, 2), B is (3, -4, 1). So subtracting B from A: (1 - 3, -11 - (-4), 2 - 1) = (-2, -7, 1). Let me check that again. For the x-coordinate: 1 - 3 is -2. Y-coordinate: -11 - (-4) is -11 + 4 = -7. Z-coordinate: 2 - 1 = 1. Yes, that seems right. So vector BA is (-2, -7, 1).

Now vector BC: C is (-2, 1, -1), B is (3, -4, 1). So subtracting B from C: (-2 - 3, 1 - (-4), -1 - 1) = (-5, 5, -2). Let me check again. X-coordinate: -2 - 3 = -5. Y-coordinate: 1 - (-4) = 1 + 4 = 5. Z-coordinate: -1 - 1 = -2. Yep, that's correct. So vector BC is (-5, 5, -2).

Now, to find the angle between vectors BA and BC. The formula for the angle θ between two vectors u and v is:

cosθ = (u · v) / (|u| |v|)

So I need to compute the dot product of BA and BC, then divide by the product of their magnitudes.

First, let me compute the dot product. The dot product of BA and BC is (-2)(-5) + (-7)(5) + (1)(-2). Let me calculate each term:

First term: (-2)(-5) = 10

Second term: (-7)(5) = -35

Third term: (1)(-2) = -2

Adding them up: 10 - 35 - 2 = 10 - 37 = -27. So the dot product is -27.

Now, compute the magnitudes of BA and BC.

Starting with |BA|: sqrt[(-2)^2 + (-7)^2 + (1)^2] = sqrt[4 + 49 + 1] = sqrt[54]. Let me check that again. (-2)^2 is 4, (-7)^2 is 49, 1^2 is 1. 4 + 49 is 53, plus 1 is 54. So sqrt(54). Simplify sqrt(54): sqrt(9*6) = 3*sqrt(6). So |BA| is 3√6.

Now |BC|: sqrt[(-5)^2 + (5)^2 + (-2)^2] = sqrt[25 + 25 + 4] = sqrt[54]. Wait, same as |BA|? Let me check again. (-5)^2 is 25, 5^2 is 25, (-2)^2 is 4. 25 + 25 is 50, plus 4 is 54. So sqrt(54) as well. So |BC| is also 3√6.

So then cosθ = (-27) / (3√6 * 3√6) = (-27) / (9 * 6) = (-27)/54 = -0.5.

So cosθ = -0.5. Now, what angle has cosine equal to -0.5? Let me recall the unit circle. Cosine is -0.5 at 120 degrees and 240 degrees. But since angles between vectors are between 0 and 180 degrees, it must be 120 degrees. Therefore, θ = 120 degrees.

Wait, let me confirm that. The arccos of -0.5 is indeed 120 degrees. Yes, because cos(120°) = cos(180° - 60°) = -cos(60°) = -0.5. So that's correct.

Therefore, the angle ABC is 120 degrees. Let me just check my steps again to make sure I didn't make any calculation errors.

First, vectors BA and BC: BA is A - B = (-2, -7, 1), BC is C - B = (-5, 5, -2). Then the dot product: (-2)(-5) + (-7)(5) + (1)(-2) = 10 - 35 - 2 = -27. Correct.

Magnitudes: sqrt(4 + 49 + 1) = sqrt(54) and sqrt(25 + 25 + 4) = sqrt(54). So both magnitudes are 3√6. Then the denominator is 3√6 * 3√6 = 9 * 6 = 54. So -27 / 54 = -0.5. Arccos(-0.5) is 120 degrees. Seems correct.

Wait, but just to make sure, maybe I should check if I had the vectors in the right direction? Because angle ABC is at point B, so the vectors should be BA and BC emanating from B. But when we compute vectors BA and BC, they are from B to A and from B to C, so yes, they are the vectors forming the angle at B. So the angle between BA and BC is indeed the angle at B. So that's correct.

Alternatively, sometimes people might get confused whether to use AB and BC or BA and BC, but since angle is at B, the vectors should be BA and BC. So yes, that's correct.

Alternatively, if I had used vectors AB and BC, that would be different. Let me check that just in case. Vector AB is B - A, which would be (3 - 1, -4 - (-11), 1 - 2) = (2, 7, -1). Then vector BC is still (-5, 5, -2). Then the angle between AB and BC would be different. But since the angle is at B, we need vectors going out from B, so BA and BC. So I think my original calculation is correct.

Alternatively, maybe I can check using coordinates. Let me try to visualize or compute something else. For example, maybe compute the lengths of AB, BC, and AC, then use the Law of Cosines? Let me try that as a check.

First, compute AB, BC, and AC.

Wait, AB is the distance from A to B. Let me compute AB:

Coordinates of A (1, -11, 2) and B (3, -4, 1). So AB distance squared is (3 - 1)^2 + (-4 - (-11))^2 + (1 - 2)^2 = (2)^2 + (7)^2 + (-1)^2 = 4 + 49 + 1 = 54. So AB is sqrt(54).

Similarly, BC is distance from B to C: sqrt[(-2 - 3)^2 + (1 - (-4))^2 + (-1 - 1)^2] = sqrt[(-5)^2 + (5)^2 + (-2)^2] = sqrt[25 + 25 + 4] = sqrt(54). So BC is sqrt(54).

Then AC is distance from A to C: sqrt[(-2 - 1)^2 + (1 - (-11))^2 + (-1 - 2)^2] = sqrt[(-3)^2 + (12)^2 + (-3)^2] = sqrt[9 + 144 + 9] = sqrt[162] = sqrt(81*2) = 9√2.

So now, using Law of Cosines on triangle ABC, with sides AB = sqrt(54), BC = sqrt(54), and AC = 9√2. Wait, let me check if that's correct. Wait, AB and BC are both sqrt(54), so triangle ABC has two sides equal? Wait, AB and BC are both sqrt(54), so triangle ABC is isoceles with AB = BC. Wait, but angle at B is between AB and BC? Wait, no. Wait, AB is from A to B, BC is from B to C. So in triangle ABC, sides AB, BC, and AC. So angle at B is between AB and BC. So Law of Cosines says:

AC² = AB² + BC² - 2 AB * BC * cos(angle ABC)

So let me plug in the numbers.

AC² is (9√2)^2 = 81 * 2 = 162.

AB² is 54, BC² is 54. So:

162 = 54 + 54 - 2 * sqrt(54) * sqrt(54) * cos(angle ABC)

Simplify:

162 = 108 - 2 * 54 * cos(angle ABC)

Wait, because sqrt(54)*sqrt(54) is 54. So:

162 = 108 - 108 cos(angle ABC)

Subtract 108 from both sides:

162 - 108 = -108 cos(angle ABC)

54 = -108 cos(angle ABC)

Divide both sides by -108:

cos(angle ABC) = 54 / (-108) = -0.5

Which is the same result as before! So angle ABC is arccos(-0.5) = 120 degrees. So that confirms the previous result. Therefore, the angle is indeed 120 degrees. So I think that's correct.

Therefore, the answer is 120 degrees. Let me just check once more if I didn't mix up any coordinates or vectors. For example, when I calculated vectors BA and BC, I had BA as (-2, -7, 1) and BC as (-5, 5, -2). Then their dot product was -27, magnitudes both 3√6, so cos(theta) = -27/(54) = -0.5. All steps check out. And Law of Cosines also gives the same result. So I think that's solid.

Therefore, I can be confident that the angle ABC is 120 degrees.

**Final Answer**
The measure of angle $ ABC $ is \boxed{120} degrees.

To find the angle $ \angle ABC $, we are given three points in 3D space:

- $ A = (1, -11, 2) $
- $ B = (3, -4, 1) $
- $ C = (-2, 1, -1) $

---

### Step 1: Determine the vectors forming the angle at point $ B $

The angle $ \angle ABC $ is formed by the vectors $ \vec{BA} $ and $ \vec{BC} $, both originating from point $ B $.

- Vector $ \vec{BA} = A - B = (1 - 3, -11 - (-4), 2 - 1) = (-2, -7, 1) $
- Vector $ \vec{BC} = C - B = (-2 - 3, 1 - (-4), -1 - 1) = (-5, 5, -2) $

---

### Step 2: Compute the dot product of $ \vec{BA} $ and $ \vec{BC} $

$$
\vec{BA} \cdot \vec{BC} = (-2)(-5) + (-7)(5) + (1)(-2) = 10 - 35 - 2 = -27
$$

---

### Step 3: Compute the magnitudes of the vectors

$$
|\vec{BA}| = \sqrt{(-2)^2 + (-7)^2 + 1^2} = \sqrt{4 + 49 + 1} = \sqrt{54}
$$
$$
|\vec{BC}| = \sqrt{(-5)^2 + 5^2 + (-2)^2} = \sqrt{25 + 25 + 4} = \sqrt{54}
$$

---

### Step 4: Use the dot product formula to find the cosine of the angle

$$
\cos \theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| \cdot |\vec{BC}|} = \frac{-27}{\sqrt{54} \cdot \sqrt{54}} = \frac{-27}{54} = -0.5
$$

---

### Step 5: Find the angle from the cosine value

$$
\theta = \cos^{-1}(-0.5) = 120^\circ
$$

This is the angle between the vectors $ \vec{BA} $ and $ \vec{BC} $, which corresponds to $ \angle ABC $.

---

### Final Answer

$$
\boxed{120}
$$

[Assistant]
Explanation:
Okay, so I need to find the angle ABC, which is the angle at point B formed by points A, B, and C. Let me recall how to approach this. I think the angle between two lines can be found using vectors. So, maybe I need to find vectors BA and BC, then use the dot product formula to find the angle between them. Let me try to remember the exact steps.

First, let me note down the coordinates of the points again to make sure I have them right. Point A is (1, -11, 2), point B is (3, -4, 1), and point C is (-2, 1, -1). So angle ABC is at point B, so the two sides of the angle are BA and BC. Therefore, I need vectors BA and BC.

To find vector BA, I subtract coordinates of A from B? Wait, no. Wait, vector BA goes from B to A, so it's A - B. Similarly, vector BC goes from B to C, so it's C - B. Let me confirm that. Yes, if you have a vector from point X to point Y, it's Y - X. So vector BA is A - B, and vector BC is C - B. Let me compute those.

First, vector BA: A is (1, -11, 2), B is (3, -4, 1). So subtracting B from A: (1 - 3, -11 - (-4), 2 - 1) = (-2, -7, 1). Let me check that again. For the x-coordinate: 1 - 3 is -2. Y-coordinate: -11 - (-4) is -11 + 4 = -7. Z-coordinate: 2 - 1 = 1. Yes, that seems right. So vector BA is (-2, -7, 1).

Now vector BC: C is (-2, 1, -1), B is (3, -4, 1). So subtracting B from C: (-2 - 3, 1 - (-4), -1 - 1) = (-5, 5, -2). Let me check again. X-coordinate: -2 - 3 = -5. Y-coordinate: 1 - (-4) = 1 + 4 = 5. Z-coordinate: -1 - 1 = -2. Yep, that's correct. So vector BC is (-5, 5, -2).

Now, to find the angle between vectors BA and BC. The formula for the angle θ between two vectors u and v is:

cosθ = (u · v) / (|u| |v|)

So I need to compute the dot product of BA and BC, then divide by the product of their magnitudes.

First, let me compute the dot product. The dot product of BA and BC is (-2)(-5) + (-7)(5) + (1)(-2). Let me calculate each term:

First term: (-2)(-5) = 10

Second term: (-7)(5) = -35

Third term: (1)(-2) = -2

Adding them up: 10 - 35 - 2 = 10 - 37 = -27. So the dot product is -27.

Now, compute the magnitudes of BA and BC.

Starting with |BA|: sqrt[(-2)^2 + (-7)^2 + (1)^2] = sqrt[4 + 49 + 1] = sqrt[54]. Let me check that again. (-2)^2 is 4, (-7)^2 is 49, 1^2 is 1. 4 + 49 is 53, plus 1 is 54. So sqrt(54). Simplify sqrt(54): sqrt(9*6) = 3*sqrt(6). So |BA| is 3√6.

Now |BC|: sqrt[(-5)^2 + (5)^2 + (-2)^2] = sqrt[25 + 25 + 4] = sqrt[54]. Wait, same as |BA|? Let me check again. (-5)^2 is 25, 5^2 is 25, (-2)^2 is 4. 25 + 25 is 50, plus 4 is 54. So sqrt(54) as well. So |BC| is also 3√6.

So then cosθ = (-27) / (3√6 * 3√6) = (-27) / (9 * 6) = (-27)/54 = -0.5.

So cosθ = -0.5. Now, what angle has cosine equal to -0.5? Let me recall the unit circle. Cosine is -0.5 at 120 degrees and 240 degrees. But since angles between vectors are between 0 and 180 degrees, it must be 120 degrees. Therefore, θ = 120 degrees.

Wait, let me confirm that. The arccos of -0.5 is indeed 120 degrees. Yes, because cos(120°) = cos(180° - 60°) = -cos(60°) = -0.5. So that's correct.

Therefore, the angle ABC is 120 degrees. Let me just check my steps again to make sure I didn't make any calculation errors.

First, vectors BA and BC: BA is A - B = (-2, -7, 1), BC is C - B = (-5, 5, -2). Then the dot product: (-2)(-5) + (-7)(5) + (1)(-2) = 10 - 35 - 2 = -27. Correct.

Magnitudes: sqrt(4 + 49 + 1) = sqrt(54) and sqrt(25 + 25 + 4) = sqrt(54). So both magnitudes are 3√6. Then the denominator is 3√6 * 3√6 = 9 * 6 = 54. So -27 / 54 = -0.5. Arccos(-0.5) is 120 degrees. Seems correct.

Wait, but just to make sure, maybe I should check if I had the vectors in the right direction? Because angle ABC is at point B, so the vectors should be BA and BC emanating from B. But when we compute vectors BA and BC, they are from B to A and from B to C, so yes, they are the vectors forming the angle at B. So the angle between BA and BC is indeed the angle at B. So that's correct.

Alternatively, sometimes people might get confused whether to use AB and BC or BA and BC, but since angle is at B, the vectors should be BA and BC. So yes, that's correct.

Alternatively, if I had used vectors AB and BC, that would be different. Let me check that just in case. Vector AB is B - A, which would be (3 - 1, -4 - (-11), 1 - 2) = (2, 7, -1). Then vector BC is still (-5, 5, -2). Then the angle between AB and BC would be different. But since the angle is at B, we need vectors going out from B, so BA and BC. So I think my original calculation is correct.

Alternatively, maybe I can check using coordinates. Let me try to visualize or compute something else. For example, maybe compute the lengths of AB, BC, and AC, then use the Law of Cosines? Let me try that as a check.

First, compute AB, BC, and AC.

Wait, AB is the distance from A to B. Let me compute AB:

Coordinates of A (1, -11, 2) and B (3, -4, 1). So AB distance squared is (3 - 1)^2 + (-4 - (-11))^2 + (1 - 2)^2 = (2)^2 + (7)^2 + (-1)^2 = 4 + 49 + 1 = 54. So AB is sqrt(54).

Similarly, BC is distance from B to C: sqrt[(-2 - 3)^2 + (1 - (-4))^2 + (-1 - 1)^2] = sqrt[(-5)^2 + (5)^2 + (-2)^2] = sqrt[25 + 25 + 4] = sqrt(54). So BC is sqrt(54).

Then AC is distance from A to C: sqrt[(-2 - 1)^2 + (1 - (-11))^2 + (-1 - 2)^2] = sqrt[(-3)^2 + (12)^2 + (-3)^2] = sqrt[9 + 144 + 9] = sqrt[162] = sqrt(81*2) = 9√2.

So now, using Law of Cosines on triangle ABC, with sides AB = sqrt(54), BC = sqrt(54), and AC = 9√2. Wait, let me check if that's correct. Wait, AB and BC are both sqrt(54), so triangle ABC has two sides equal? Wait, AB and BC are both sqrt(54), so triangle ABC is isoceles with AB = BC. Wait, but angle at B is between AB and BC? Wait, no. Wait, AB is from A to B, BC is from B to C. So in triangle ABC, sides AB, BC, and AC. So angle at B is between AB and BC. So Law of Cosines says:

AC² = AB² + BC² - 2 AB * BC * cos(angle ABC)

So let me plug in the numbers.

AC² is (9√2)^2 = 81 * 2 = 162.

AB² is 54, BC² is 54. So:

162 = 54 + 54 - 2 * sqrt(54) * sqrt(54) * cos(angle ABC)

Simplify:

162 = 108 - 2 * 54 * cos(angle ABC)

Wait, because sqrt(54)*sqrt(54) is 54. So:

162 = 108 - 108 cos(angle ABC)

Subtract 108 from both sides:

162 - 108 = -108 cos(angle ABC)

54 = -108 cos(angle ABC)

Divide both sides by -108:

cos(angle ABC) = 54 / (-108) = -0.5

Which is the same result as before! So angle ABC is arccos(-0.5) = 120 degrees. So that confirms the previous result. Therefore, the angle is indeed 120 degrees. So I think that's correct.

Therefore, the answer is 120 degrees. Let me just check once more if I didn't mix up any coordinates or vectors. For example, when I calculated vectors BA and BC, I had BA as (-2, -7, 1) and BC as (-5, 5, -2). Then their dot product was -27, magnitudes both 3√6, so cos(theta) = -27/(54) = -0.5. All steps check out. And Law of Cosines also gives the same result. So I think that's solid.

Therefore, I can be confident that the angle ABC is 120 degrees.

**Final Answer**
The measure of angle $ ABC $ is \boxed{120} degrees.

To find the angle $ \angle ABC $, we are given three points in 3D space:

- $ A = (1, -11, 2) $
- $ B = (3, -4, 1) $
- $ C = (-2, 1, -1) $

---

### Step 1: Determine the vectors forming the angle at point $ B $

The angle $ \angle ABC $ is formed by the vectors $ \vec{BA} $ and $ \vec{BC} $, both originating from point $ B $.

- Vector $ \vec{BA} = A - B = (1 - 3, -11 - (-4), 2 - 1) = (-2, -7, 1) $
- Vector $ \vec{BC} = C - B = (-2 - 3, 1 - (-4), -1 - 1) = (-5, 5, -2) $

---

### Step 2: Compute the dot product of $ \vec{BA} $ and $ \vec{BC} $

$$
\vec{BA} \cdot \vec{BC} = (-2)(-5) + (-7)(5) + (1)(-2) = 10 - 35 - 2 = -27
$$

---

### Step 3: Compute the magnitudes of the vectors

$$
|\vec{BA}| = \sqrt{(-2)^2 + (-7)^2 + 1^2} = \sqrt{4 + 49 + 1} = \sqrt{54}
$$
$$
|\vec{BC}| = \sqrt{(-5)^2 + 5^2 + (-2)^2} = \sqrt{25 + 25 + 4} = \sqrt{54}
$$

---

### Step 4: Use the dot product formula to find the cosine of the angle

$$
\cos \theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| \cdot |\vec{BC}|} = \frac{-27}{\sqrt{54} \cdot \sqrt{54}} = \frac{-27}{54} = -0.5
$$

---

### Step 5: Find the angle from the cosine value

$$
\theta = \cos^{-1}(-0.5) = 120^\circ
$$

This is the angle between the vectors $ \vec{BA} $ and $ \vec{BC} $, which corresponds to $ \angle ABC $.

---

### Final Answer

$$
\boxed{120}
$$

Solution:
From the distance formula, we compute that $AB = 3 \sqrt{6},$ $AC = 9 \sqrt{2},$ and $BC = 3 \sqrt{6}.$  Then from the Law of Cosines,
\[\cos \angle ABC = \frac{(3 \sqrt{6})^2 + (3 \sqrt{6})^2 - (9 \sqrt{2})^2}{2 \cdot 3 \sqrt{6} \cdot 3 \sqrt{6}} = -\frac{1}{2}.\]Therefore, $\angle ABC = \boxed{120^\circ}.$


Explanation:
Okay, so I need to find the measure of angle B in triangle ABC, where angle A is 14 degrees, and AB is the diameter of the circle in which the triangle is inscribed. Hmm, let me think about how to approach this.

First, I remember that when a triangle is inscribed in a circle and one of its sides is the diameter, there's a special property related to right angles. Is it the Thales' theorem? Yeah, Thales' theorem states that if you have a triangle inscribed in a circle where one side is the diameter, then the angle opposite that diameter is a right angle. So, that means angle C must be 90 degrees because AB is the diameter. Wait, let me confirm that. If AB is the diameter, then the triangle ABC has AB as its hypotenuse, and angle C is the right angle. So, angle C is 90 degrees. Got it.

So, triangle ABC is a right-angled triangle at C. Therefore, the sum of angles A and B should be 90 degrees because the total sum of angles in a triangle is 180 degrees. Since angle A is given as 14 degrees, angle B should be 90 - 14 = 76 degrees. Is that all? Wait, maybe I need to check if there's more to it?

Wait, let me make sure I didn't skip any steps. Let me visualize the triangle. AB is the diameter of the circle, so points A and B are at the ends of the diameter. Point C is somewhere on the circumference, forming triangle ABC. According to Thales' theorem, angle ACB is 90 degrees. Therefore, triangle ABC is right-angled at C. Therefore, angles at A and B must add up to 90 degrees. Since angle A is 14 degrees, angle B is 90 - 14 = 76 degrees. That seems straightforward.

But maybe I need to verify using other methods? Let me try using the Law of Sines or something else to confirm.

In any triangle, the Law of Sines states that a/sin A = b/sin B = c/sin C. But since we know angle C is 90 degrees, maybe we can use that. Let me denote the sides. Let me call AB as the diameter, so AB = 2R, where R is the radius of the circle. Then, AC and BC are the other sides.

But maybe I can assign some variables. Let me denote AB as the diameter, so AB = 2R. Then, angle at C is 90 degrees. Let me denote angle at A as 14 degrees, angle at B as what we need to find, let's call it x. Then, as before, 14 + x + 90 = 180, so x = 76 degrees. So, that seems consistent.

Alternatively, maybe using coordinates? Let me try that. Let me place the circle with diameter AB on the coordinate plane. Let me set point A at (-1, 0) and point B at (1, 0), so that AB is the diameter of length 2 units, and the center is at the origin (0, 0). Then, point C is somewhere on the circle. The coordinates of C can be represented as (cos θ, sin θ) for some angle θ. Then, angle at A is 14 degrees. Wait, how do I relate that?

Alternatively, maybe using vectors or slopes? Hmm, maybe that's more complicated. Alternatively, since angle at A is 14 degrees, maybe I can compute the coordinates of point C such that angle BAC is 14 degrees. Let me think.

In triangle ABC, angle at A is 14 degrees, angle at C is 90 degrees. So, angle at B is 76 degrees. But maybe I can check using coordinates. Let me try.

Let me consider AB as diameter from (-1, 0) to (1, 0). Let me find coordinates of point C such that angle BAC is 14 degrees. Let me denote point C as (x, y) on the circle x² + y² = 1. Then, angle BAC is the angle between vectors AB and AC. Wait, AB is from A to B, which is (2, 0). AC is from A to C, which is (x - (-1), y - 0) = (x + 1, y). The angle between AB and AC is 14 degrees. So, using the dot product formula:

cos(theta) = (AB · AC) / (|AB| |AC|)

AB vector is (2, 0), AC vector is (x + 1, y). The dot product is 2*(x + 1) + 0*y = 2(x + 1). The magnitude of AB is 2, the magnitude of AC is sqrt((x + 1)^2 + y^2). Therefore:

cos(14°) = [2(x + 1)] / [2 * sqrt((x + 1)^2 + y^2)]

Simplify:

cos(14°) = (x + 1)/sqrt((x + 1)^2 + y^2)

But since point C lies on the circle x² + y² = 1, so y² = 1 - x². Substitute that into the equation:

cos(14°) = (x + 1)/sqrt((x + 1)^2 + 1 - x²)

Simplify denominator:

(x + 1)^2 + 1 - x² = x² + 2x + 1 + 1 - x² = 2x + 2 = 2(x + 1)

Therefore, denominator is sqrt(2(x + 1)).

So, cos(14°) = (x + 1)/sqrt(2(x + 1)) = sqrt(x + 1)/sqrt(2)

Therefore:

cos(14°) = sqrt( (x + 1)/2 )

Square both sides:

cos²(14°) = (x + 1)/2

Therefore:

x + 1 = 2 cos²(14°)

So:

x = 2 cos²(14°) - 1

But 2 cos²(theta) - 1 is equal to cos(2 theta). Therefore:

x = cos(28°)

So, x-coordinate of point C is cos(28°). Therefore, coordinates of C are (cos(28°), sin(28°)).

Wait, but then angle at B? Let me compute angle at B. Let me think. Alternatively, since we know coordinates of points A, B, C, we can compute angle at B.

Point B is at (1, 0), point A is at (-1, 0), point C is at (cos(28°), sin(28°)). Let me compute vectors BA and BC.

Vector BA is A - B = (-1 - 1, 0 - 0) = (-2, 0)

Vector BC is C - B = (cos(28°) - 1, sin(28°) - 0) = (cos(28°) - 1, sin(28°))

Then, angle at B is the angle between vectors BA and BC. Let me compute that angle using the dot product.

Dot product of BA and BC is (-2)(cos(28°) - 1) + 0*(sin(28°)) = -2 cos(28°) + 2

The magnitude of BA is sqrt((-2)^2 + 0^2) = 2

The magnitude of BC is sqrt( (cos(28°) - 1)^2 + (sin(28°))^2 )

Let me compute that:

(cos(28°) - 1)^2 + sin²(28°) = cos²(28°) - 2 cos(28°) + 1 + sin²(28°) = [cos²(28°) + sin²(28°)] - 2 cos(28°) + 1 = 1 - 2 cos(28°) + 1 = 2 - 2 cos(28°) = 2(1 - cos(28°))

Therefore, magnitude of BC is sqrt(2(1 - cos(28°)))

Therefore, the cosine of angle at B is:

[ -2 cos(28°) + 2 ] / [ 2 * sqrt(2(1 - cos(28°))) ] = [ 2(1 - cos(28°)) ] / [ 2 * sqrt(2(1 - cos(28°))) ] = [ (1 - cos(28°)) ] / [ sqrt(2(1 - cos(28°))) ] = sqrt(1 - cos(28°)) / sqrt(2)

Which is equal to sqrt( (1 - cos(28°))/2 ) = sin(14°), since 1 - cos(theta) = 2 sin²(theta/2), so sqrt( (2 sin²(14°))/2 ) = sin(14°). Therefore, cos(angle at B) = sin(14°). Wait, but cos(angle at B) = sin(14°). Let me recall that sin(theta) = cos(90° - theta). Therefore, sin(14°) = cos(76°). Therefore, cos(angle at B) = cos(76°). Therefore, angle at B is 76°, since it's between 0 and 90 degrees. Therefore, angle B is 76 degrees. So that confirms the earlier result.

Therefore, angle B is 76 degrees. So, that seems consistent. Therefore, the answer is 76 degrees.

But let me check once again. Since angle A is 14°, angle C is 90°, so angle B is 180 - 14 - 90 = 76°, which is straightforward. So, maybe there's no need for all that coordinate geometry. But it's good that I verified it with coordinates. So, seems like 76 degrees is the answer.

Wait, but let me think again. Is there a possibility that angle B could be something else? For example, if the triangle is not right-angled? But no, because AB is the diameter, so by Thales' theorem, angle C is right. Therefore, triangle is right-angled at C, so angles at A and B must add to 90°, so angle B is 90° - 14° = 76°. So, that's straightforward.

Alternatively, maybe using the inscribed angle theorem? Let me think. The inscribed angle theorem says that an angle subtended by an arc at the circumference is half the measure of the central angle subtended by the same arc. But since AB is the diameter, the central angle for arc AB is 180°, so any inscribed angle subtended by AB would be half that, which is 90°, which is angle C. So, that again confirms angle C is 90°, so angles A and B add to 90°, so angle B is 76°. So, all methods lead to the same answer.

Therefore, I think the answer is 76 degrees. I don't see any mistakes in my reasoning. So, I think that's the correct answer.

**Final Answer**
The measure of angle $ \angle B $ is \boxed{76} degrees.

To determine the measure of angle $ \angle B $ in triangle $ \triangle ABC $, we are given the following:

- $ \angle A = 14^\circ $
- $ AB $ is the diameter of the circle in which the triangle is inscribed

---

### Step 1: Apply Thales' Theorem

Thales' Theorem states that if a triangle is inscribed in a circle and one of its sides is the diameter of the circle, then the angle opposite that diameter is a right angle.

In this case, since $ AB $ is the diameter, the angle opposite to it, which is $ \angle C $, must be a right angle:

$$
\angle C = 90^\circ
$$

---

### Step 2: Use the Triangle Angle Sum Theorem

In any triangle, the sum of the interior angles is $ 180^\circ $. Therefore, in triangle $ \triangle ABC $:

$$
\angle A + \angle B + \angle C = 180^\circ
$$

Substitute the known values:

$$
14^\circ + \angle B + 90^\circ = 180^\circ
$$

Solving for $ \angle B $:

$$
\angle B = 180^\circ - 14^\circ - 90^\circ = 76^\circ
$$

---

### Step 3: Verification via Coordinate Geometry (Optional)

To confirm, we can place the triangle on a coordinate plane with $ A = (-1, 0) $, $ B = (1, 0) $, and $ C $ on the unit circle. Using vector analysis and the dot product, we can compute the angle at $ B $, and it again confirms that $ \angle B = 76^\circ $.

---

### Final Answer

$$
\boxed{76}
$$

[Assistant]
Explanation:
Okay, so I need to find the measure of angle B in triangle ABC, where angle A is 14 degrees, and AB is the diameter of the circle in which the triangle is inscribed. Hmm, let me think about how to approach this.

First, I remember that when a triangle is inscribed in a circle and one of its sides is the diameter, there's a special property related to right angles. Is it the Thales' theorem? Yeah, Thales' theorem states that if you have a triangle inscribed in a circle where one side is the diameter, then the angle opposite that diameter is a right angle. So, that means angle C must be 90 degrees because AB is the diameter. Wait, let me confirm that. If AB is the diameter, then the triangle ABC has AB as its hypotenuse, and angle C is the right angle. So, angle C is 90 degrees. Got it.

So, triangle ABC is a right-angled triangle at C. Therefore, the sum of angles A and B should be 90 degrees because the total sum of angles in a triangle is 180 degrees. Since angle A is given as 14 degrees, angle B should be 90 - 14 = 76 degrees. Is that all? Wait, maybe I need to check if there's more to it?

Wait, let me make sure I didn't skip any steps. Let me visualize the triangle. AB is the diameter of the circle, so points A and B are at the ends of the diameter. Point C is somewhere on the circumference, forming triangle ABC. According to Thales' theorem, angle ACB is 90 degrees. Therefore, triangle ABC is right-angled at C. Therefore, angles at A and B must add up to 90 degrees. Since angle A is 14 degrees, angle B is 90 - 14 = 76 degrees. That seems straightforward.

But maybe I need to verify using other methods? Let me try using the Law of Sines or something else to confirm.

In any triangle, the Law of Sines states that a/sin A = b/sin B = c/sin C. But since we know angle C is 90 degrees, maybe we can use that. Let me denote the sides. Let me call AB as the diameter, so AB = 2R, where R is the radius of the circle. Then, AC and BC are the other sides.

But maybe I can assign some variables. Let me denote AB as the diameter, so AB = 2R. Then, angle at C is 90 degrees. Let me denote angle at A as 14 degrees, angle at B as what we need to find, let's call it x. Then, as before, 14 + x + 90 = 180, so x = 76 degrees. So, that seems consistent.

Alternatively, maybe using coordinates? Let me try that. Let me place the circle with diameter AB on the coordinate plane. Let me set point A at (-1, 0) and point B at (1, 0), so that AB is the diameter of length 2 units, and the center is at the origin (0, 0). Then, point C is somewhere on the circle. The coordinates of C can be represented as (cos θ, sin θ) for some angle θ. Then, angle at A is 14 degrees. Wait, how do I relate that?

Alternatively, maybe using vectors or slopes? Hmm, maybe that's more complicated. Alternatively, since angle at A is 14 degrees, maybe I can compute the coordinates of point C such that angle BAC is 14 degrees. Let me think.

In triangle ABC, angle at A is 14 degrees, angle at C is 90 degrees. So, angle at B is 76 degrees. But maybe I can check using coordinates. Let me try.

Let me consider AB as diameter from (-1, 0) to (1, 0). Let me find coordinates of point C such that angle BAC is 14 degrees. Let me denote point C as (x, y) on the circle x² + y² = 1. Then, angle BAC is the angle between vectors AB and AC. Wait, AB is from A to B, which is (2, 0). AC is from A to C, which is (x - (-1), y - 0) = (x + 1, y). The angle between AB and AC is 14 degrees. So, using the dot product formula:

cos(theta) = (AB · AC) / (|AB| |AC|)

AB vector is (2, 0), AC vector is (x + 1, y). The dot product is 2*(x + 1) + 0*y = 2(x + 1). The magnitude of AB is 2, the magnitude of AC is sqrt((x + 1)^2 + y^2). Therefore:

cos(14°) = [2(x + 1)] / [2 * sqrt((x + 1)^2 + y^2)]

Simplify:

cos(14°) = (x + 1)/sqrt((x + 1)^2 + y^2)

But since point C lies on the circle x² + y² = 1, so y² = 1 - x². Substitute that into the equation:

cos(14°) = (x + 1)/sqrt((x + 1)^2 + 1 - x²)

Simplify denominator:

(x + 1)^2 + 1 - x² = x² + 2x + 1 + 1 - x² = 2x + 2 = 2(x + 1)

Therefore, denominator is sqrt(2(x + 1)).

So, cos(14°) = (x + 1)/sqrt(2(x + 1)) = sqrt(x + 1)/sqrt(2)

Therefore:

cos(14°) = sqrt( (x + 1)/2 )

Square both sides:

cos²(14°) = (x + 1)/2

Therefore:

x + 1 = 2 cos²(14°)

So:

x = 2 cos²(14°) - 1

But 2 cos²(theta) - 1 is equal to cos(2 theta). Therefore:

x = cos(28°)

So, x-coordinate of point C is cos(28°). Therefore, coordinates of C are (cos(28°), sin(28°)).

Wait, but then angle at B? Let me compute angle at B. Let me think. Alternatively, since we know coordinates of points A, B, C, we can compute angle at B.

Point B is at (1, 0), point A is at (-1, 0), point C is at (cos(28°), sin(28°)). Let me compute vectors BA and BC.

Vector BA is A - B = (-1 - 1, 0 - 0) = (-2, 0)

Vector BC is C - B = (cos(28°) - 1, sin(28°) - 0) = (cos(28°) - 1, sin(28°))

Then, angle at B is the angle between vectors BA and BC. Let me compute that angle using the dot product.

Dot product of BA and BC is (-2)(cos(28°) - 1) + 0*(sin(28°)) = -2 cos(28°) + 2

The magnitude of BA is sqrt((-2)^2 + 0^2) = 2

The magnitude of BC is sqrt( (cos(28°) - 1)^2 + (sin(28°))^2 )

Let me compute that:

(cos(28°) - 1)^2 + sin²(28°) = cos²(28°) - 2 cos(28°) + 1 + sin²(28°) = [cos²(28°) + sin²(28°)] - 2 cos(28°) + 1 = 1 - 2 cos(28°) + 1 = 2 - 2 cos(28°) = 2(1 - cos(28°))

Therefore, magnitude of BC is sqrt(2(1 - cos(28°)))

Therefore, the cosine of angle at B is:

[ -2 cos(28°) + 2 ] / [ 2 * sqrt(2(1 - cos(28°))) ] = [ 2(1 - cos(28°)) ] / [ 2 * sqrt(2(1 - cos(28°))) ] = [ (1 - cos(28°)) ] / [ sqrt(2(1 - cos(28°))) ] = sqrt(1 - cos(28°)) / sqrt(2)

Which is equal to sqrt( (1 - cos(28°))/2 ) = sin(14°), since 1 - cos(theta) = 2 sin²(theta/2), so sqrt( (2 sin²(14°))/2 ) = sin(14°). Therefore, cos(angle at B) = sin(14°). Wait, but cos(angle at B) = sin(14°). Let me recall that sin(theta) = cos(90° - theta). Therefore, sin(14°) = cos(76°). Therefore, cos(angle at B) = cos(76°). Therefore, angle at B is 76°, since it's between 0 and 90 degrees. Therefore, angle B is 76 degrees. So that confirms the earlier result.

Therefore, angle B is 76 degrees. So, that seems consistent. Therefore, the answer is 76 degrees.

But let me check once again. Since angle A is 14°, angle C is 90°, so angle B is 180 - 14 - 90 = 76°, which is straightforward. So, maybe there's no need for all that coordinate geometry. But it's good that I verified it with coordinates. So, seems like 76 degrees is the answer.

Wait, but let me think again. Is there a possibility that angle B could be something else? For example, if the triangle is not right-angled? But no, because AB is the diameter, so by Thales' theorem, angle C is right. Therefore, triangle is right-angled at C, so angles at A and B must add to 90°, so angle B is 90° - 14° = 76°. So, that's straightforward.

Alternatively, maybe using the inscribed angle theorem? Let me think. The inscribed angle theorem says that an angle subtended by an arc at the circumference is half the measure of the central angle subtended by the same arc. But since AB is the diameter, the central angle for arc AB is 180°, so any inscribed angle subtended by AB would be half that, which is 90°, which is angle C. So, that again confirms angle C is 90°, so angles A and B add to 90°, so angle B is 76°. So, all methods lead to the same answer.

Therefore, I think the answer is 76 degrees. I don't see any mistakes in my reasoning. So, I think that's the correct answer.

**Final Answer**
The measure of angle $ \angle B $ is \boxed{76} degrees.

To determine the measure of angle $ \angle B $ in triangle $ \triangle ABC $, we are given the following:

- $ \angle A = 14^\circ $
- $ AB $ is the diameter of the circle in which the triangle is inscribed

---

### Step 1: Apply Thales' Theorem

Thales' Theorem states that if a triangle is inscribed in a circle and one of its sides is the diameter of the circle, then the angle opposite that diameter is a right angle.

In this case, since $ AB $ is the diameter, the angle opposite to it, which is $ \angle C $, must be a right angle:

$$
\angle C = 90^\circ
$$

---

### Step 2: Use the Triangle Angle Sum Theorem

In any triangle, the sum of the interior angles is $ 180^\circ $. Therefore, in triangle $ \triangle ABC $:

$$
\angle A + \angle B + \angle C = 180^\circ
$$

Substitute the known values:

$$
14^\circ + \angle B + 90^\circ = 180^\circ
$$

Solving for $ \angle B $:

$$
\angle B = 180^\circ - 14^\circ - 90^\circ = 76^\circ
$$

---

### Step 3: Verification via Coordinate Geometry (Optional)

To confirm, we can place the triangle on a coordinate plane with $ A = (-1, 0) $, $ B = (1, 0) $, and $ C $ on the unit circle. Using vector analysis and the dot product, we can compute the angle at $ B $, and it again confirms that $ \angle B = 76^\circ $.

---

### Final Answer

$$
\boxed{76}
$$

Solution:
If $AB$ is a diameter, that means the triangle must have a right angle at $C.$ Therefore, we have that \begin{align*}
\angle B &= 180^\circ - (\angle A + \angle C) \\
&= 180^\circ - (14^\circ + 90^\circ) = \boxed{76^\circ}.
\end{align*}


Explanation:
Okay, so I need to find the probability that a randomly selected number from 1 to 100 is a multiple of 3. Let me think about how to approach this.

First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. In this case, the total number of possible outcomes is 100 because we're selecting a number from 1 through 100 inclusive. That part seems straightforward.

Now, the tricky part is figuring out how many numbers between 1 and 100 are multiples of 3. Let me recall that a multiple of 3 is any number that can be expressed as 3 times some integer. So, numbers like 3, 6, 9, ..., up to the largest multiple of 3 that's less than or equal to 100.

Hmm, how do I find how many such numbers there are? Maybe I can divide 100 by 3 and take the floor of that? Let me check. If I divide 100 by 3, I get approximately 33.333... So, does that mean there are 33 multiples of 3 between 1 and 100? Wait, let me verify that.

Let me list some multiples of 3 and see. The first multiple is 3*1=3, then 3*2=6, 3*3=9, ..., up to 3*33=99. Because 3*33 is 99, which is less than 100. Then the next multiple would be 3*34=102, which is over 100. So, yes, there are 33 multiples of 3 between 1 and 100. Therefore, the number of favorable outcomes is 33.

Wait, but let me make sure I didn't miss anything. Let me think of another way. If I take the number 100 and divide by 3, the quotient is 33 with a remainder of 1. So, that would mean that there are 33 full multiples of 3 in that range. So that seems to confirm it.

Alternatively, maybe I can use the formula for the number of terms in an arithmetic sequence. The multiples of 3 form an arithmetic sequence where the first term a1 is 3, the common difference d is 3, and the last term an is the largest multiple of 3 less than or equal to 100.

So, to find the number of terms, n, in this sequence, I can use the formula:

an = a1 + (n - 1)d

Plugging in the known values:

99 = 3 + (n - 1)*3

Subtract 3 from both sides:

96 = (n - 1)*3

Divide both sides by 3:

32 = n - 1

Add 1 to both sides:

n = 33

So that gives me 33 terms, which matches my previous calculation. Therefore, there are 33 multiples of 3 between 1 and 100.

Therefore, the probability would be the number of favorable outcomes (33) divided by the total number of possible outcomes (100). So, probability = 33/100.

But wait, let me check if there's a possibility that 100 is a multiple of 3. 100 divided by 3 is 33.333..., so no, 100 isn't a multiple of 3. Therefore, the last multiple is indeed 99.

Alternatively, if I wanted to check for numbers from 1 to N inclusive, the number of multiples of k is floor(N/k). In this case, floor(100/3) = 33. So that formula also gives me 33.

Therefore, all methods point to 33 multiples of 3.

But just to be thorough, let me count them in another way. Let's think about how many numbers are divisible by 3. For every 3 numbers, there is one multiple of 3. So, in 1-3: 1 multiple (3); 4-6: 1 multiple (6); 7-9: 1 multiple (9); and so on.

So, how many such groups of 3 are there in 100 numbers? Well, 100 divided by 3 is approximately 33.333. So, there are 33 full groups, each contributing 1 multiple, so 33 multiples. Then, there is a remainder of 1 number (since 33*3=99, so 100-99=1). Therefore, that last group (97-100) only has 1 number, which is 100, which isn't a multiple of 3. Therefore, again, 33 multiples.

Therefore, I think I can be confident that there are 33 multiples of 3 between 1 and 100.

Therefore, the probability is 33/100.

But let me check if 33/100 can be simplified. The greatest common divisor of 33 and 100 is 1, since 33 factors are 3 and 11, and 100 factors are 2 and 5. So, no common factors. Therefore, 33/100 is already in simplest form.

Alternatively, if I wanted to write it as a decimal, it would be 0.33, but the question doesn't specify the form, so fractional form is probably better.

Wait, but let me check once more. Maybe I made a mistake in counting? Let me try to count the number of multiples of 3 between 1 and 100 by another method.

Let me think of the numbers from 1 to 100. The first multiple of 3 is 3, and the last is 99. So, the number of terms is ((last term - first term)/common difference) + 1.

So, ((99 - 3)/3) + 1 = (96/3) + 1 = 32 + 1 = 33. Yep, same result.

Therefore, that formula also gives 33.

Therefore, I think I can be confident that the number of favorable outcomes is 33.

Therefore, the probability is 33/100.

But just to make sure, let me think of a smaller example. Let's say numbers from 1 to 6. How many multiples of 3? 3 and 6, so 2. According to the formula, floor(6/3)=2. Correct. If I use the formula ((last - first)/d) +1, ((6-3)/3)+1= (3/3)+1=1+1=2. Correct.

Another example: numbers from 1 to 7. Multiples of 3: 3,6. So 2. floor(7/3)=2. Correct.

Another example: numbers from 1 to 9. Multiples of 3: 3,6,9. So 3. floor(9/3)=3. Correct.

So, seems like the formula works. Therefore, applying it to 1-100, floor(100/3)=33.

Therefore, probability is 33/100.

Alternatively, if I think of the probability as 1/3, since every third number is a multiple of 3. But since 100 isn't a multiple of 3, the exact probability is slightly less than 1/3.

Indeed, 33/100 is 0.33, which is 1/3 approximately. But since 1/3 is approximately 0.3333..., so 33/100 is 0.33, which is slightly less.

But since the question asks for the probability, and we have an exact value, which is 33/100.

Therefore, I think that is the answer.

But just to make sure, let me check with actual counting. Let me try to count how many multiples of 3 there are between 1 and 100.

Starting from 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99.

Let me count these. Let's see:

Starting from 3:

1. 3
2. 6
3. 9
4. 12
5. 15
6. 18
7. 21
8. 24
9. 27
10. 30
11. 33
12. 36
13. 39
14. 42
15. 45
16. 48
17. 51
18. 54
19. 57
20. 60
21. 63
22. 66
23. 69
24. 72
25. 75
26. 78
27. 81
28. 84
29. 87
30. 90
31. 93
32. 96
33. 99

Yes, that's 33 numbers. So, that's 33 multiples of 3. Therefore, the count is correct.

Therefore, the probability is 33/100.

So, after all these checks and methods, I can be confident that the probability is 33/100.

**Final Answer**
The probability is \boxed{\dfrac{33}{100}}.

To determine the probability that a randomly selected number from 1 through 100 is a multiple of 3, we follow a structured approach based on the definition of probability:

---

### **Step 1: Total Number of Outcomes**

We are selecting a number at random from the set of integers from 1 to 100, inclusive. Therefore, the total number of possible outcomes is:

$$
\text{Total outcomes} = 100
$$

---

### **Step 2: Count the Favorable Outcomes**

We are interested in the numbers that are **multiples of 3** within this range. A number is a multiple of 3 if it can be written in the form $3k$, where $k$ is an integer.

To find how many such numbers exist between 1 and 100, we can use the formula for the number of terms in an arithmetic sequence:

$$
\text{Number of multiples of 3} = \left\lfloor \frac{100}{3} \right\rfloor
$$

$$
\left\lfloor \frac{100}{3} \right\rfloor = \left\lfloor 33.333\ldots \right\rfloor = 33
$$

This means there are **33** multiples of 3 between 1 and 100.

---

### **Step 3: Compute the Probability**

The probability of selecting a multiple of 3 is the ratio of favorable outcomes to total outcomes:

$$
\text{Probability} = \frac{\text{Number of multiples of 3}}{\text{Total numbers}} = \frac{33}{100}
$$

This fraction is already in its simplest form, as 33 and 100 have no common factors other than 1.

---

### **Final Answer**

$$
\boxed{\dfrac{33}{100}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that a randomly selected number from 1 to 100 is a multiple of 3. Let me think about how to approach this.

First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. In this case, the total number of possible outcomes is 100 because we're selecting a number from 1 through 100 inclusive. That part seems straightforward.

Now, the tricky part is figuring out how many numbers between 1 and 100 are multiples of 3. Let me recall that a multiple of 3 is any number that can be expressed as 3 times some integer. So, numbers like 3, 6, 9, ..., up to the largest multiple of 3 that's less than or equal to 100.

Hmm, how do I find how many such numbers there are? Maybe I can divide 100 by 3 and take the floor of that? Let me check. If I divide 100 by 3, I get approximately 33.333... So, does that mean there are 33 multiples of 3 between 1 and 100? Wait, let me verify that.

Let me list some multiples of 3 and see. The first multiple is 3*1=3, then 3*2=6, 3*3=9, ..., up to 3*33=99. Because 3*33 is 99, which is less than 100. Then the next multiple would be 3*34=102, which is over 100. So, yes, there are 33 multiples of 3 between 1 and 100. Therefore, the number of favorable outcomes is 33.

Wait, but let me make sure I didn't miss anything. Let me think of another way. If I take the number 100 and divide by 3, the quotient is 33 with a remainder of 1. So, that would mean that there are 33 full multiples of 3 in that range. So that seems to confirm it.

Alternatively, maybe I can use the formula for the number of terms in an arithmetic sequence. The multiples of 3 form an arithmetic sequence where the first term a1 is 3, the common difference d is 3, and the last term an is the largest multiple of 3 less than or equal to 100.

So, to find the number of terms, n, in this sequence, I can use the formula:

an = a1 + (n - 1)d

Plugging in the known values:

99 = 3 + (n - 1)*3

Subtract 3 from both sides:

96 = (n - 1)*3

Divide both sides by 3:

32 = n - 1

Add 1 to both sides:

n = 33

So that gives me 33 terms, which matches my previous calculation. Therefore, there are 33 multiples of 3 between 1 and 100.

Therefore, the probability would be the number of favorable outcomes (33) divided by the total number of possible outcomes (100). So, probability = 33/100.

But wait, let me check if there's a possibility that 100 is a multiple of 3. 100 divided by 3 is 33.333..., so no, 100 isn't a multiple of 3. Therefore, the last multiple is indeed 99.

Alternatively, if I wanted to check for numbers from 1 to N inclusive, the number of multiples of k is floor(N/k). In this case, floor(100/3) = 33. So that formula also gives me 33.

Therefore, all methods point to 33 multiples of 3.

But just to be thorough, let me count them in another way. Let's think about how many numbers are divisible by 3. For every 3 numbers, there is one multiple of 3. So, in 1-3: 1 multiple (3); 4-6: 1 multiple (6); 7-9: 1 multiple (9); and so on.

So, how many such groups of 3 are there in 100 numbers? Well, 100 divided by 3 is approximately 33.333. So, there are 33 full groups, each contributing 1 multiple, so 33 multiples. Then, there is a remainder of 1 number (since 33*3=99, so 100-99=1). Therefore, that last group (97-100) only has 1 number, which is 100, which isn't a multiple of 3. Therefore, again, 33 multiples.

Therefore, I think I can be confident that there are 33 multiples of 3 between 1 and 100.

Therefore, the probability is 33/100.

But let me check if 33/100 can be simplified. The greatest common divisor of 33 and 100 is 1, since 33 factors are 3 and 11, and 100 factors are 2 and 5. So, no common factors. Therefore, 33/100 is already in simplest form.

Alternatively, if I wanted to write it as a decimal, it would be 0.33, but the question doesn't specify the form, so fractional form is probably better.

Wait, but let me check once more. Maybe I made a mistake in counting? Let me try to count the number of multiples of 3 between 1 and 100 by another method.

Let me think of the numbers from 1 to 100. The first multiple of 3 is 3, and the last is 99. So, the number of terms is ((last term - first term)/common difference) + 1.

So, ((99 - 3)/3) + 1 = (96/3) + 1 = 32 + 1 = 33. Yep, same result.

Therefore, that formula also gives 33.

Therefore, I think I can be confident that the number of favorable outcomes is 33.

Therefore, the probability is 33/100.

But just to make sure, let me think of a smaller example. Let's say numbers from 1 to 6. How many multiples of 3? 3 and 6, so 2. According to the formula, floor(6/3)=2. Correct. If I use the formula ((last - first)/d) +1, ((6-3)/3)+1= (3/3)+1=1+1=2. Correct.

Another example: numbers from 1 to 7. Multiples of 3: 3,6. So 2. floor(7/3)=2. Correct.

Another example: numbers from 1 to 9. Multiples of 3: 3,6,9. So 3. floor(9/3)=3. Correct.

So, seems like the formula works. Therefore, applying it to 1-100, floor(100/3)=33.

Therefore, probability is 33/100.

Alternatively, if I think of the probability as 1/3, since every third number is a multiple of 3. But since 100 isn't a multiple of 3, the exact probability is slightly less than 1/3.

Indeed, 33/100 is 0.33, which is 1/3 approximately. But since 1/3 is approximately 0.3333..., so 33/100 is 0.33, which is slightly less.

But since the question asks for the probability, and we have an exact value, which is 33/100.

Therefore, I think that is the answer.

But just to make sure, let me check with actual counting. Let me try to count how many multiples of 3 there are between 1 and 100.

Starting from 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99.

Let me count these. Let's see:

Starting from 3:

1. 3
2. 6
3. 9
4. 12
5. 15
6. 18
7. 21
8. 24
9. 27
10. 30
11. 33
12. 36
13. 39
14. 42
15. 45
16. 48
17. 51
18. 54
19. 57
20. 60
21. 63
22. 66
23. 69
24. 72
25. 75
26. 78
27. 81
28. 84
29. 87
30. 90
31. 93
32. 96
33. 99

Yes, that's 33 numbers. So, that's 33 multiples of 3. Therefore, the count is correct.

Therefore, the probability is 33/100.

So, after all these checks and methods, I can be confident that the probability is 33/100.

**Final Answer**
The probability is \boxed{\dfrac{33}{100}}.

To determine the probability that a randomly selected number from 1 through 100 is a multiple of 3, we follow a structured approach based on the definition of probability:

---

### **Step 1: Total Number of Outcomes**

We are selecting a number at random from the set of integers from 1 to 100, inclusive. Therefore, the total number of possible outcomes is:

$$
\text{Total outcomes} = 100
$$

---

### **Step 2: Count the Favorable Outcomes**

We are interested in the numbers that are **multiples of 3** within this range. A number is a multiple of 3 if it can be written in the form $3k$, where $k$ is an integer.

To find how many such numbers exist between 1 and 100, we can use the formula for the number of terms in an arithmetic sequence:

$$
\text{Number of multiples of 3} = \left\lfloor \frac{100}{3} \right\rfloor
$$

$$
\left\lfloor \frac{100}{3} \right\rfloor = \left\lfloor 33.333\ldots \right\rfloor = 33
$$

This means there are **33** multiples of 3 between 1 and 100.

---

### **Step 3: Compute the Probability**

The probability of selecting a multiple of 3 is the ratio of favorable outcomes to total outcomes:

$$
\text{Probability} = \frac{\text{Number of multiples of 3}}{\text{Total numbers}} = \frac{33}{100}
$$

This fraction is already in its simplest form, as 33 and 100 have no common factors other than 1.

---

### **Final Answer**

$$
\boxed{\dfrac{33}{100}}
$$

Solution: There are 100 numbers possible between 1 and 100. There are 33 multiples of 3 between 1 and 100: $(3,6,9,\ldots,99)=(1\times 3,2\times 3,3\times 3,\ldots,33\times 3)$.  So the probability that a randomly selected number is a multiple of 3 is $\boxed{\dfrac{33}{100}}$.


Explanation:
Okay, so I need to figure out how many distinct arrangements there are for putting 2 indistinguishable IMO gold medals and 2 indistinguishable IPhO gold medals in a row. Let me think about this step by step.

First, let me make sure I understand the problem correctly. There are four medals in total: two of one type (IMO gold) and two of another type (IPhO gold). The medals of the same type are indistinguishable, meaning that swapping two IMO medals doesn't create a new arrangement. Similarly for the IPhO medals. The question is asking for the number of distinct ways to arrange these four medals in a row.

Hmm, so this seems like a permutations problem with identical items. I remember that when you have items that are not all distinct, the number of distinct arrangements is given by the factorial of the total number of items divided by the product of the factorials of the numbers of each identical item. The formula is something like:

Number of arrangements = n! / (n1! * n2! * ... * nk!)

Where n is the total number of items, and n1, n2, ..., nk are the numbers of each identical item.

In this case, we have 4 medals total. There are two types: IMO and IPhO. Each has 2 medals. So applying the formula, it should be 4! divided by (2! * 2!). Let me check that.

Calculating 4! is 24. Then 2! is 2, so 2! * 2! is 2 * 2 = 4. Then 24 divided by 4 is 6. So does that mean there are 6 distinct arrangements?

Wait, let me verify this by actually listing all possible arrangements to make sure I didn't make a mistake. Sometimes with small numbers, it's easy to list them out and count.

Let me denote the IMO medals as I and the IPhO medals as P. Since they are indistinct, we can't tell the difference between the two I's or the two P's.

So the possible arrangements are all the different ways to arrange two I's and two P's. Let me list them:

1. I, I, P, P
2. I, P, I, P
3. I, P, P, I
4. P, I, I, P
5. P, I, P, I
6. P, P, I, I

Is that all? Let me check if there are more. Let's think of it as permutations of the letters I and P with two of each. The number of unique permutations is indeed 4! / (2!2!) = 6, which matches the list above. So that seems correct.

But wait, let me think again. Is there a possibility that the problem is different? For example, maybe the medals are placed in a row, but there's some other constraint? The problem says "put 2 of his indistinguishable IMO gold medals and 2 of his indistinguishable IPhO gold medals in one row." So it's just arranging four medals, two of each type, in a row. So yeah, the answer should be 6.

But let me think again if there's another way to approach this. Maybe using combinations? For example, choosing positions for the IMO medals and the rest will be IPhO.

So, there are 4 positions. We need to choose 2 positions to place the IMO medals, and the remaining 2 will be IPhO. The number of ways to choose 2 positions out of 4 is C(4,2), which is 6. That's the same answer as before. So that confirms it again.

Alternatively, if I think of it as permutations with identical elements, the formula is the same. So both methods give me 6. Therefore, I think the answer is 6.

But just to be thorough, let me think if there's any chance that the medals are considered different in some way. The problem says they are indistinguishable. So, for example, if the IMO medals are different from each other, but the problem says they are indistinguishable. Similarly for IPhO. So no, there's no difference between the medals of the same type. Therefore, the answer is indeed 6.

Wait, but maybe the problem is in another language? The original problem mentions IMO and IPhO. But the user wrote the problem in English, so I think that's fine. The key is that the medals are indistinct within their types.

Alternatively, maybe there's a trick here? For example, if the medals are placed in a row, but maybe the row has some symmetry? But no, the problem doesn't mention anything about rotations or reflections being considered the same. It just says "in one row," so I think linear arrangements where the order matters, and the only thing that matters is the sequence of medals.

Therefore, I think my reasoning is correct. The answer is 6. Therefore, the number of distinct arrangements is 6.

But just to make sure, let me think of another approach. Suppose I try to count the number of permutations step by step.

First, the first position can be either I or P. Let's consider cases.

Case 1: First position is I.

Then, we have one I and two P's left. The remaining three positions need to be filled with one I and two P's. The number of arrangements for this case is the number of ways to arrange one I and two P's in three positions, which is C(3,1) = 3.

Case 2: First position is P.

Similarly, then we have two I's and one P left. The number of arrangements for this case is the number of ways to arrange two I's and one P in three positions, which is C(3,2) = 3.

Therefore, total number of arrangements is 3 + 3 = 6. Yep, same answer again. So that's consistent.

Alternatively, if I think recursively, but that might be overcomplicating. But just to check, suppose I think of the first position: 2 choices (I or P). Then, depending on that, the next positions have different numbers. But as above, it still leads to 6.

Alternatively, maybe using generating functions? Although that might be more complex. The exponential generating function for each type is (1 + x)^2, but since we have two types, maybe not. Wait, actually, generating functions for permutations with identical items... Maybe not necessary here.

Alternatively, think of the problem as arranging letters. If we have the word "IIP", but with two I's and two P's. The number of distinct permutations is 4! / (2!2!) = 6. Yep, same thing.

Therefore, all methods lead to 6. Therefore, I think the answer is 6. Therefore, the number of distinct arrangements is 6.

But just to make sure, let me list all the possibilities again:

1. II PP
2. I P I P
3. I PP I
4. P II P
5. P I P I
6. PP II

Yes, that's six. Each of these is unique. For example, the first one is two I's followed by two P's. The second is alternating I and P starting with I. The third is I, then two P's, then I. The fourth is P, then two I's, then P. The fifth is alternating starting with P. The sixth is two P's followed by two I's. So that's all six. There's no other arrangement possible. For example, if I try to make something like I P P I, that's already listed as number 3. Similarly, P I I P is number 4. So yeah, all possibilities are covered.

Therefore, I think I can confidently say that the answer is 6. Therefore, the number of distinct arrangements is 6.

**Final Answer**
The number of distinct arrangements is \boxed{6}.

To determine the number of distinct arrangements of 2 indistinguishable IMO gold medals and 2 indistinguishable IPhO gold medals in a row, we approach the problem using the concept of **permutations with identical items**.

---

### Step 1: Understanding the Problem

We are arranging **4 medals** in a row:
- 2 of them are **indistinguishable IMO gold medals** (denoted as `I`)
- 2 of them are **indistinguishable IPhO gold medals** (denoted as `P`)

Since the medals of the same type are **indistinguishable**, swapping two `I`s or two `P`s does **not** result in a new arrangement.

---

### Step 2: Applying the Permutation Formula

The general formula for the number of distinct permutations of a multiset is:

$$
\text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
$$

Where:
- $ n $ is the total number of items (here, $ n = 4 $)
- $ n_1, n_2, \ldots, n_k $ are the counts of each type of indistinguishable item

In this case:
- $ n = 4 $
- $ n_1 = 2 $ (number of `I`s)
- $ n_2 = 2 $ (number of `P`s)

So the number of distinct arrangements is:

$$
\frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = \frac{24}{4} = 6
$$

---

### Step 3: Verifying by Enumeration

To confirm, we can list all the distinct arrangements of the letters `I` and `P` with two of each:

1. `I I P P`
2. `I P I P`
3. `I P P I`
4. `P I I P`
5. `P I P I`
6. `P P I I`

These are all the unique permutations, and there are exactly **6** of them.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many distinct arrangements there are for putting 2 indistinguishable IMO gold medals and 2 indistinguishable IPhO gold medals in a row. Let me think about this step by step.

First, let me make sure I understand the problem correctly. There are four medals in total: two of one type (IMO gold) and two of another type (IPhO gold). The medals of the same type are indistinguishable, meaning that swapping two IMO medals doesn't create a new arrangement. Similarly for the IPhO medals. The question is asking for the number of distinct ways to arrange these four medals in a row.

Hmm, so this seems like a permutations problem with identical items. I remember that when you have items that are not all distinct, the number of distinct arrangements is given by the factorial of the total number of items divided by the product of the factorials of the numbers of each identical item. The formula is something like:

Number of arrangements = n! / (n1! * n2! * ... * nk!)

Where n is the total number of items, and n1, n2, ..., nk are the numbers of each identical item.

In this case, we have 4 medals total. There are two types: IMO and IPhO. Each has 2 medals. So applying the formula, it should be 4! divided by (2! * 2!). Let me check that.

Calculating 4! is 24. Then 2! is 2, so 2! * 2! is 2 * 2 = 4. Then 24 divided by 4 is 6. So does that mean there are 6 distinct arrangements?

Wait, let me verify this by actually listing all possible arrangements to make sure I didn't make a mistake. Sometimes with small numbers, it's easy to list them out and count.

Let me denote the IMO medals as I and the IPhO medals as P. Since they are indistinct, we can't tell the difference between the two I's or the two P's.

So the possible arrangements are all the different ways to arrange two I's and two P's. Let me list them:

1. I, I, P, P
2. I, P, I, P
3. I, P, P, I
4. P, I, I, P
5. P, I, P, I
6. P, P, I, I

Is that all? Let me check if there are more. Let's think of it as permutations of the letters I and P with two of each. The number of unique permutations is indeed 4! / (2!2!) = 6, which matches the list above. So that seems correct.

But wait, let me think again. Is there a possibility that the problem is different? For example, maybe the medals are placed in a row, but there's some other constraint? The problem says "put 2 of his indistinguishable IMO gold medals and 2 of his indistinguishable IPhO gold medals in one row." So it's just arranging four medals, two of each type, in a row. So yeah, the answer should be 6.

But let me think again if there's another way to approach this. Maybe using combinations? For example, choosing positions for the IMO medals and the rest will be IPhO.

So, there are 4 positions. We need to choose 2 positions to place the IMO medals, and the remaining 2 will be IPhO. The number of ways to choose 2 positions out of 4 is C(4,2), which is 6. That's the same answer as before. So that confirms it again.

Alternatively, if I think of it as permutations with identical elements, the formula is the same. So both methods give me 6. Therefore, I think the answer is 6.

But just to be thorough, let me think if there's any chance that the medals are considered different in some way. The problem says they are indistinguishable. So, for example, if the IMO medals are different from each other, but the problem says they are indistinguishable. Similarly for IPhO. So no, there's no difference between the medals of the same type. Therefore, the answer is indeed 6.

Wait, but maybe the problem is in another language? The original problem mentions IMO and IPhO. But the user wrote the problem in English, so I think that's fine. The key is that the medals are indistinct within their types.

Alternatively, maybe there's a trick here? For example, if the medals are placed in a row, but maybe the row has some symmetry? But no, the problem doesn't mention anything about rotations or reflections being considered the same. It just says "in one row," so I think linear arrangements where the order matters, and the only thing that matters is the sequence of medals.

Therefore, I think my reasoning is correct. The answer is 6. Therefore, the number of distinct arrangements is 6.

But just to make sure, let me think of another approach. Suppose I try to count the number of permutations step by step.

First, the first position can be either I or P. Let's consider cases.

Case 1: First position is I.

Then, we have one I and two P's left. The remaining three positions need to be filled with one I and two P's. The number of arrangements for this case is the number of ways to arrange one I and two P's in three positions, which is C(3,1) = 3.

Case 2: First position is P.

Similarly, then we have two I's and one P left. The number of arrangements for this case is the number of ways to arrange two I's and one P in three positions, which is C(3,2) = 3.

Therefore, total number of arrangements is 3 + 3 = 6. Yep, same answer again. So that's consistent.

Alternatively, if I think recursively, but that might be overcomplicating. But just to check, suppose I think of the first position: 2 choices (I or P). Then, depending on that, the next positions have different numbers. But as above, it still leads to 6.

Alternatively, maybe using generating functions? Although that might be more complex. The exponential generating function for each type is (1 + x)^2, but since we have two types, maybe not. Wait, actually, generating functions for permutations with identical items... Maybe not necessary here.

Alternatively, think of the problem as arranging letters. If we have the word "IIP", but with two I's and two P's. The number of distinct permutations is 4! / (2!2!) = 6. Yep, same thing.

Therefore, all methods lead to 6. Therefore, I think the answer is 6. Therefore, the number of distinct arrangements is 6.

But just to make sure, let me list all the possibilities again:

1. II PP
2. I P I P
3. I PP I
4. P II P
5. P I P I
6. PP II

Yes, that's six. Each of these is unique. For example, the first one is two I's followed by two P's. The second is alternating I and P starting with I. The third is I, then two P's, then I. The fourth is P, then two I's, then P. The fifth is alternating starting with P. The sixth is two P's followed by two I's. So that's all six. There's no other arrangement possible. For example, if I try to make something like I P P I, that's already listed as number 3. Similarly, P I I P is number 4. So yeah, all possibilities are covered.

Therefore, I think I can confidently say that the answer is 6. Therefore, the number of distinct arrangements is 6.

**Final Answer**
The number of distinct arrangements is \boxed{6}.

To determine the number of distinct arrangements of 2 indistinguishable IMO gold medals and 2 indistinguishable IPhO gold medals in a row, we approach the problem using the concept of **permutations with identical items**.

---

### Step 1: Understanding the Problem

We are arranging **4 medals** in a row:
- 2 of them are **indistinguishable IMO gold medals** (denoted as `I`)
- 2 of them are **indistinguishable IPhO gold medals** (denoted as `P`)

Since the medals of the same type are **indistinguishable**, swapping two `I`s or two `P`s does **not** result in a new arrangement.

---

### Step 2: Applying the Permutation Formula

The general formula for the number of distinct permutations of a multiset is:

$$
\text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
$$

Where:
- $ n $ is the total number of items (here, $ n = 4 $)
- $ n_1, n_2, \ldots, n_k $ are the counts of each type of indistinguishable item

In this case:
- $ n = 4 $
- $ n_1 = 2 $ (number of `I`s)
- $ n_2 = 2 $ (number of `P`s)

So the number of distinct arrangements is:

$$
\frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = \frac{24}{4} = 6
$$

---

### Step 3: Verifying by Enumeration

To confirm, we can list all the distinct arrangements of the letters `I` and `P` with two of each:

1. `I I P P`
2. `I P I P`
3. `I P P I`
4. `P I I P`
5. `P I P I`
6. `P P I I`

These are all the unique permutations, and there are exactly **6** of them.

---

### Final Answer

$$
\boxed{6}
$$

Solution: There are 4 slots in the row.  He can choose two of them for his IMO medals in $\binom{4}{2} = \boxed{6}$ ways.


Explanation:
Okay, so I need to figure out what the largest eight-digit number is in base 2, and then convert that to base 10. Let me start by recalling what an eight-digit number in base 2 would look like.

In any base system, the number of digits determines the range of values. For example, in base 10, an eight-digit number ranges from 10,000,000 to 99,999,999. Similarly, in base 2, an eight-digit number would start with 1 followed by seven more digits. But since it's base 2, each digit can only be 0 or 1.

So, the largest eight-digit base 2 number would be the one where all the digits are 1s. That makes sense because each digit represents a power of 2, and having a 1 in the highest place value gives the maximum value. Let me confirm that.

An eight-digit binary number has positions from the 7th power of 2 down to the 0th power. Wait, let me think. The rightmost digit is 2^0, then moving left, it's 2^1, 2^2, up to 2^7 for the leftmost digit. So, an eight-digit binary number would be from 2^7 down to 2^0.

Therefore, the largest eight-digit binary number would be 11111111 in binary. Let me check that. If I have eight 1s, that's the maximum possible value for an eight-digit binary number.

Now, I need to convert this binary number to decimal (base 10). How do I do that? Well, each digit represents a power of 2. Starting from the right, the first digit is 2^0, then 2^1, up to 2^7 for the leftmost digit.

So, let's write out the binary number 11111111 and convert each digit to its decimal equivalent.

Breaking it down:

From left to right, the digits are:

1 (2^7), 1 (2^6), 1 (2^5), 1 (2^4), 1 (2^3), 1 (2^2), 1 (2^1), 1 (2^0).

So, the decimal value would be the sum of all these powers of 2.

Calculating each term:

2^7 = 128

2^6 = 64

2^5 = 32

2^4 = 16

2^3 = 8

2^2 = 4

2^1 = 2

2^0 = 1

Adding them all up: 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1.

Let me add them step by step:

Start with 128 + 64 = 192

192 + 32 = 224

224 + 16 = 240

240 + 8 = 248

248 + 4 = 252

252 + 2 = 254

254 + 1 = 255

So, the total is 255.

Wait, so the largest eight-digit binary number is 255 in decimal? Let me check if that's correct.

Alternatively, I remember that 2^n - 1 is the value of a binary number with n ones. For example, 2^8 - 1 = 256 - 1 = 255. But wait, hold on. If we have eight digits, does that correspond to n=8? Let me think.

The number of digits in binary is related to the exponent. For example, the number 10000000 in binary is 2^7, which is 128 in decimal. So, an eight-digit binary number ranges from 2^7 (which is 10000000) up to 2^8 - 1 (which is 11111111). Therefore, 2^8 - 1 is indeed 256 - 1 = 255.

Therefore, the largest eight-digit binary number is 255 in decimal.

But let me verify once more. Let's take a smaller example. Let's say the largest two-digit binary number is 11, which is 3 in decimal. According to the formula, 2^2 - 1 = 3. Correct. Similarly, the largest three-digit binary number is 111, which is 7, and 2^3 - 1 = 7. So, yes, the pattern holds.

Therefore, for eight digits, it's 2^8 - 1 = 255.

But wait, hold on. Wait, 2^8 is 256. So, 2^8 - 1 is 255. But the number of digits in binary for 255 is... Let me check. How many digits does 255 have in binary?

Well, 2^8 is 256, so 255 is one less than that, which would be 11111111 in binary, which is eight 1s. Therefore, eight digits. So that's correct.

Therefore, the answer is 255.

But let me think again. Maybe I confused the number of digits. For example, in base 10, the number of digits is determined by the highest power. For example, 10^7 is 10,000,000 which is an eight-digit number. Similarly, in base 2, the number 2^7 is 10000000, which is an eight-digit binary number. Therefore, the maximum eight-digit binary number is 2^8 - 1, which is 255.

Alternatively, if I think of the number of digits, the first digit is 2^(n-1) where n is the number of digits. So, for eight digits, the first digit is 2^7, and the maximum number is 2^7 + 2^6 + ... + 2^0 = 2^8 - 1.

Yes, that's the formula for a geometric series. The sum of a geometric series from k=0 to n-1 of 2^k is 2^n - 1.

Therefore, for eight digits, n=8, so sum is 2^8 - 1 = 255.

Therefore, the answer is 255.

But just to make sure, let me convert 11111111 binary to decimal step by step again.

Starting from the right:

Positions (from right, starting at 0):

Position 7 6 5 4 3 2 1 0

Digits:        1 1 1 1 1 1 1 1

So, each digit is 1, so the value is:

2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0

Which is 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1.

Adding them up:

128 + 64 = 192

192 + 32 = 224

224 + 16 = 240

240 + 8 = 248

248 + 4 = 252

252 + 2 = 254

254 + 1 = 255.

Yep, same result.

Therefore, I can be confident that the largest eight-digit base 2 integer is 255 in base 10.

But just to make sure there isn't a misunderstanding about what an eight-digit number in base 2 is. For example, does leading zeros count? But no, in any numbering system, leading zeros are not counted as digits. So, an eight-digit number in base 2 must start with a 1, followed by seven more digits. Therefore, the maximum is indeed 11111111.

Alternatively, if someone thought that eight digits could include leading zeros, then the maximum would still be 11111111, because leading zeros don't add value. For example, 01111111 is just 127, which is less than 255.

Therefore, the answer is definitely 255.

Another way to think about it: the number of unique numbers that can be represented with n digits in base b is b^n. But here, we are talking about the maximum value. For binary, the maximum value with n digits is 2^n - 1.

So, for n=8, it's 2^8 - 1 = 255.

Therefore, all methods point to 255.

I think that's solid. I can't see any mistakes in my reasoning.

**Final Answer**
The largest eight-digit base 2 integer expressed in base 10 is \boxed{255}.

To determine the largest eight-digit base 2 (binary) integer and express it in base 10, we begin by understanding the structure of binary numbers.

---

### Understanding Binary Digits

In base 2, each digit represents a power of 2, starting from $2^0$ on the right. An eight-digit binary number has positions corresponding to powers from $2^0$ to $2^7$. The leftmost digit is the most significant, and the rightmost is the least significant.

The largest possible eight-digit binary number is the one where **all digits are 1**, i.e., `11111111`.

---

### Converting Binary to Decimal

To convert `11111111` from binary to decimal, we sum the values of each bit:

$$
1 \cdot 2^7 + 1 \cdot 2^6 + 1 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0
$$

$$
= 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
$$

---

### General Formula

This is a special case of a geometric series. The sum of $n$ ones in binary is:

$$
\sum_{k=0}^{n-1} 2^k = 2^n - 1
$$

For $n = 8$, this gives:

$$
2^8 - 1 = 256 - 1 = 255
$$

---

### Final Answer

Thus, the largest eight-digit base 2 integer is:

$$
\boxed{255}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what the largest eight-digit number is in base 2, and then convert that to base 10. Let me start by recalling what an eight-digit number in base 2 would look like.

In any base system, the number of digits determines the range of values. For example, in base 10, an eight-digit number ranges from 10,000,000 to 99,999,999. Similarly, in base 2, an eight-digit number would start with 1 followed by seven more digits. But since it's base 2, each digit can only be 0 or 1.

So, the largest eight-digit base 2 number would be the one where all the digits are 1s. That makes sense because each digit represents a power of 2, and having a 1 in the highest place value gives the maximum value. Let me confirm that.

An eight-digit binary number has positions from the 7th power of 2 down to the 0th power. Wait, let me think. The rightmost digit is 2^0, then moving left, it's 2^1, 2^2, up to 2^7 for the leftmost digit. So, an eight-digit binary number would be from 2^7 down to 2^0.

Therefore, the largest eight-digit binary number would be 11111111 in binary. Let me check that. If I have eight 1s, that's the maximum possible value for an eight-digit binary number.

Now, I need to convert this binary number to decimal (base 10). How do I do that? Well, each digit represents a power of 2. Starting from the right, the first digit is 2^0, then 2^1, up to 2^7 for the leftmost digit.

So, let's write out the binary number 11111111 and convert each digit to its decimal equivalent.

Breaking it down:

From left to right, the digits are:

1 (2^7), 1 (2^6), 1 (2^5), 1 (2^4), 1 (2^3), 1 (2^2), 1 (2^1), 1 (2^0).

So, the decimal value would be the sum of all these powers of 2.

Calculating each term:

2^7 = 128

2^6 = 64

2^5 = 32

2^4 = 16

2^3 = 8

2^2 = 4

2^1 = 2

2^0 = 1

Adding them all up: 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1.

Let me add them step by step:

Start with 128 + 64 = 192

192 + 32 = 224

224 + 16 = 240

240 + 8 = 248

248 + 4 = 252

252 + 2 = 254

254 + 1 = 255

So, the total is 255.

Wait, so the largest eight-digit binary number is 255 in decimal? Let me check if that's correct.

Alternatively, I remember that 2^n - 1 is the value of a binary number with n ones. For example, 2^8 - 1 = 256 - 1 = 255. But wait, hold on. If we have eight digits, does that correspond to n=8? Let me think.

The number of digits in binary is related to the exponent. For example, the number 10000000 in binary is 2^7, which is 128 in decimal. So, an eight-digit binary number ranges from 2^7 (which is 10000000) up to 2^8 - 1 (which is 11111111). Therefore, 2^8 - 1 is indeed 256 - 1 = 255.

Therefore, the largest eight-digit binary number is 255 in decimal.

But let me verify once more. Let's take a smaller example. Let's say the largest two-digit binary number is 11, which is 3 in decimal. According to the formula, 2^2 - 1 = 3. Correct. Similarly, the largest three-digit binary number is 111, which is 7, and 2^3 - 1 = 7. So, yes, the pattern holds.

Therefore, for eight digits, it's 2^8 - 1 = 255.

But wait, hold on. Wait, 2^8 is 256. So, 2^8 - 1 is 255. But the number of digits in binary for 255 is... Let me check. How many digits does 255 have in binary?

Well, 2^8 is 256, so 255 is one less than that, which would be 11111111 in binary, which is eight 1s. Therefore, eight digits. So that's correct.

Therefore, the answer is 255.

But let me think again. Maybe I confused the number of digits. For example, in base 10, the number of digits is determined by the highest power. For example, 10^7 is 10,000,000 which is an eight-digit number. Similarly, in base 2, the number 2^7 is 10000000, which is an eight-digit binary number. Therefore, the maximum eight-digit binary number is 2^8 - 1, which is 255.

Alternatively, if I think of the number of digits, the first digit is 2^(n-1) where n is the number of digits. So, for eight digits, the first digit is 2^7, and the maximum number is 2^7 + 2^6 + ... + 2^0 = 2^8 - 1.

Yes, that's the formula for a geometric series. The sum of a geometric series from k=0 to n-1 of 2^k is 2^n - 1.

Therefore, for eight digits, n=8, so sum is 2^8 - 1 = 255.

Therefore, the answer is 255.

But just to make sure, let me convert 11111111 binary to decimal step by step again.

Starting from the right:

Positions (from right, starting at 0):

Position 7 6 5 4 3 2 1 0

Digits:        1 1 1 1 1 1 1 1

So, each digit is 1, so the value is:

2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0

Which is 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1.

Adding them up:

128 + 64 = 192

192 + 32 = 224

224 + 16 = 240

240 + 8 = 248

248 + 4 = 252

252 + 2 = 254

254 + 1 = 255.

Yep, same result.

Therefore, I can be confident that the largest eight-digit base 2 integer is 255 in base 10.

But just to make sure there isn't a misunderstanding about what an eight-digit number in base 2 is. For example, does leading zeros count? But no, in any numbering system, leading zeros are not counted as digits. So, an eight-digit number in base 2 must start with a 1, followed by seven more digits. Therefore, the maximum is indeed 11111111.

Alternatively, if someone thought that eight digits could include leading zeros, then the maximum would still be 11111111, because leading zeros don't add value. For example, 01111111 is just 127, which is less than 255.

Therefore, the answer is definitely 255.

Another way to think about it: the number of unique numbers that can be represented with n digits in base b is b^n. But here, we are talking about the maximum value. For binary, the maximum value with n digits is 2^n - 1.

So, for n=8, it's 2^8 - 1 = 255.

Therefore, all methods point to 255.

I think that's solid. I can't see any mistakes in my reasoning.

**Final Answer**
The largest eight-digit base 2 integer expressed in base 10 is \boxed{255}.

To determine the largest eight-digit base 2 (binary) integer and express it in base 10, we begin by understanding the structure of binary numbers.

---

### Understanding Binary Digits

In base 2, each digit represents a power of 2, starting from $2^0$ on the right. An eight-digit binary number has positions corresponding to powers from $2^0$ to $2^7$. The leftmost digit is the most significant, and the rightmost is the least significant.

The largest possible eight-digit binary number is the one where **all digits are 1**, i.e., `11111111`.

---

### Converting Binary to Decimal

To convert `11111111` from binary to decimal, we sum the values of each bit:

$$
1 \cdot 2^7 + 1 \cdot 2^6 + 1 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0
$$

$$
= 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
$$

---

### General Formula

This is a special case of a geometric series. The sum of $n$ ones in binary is:

$$
\sum_{k=0}^{n-1} 2^k = 2^n - 1
$$

For $n = 8$, this gives:

$$
2^8 - 1 = 256 - 1 = 255
$$

---

### Final Answer

Thus, the largest eight-digit base 2 integer is:

$$
\boxed{255}
$$

Solution: The largest eight-digit base 2 integer is 1 less than the smallest nine-digit base 2 integer, which is $$ 100000000_{2} = 1 \cdot 2^8 = 256. $$Thus, the largest eight-digit base 2 integer is $256 - 1 = \boxed{255}$.