Explanation:
Okay, so I need to find three consecutive integers whose product is 120, and when you divide that product by the mean of the three integers, you get 24. Then, I have to find the largest of those three integers. Let me try to break this down step by step.

First, let me recall that consecutive integers are numbers that follow each other in order without gaps. For example, 3, 4, 5 or -2, -1, 0. Since the problem mentions three consecutive integers, I can represent them algebraically. Let me think... If I let the middle integer be 'n', then the three consecutive integers would be n-1, n, and n+1. That seems like a good approach because it centers the numbers around 'n', which might make calculations easier.

So, the three integers are n-1, n, n+1. Their product is given as 120. So, I can write the equation:

(n - 1) * n * (n + 1) = 120

Hmm, that's the product. Let me compute that. First, maybe I can simplify the left side. Multiplying (n - 1)(n + 1) first, which is a difference of squares. That would be n² - 1. Then multiply by n:

n(n² - 1) = n³ - n

So, the equation becomes:

n³ - n = 120

Therefore:

n³ - n - 120 = 0

Okay, so I need to solve this cubic equation. Let me see if I can find an integer solution here. Since we're dealing with integers, maybe there's a whole number that satisfies this equation. Let me try plugging in some integers for n.

Let me start with n=5: 5³ - 5 - 120 = 125 - 5 - 120 = 0. Oh! That works! So n=5 is a solution. Let me check n=4: 64 - 4 - 120 = -60, which is not zero. n=6: 216 - 6 - 120 = 90, not zero. So n=5 is the solution here. Therefore, the three consecutive integers are 4, 5, 6. Let me check their product: 4*5*6 = 120. Yes, that's correct. So that seems to satisfy the first condition.

But wait, the problem also says that when you divide the product by the mean of the three integers, you get 24. Let me check if that's true here. The mean of the three integers is (4 + 5 + 6)/3 = 15/3 = 5. Then, the product is 120, so 120 divided by 5 is 24. Yes! That matches the second condition. So, the three integers are indeed 4, 5, 6. Therefore, the largest of them is 6. So, the answer is 6?

Wait, but let me make sure there are no other possible solutions. Because sometimes cubic equations can have more than one real root. Let me check if there are other integer solutions. For example, negative numbers. Let me try n=-5: (-5)^3 - (-5) - 120 = -125 +5 -120 = -240, which is not zero. How about n= -4: (-64) - (-4) -120 = -64 +4 -120 = -180. Not zero. n= -3: -27 - (-3) -120 = -27 +3 -120 = -144. Still not. n= -2: -8 - (-2) -120 = -8 +2 -120 = -126. n=-1: -1 - (-1) -120 = -1 +1 -120 = -120. n=0: 0 -0 -120 = -120. n=1: 1 -1 -120 = -120. n=2: 8 -2 -120 = -114. n=3: 27 -3 -120 = -96. n=4: 64 -4 -120 = -60. n=5: 125 -5 -120 = 0. n=6: 216 -6 -120 = 90. So, only n=5 gives zero. Therefore, the only integer solution is n=5, so the three integers are 4,5,6. Therefore, the largest is 6.

But let me think again. The problem says "three consecutive integers," so they could be negative as well. But in that case, their product would be negative if one of them is negative. Wait, the product is 120, which is positive. So, either all three integers are positive or all three are negative. If all three are negative, then their product would be negative because multiplying three negative numbers: (-a)*(-b)*(-c) = -abc. So, if the product is positive 120, then they must all be positive. Therefore, there are no negative solutions. Therefore, the only possible solution is 4,5,6.

But let me check if there's another set of three consecutive integers with a different middle number. Wait, the equation n³ - n - 120 = 0 only has one real root? Wait, cubic equations have at least one real root, but they can have three. However, since we found n=5 is a root, we can factor the cubic equation. Let me try to factor it.

If n=5 is a root, then (n - 5) is a factor. Let me perform polynomial division or use synthetic division. Let me do synthetic division with root 5:

Coefficients of the polynomial n³ - n - 120. Let me write it as n³ + 0n² - n - 120.

So, coefficients: 1 (for n³), 0 (for n²), -1 (for n), -120 (constant term).

Using synthetic division with 5:

Bring down the 1.

Multiply 1 by 5: 5. Add to next coefficient: 0 + 5 = 5.

Multiply 5 by 5: 25. Add to next coefficient: -1 + 25 = 24.

Multiply 24 by 5: 120. Add to next coefficient: -120 + 120 = 0. Perfect, no remainder.

So, the polynomial factors as (n - 5)(n² + 5n + 24).

So, the equation is (n - 5)(n² + 5n + 24) = 0.

So, the other roots are solutions to n² + 5n + 24 = 0. Let me compute the discriminant: D = 25 - 96 = -71. Since discriminant is negative, there are no real roots here. Therefore, the only real solution is n=5. Therefore, the only set of three consecutive integers is 4,5,6.

Therefore, the largest is 6.

But let me check again the second condition. The problem says "the product divided by the mean of the three integers is 24." So, product is 120, mean is (4+5+6)/3 = 15/3 = 5. 120 divided by 5 is 24. Correct. So that's exactly the condition given. Therefore, all conditions are satisfied.

Therefore, the answer is 6.

But just to make sure, let me think if there's another way to approach this problem. Maybe by letting the three consecutive integers be x, x+1, x+2. Let me try that approach as well.

So, product is x(x+1)(x+2) = 120. Then, the mean is (x + (x+1) + (x+2))/3 = (3x + 3)/3 = x + 1. Therefore, the product divided by the mean is [x(x+1)(x+2)] / (x + 1) = x(x+2) = 24.

Wait, that's interesting. So, if I do that, then the product divided by the mean is x(x+2) = 24.

So, we have two equations:

1. x(x+1)(x+2) = 120

2. x(x+2) = 24

So, maybe we can use equation 2 to find x. Let me solve equation 2 first.

x(x + 2) = 24

Which is x² + 2x - 24 = 0

Solving this quadratic equation:

x = [-2 ± sqrt(4 + 96)] / 2 = [-2 ± sqrt(100)] / 2 = [-2 ± 10]/2

So, two solutions:

(-2 + 10)/2 = 8/2 = 4

(-2 -10)/2 = -12/2 = -6

So, x = 4 or x = -6

Therefore, if x = 4, then the three integers are 4, 5, 6. If x = -6, then the three integers are -6, -5, -4. Let me check if these satisfy the first equation.

First, for x=4: product is 4*5*6 = 120. Correct. For x=-6: (-6)*(-5)*(-4) = (-6)*(-5) is 30, times (-4) is -120. Which is not 120. Therefore, x=-6 is not a valid solution. Therefore, only x=4 is valid. Therefore, the three integers are 4,5,6. Therefore, the largest is 6.

So, this approach also leads to the same conclusion. Therefore, the answer is 6.

But why did the second approach give us x=-6? Because when we divided the product by the mean, we canceled out the (x+1) term. However, if x=-6, then the mean is x + 1 = -5. Then, the product divided by the mean would be (-6)*(-5)*(-4)/(-5) = (-120)/(-5) = 24. Wait, that's also 24. But the product is -120, not 120. However, the problem says the product is 120. Therefore, even though dividing by the mean gives 24, the product is -120, which does not satisfy the first condition. Therefore, x=-6 is invalid.

Therefore, only x=4 is valid. Therefore, the three integers are 4,5,6.

So, both methods lead to the same conclusion. Therefore, the answer is 6.

But let me check once again. If the three integers were -6, -5, -4, then their product is -120, which is not 120. Therefore, that's invalid. Therefore, only 4,5,6 works.

Alternatively, if I didn't think of the second approach, but just solved the first equation, I still get the same answer.

Therefore, I think the answer is definitely 6.

But just to make sure, let me check if there's any other possible way. For example, maybe the three consecutive integers are not centered around n? Wait, but consecutive integers are always like x, x+1, x+2. So, whether you take them as n-1, n, n+1 or x, x+1, x+2, it's the same thing.

Alternatively, maybe someone might think of the three numbers as a, a+1, a+2. So, same as above.

Alternatively, someone might think of them as n, n+1, n+2. So, same thing.

Therefore, regardless of how you represent them, you end up with similar equations.

Therefore, I think the answer is 6.

But let me check once again if there's a possibility of another set of numbers. For example, maybe non-integer numbers? But the problem says "three consecutive integers," so they have to be integers. Therefore, no.

Therefore, I think I can be confident that the answer is 6.

**Final Answer**
The largest of the three consecutive integers is \boxed{6}.

To solve the problem, we are given two key pieces of information:

1. The **product** of three consecutive integers is **120**.
2. When this product is **divided by the mean** of the three integers, the result is **24**.

---

### Step 1: Represent the Integers Algebraically

Let the three consecutive integers be:

$$
x, \quad x+1, \quad x+2
$$

Their **product** is:

$$
x(x+1)(x+2) = 120
$$

Their **mean** is:

$$
\frac{x + (x+1) + (x+2)}{3} = \frac{3x + 3}{3} = x + 1
$$

So, the second condition becomes:

$$
\frac{x(x+1)(x+2)}{x+1} = 24
$$

Simplifying the left-hand side:

$$
\frac{x(x+1)(x+2)}{x+1} = x(x+2)
$$

So we now have the equation:

$$
x(x+2) = 24
$$

---

### Step 2: Solve the Equation

We solve the quadratic:

$$
x(x+2) = 24 \Rightarrow x^2 + 2x - 24 = 0
$$

Use the quadratic formula:

$$
x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2}
$$

$$
x = \frac{-2 \pm 10}{2}
$$

So the two possible values for $ x $ are:

$$
x = \frac{8}{2} = 4 \quad \text{or} \quad x = \frac{-12}{2} = -6
$$

---

### Step 3: Check Which Value Satisfies the Original Product Condition

- If $ x = 4 $, the integers are $ 4, 5, 6 $, and their product is:

$$
4 \times 5 \times 6 = 120
$$

- If $ x = -6 $, the integers are $ -6, -5, -4 $, and their product is:

$$
(-6) \times (-5) \times (-4) = -120
$$

This does **not** satisfy the first condition (product is 120), so we discard $ x = -6 $.

---

### Step 4: Confirm the Second Condition

For $ x = 4 $, the integers are $ 4, 5, 6 $, and the mean is:

$$
\frac{4 + 5 + 6}{3} = \frac{15}{3} = 5
$$

Then:

$$
\frac{120}{5} = 24
$$

This matches the second condition.

---

### Final Answer

The three consecutive integers are $ 4, 5, 6 $, and the **largest** of them is:

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find three consecutive integers whose product is 120, and when you divide that product by the mean of the three integers, you get 24. Then, I have to find the largest of those three integers. Let me try to break this down step by step.

First, let me recall that consecutive integers are numbers that follow each other in order without gaps. For example, 3, 4, 5 or -2, -1, 0. Since the problem mentions three consecutive integers, I can represent them algebraically. Let me think... If I let the middle integer be 'n', then the three consecutive integers would be n-1, n, and n+1. That seems like a good approach because it centers the numbers around 'n', which might make calculations easier.

So, the three integers are n-1, n, n+1. Their product is given as 120. So, I can write the equation:

(n - 1) * n * (n + 1) = 120

Hmm, that's the product. Let me compute that. First, maybe I can simplify the left side. Multiplying (n - 1)(n + 1) first, which is a difference of squares. That would be n² - 1. Then multiply by n:

n(n² - 1) = n³ - n

So, the equation becomes:

n³ - n = 120

Therefore:

n³ - n - 120 = 0

Okay, so I need to solve this cubic equation. Let me see if I can find an integer solution here. Since we're dealing with integers, maybe there's a whole number that satisfies this equation. Let me try plugging in some integers for n.

Let me start with n=5: 5³ - 5 - 120 = 125 - 5 - 120 = 0. Oh! That works! So n=5 is a solution. Let me check n=4: 64 - 4 - 120 = -60, which is not zero. n=6: 216 - 6 - 120 = 90, not zero. So n=5 is the solution here. Therefore, the three consecutive integers are 4, 5, 6. Let me check their product: 4*5*6 = 120. Yes, that's correct. So that seems to satisfy the first condition.

But wait, the problem also says that when you divide the product by the mean of the three integers, you get 24. Let me check if that's true here. The mean of the three integers is (4 + 5 + 6)/3 = 15/3 = 5. Then, the product is 120, so 120 divided by 5 is 24. Yes! That matches the second condition. So, the three integers are indeed 4, 5, 6. Therefore, the largest of them is 6. So, the answer is 6?

Wait, but let me make sure there are no other possible solutions. Because sometimes cubic equations can have more than one real root. Let me check if there are other integer solutions. For example, negative numbers. Let me try n=-5: (-5)^3 - (-5) - 120 = -125 +5 -120 = -240, which is not zero. How about n= -4: (-64) - (-4) -120 = -64 +4 -120 = -180. Not zero. n= -3: -27 - (-3) -120 = -27 +3 -120 = -144. Still not. n= -2: -8 - (-2) -120 = -8 +2 -120 = -126. n=-1: -1 - (-1) -120 = -1 +1 -120 = -120. n=0: 0 -0 -120 = -120. n=1: 1 -1 -120 = -120. n=2: 8 -2 -120 = -114. n=3: 27 -3 -120 = -96. n=4: 64 -4 -120 = -60. n=5: 125 -5 -120 = 0. n=6: 216 -6 -120 = 90. So, only n=5 gives zero. Therefore, the only integer solution is n=5, so the three integers are 4,5,6. Therefore, the largest is 6.

But let me think again. The problem says "three consecutive integers," so they could be negative as well. But in that case, their product would be negative if one of them is negative. Wait, the product is 120, which is positive. So, either all three integers are positive or all three are negative. If all three are negative, then their product would be negative because multiplying three negative numbers: (-a)*(-b)*(-c) = -abc. So, if the product is positive 120, then they must all be positive. Therefore, there are no negative solutions. Therefore, the only possible solution is 4,5,6.

But let me check if there's another set of three consecutive integers with a different middle number. Wait, the equation n³ - n - 120 = 0 only has one real root? Wait, cubic equations have at least one real root, but they can have three. However, since we found n=5 is a root, we can factor the cubic equation. Let me try to factor it.

If n=5 is a root, then (n - 5) is a factor. Let me perform polynomial division or use synthetic division. Let me do synthetic division with root 5:

Coefficients of the polynomial n³ - n - 120. Let me write it as n³ + 0n² - n - 120.

So, coefficients: 1 (for n³), 0 (for n²), -1 (for n), -120 (constant term).

Using synthetic division with 5:

Bring down the 1.

Multiply 1 by 5: 5. Add to next coefficient: 0 + 5 = 5.

Multiply 5 by 5: 25. Add to next coefficient: -1 + 25 = 24.

Multiply 24 by 5: 120. Add to next coefficient: -120 + 120 = 0. Perfect, no remainder.

So, the polynomial factors as (n - 5)(n² + 5n + 24).

So, the equation is (n - 5)(n² + 5n + 24) = 0.

So, the other roots are solutions to n² + 5n + 24 = 0. Let me compute the discriminant: D = 25 - 96 = -71. Since discriminant is negative, there are no real roots here. Therefore, the only real solution is n=5. Therefore, the only set of three consecutive integers is 4,5,6.

Therefore, the largest is 6.

But let me check again the second condition. The problem says "the product divided by the mean of the three integers is 24." So, product is 120, mean is (4+5+6)/3 = 15/3 = 5. 120 divided by 5 is 24. Correct. So that's exactly the condition given. Therefore, all conditions are satisfied.

Therefore, the answer is 6.

But just to make sure, let me think if there's another way to approach this problem. Maybe by letting the three consecutive integers be x, x+1, x+2. Let me try that approach as well.

So, product is x(x+1)(x+2) = 120. Then, the mean is (x + (x+1) + (x+2))/3 = (3x + 3)/3 = x + 1. Therefore, the product divided by the mean is [x(x+1)(x+2)] / (x + 1) = x(x+2) = 24.

Wait, that's interesting. So, if I do that, then the product divided by the mean is x(x+2) = 24.

So, we have two equations:

1. x(x+1)(x+2) = 120

2. x(x+2) = 24

So, maybe we can use equation 2 to find x. Let me solve equation 2 first.

x(x + 2) = 24

Which is x² + 2x - 24 = 0

Solving this quadratic equation:

x = [-2 ± sqrt(4 + 96)] / 2 = [-2 ± sqrt(100)] / 2 = [-2 ± 10]/2

So, two solutions:

(-2 + 10)/2 = 8/2 = 4

(-2 -10)/2 = -12/2 = -6

So, x = 4 or x = -6

Therefore, if x = 4, then the three integers are 4, 5, 6. If x = -6, then the three integers are -6, -5, -4. Let me check if these satisfy the first equation.

First, for x=4: product is 4*5*6 = 120. Correct. For x=-6: (-6)*(-5)*(-4) = (-6)*(-5) is 30, times (-4) is -120. Which is not 120. Therefore, x=-6 is not a valid solution. Therefore, only x=4 is valid. Therefore, the three integers are 4,5,6. Therefore, the largest is 6.

So, this approach also leads to the same conclusion. Therefore, the answer is 6.

But why did the second approach give us x=-6? Because when we divided the product by the mean, we canceled out the (x+1) term. However, if x=-6, then the mean is x + 1 = -5. Then, the product divided by the mean would be (-6)*(-5)*(-4)/(-5) = (-120)/(-5) = 24. Wait, that's also 24. But the product is -120, not 120. However, the problem says the product is 120. Therefore, even though dividing by the mean gives 24, the product is -120, which does not satisfy the first condition. Therefore, x=-6 is invalid.

Therefore, only x=4 is valid. Therefore, the three integers are 4,5,6.

So, both methods lead to the same conclusion. Therefore, the answer is 6.

But let me check once again. If the three integers were -6, -5, -4, then their product is -120, which is not 120. Therefore, that's invalid. Therefore, only 4,5,6 works.

Alternatively, if I didn't think of the second approach, but just solved the first equation, I still get the same answer.

Therefore, I think the answer is definitely 6.

But just to make sure, let me check if there's any other possible way. For example, maybe the three consecutive integers are not centered around n? Wait, but consecutive integers are always like x, x+1, x+2. So, whether you take them as n-1, n, n+1 or x, x+1, x+2, it's the same thing.

Alternatively, maybe someone might think of the three numbers as a, a+1, a+2. So, same as above.

Alternatively, someone might think of them as n, n+1, n+2. So, same thing.

Therefore, regardless of how you represent them, you end up with similar equations.

Therefore, I think the answer is 6.

But let me check once again if there's a possibility of another set of numbers. For example, maybe non-integer numbers? But the problem says "three consecutive integers," so they have to be integers. Therefore, no.

Therefore, I think I can be confident that the answer is 6.

**Final Answer**
The largest of the three consecutive integers is \boxed{6}.

To solve the problem, we are given two key pieces of information:

1. The **product** of three consecutive integers is **120**.
2. When this product is **divided by the mean** of the three integers, the result is **24**.

---

### Step 1: Represent the Integers Algebraically

Let the three consecutive integers be:

$$
x, \quad x+1, \quad x+2
$$

Their **product** is:

$$
x(x+1)(x+2) = 120
$$

Their **mean** is:

$$
\frac{x + (x+1) + (x+2)}{3} = \frac{3x + 3}{3} = x + 1
$$

So, the second condition becomes:

$$
\frac{x(x+1)(x+2)}{x+1} = 24
$$

Simplifying the left-hand side:

$$
\frac{x(x+1)(x+2)}{x+1} = x(x+2)
$$

So we now have the equation:

$$
x(x+2) = 24
$$

---

### Step 2: Solve the Equation

We solve the quadratic:

$$
x(x+2) = 24 \Rightarrow x^2 + 2x - 24 = 0
$$

Use the quadratic formula:

$$
x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2}
$$

$$
x = \frac{-2 \pm 10}{2}
$$

So the two possible values for $ x $ are:

$$
x = \frac{8}{2} = 4 \quad \text{or} \quad x = \frac{-12}{2} = -6
$$

---

### Step 3: Check Which Value Satisfies the Original Product Condition

- If $ x = 4 $, the integers are $ 4, 5, 6 $, and their product is:

$$
4 \times 5 \times 6 = 120
$$

- If $ x = -6 $, the integers are $ -6, -5, -4 $, and their product is:

$$
(-6) \times (-5) \times (-4) = -120
$$

This does **not** satisfy the first condition (product is 120), so we discard $ x = -6 $.

---

### Step 4: Confirm the Second Condition

For $ x = 4 $, the integers are $ 4, 5, 6 $, and the mean is:

$$
\frac{4 + 5 + 6}{3} = \frac{15}{3} = 5
$$

Then:

$$
\frac{120}{5} = 24
$$

This matches the second condition.

---

### Final Answer

The three consecutive integers are $ 4, 5, 6 $, and the **largest** of them is:

$$
\boxed{6}
$$

Solution: Call the integers $n-1$, $n$, and $n+1$. Their mean is $n$; their product is $(n-1)(n)(n+1)=120$ and their product divided by their mean is $(n-1)(n+1)=24$. Dividing the first equation by the second, we get $n=5$. The largest of the three is $n+1=\boxed{6}$.


Explanation:
Okay, so I need to figure out what 999 in base 10 is in base six. Hmm, let me remember how to convert numbers from base 10 to another base. I think the standard method is to divide the number by the base and keep track of the remainders. Let me try to recall the exact steps.

First, I know that base six means each digit represents a power of six. So, the rightmost digit is 6^0 (which is 1), then 6^1 (which is 6), then 6^2 (which is 36), then 6^3 (which is 216), and so on. So, to convert 999 into base six, I need to find out how many times each power of six fits into 999, starting from the highest power possible.

But maybe the division method is more straightforward. Let me try that. The process is to divide the number by 6, record the remainder, then divide the quotient by 6 again, and repeat until the quotient is zero. Then, the base six number is the remainders read in reverse order. Let me try that.

So, starting with 999 divided by 6. Let me calculate that. 6 times 166 is 996, right? Because 6*160 is 960, and 6*6 is 36, so 960+36=996. Then 999 minus 996 is 3. So the first remainder is 3. So that's the least significant digit (rightmost digit) in base six.

Now, take the quotient, which is 166, and divide that by 6. Let me do 166 divided by 6. 6*27 is 162, so 166-162=4. So the remainder is 4. So the next digit is 4.

Now take the quotient 27 and divide by 6. 6*4 is 24, so 27-24=3. Remainder is 3. Next digit is 3.

Take the quotient 4 and divide by 6. Well, 4 divided by 6 is 0 with a remainder of 4. So the next digit is 4.

Now the quotient is 0, so we stop. Then, reading the remainders from last to first, we have 4, 3, 4, 3. Wait, let me check again.

Wait, let me write down each step properly:

Step 1: 999 ÷ 6 = 166 with remainder 3.

Step 2: 166 ÷ 6 = 27 with remainder 4.

Step 3: 27 ÷ 6 = 4 with remainder 3.

Step 4: 4 ÷ 6 = 0 with remainder 4.

So the remainders, in order from last to first, are 4, 3, 4, 3. Therefore, the base six number is 4343? Let me check if that's correct.

Wait, let me verify by converting 4343 base six back to decimal. Let's see:

Starting from the right, the digits are 4, 3, 4, 3. Wait, actually, in base six, the rightmost digit is 6^0, then 6^1, 6^2, 6^3. So the number 4343 in base six would be:

4*6^3 + 3*6^2 + 4*6^1 + 3*6^0.

Calculating each term:

6^0 = 1, so 3*1 = 3.

6^1 = 6, so 4*6 = 24.

6^2 = 36, so 3*36 = 108.

6^3 = 216, so 4*216 = 864.

Adding them up: 864 + 108 = 972; 972 + 24 = 996; 996 + 3 = 999. Perfect! That matches the original number. So 999 in base 10 is 4343 in base 6. Therefore, the answer is 4343 base six.

But let me just make sure I didn't make any mistakes in the division steps. Let me go through them again.

First division: 999 divided by 6. 6*160 is 960, 999-960=39. Then 39 divided by 6 is 6*6=36, so total quotient is 160+6=166, remainder 3. Correct.

Second division: 166 divided by 6. 6*20=120, 166-120=46. 46 divided by 6 is 7*6=42, remainder 4. So total quotient is 20+7=27, remainder 4. Correct.

Third division: 27 divided by 6. 6*4=24, remainder 3. Quotient 4, remainder 3. Correct.

Fourth division: 4 divided by 6 is 0 with remainder 4. Correct.

So the remainders are 3, 4, 3, 4. Wait, but when we read them in reverse order, it's 4, 3, 4, 3. So 4343. Yes, that's correct. So that seems right.

Alternatively, maybe I can try another method to cross-verify. Let me try expanding the number in terms of powers of six.

What's the highest power of 6 less than or equal to 999?

Compute 6^0 = 1

6^1 = 6

6^2 = 36

6^3 = 216

6^4 = 1296

Okay, 6^4 is 1296 which is larger than 999, so the highest power is 6^3 = 216.

So how many times does 216 go into 999?

Let me calculate 999 divided by 216. 216*4 = 864, 216*5=1080 which is too big. So 4 times. So the coefficient for 6^3 is 4. Then subtract 4*216=864 from 999: 999 - 864 = 135.

Now move to the next lower power, which is 6^2 = 36. How many times does 36 go into 135?

36*3=108, 36*4=144 which is too big. So coefficient is 3. Subtract 108 from 135: 135 - 108 = 27.

Next power is 6^1 = 6. How many times does 6 go into 27? 6*4=24, 6*5=30 which is too big. So coefficient is 4. Subtract 24 from 27: 3.

Next power is 6^0 = 1. How many times does 1 go into 3? 3 times. So coefficient is 3.

Therefore, the coefficients are 4 (for 6^3), 3 (for 6^2), 4 (for 6^1), 3 (for 6^0). So the number is 4343 in base six. Same result as before. So that confirms it.

Therefore, I think the answer is 4343 base six. Let me check once more with another approach.

Alternatively, I can use the fact that 6^3 is 216, and 6^4 is 1296. Since 999 is between 6^3 and 6^4, the base six number will have four digits. Let me think of how to compute each digit.

First digit (from left): how many 216s in 999? As before, 4. Then 4*216=864. 999-864=135.

Second digit: how many 36s in 135? 3*36=108. 135-108=27.

Third digit: how many 6s in 27? 4*6=24. 27-24=3.

Fourth digit: how many 1s in 3? 3*1=3. So digits are 4, 3, 4, 3. Yep, same as before.

So all methods lead to 4343. Therefore, I think that's the correct answer. I don't see any mistakes in my reasoning. Let me just check once again by converting 4343 base six to decimal:

Starting from the right:

Digits: 4 (position 3), 3 (position 2), 4 (position 1), 3 (position 0).

Wait, positions are from right starting at 0. Wait, actually, the leftmost digit is the highest power. So 4 is in the 6^3 place, then 3 in 6^2, 4 in 6^1, 3 in 6^0. So:

4*6^3 = 4*216 = 864

3*6^2 = 3*36 = 108

4*6^1 = 4*6 = 24

3*6^0 = 3*1 = 3

Adding them up: 864 + 108 = 972; 972 + 24 = 996; 996 + 3 = 999. Perfect, that's exactly the original number. So that's correct.

Therefore, I can be confident that 999 in base 10 is 4343 in base 6.

**Final Answer**
The base six equivalent of $999_{10}$ is \boxed{4343}.

To convert the decimal number $ 999_{10} $ into base six, we use the **division-remainder method**, which is a standard approach for converting numbers from base 10 to any other base.

---

### Step-by-Step Conversion

We repeatedly divide the number by 6 and record the remainders. The remainders, read in reverse order, give the digits of the number in base six.

1. **Divide 999 by 6:**
   $$
   999 \div 6 = 166 \text{ remainder } 3
   $$
   - Remainder: 3

2. **Divide 166 by 6:**
   $$
   166 \div 6 = 27 \text{ remainder } 4
   $$
   - Remainder: 4

3. **Divide 27 by 6:**
   $$
   27 \div 6 = 4 \text{ remainder } 3
   $$
   - Remainder: 3

4. **Divide 4 by 6:**
   $$
   4 \div 6 = 0 \text{ remainder } 4
   $$
   - Remainder: 4

Now, reading the remainders from **last to first**, we get the base six number:
$$
4\ 3\ 4\ 3
$$

---

### Verification

To confirm, we convert $ 4343_6 $ back to base 10:

$$
4 \times 6^3 + 3 \times 6^2 + 4 \times 6^1 + 3 \times 6^0
= 4 \times 216 + 3 \times 36 + 4 \times 6 + 3 \times 1
= 864 + 108 + 24 + 3 = 999
$$

This matches the original number, confirming the conversion is correct.

---

### Final Answer

$$
\boxed{4343}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what 999 in base 10 is in base six. Hmm, let me remember how to convert numbers from base 10 to another base. I think the standard method is to divide the number by the base and keep track of the remainders. Let me try to recall the exact steps.

First, I know that base six means each digit represents a power of six. So, the rightmost digit is 6^0 (which is 1), then 6^1 (which is 6), then 6^2 (which is 36), then 6^3 (which is 216), and so on. So, to convert 999 into base six, I need to find out how many times each power of six fits into 999, starting from the highest power possible.

But maybe the division method is more straightforward. Let me try that. The process is to divide the number by 6, record the remainder, then divide the quotient by 6 again, and repeat until the quotient is zero. Then, the base six number is the remainders read in reverse order. Let me try that.

So, starting with 999 divided by 6. Let me calculate that. 6 times 166 is 996, right? Because 6*160 is 960, and 6*6 is 36, so 960+36=996. Then 999 minus 996 is 3. So the first remainder is 3. So that's the least significant digit (rightmost digit) in base six.

Now, take the quotient, which is 166, and divide that by 6. Let me do 166 divided by 6. 6*27 is 162, so 166-162=4. So the remainder is 4. So the next digit is 4.

Now take the quotient 27 and divide by 6. 6*4 is 24, so 27-24=3. Remainder is 3. Next digit is 3.

Take the quotient 4 and divide by 6. Well, 4 divided by 6 is 0 with a remainder of 4. So the next digit is 4.

Now the quotient is 0, so we stop. Then, reading the remainders from last to first, we have 4, 3, 4, 3. Wait, let me check again.

Wait, let me write down each step properly:

Step 1: 999 ÷ 6 = 166 with remainder 3.

Step 2: 166 ÷ 6 = 27 with remainder 4.

Step 3: 27 ÷ 6 = 4 with remainder 3.

Step 4: 4 ÷ 6 = 0 with remainder 4.

So the remainders, in order from last to first, are 4, 3, 4, 3. Therefore, the base six number is 4343? Let me check if that's correct.

Wait, let me verify by converting 4343 base six back to decimal. Let's see:

Starting from the right, the digits are 4, 3, 4, 3. Wait, actually, in base six, the rightmost digit is 6^0, then 6^1, 6^2, 6^3. So the number 4343 in base six would be:

4*6^3 + 3*6^2 + 4*6^1 + 3*6^0.

Calculating each term:

6^0 = 1, so 3*1 = 3.

6^1 = 6, so 4*6 = 24.

6^2 = 36, so 3*36 = 108.

6^3 = 216, so 4*216 = 864.

Adding them up: 864 + 108 = 972; 972 + 24 = 996; 996 + 3 = 999. Perfect! That matches the original number. So 999 in base 10 is 4343 in base 6. Therefore, the answer is 4343 base six.

But let me just make sure I didn't make any mistakes in the division steps. Let me go through them again.

First division: 999 divided by 6. 6*160 is 960, 999-960=39. Then 39 divided by 6 is 6*6=36, so total quotient is 160+6=166, remainder 3. Correct.

Second division: 166 divided by 6. 6*20=120, 166-120=46. 46 divided by 6 is 7*6=42, remainder 4. So total quotient is 20+7=27, remainder 4. Correct.

Third division: 27 divided by 6. 6*4=24, remainder 3. Quotient 4, remainder 3. Correct.

Fourth division: 4 divided by 6 is 0 with remainder 4. Correct.

So the remainders are 3, 4, 3, 4. Wait, but when we read them in reverse order, it's 4, 3, 4, 3. So 4343. Yes, that's correct. So that seems right.

Alternatively, maybe I can try another method to cross-verify. Let me try expanding the number in terms of powers of six.

What's the highest power of 6 less than or equal to 999?

Compute 6^0 = 1

6^1 = 6

6^2 = 36

6^3 = 216

6^4 = 1296

Okay, 6^4 is 1296 which is larger than 999, so the highest power is 6^3 = 216.

So how many times does 216 go into 999?

Let me calculate 999 divided by 216. 216*4 = 864, 216*5=1080 which is too big. So 4 times. So the coefficient for 6^3 is 4. Then subtract 4*216=864 from 999: 999 - 864 = 135.

Now move to the next lower power, which is 6^2 = 36. How many times does 36 go into 135?

36*3=108, 36*4=144 which is too big. So coefficient is 3. Subtract 108 from 135: 135 - 108 = 27.

Next power is 6^1 = 6. How many times does 6 go into 27? 6*4=24, 6*5=30 which is too big. So coefficient is 4. Subtract 24 from 27: 3.

Next power is 6^0 = 1. How many times does 1 go into 3? 3 times. So coefficient is 3.

Therefore, the coefficients are 4 (for 6^3), 3 (for 6^2), 4 (for 6^1), 3 (for 6^0). So the number is 4343 in base six. Same result as before. So that confirms it.

Therefore, I think the answer is 4343 base six. Let me check once more with another approach.

Alternatively, I can use the fact that 6^3 is 216, and 6^4 is 1296. Since 999 is between 6^3 and 6^4, the base six number will have four digits. Let me think of how to compute each digit.

First digit (from left): how many 216s in 999? As before, 4. Then 4*216=864. 999-864=135.

Second digit: how many 36s in 135? 3*36=108. 135-108=27.

Third digit: how many 6s in 27? 4*6=24. 27-24=3.

Fourth digit: how many 1s in 3? 3*1=3. So digits are 4, 3, 4, 3. Yep, same as before.

So all methods lead to 4343. Therefore, I think that's the correct answer. I don't see any mistakes in my reasoning. Let me just check once again by converting 4343 base six to decimal:

Starting from the right:

Digits: 4 (position 3), 3 (position 2), 4 (position 1), 3 (position 0).

Wait, positions are from right starting at 0. Wait, actually, the leftmost digit is the highest power. So 4 is in the 6^3 place, then 3 in 6^2, 4 in 6^1, 3 in 6^0. So:

4*6^3 = 4*216 = 864

3*6^2 = 3*36 = 108

4*6^1 = 4*6 = 24

3*6^0 = 3*1 = 3

Adding them up: 864 + 108 = 972; 972 + 24 = 996; 996 + 3 = 999. Perfect, that's exactly the original number. So that's correct.

Therefore, I can be confident that 999 in base 10 is 4343 in base 6.

**Final Answer**
The base six equivalent of $999_{10}$ is \boxed{4343}.

To convert the decimal number $ 999_{10} $ into base six, we use the **division-remainder method**, which is a standard approach for converting numbers from base 10 to any other base.

---

### Step-by-Step Conversion

We repeatedly divide the number by 6 and record the remainders. The remainders, read in reverse order, give the digits of the number in base six.

1. **Divide 999 by 6:**
   $$
   999 \div 6 = 166 \text{ remainder } 3
   $$
   - Remainder: 3

2. **Divide 166 by 6:**
   $$
   166 \div 6 = 27 \text{ remainder } 4
   $$
   - Remainder: 4

3. **Divide 27 by 6:**
   $$
   27 \div 6 = 4 \text{ remainder } 3
   $$
   - Remainder: 3

4. **Divide 4 by 6:**
   $$
   4 \div 6 = 0 \text{ remainder } 4
   $$
   - Remainder: 4

Now, reading the remainders from **last to first**, we get the base six number:
$$
4\ 3\ 4\ 3
$$

---

### Verification

To confirm, we convert $ 4343_6 $ back to base 10:

$$
4 \times 6^3 + 3 \times 6^2 + 4 \times 6^1 + 3 \times 6^0
= 4 \times 216 + 3 \times 36 + 4 \times 6 + 3 \times 1
= 864 + 108 + 24 + 3 = 999
$$

This matches the original number, confirming the conversion is correct.

---

### Final Answer

$$
\boxed{4343}
$$

Solution: We know that $6^{4}>999_{10}>6^{3}$. So, we can tell that $999_{10}$ in base six will have four digits. $6^{3}=216$, which can go into 999 four times at most, leaving $999-4\cdot216 = 135$ for the next three digits. $6^{2}=36$ goes into 135 three times at most, leaving us with $135-3\cdot36 = 27$. Then, $6^{1}=6$ goes into 27 four times at most, leaving $27-4\cdot6 = 3$ for the ones digit. All together, the base six equivalent of $999_{10}$ is $\boxed{4343_6}$.


Explanation:
Okay, so I need to find the sum a + b + c + d + e + f in terms of k, where the original parabola is y = ax² + bx + c with vertex (h, k), and after reflecting it over the line y = k, we get another parabola y = dx² + ex + f.

First, let me recall that reflecting a graph over a horizontal line like y = k would invert the vertical distances from that line. Since the original parabola has its vertex at (h, k), reflecting it over y = k should result in another parabola that's symmetric with respect to y = k. But since the vertex is on the line of reflection, the reflection of the vertex should be itself. So the reflected parabola should also have its vertex at (h, k)? Wait, that seems odd. If you reflect a parabola over its own vertex's horizontal line, does it stay the same? Or does it flip?

Wait, let me think. Suppose I have a parabola opening upwards with vertex at (h, k). If I reflect it over the line y = k, then points above y = k would be mirrored below y = k and vice versa. However, since the vertex is on the line of reflection, it remains the same. But the shape of the parabola would be flipped. So instead of opening upwards, it would open downwards? But the original parabola is y = ax² + bx + c. If we reflect it over y = k, then the new parabola would have the same vertex but open in the opposite direction? So if the original parabola opens upwards (a > 0), the reflected one would open downwards (d < 0), and vice versa.

But how do I find the equation of the reflected parabola? Maybe I can use the vertex form of a parabola. The original parabola can be written as y = a(x - h)² + k. Because the vertex is (h, k). Then reflecting over y = k would mean that for any point (x, y) on the original parabola, the reflected point (x, y') would satisfy that the distance from y to k is the same as from y' to k, but on the opposite side. So y' = 2k - y.

Therefore, if the original equation is y = a(x - h)² + k, then the reflected parabola would have equation y' = 2k - y. Substituting y from the original equation, we get y' = 2k - [a(x - h)² + k] = 2k - a(x - h)² - k = k - a(x - h)².

So the reflected parabola is y = -a(x - h)² + k. Therefore, comparing this to the standard form y = dx² + ex + f, we need to expand this equation.

Let me expand y = -a(x - h)² + k. First, expand (x - h)²: x² - 2hx + h². Multiply by -a: -a x² + 2a h x - a h². Then add k: y = -a x² + 2a h x - a h² + k.

Therefore, the coefficients for the reflected parabola are d = -a, e = 2a h, f = -a h² + k.

Now, the original parabola was y = a x² + b x + c. Let me write that in vertex form to find relations between a, b, c and h, k.

The vertex form is y = a(x - h)² + k. Expanding this, we get y = a x² - 2a h x + a h² + k. Comparing with y = a x² + b x + c, we can equate coefficients:

Original coefficients:
a = a (same)
b = -2a h
c = a h² + k

So from this, we can express b and c in terms of a, h, and k.

Now, the reflected parabola has coefficients:
d = -a
e = 2a h
f = -a h² + k

So now, we need to compute a + b + c + d + e + f. Let's compute each term:

First, a is just a.

Then, b = -2a h.

c = a h² + k.

d = -a.

e = 2a h.

f = -a h² + k.

So adding them all up:

a + (-2a h) + (a h² + k) + (-a) + (2a h) + (-a h² + k)

Let me compute term by term:

Start with a.

Then -2a h.

Then + a h² + k.

Then -a.

Then + 2a h.

Then -a h² + k.

Now, let's combine like terms.

First, the a terms: a - a = 0.

Then the terms with -2a h and +2a h: -2a h + 2a h = 0.

Then the terms with a h²: +a h² - a h² = 0.

Then the constants: k + k = 2k.

So all the other terms cancel out, and we are left with 2k.

Wait, so the sum a + b + c + d + e + f equals 2k? That seems surprisingly simple. Let me check my steps again to make sure I didn't make a mistake.

Original parabola: y = a(x - h)^2 + k. Expanded to y = a x² - 2a h x + a h² + k. Therefore, original coefficients: a, b = -2a h, c = a h² + k. Correct.

Reflected parabola: y = -a(x - h)^2 + k. Expanded to y = -a x² + 2a h x - a h² + k. Therefore, coefficients: d = -a, e = 2a h, f = -a h² + k. Correct.

Sum: a + b + c + d + e + f.

Substituting:

a + (-2a h) + (a h² + k) + (-a) + (2a h) + (-a h² + k)

Yes, as above. The a and -a cancel. The -2a h and +2a h cancel. The a h² and -a h² cancel. Then k + k = 2k. So total sum is 2k.

Therefore, regardless of the values of a, h, or the original coefficients, the sum a + b + c + d + e + f is equal to 2k.

But let me check with a concrete example to make sure. Let's take a specific parabola, reflect it, compute the sum, and see if it equals 2k.

Let me choose a simple parabola. Let me take the vertex at (h, k) = (0, 0). So original parabola is y = a x² + 0x + c. But since vertex is (0, 0), then in vertex form it's y = a(x - 0)^2 + 0, so y = a x². Therefore, original equation is y = a x². Therefore, coefficients: a = a, b = 0, c = 0.

Reflecting over y = 0 (the x-axis). The reflection of y = a x² over y = 0 is y = -a x². Therefore, the reflected parabola is y = -a x² + 0x + 0. Therefore, coefficients d = -a, e = 0, f = 0.

Sum a + b + c + d + e + f = a + 0 + 0 + (-a) + 0 + 0 = 0.

But according to our formula, 2k. Here, k = 0, so 2k = 0. Correct.

Another example. Let me take vertex at (h, k) = (1, 2). Let me choose a = 1. Then original parabola is y = (x - 1)^2 + 2. Expanding: y = x² - 2x + 1 + 2 = x² - 2x + 3. Therefore, a = 1, b = -2, c = 3.

Reflecting over y = 2. The reflected parabola is y = - (x - 1)^2 + 2. Expanding: y = -x² + 2x - 1 + 2 = -x² + 2x + 1. Therefore, coefficients d = -1, e = 2, f = 1.

Now compute a + b + c + d + e + f: 1 + (-2) + 3 + (-1) + 2 + 1 = (1 - 2 + 3) + (-1 + 2 + 1) = (2) + (2) = 4.

Now, k = 2, so 2k = 4. Correct.

Another example. Let me take vertex (h, k) = (2, 3), a = -1. Original parabola: y = -1(x - 2)^2 + 3. Expanding: y = - (x² - 4x + 4) + 3 = -x² + 4x - 4 + 3 = -x² + 4x - 1. Therefore, a = -1, b = 4, c = -1.

Reflecting over y = 3: the reflected parabola is y = - [ - (x - 2)^2 + 3 ] + 3? Wait, no. Wait, original equation is y = - (x - 2)^2 + 3. Reflecting over y = 3 would be y' = 2*3 - y = 6 - y. Therefore, substituting y: y' = 6 - [ - (x - 2)^2 + 3 ] = 6 + (x - 2)^2 - 3 = 3 + (x - 2)^2.

Expanding that: (x - 2)^2 + 3 = x² - 4x + 4 + 3 = x² - 4x + 7. Therefore, coefficients d = 1, e = -4, f = 7.

Now compute a + b + c + d + e + f: (-1) + 4 + (-1) + 1 + (-4) + 7. Let's compute step by step: (-1 + 4) = 3; (3 + (-1)) = 2; (2 + 1) = 3; (3 + (-4)) = -1; (-1 + 7) = 6.

Now, k = 3, so 2k = 6. Correct.

So the formula seems to hold. Therefore, regardless of the values of a, h, or the original coefficients, the sum a + b + c + d + e + f equals 2k.

Therefore, the answer is 2k.

But let me check another case where the original parabola is not in vertex form. Suppose I have a parabola with vertex (h, k) but written in standard form. Let me take vertex (1, 2) again, but with a different a. Let me take a = 2. Original equation: y = 2(x - 1)^2 + 2. Expanding: 2(x² - 2x + 1) + 2 = 2x² - 4x + 2 + 2 = 2x² - 4x + 4. Therefore, a = 2, b = -4, c = 4.

Reflecting over y = 2: y' = 2*2 - y = 4 - y. So substituting y: 4 - (2x² - 4x + 4) = 4 - 2x² + 4x - 4 = -2x² + 4x. Therefore, coefficients d = -2, e = 4, f = 0.

Sum a + b + c + d + e + f: 2 + (-4) + 4 + (-2) + 4 + 0 = (2 - 4 + 4) + (-2 + 4 + 0) = (2) + (2) = 4. Which is 2k = 2*2 = 4. Correct.

Therefore, all examples check out. Therefore, the answer is 2k.

But let me think again if there's another way to approach this problem, maybe more algebraically, to confirm.

Alternatively, since the sum a + b + c is the value of the original parabola at x = 1, because substituting x = 1 into y = ax² + bx + c gives y = a + b + c. Similarly, the sum d + e + f is the value of the reflected parabola at x = 1. Therefore, the total sum is y_original(1) + y_reflected(1).

So if I can compute y_original(1) + y_reflected(1), that would be equal to a + b + c + d + e + f.

Let me check this.

Original parabola: y = ax² + bx + c. At x = 1, y = a + b + c.

Reflected parabola: y = dx² + ex + f. At x = 1, y = d + e + f.

Therefore, total sum is y_original(1) + y_reflected(1).

But what is y_reflected(1)?

Alternatively, since reflecting over y = k, the reflection of the original parabola. So for any point (x, y) on the original parabola, the reflected point is (x, 2k - y). Therefore, if we consider the reflected parabola, then for x = 1, the y-coordinate is 2k - y_original(1). Therefore, y_reflected(1) = 2k - y_original(1).

Therefore, the total sum is y_original(1) + y_reflected(1) = y_original(1) + 2k - y_original(1) = 2k.

Therefore, regardless of the original parabola, the sum a + b + c + d + e + f equals 2k.

This is a much more straightforward way to see it. Since the sum is y_original(1) + y_reflected(1), and because reflecting over y = k maps y_original(1) to 2k - y_original(1), their sum is 2k.

This confirms the previous result.

Therefore, the answer is 2k.

**Final Answer**
\boxed{2k}

To solve the problem, we are given a parabola with equation:

$$
y = ax^2 + bx + c
$$

and vertex $(h, k)$. This parabola is reflected over the horizontal line $y = k$, resulting in a new parabola with equation:

$$
y = dx^2 + ex + f
$$

We are asked to compute the sum:

$$
a + b + c + d + e + f
$$

in terms of $k$.

---

### Step 1: Express the Original Parabola in Vertex Form

The vertex form of a parabola is:

$$
y = a(x - h)^2 + k
$$

Expanding this:

$$
y = a(x^2 - 2hx + h^2) + k = ax^2 - 2ahx + ah^2 + k
$$

So, comparing with the standard form $y = ax^2 + bx + c$, we identify:

- $a = a$
- $b = -2ah$
- $c = ah^2 + k$

---

### Step 2: Reflect the Parabola Over the Line $y = k$

Reflecting a point $(x, y)$ over the line $y = k$ gives the point $(x, 2k - y)$. So, the reflected parabola has the equation:

$$
y = 2k - [a(x - h)^2 + k] = k - a(x - h)^2
$$

Expanding this:

$$
y = -a(x - h)^2 + k = -a(x^2 - 2hx + h^2) + k = -ax^2 + 2ahx - ah^2 + k
$$

So, the reflected parabola has coefficients:

- $d = -a$
- $e = 2ah$
- $f = -ah^2 + k$

---

### Step 3: Compute the Sum $a + b + c + d + e + f$

Now, substitute the values:

- $a = a$
- $b = -2ah$
- $c = ah^2 + k$
- $d = -a$
- $e = 2ah$
- $f = -ah^2 + k$

Add them up:

$$
a + (-2ah) + (ah^2 + k) + (-a) + (2ah) + (-ah^2 + k)
$$

Group like terms:

- $a - a = 0$
- $-2ah + 2ah = 0$
- $ah^2 - ah^2 = 0$
- $k + k = 2k$

So, the total sum is:

$$
a + b + c + d + e + f = 2k
$$

---

### Final Answer

$$
\boxed{2k}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum a + b + c + d + e + f in terms of k, where the original parabola is y = ax² + bx + c with vertex (h, k), and after reflecting it over the line y = k, we get another parabola y = dx² + ex + f.

First, let me recall that reflecting a graph over a horizontal line like y = k would invert the vertical distances from that line. Since the original parabola has its vertex at (h, k), reflecting it over y = k should result in another parabola that's symmetric with respect to y = k. But since the vertex is on the line of reflection, the reflection of the vertex should be itself. So the reflected parabola should also have its vertex at (h, k)? Wait, that seems odd. If you reflect a parabola over its own vertex's horizontal line, does it stay the same? Or does it flip?

Wait, let me think. Suppose I have a parabola opening upwards with vertex at (h, k). If I reflect it over the line y = k, then points above y = k would be mirrored below y = k and vice versa. However, since the vertex is on the line of reflection, it remains the same. But the shape of the parabola would be flipped. So instead of opening upwards, it would open downwards? But the original parabola is y = ax² + bx + c. If we reflect it over y = k, then the new parabola would have the same vertex but open in the opposite direction? So if the original parabola opens upwards (a > 0), the reflected one would open downwards (d < 0), and vice versa.

But how do I find the equation of the reflected parabola? Maybe I can use the vertex form of a parabola. The original parabola can be written as y = a(x - h)² + k. Because the vertex is (h, k). Then reflecting over y = k would mean that for any point (x, y) on the original parabola, the reflected point (x, y') would satisfy that the distance from y to k is the same as from y' to k, but on the opposite side. So y' = 2k - y.

Therefore, if the original equation is y = a(x - h)² + k, then the reflected parabola would have equation y' = 2k - y. Substituting y from the original equation, we get y' = 2k - [a(x - h)² + k] = 2k - a(x - h)² - k = k - a(x - h)².

So the reflected parabola is y = -a(x - h)² + k. Therefore, comparing this to the standard form y = dx² + ex + f, we need to expand this equation.

Let me expand y = -a(x - h)² + k. First, expand (x - h)²: x² - 2hx + h². Multiply by -a: -a x² + 2a h x - a h². Then add k: y = -a x² + 2a h x - a h² + k.

Therefore, the coefficients for the reflected parabola are d = -a, e = 2a h, f = -a h² + k.

Now, the original parabola was y = a x² + b x + c. Let me write that in vertex form to find relations between a, b, c and h, k.

The vertex form is y = a(x - h)² + k. Expanding this, we get y = a x² - 2a h x + a h² + k. Comparing with y = a x² + b x + c, we can equate coefficients:

Original coefficients:
a = a (same)
b = -2a h
c = a h² + k

So from this, we can express b and c in terms of a, h, and k.

Now, the reflected parabola has coefficients:
d = -a
e = 2a h
f = -a h² + k

So now, we need to compute a + b + c + d + e + f. Let's compute each term:

First, a is just a.

Then, b = -2a h.

c = a h² + k.

d = -a.

e = 2a h.

f = -a h² + k.

So adding them all up:

a + (-2a h) + (a h² + k) + (-a) + (2a h) + (-a h² + k)

Let me compute term by term:

Start with a.

Then -2a h.

Then + a h² + k.

Then -a.

Then + 2a h.

Then -a h² + k.

Now, let's combine like terms.

First, the a terms: a - a = 0.

Then the terms with -2a h and +2a h: -2a h + 2a h = 0.

Then the terms with a h²: +a h² - a h² = 0.

Then the constants: k + k = 2k.

So all the other terms cancel out, and we are left with 2k.

Wait, so the sum a + b + c + d + e + f equals 2k? That seems surprisingly simple. Let me check my steps again to make sure I didn't make a mistake.

Original parabola: y = a(x - h)^2 + k. Expanded to y = a x² - 2a h x + a h² + k. Therefore, original coefficients: a, b = -2a h, c = a h² + k. Correct.

Reflected parabola: y = -a(x - h)^2 + k. Expanded to y = -a x² + 2a h x - a h² + k. Therefore, coefficients: d = -a, e = 2a h, f = -a h² + k. Correct.

Sum: a + b + c + d + e + f.

Substituting:

a + (-2a h) + (a h² + k) + (-a) + (2a h) + (-a h² + k)

Yes, as above. The a and -a cancel. The -2a h and +2a h cancel. The a h² and -a h² cancel. Then k + k = 2k. So total sum is 2k.

Therefore, regardless of the values of a, h, or the original coefficients, the sum a + b + c + d + e + f is equal to 2k.

But let me check with a concrete example to make sure. Let's take a specific parabola, reflect it, compute the sum, and see if it equals 2k.

Let me choose a simple parabola. Let me take the vertex at (h, k) = (0, 0). So original parabola is y = a x² + 0x + c. But since vertex is (0, 0), then in vertex form it's y = a(x - 0)^2 + 0, so y = a x². Therefore, original equation is y = a x². Therefore, coefficients: a = a, b = 0, c = 0.

Reflecting over y = 0 (the x-axis). The reflection of y = a x² over y = 0 is y = -a x². Therefore, the reflected parabola is y = -a x² + 0x + 0. Therefore, coefficients d = -a, e = 0, f = 0.

Sum a + b + c + d + e + f = a + 0 + 0 + (-a) + 0 + 0 = 0.

But according to our formula, 2k. Here, k = 0, so 2k = 0. Correct.

Another example. Let me take vertex at (h, k) = (1, 2). Let me choose a = 1. Then original parabola is y = (x - 1)^2 + 2. Expanding: y = x² - 2x + 1 + 2 = x² - 2x + 3. Therefore, a = 1, b = -2, c = 3.

Reflecting over y = 2. The reflected parabola is y = - (x - 1)^2 + 2. Expanding: y = -x² + 2x - 1 + 2 = -x² + 2x + 1. Therefore, coefficients d = -1, e = 2, f = 1.

Now compute a + b + c + d + e + f: 1 + (-2) + 3 + (-1) + 2 + 1 = (1 - 2 + 3) + (-1 + 2 + 1) = (2) + (2) = 4.

Now, k = 2, so 2k = 4. Correct.

Another example. Let me take vertex (h, k) = (2, 3), a = -1. Original parabola: y = -1(x - 2)^2 + 3. Expanding: y = - (x² - 4x + 4) + 3 = -x² + 4x - 4 + 3 = -x² + 4x - 1. Therefore, a = -1, b = 4, c = -1.

Reflecting over y = 3: the reflected parabola is y = - [ - (x - 2)^2 + 3 ] + 3? Wait, no. Wait, original equation is y = - (x - 2)^2 + 3. Reflecting over y = 3 would be y' = 2*3 - y = 6 - y. Therefore, substituting y: y' = 6 - [ - (x - 2)^2 + 3 ] = 6 + (x - 2)^2 - 3 = 3 + (x - 2)^2.

Expanding that: (x - 2)^2 + 3 = x² - 4x + 4 + 3 = x² - 4x + 7. Therefore, coefficients d = 1, e = -4, f = 7.

Now compute a + b + c + d + e + f: (-1) + 4 + (-1) + 1 + (-4) + 7. Let's compute step by step: (-1 + 4) = 3; (3 + (-1)) = 2; (2 + 1) = 3; (3 + (-4)) = -1; (-1 + 7) = 6.

Now, k = 3, so 2k = 6. Correct.

So the formula seems to hold. Therefore, regardless of the values of a, h, or the original coefficients, the sum a + b + c + d + e + f equals 2k.

Therefore, the answer is 2k.

But let me check another case where the original parabola is not in vertex form. Suppose I have a parabola with vertex (h, k) but written in standard form. Let me take vertex (1, 2) again, but with a different a. Let me take a = 2. Original equation: y = 2(x - 1)^2 + 2. Expanding: 2(x² - 2x + 1) + 2 = 2x² - 4x + 2 + 2 = 2x² - 4x + 4. Therefore, a = 2, b = -4, c = 4.

Reflecting over y = 2: y' = 2*2 - y = 4 - y. So substituting y: 4 - (2x² - 4x + 4) = 4 - 2x² + 4x - 4 = -2x² + 4x. Therefore, coefficients d = -2, e = 4, f = 0.

Sum a + b + c + d + e + f: 2 + (-4) + 4 + (-2) + 4 + 0 = (2 - 4 + 4) + (-2 + 4 + 0) = (2) + (2) = 4. Which is 2k = 2*2 = 4. Correct.

Therefore, all examples check out. Therefore, the answer is 2k.

But let me think again if there's another way to approach this problem, maybe more algebraically, to confirm.

Alternatively, since the sum a + b + c is the value of the original parabola at x = 1, because substituting x = 1 into y = ax² + bx + c gives y = a + b + c. Similarly, the sum d + e + f is the value of the reflected parabola at x = 1. Therefore, the total sum is y_original(1) + y_reflected(1).

So if I can compute y_original(1) + y_reflected(1), that would be equal to a + b + c + d + e + f.

Let me check this.

Original parabola: y = ax² + bx + c. At x = 1, y = a + b + c.

Reflected parabola: y = dx² + ex + f. At x = 1, y = d + e + f.

Therefore, total sum is y_original(1) + y_reflected(1).

But what is y_reflected(1)?

Alternatively, since reflecting over y = k, the reflection of the original parabola. So for any point (x, y) on the original parabola, the reflected point is (x, 2k - y). Therefore, if we consider the reflected parabola, then for x = 1, the y-coordinate is 2k - y_original(1). Therefore, y_reflected(1) = 2k - y_original(1).

Therefore, the total sum is y_original(1) + y_reflected(1) = y_original(1) + 2k - y_original(1) = 2k.

Therefore, regardless of the original parabola, the sum a + b + c + d + e + f equals 2k.

This is a much more straightforward way to see it. Since the sum is y_original(1) + y_reflected(1), and because reflecting over y = k maps y_original(1) to 2k - y_original(1), their sum is 2k.

This confirms the previous result.

Therefore, the answer is 2k.

**Final Answer**
\boxed{2k}

To solve the problem, we are given a parabola with equation:

$$
y = ax^2 + bx + c
$$

and vertex $(h, k)$. This parabola is reflected over the horizontal line $y = k$, resulting in a new parabola with equation:

$$
y = dx^2 + ex + f
$$

We are asked to compute the sum:

$$
a + b + c + d + e + f
$$

in terms of $k$.

---

### Step 1: Express the Original Parabola in Vertex Form

The vertex form of a parabola is:

$$
y = a(x - h)^2 + k
$$

Expanding this:

$$
y = a(x^2 - 2hx + h^2) + k = ax^2 - 2ahx + ah^2 + k
$$

So, comparing with the standard form $y = ax^2 + bx + c$, we identify:

- $a = a$
- $b = -2ah$
- $c = ah^2 + k$

---

### Step 2: Reflect the Parabola Over the Line $y = k$

Reflecting a point $(x, y)$ over the line $y = k$ gives the point $(x, 2k - y)$. So, the reflected parabola has the equation:

$$
y = 2k - [a(x - h)^2 + k] = k - a(x - h)^2
$$

Expanding this:

$$
y = -a(x - h)^2 + k = -a(x^2 - 2hx + h^2) + k = -ax^2 + 2ahx - ah^2 + k
$$

So, the reflected parabola has coefficients:

- $d = -a$
- $e = 2ah$
- $f = -ah^2 + k$

---

### Step 3: Compute the Sum $a + b + c + d + e + f$

Now, substitute the values:

- $a = a$
- $b = -2ah$
- $c = ah^2 + k$
- $d = -a$
- $e = 2ah$
- $f = -ah^2 + k$

Add them up:

$$
a + (-2ah) + (ah^2 + k) + (-a) + (2ah) + (-ah^2 + k)
$$

Group like terms:

- $a - a = 0$
- $-2ah + 2ah = 0$
- $ah^2 - ah^2 = 0$
- $k + k = 2k$

So, the total sum is:

$$
a + b + c + d + e + f = 2k
$$

---

### Final Answer

$$
\boxed{2k}
$$

Solution:
The equation of the original parabola can be written as
\[y = a(x - h)^2 + k.\]The equation of the reflected parabola is then
\[y = -a(x - h)^2 + k.\]Thus,
\[ax^2 + bx + c + dx^2 + ex + f = 2k.\]Setting $x = 1,$ we get $a + b + c + d + e + f = \boxed{2k}.$