Explanation:
Okay, so I need to find the value of f(-2) + f(-1) + f(0) where the function f(x) is given as (3x - 2)/(x - 2). Let me start by recalling how to evaluate a function at specific points.

First, I should remember that to find f(a), I just substitute x with a in the function. So, for each of these values (-2, -1, 0), I need to plug them into the function and then add the results together. Let me write that down step by step.

Starting with f(-2). Let me substitute x = -2 into the function:

f(-2) = [3*(-2) - 2] / [(-2) - 2]

Let me compute the numerator first: 3*(-2) is -6, then subtract 2, so that's -6 - 2 = -8.

Now the denominator: (-2) - 2 is -4.

So f(-2) = -8 / -4. Dividing two negatives gives a positive, so that's 2. Okay, so f(-2) is 2. Got that.

Next, let me compute f(-1). Again, substitute x = -1 into the function:

f(-1) = [3*(-1) - 2] / [(-1) - 2]

Calculating numerator: 3*(-1) is -3, then subtract 2, so that's -3 - 2 = -5.

Denominator: (-1) - 2 is -3.

So f(-1) = -5 / -3. Again, negatives cancel, so that's 5/3. Hmm, 5 over 3. Let me check that again. Yes, numerator is -5, denominator is -3, so positive 5/3. Okay.

Now, moving on to f(0). Substitute x = 0 into the function:

f(0) = [3*(0) - 2] / [0 - 2]

Numerator: 3*0 is 0, then subtract 2, so that's 0 - 2 = -2.

Denominator: 0 - 2 is -2.

Therefore, f(0) = (-2)/(-2) = 1. That's straightforward.

So now, I have f(-2) = 2, f(-1) = 5/3, and f(0) = 1. The question asks for the sum of these three values: 2 + 5/3 + 1.

Let me compute that. First, adding 2 and 1 gives 3. Then adding 5/3 to that. So 3 + 5/3. To add these, I need a common denominator. 3 can be written as 9/3, so 9/3 + 5/3 = 14/3.

Wait, let me check that again. 3 is equal to 9/3? Wait, no. Wait, 3 is 3/1. To convert to thirds, multiply numerator and denominator by 3: 3 * 3 / 1 * 3 = 9/3. Yes, that's correct. So 9/3 + 5/3 is indeed 14/3.

So the total sum is 14/3. Let me just verify each step again to make sure I didn't make any mistakes.

Starting with f(-2): numerator 3*(-2) -2 = -6 -2 = -8. Denominator -2 -2 = -4. -8/-4 = 2. Correct.

f(-1): numerator 3*(-1) -2 = -3 -2 = -5. Denominator -1 -2 = -3. -5/-3 = 5/3. Correct.

f(0): numerator 0 -2 = -2. Denominator 0 -2 = -2. -2/-2 = 1. Correct.

Adding them up: 2 + 5/3 + 1. 2 + 1 is 3. 3 + 5/3. Converting 3 to thirds: 9/3 + 5/3 = 14/3. Yes, that seems right.

Wait, but let me check if there's another way to approach this problem. Maybe there's a shortcut or something? Sometimes with functions, there might be symmetry or properties that can be used. Let me think.

The function is f(x) = (3x - 2)/(x - 2). Is there a way to simplify this function or find a pattern? Let me try to perform the division or see if it can be rewritten.

Let me try to divide 3x - 2 by x - 2. Let me do polynomial long division. Dividing 3x - 2 by x - 2.

How many times does x - 2 go into 3x - 2? The leading term is 3x divided by x, which is 3. Multiply 3*(x - 2) = 3x - 6. Subtract that from 3x - 2:

(3x - 2) - (3x - 6) = 3x - 2 - 3x + 6 = 4. So the division gives 3 with a remainder of 4. Therefore, f(x) can be written as 3 + 4/(x - 2).

So f(x) = 3 + 4/(x - 2). Maybe this form is easier for evaluating? Let me check with one of the previous values. For example, f(-2):

Using the original expression: (3*(-2) -2)/(-2 -2) = (-8)/(-4) = 2. Using the new expression: 3 + 4/(-2 -2) = 3 + 4/(-4) = 3 - 1 = 2. Correct. Similarly, f(-1): original was 5/3. New expression: 3 + 4/(-1 -2) = 3 + 4/(-3) = 3 - 4/3 = 9/3 - 4/3 = 5/3. Correct. And f(0): original was 1. New expression: 3 + 4/(0 - 2) = 3 + 4/(-2) = 3 - 2 = 1. Correct. So this decomposition is valid.

So maybe using this form can make the addition easier? Let me try.

So f(-2) + f(-1) + f(0) = [3 + 4/(-2 -2)] + [3 + 4/(-1 -2)] + [3 + 4/(0 -2)].

Which is 3 + 3 + 3 + [4/(-4) + 4/(-3) + 4/(-2)].

So that's 9 + [ -1 - 4/3 - 2 ].

Wait, let me compute each term:

First, the constants: 3 + 3 + 3 = 9.

Then the fractions: 4/(-4) is -1, 4/(-3) is -4/3, and 4/(-2) is -2. So adding those together: -1 - 4/3 - 2. Let me compute that:

First, combine -1 and -2: that's -3. Then -3 - 4/3. To combine, convert -3 to thirds: -9/3 - 4/3 = -13/3.

Therefore, total sum is 9 + (-13/3). Convert 9 to thirds: 27/3 - 13/3 = 14/3. Which matches the previous result. So that's 14/3.

So either way, whether I compute each term separately and add them, or decompose the function first, I get 14/3. So that seems to confirm that the answer is 14/3.

But just to be thorough, let me check once again each step.

Original function: f(x) = (3x - 2)/(x - 2).

Calculating f(-2):

Numerator: 3*(-2) - 2 = -6 - 2 = -8. Denominator: -2 - 2 = -4. So -8/-4 = 2. Correct.

f(-1):

Numerator: 3*(-1) - 2 = -3 - 2 = -5. Denominator: -1 - 2 = -3. So -5/-3 = 5/3. Correct.

f(0):

Numerator: 3*0 - 2 = 0 - 2 = -2. Denominator: 0 - 2 = -2. So -2/-2 = 1. Correct.

Adding them up: 2 + 5/3 + 1.

Convert all to thirds:

2 is 6/3, 5/3 is 5/3, 1 is 3/3. So 6/3 + 5/3 + 3/3 = (6 + 5 + 3)/3 = 14/3.

Yes, same result.

Alternatively, adding 2 + 1 = 3, then 3 + 5/3. 3 is 9/3, so 9/3 + 5/3 = 14/3.

So all methods lead to 14/3. Therefore, I can be confident that the answer is 14/3.

But just to make sure, let me check if there's any possible mistake in arithmetic. For example, when I did the polynomial division, I had f(x) = 3 + 4/(x - 2). Let me verify that again.

Dividing 3x - 2 by x - 2.

Divide 3x by x, get 3. Multiply 3*(x - 2) = 3x - 6. Subtract that from 3x - 2: (3x - 2) - (3x - 6) = 3x - 2 - 3x + 6 = 4. So yes, the remainder is 4, so f(x) = 3 + 4/(x - 2). Correct.

So using that expression, when I calculated the sum, I had 9 - 1 - 4/3 - 2. Let me check that again.

Original decomposition:

Each term is 3 + 4/(x - 2). So for x = -2, -1, 0, the sum is 3 + 3 + 3 + [4/(-4) + 4/(-3) + 4/(-2)].

Which is 9 + [ -1 - 4/3 - 2 ].

Wait, 4/(-4) is -1, 4/(-3) is -4/3, 4/(-2) is -2. So adding those three terms: -1 - 4/3 - 2.

Let me compute that again:

Start with -1 - 2 = -3. Then -3 - 4/3.

Convert -3 to thirds: -9/3. Then -9/3 - 4/3 = -13/3.

So total sum is 9 - 13/3.

Convert 9 to thirds: 27/3 - 13/3 = 14/3. Correct.

So that's consistent.

Alternatively, if I think about the sum f(-2) + f(-1) + f(0) = sum [3 + 4/(x - 2)] = 3*3 + 4*[1/(-2 - 2) + 1/(-1 - 2) + 1/(0 - 2)] = 9 + 4*[ (-1/4) + (-1/3) + (-1/2) ].

Wait, hold on. Wait, the decomposition is 3 + 4/(x - 2). So when I sum over x = -2, -1, 0, the sum is 3*3 + 4*[1/(-2 - 2) + 1/(-1 - 2) + 1/(0 - 2)].

Which is 9 + 4*[ (-1/4) + (-1/3) + (-1/2) ].

Let me compute that.

First, compute the terms inside the brackets:

-1/4 -1/3 -1/2. Let me find a common denominator. The denominators are 4, 3, 2. The least common denominator is 12.

Convert each term:

-1/4 = -3/12

-1/3 = -4/12

-1/2 = -6/12

Adding them together: -3/12 -4/12 -6/12 = (-3 -4 -6)/12 = (-13)/12.

Therefore, the entire expression is 9 + 4*(-13/12) = 9 - (52/12). Simplify 52/12: divide numerator and denominator by 4: 13/3. So 9 - 13/3.

Convert 9 to thirds: 27/3 - 13/3 = 14/3. Same result.

So that's another way to get there.

Therefore, regardless of the method, the answer is 14/3.

I think that's thorough enough. I don't see any mistakes in my calculations. All methods converge to 14/3, so I can be confident that the answer is correct.

**Final Answer**
\boxed{\dfrac{14}{3}}

To evaluate the expression $ f(-2) + f(-1) + f(0) $, we are given the function:

$$
f(x) = \frac{3x - 2}{x - 2}
$$

---

### Step 1: Evaluate $ f(-2) $

$$
f(-2) = \frac{3(-2) - 2}{-2 - 2} = \frac{-6 - 2}{-4} = \frac{-8}{-4} = 2
$$

---

### Step 2: Evaluate $ f(-1) $

$$
f(-1) = \frac{3(-1) - 2}{-1 - 2} = \frac{-3 - 2}{-3} = \frac{-5}{-3} = \frac{5}{3}
$$

---

### Step 3: Evaluate $ f(0) $

$$
f(0) = \frac{3(0) - 2}{0 - 2} = \frac{-2}{-2} = 1
$$

---

### Step 4: Add the results

$$
f(-2) + f(-1) + f(0) = 2 + \frac{5}{3} + 1
$$

Convert all terms to have a common denominator of 3:

$$
2 = \frac{6}{3}, \quad 1 = \frac{3}{3}
$$

$$
\frac{6}{3} + \frac{5}{3} + \frac{3}{3} = \frac{14}{3}
$$

---

### Final Answer

$$
\boxed{\dfrac{14}{3}}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of f(-2) + f(-1) + f(0) where the function f(x) is given as (3x - 2)/(x - 2). Let me start by recalling how to evaluate a function at specific points.

First, I should remember that to find f(a), I just substitute x with a in the function. So, for each of these values (-2, -1, 0), I need to plug them into the function and then add the results together. Let me write that down step by step.

Starting with f(-2). Let me substitute x = -2 into the function:

f(-2) = [3*(-2) - 2] / [(-2) - 2]

Let me compute the numerator first: 3*(-2) is -6, then subtract 2, so that's -6 - 2 = -8.

Now the denominator: (-2) - 2 is -4.

So f(-2) = -8 / -4. Dividing two negatives gives a positive, so that's 2. Okay, so f(-2) is 2. Got that.

Next, let me compute f(-1). Again, substitute x = -1 into the function:

f(-1) = [3*(-1) - 2] / [(-1) - 2]

Calculating numerator: 3*(-1) is -3, then subtract 2, so that's -3 - 2 = -5.

Denominator: (-1) - 2 is -3.

So f(-1) = -5 / -3. Again, negatives cancel, so that's 5/3. Hmm, 5 over 3. Let me check that again. Yes, numerator is -5, denominator is -3, so positive 5/3. Okay.

Now, moving on to f(0). Substitute x = 0 into the function:

f(0) = [3*(0) - 2] / [0 - 2]

Numerator: 3*0 is 0, then subtract 2, so that's 0 - 2 = -2.

Denominator: 0 - 2 is -2.

Therefore, f(0) = (-2)/(-2) = 1. That's straightforward.

So now, I have f(-2) = 2, f(-1) = 5/3, and f(0) = 1. The question asks for the sum of these three values: 2 + 5/3 + 1.

Let me compute that. First, adding 2 and 1 gives 3. Then adding 5/3 to that. So 3 + 5/3. To add these, I need a common denominator. 3 can be written as 9/3, so 9/3 + 5/3 = 14/3.

Wait, let me check that again. 3 is equal to 9/3? Wait, no. Wait, 3 is 3/1. To convert to thirds, multiply numerator and denominator by 3: 3 * 3 / 1 * 3 = 9/3. Yes, that's correct. So 9/3 + 5/3 is indeed 14/3.

So the total sum is 14/3. Let me just verify each step again to make sure I didn't make any mistakes.

Starting with f(-2): numerator 3*(-2) -2 = -6 -2 = -8. Denominator -2 -2 = -4. -8/-4 = 2. Correct.

f(-1): numerator 3*(-1) -2 = -3 -2 = -5. Denominator -1 -2 = -3. -5/-3 = 5/3. Correct.

f(0): numerator 0 -2 = -2. Denominator 0 -2 = -2. -2/-2 = 1. Correct.

Adding them up: 2 + 5/3 + 1. 2 + 1 is 3. 3 + 5/3. Converting 3 to thirds: 9/3 + 5/3 = 14/3. Yes, that seems right.

Wait, but let me check if there's another way to approach this problem. Maybe there's a shortcut or something? Sometimes with functions, there might be symmetry or properties that can be used. Let me think.

The function is f(x) = (3x - 2)/(x - 2). Is there a way to simplify this function or find a pattern? Let me try to perform the division or see if it can be rewritten.

Let me try to divide 3x - 2 by x - 2. Let me do polynomial long division. Dividing 3x - 2 by x - 2.

How many times does x - 2 go into 3x - 2? The leading term is 3x divided by x, which is 3. Multiply 3*(x - 2) = 3x - 6. Subtract that from 3x - 2:

(3x - 2) - (3x - 6) = 3x - 2 - 3x + 6 = 4. So the division gives 3 with a remainder of 4. Therefore, f(x) can be written as 3 + 4/(x - 2).

So f(x) = 3 + 4/(x - 2). Maybe this form is easier for evaluating? Let me check with one of the previous values. For example, f(-2):

Using the original expression: (3*(-2) -2)/(-2 -2) = (-8)/(-4) = 2. Using the new expression: 3 + 4/(-2 -2) = 3 + 4/(-4) = 3 - 1 = 2. Correct. Similarly, f(-1): original was 5/3. New expression: 3 + 4/(-1 -2) = 3 + 4/(-3) = 3 - 4/3 = 9/3 - 4/3 = 5/3. Correct. And f(0): original was 1. New expression: 3 + 4/(0 - 2) = 3 + 4/(-2) = 3 - 2 = 1. Correct. So this decomposition is valid.

So maybe using this form can make the addition easier? Let me try.

So f(-2) + f(-1) + f(0) = [3 + 4/(-2 -2)] + [3 + 4/(-1 -2)] + [3 + 4/(0 -2)].

Which is 3 + 3 + 3 + [4/(-4) + 4/(-3) + 4/(-2)].

So that's 9 + [ -1 - 4/3 - 2 ].

Wait, let me compute each term:

First, the constants: 3 + 3 + 3 = 9.

Then the fractions: 4/(-4) is -1, 4/(-3) is -4/3, and 4/(-2) is -2. So adding those together: -1 - 4/3 - 2. Let me compute that:

First, combine -1 and -2: that's -3. Then -3 - 4/3. To combine, convert -3 to thirds: -9/3 - 4/3 = -13/3.

Therefore, total sum is 9 + (-13/3). Convert 9 to thirds: 27/3 - 13/3 = 14/3. Which matches the previous result. So that's 14/3.

So either way, whether I compute each term separately and add them, or decompose the function first, I get 14/3. So that seems to confirm that the answer is 14/3.

But just to be thorough, let me check once again each step.

Original function: f(x) = (3x - 2)/(x - 2).

Calculating f(-2):

Numerator: 3*(-2) - 2 = -6 - 2 = -8. Denominator: -2 - 2 = -4. So -8/-4 = 2. Correct.

f(-1):

Numerator: 3*(-1) - 2 = -3 - 2 = -5. Denominator: -1 - 2 = -3. So -5/-3 = 5/3. Correct.

f(0):

Numerator: 3*0 - 2 = 0 - 2 = -2. Denominator: 0 - 2 = -2. So -2/-2 = 1. Correct.

Adding them up: 2 + 5/3 + 1.

Convert all to thirds:

2 is 6/3, 5/3 is 5/3, 1 is 3/3. So 6/3 + 5/3 + 3/3 = (6 + 5 + 3)/3 = 14/3.

Yes, same result.

Alternatively, adding 2 + 1 = 3, then 3 + 5/3. 3 is 9/3, so 9/3 + 5/3 = 14/3.

So all methods lead to 14/3. Therefore, I can be confident that the answer is 14/3.

But just to make sure, let me check if there's any possible mistake in arithmetic. For example, when I did the polynomial division, I had f(x) = 3 + 4/(x - 2). Let me verify that again.

Dividing 3x - 2 by x - 2.

Divide 3x by x, get 3. Multiply 3*(x - 2) = 3x - 6. Subtract that from 3x - 2: (3x - 2) - (3x - 6) = 3x - 2 - 3x + 6 = 4. So yes, the remainder is 4, so f(x) = 3 + 4/(x - 2). Correct.

So using that expression, when I calculated the sum, I had 9 - 1 - 4/3 - 2. Let me check that again.

Original decomposition:

Each term is 3 + 4/(x - 2). So for x = -2, -1, 0, the sum is 3 + 3 + 3 + [4/(-4) + 4/(-3) + 4/(-2)].

Which is 9 + [ -1 - 4/3 - 2 ].

Wait, 4/(-4) is -1, 4/(-3) is -4/3, 4/(-2) is -2. So adding those three terms: -1 - 4/3 - 2.

Let me compute that again:

Start with -1 - 2 = -3. Then -3 - 4/3.

Convert -3 to thirds: -9/3. Then -9/3 - 4/3 = -13/3.

So total sum is 9 - 13/3.

Convert 9 to thirds: 27/3 - 13/3 = 14/3. Correct.

So that's consistent.

Alternatively, if I think about the sum f(-2) + f(-1) + f(0) = sum [3 + 4/(x - 2)] = 3*3 + 4*[1/(-2 - 2) + 1/(-1 - 2) + 1/(0 - 2)] = 9 + 4*[ (-1/4) + (-1/3) + (-1/2) ].

Wait, hold on. Wait, the decomposition is 3 + 4/(x - 2). So when I sum over x = -2, -1, 0, the sum is 3*3 + 4*[1/(-2 - 2) + 1/(-1 - 2) + 1/(0 - 2)].

Which is 9 + 4*[ (-1/4) + (-1/3) + (-1/2) ].

Let me compute that.

First, compute the terms inside the brackets:

-1/4 -1/3 -1/2. Let me find a common denominator. The denominators are 4, 3, 2. The least common denominator is 12.

Convert each term:

-1/4 = -3/12

-1/3 = -4/12

-1/2 = -6/12

Adding them together: -3/12 -4/12 -6/12 = (-3 -4 -6)/12 = (-13)/12.

Therefore, the entire expression is 9 + 4*(-13/12) = 9 - (52/12). Simplify 52/12: divide numerator and denominator by 4: 13/3. So 9 - 13/3.

Convert 9 to thirds: 27/3 - 13/3 = 14/3. Same result.

So that's another way to get there.

Therefore, regardless of the method, the answer is 14/3.

I think that's thorough enough. I don't see any mistakes in my calculations. All methods converge to 14/3, so I can be confident that the answer is correct.

**Final Answer**
\boxed{\dfrac{14}{3}}

To evaluate the expression $ f(-2) + f(-1) + f(0) $, we are given the function:

$$
f(x) = \frac{3x - 2}{x - 2}
$$

---

### Step 1: Evaluate $ f(-2) $

$$
f(-2) = \frac{3(-2) - 2}{-2 - 2} = \frac{-6 - 2}{-4} = \frac{-8}{-4} = 2
$$

---

### Step 2: Evaluate $ f(-1) $

$$
f(-1) = \frac{3(-1) - 2}{-1 - 2} = \frac{-3 - 2}{-3} = \frac{-5}{-3} = \frac{5}{3}
$$

---

### Step 3: Evaluate $ f(0) $

$$
f(0) = \frac{3(0) - 2}{0 - 2} = \frac{-2}{-2} = 1
$$

---

### Step 4: Add the results

$$
f(-2) + f(-1) + f(0) = 2 + \frac{5}{3} + 1
$$

Convert all terms to have a common denominator of 3:

$$
2 = \frac{6}{3}, \quad 1 = \frac{3}{3}
$$

$$
\frac{6}{3} + \frac{5}{3} + \frac{3}{3} = \frac{14}{3}
$$

---

### Final Answer

$$
\boxed{\dfrac{14}{3}}
$$

Solution: $f(-2)+f(-1)+f(0)=\frac{3(-2)-2}{-2-2}+\frac{3(-1)-2}{-1-2}+\frac{3(0)-2}{0-2}=\frac{-8}{-4}+\frac{-5}{-3}+\frac{-2}{-2}=2+\frac{5}{3}+1=\boxed{\frac{14}{3}}$


Explanation:
Okay, so I need to figure out how many positive whole-number divisors 196 has. Hmm, divisors... right. Divisors are numbers that divide into 196 without leaving a remainder. So, I guess I need to find all the numbers that can multiply together to make 196. Let me think about how to approach this systematically.

First, I remember that to find the number of divisors of a number, it's helpful to start by finding its prime factorization. Once I have the prime factors, there's a formula to calculate the number of divisors. Let me recall... I think it's something like adding one to each of the exponents in the prime factorization and then multiplying those together. Let me check if that's right. For example, if a number is factored into primes as $ p^a \times q^b \times r^c $, then the number of divisors is $ (a+1)(b+1)(c+1) $. Yeah, that sounds familiar. So, if I can find the prime factors of 196, then I can apply this formula.

Alright, let's start by factoring 196. Let me try dividing by small primes. First, check if it's even. 196 ends with a 6, which is even, so it's divisible by 2. Let me divide 196 by 2. 196 divided by 2 is 98. Okay, so 2 is a prime factor. Now, 98 is also even, so divide by 2 again. 98 divided by 2 is 49. So now we have 2^2 times 49. Now, 49... I know that's 7 squared. Let me confirm: 7 times 7 is 49. So, 49 is 7^2. Therefore, putting it all together, the prime factorization of 196 is 2 squared times 7 squared. So, $ 196 = 2^2 \times 7^2 $.

Now, applying the formula for the number of divisors. The exponents here are 2 and 2. So, adding one to each exponent gives 2 + 1 = 3 and 2 + 1 = 3. Then multiplying those together: 3 times 3 is 9. Therefore, 196 should have 9 positive divisors. Wait, let me make sure I didn't skip any steps or make a mistake in the prime factorization.

Let me verify the prime factorization again. Starting with 196. Divided by 2 is 98, divided by 2 is 49. Then 49 divided by 7 is 7, and divided by 7 is 1. So, yes, that's 2^2 * 7^2. So that seems correct. So exponents are 2 and 2. Then (2+1)(2+1) = 9. So, 9 divisors. Let me try listing them out to check if that's accurate.

Starting with 1, since 1 divides every number. Then 2, because 196 divided by 2 is 98. Then 4, since 196 divided by 4 is 49. Then 7, because 196 divided by 7 is 28. Then 14, since 196 divided by 14 is 14. Then 28, which is 196 divided by 7. Wait, but I already have 7 and 14 and 28. Let me think. Maybe I need to list them systematically.

Alternatively, since the prime factors are 2^2 and 7^2, the divisors can be formed by taking exponents of 2 from 0 to 2 and exponents of 7 from 0 to 2. So, for each combination, we can have:

- 2^0 * 7^0 = 1
- 2^1 * 7^0 = 2
- 2^2 * 7^0 = 4
- 2^0 * 7^1 = 7
- 2^1 * 7^1 = 14
- 2^2 * 7^1 = 28
- 2^0 * 7^2 = 49
- 2^1 * 7^2 = 98
- 2^2 * 7^2 = 196

So that's 9 divisors: 1, 2, 4, 7, 14, 28, 49, 98, 196. Let me count them: 1, 2, 3, 4, 5, 6, 7, 8, 9. Yep, that's nine divisors. So that matches the formula result. Therefore, the number of positive whole-number divisors of 196 is 9.

But just to be thorough, let me check if there are any other divisors that I might have missed. Let me try dividing 196 by numbers from 1 upwards and see which ones divide evenly.

Starting with 1: 196 / 1 = 196, which is an integer. So 1 is a divisor.

2: 196 / 2 = 98, integer. So 2 is a divisor.

3: 196 divided by 3 is approximately 65.333... Not an integer. So 3 is not a divisor.

4: 196 / 4 = 49, which is integer. So 4 is a divisor.

5: 196 ends with 6, so not divisible by 5. 196 / 5 = 39.2, not integer.

6: Since it's divisible by 2 but not by 3, so not divisible by 6.

7: 196 / 7 = 28, integer. So 7 is a divisor.

8: 196 divided by 8 is 24.5, not integer.

9: 196 divided by 9 is about 21.777... Not integer.

10: Ends with 6, not divisible by 10.

11: Let me check. 11 times 17 is 187, 11 times 18 is 198, which is more than 196. So no.

12: Not divisible by 3, so no.

13: 13*15 is 195, 13*15 +1 = 196, so no.

14: 196 /14 = 14, which is integer. So 14 is a divisor.

15: Not divisible by 3 or 5.

16: 16*12 is 192, 16*12 +4 = 196, so no.

17: 17*11 is 187, 17*12 is 204, so no.

18: Not divisible by 3? Wait, 18 is 2*9, so since 196 isn't divisible by 9, no.

19: 19*10 is 190, 19*10 +6=196, so no.

20: Ends with 6, not divisible by 5.

21: 21*9 is 189, 21*9 +7=196, so no.

22: 22*8 is 176, 22*9 is 198, so no.

23: 23*8 is 184, 23*8 +12=196, nope.

24: Not divisible by 3?

Wait, 24 is 8*3. Since 196 isn't divisible by 3, no.

25: Ends with 6, not divisible by 5.

26: 26*7 is 182, 26*7 +14=196, so no.

27: Not divisible by 3? Wait, 27 is 3^3. Since 196 isn't divisible by 3, no.

28: 196 /28 = 7, which is integer. So 28 is a divisor.

29: 29*6 is 174, 29*7 is 203, so nope.

30: Not divisible by 5 or 3.

...and so on. So up to 28, we have already found all the divisors. The next ones would be 49, 98, 196. Let me check 49: 196 /49 = 4, which is integer. 98: 196 /98 = 2, integer. 196: 196 /196 = 1, integer. So those are the remaining ones. So total divisors are 1, 2, 4, 7, 14, 28, 49, 98, 196. That's nine numbers. So that's consistent with the earlier count.

Therefore, I think my answer is correct. The number of positive whole-number divisors of 196 is 9. So I can confidently say the answer is 9.

**Final Answer**
The number of positive whole-number divisors of 196 is \boxed{9}.

To determine how many **positive whole-number divisors** the number **196** has, we can follow a systematic approach based on **prime factorization** and a known formula for counting divisors.

---

### Step 1: Prime Factorization of 196

We begin by breaking down 196 into its prime factors:

- 196 is even, so divide by 2:
  $ 196 \div 2 = 98 $

- 98 is also even:
  $ 98 \div 2 = 49 $

- 49 is not divisible by 2, but it is divisible by 7:
  $ 49 \div 7 = 7 $

- 7 is a prime number.

So, the prime factorization of 196 is:

$$
196 = 2^2 \times 7^2
$$

---

### Step 2: Use the Divisor Formula

If a number $ N $ has a prime factorization of the form:

$$
N = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k}
$$

Then the total number of **positive divisors** of $ N $ is given by:

$$
(a_1 + 1)(a_2 + 1)\dots(a_k + 1)
$$

In our case:

- $ a_1 = 2 $ (for the prime factor 2)
- $ a_2 = 2 $ (for the prime factor 7)

So the number of divisors is:

$$
(2 + 1)(2 + 1) = 3 \times 3 = 9
$$

---

### Step 3: Verify by Listing the Divisors

To double-check, we can list all the divisors of 196 by considering all combinations of the prime factors $ 2^a \times 7^b $, where $ a = 0, 1, 2 $ and $ b = 0, 1, 2 $:

- $ 2^0 \times 7^0 = 1 $
- $ 2^1 \times 7^0 = 2 $
- $ 2^2 \times 7^0 = 4 $
- $ 2^0 \times 7^1 = 7 $
- $ 2^1 \times 7^1 = 14 $
- $ 2^2 \times 7^1 = 28 $
- $ 2^0 \times 7^2 = 49 $
- $ 2^1 \times 7^2 = 98 $
- $ 2^2 \times 7^2 = 196 $

These are the 9 positive divisors:
$$
1, 2, 4, 7, 14, 28, 49, 98, 196
$$

---

### Final Answer

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many positive whole-number divisors 196 has. Hmm, divisors... right. Divisors are numbers that divide into 196 without leaving a remainder. So, I guess I need to find all the numbers that can multiply together to make 196. Let me think about how to approach this systematically.

First, I remember that to find the number of divisors of a number, it's helpful to start by finding its prime factorization. Once I have the prime factors, there's a formula to calculate the number of divisors. Let me recall... I think it's something like adding one to each of the exponents in the prime factorization and then multiplying those together. Let me check if that's right. For example, if a number is factored into primes as $ p^a \times q^b \times r^c $, then the number of divisors is $ (a+1)(b+1)(c+1) $. Yeah, that sounds familiar. So, if I can find the prime factors of 196, then I can apply this formula.

Alright, let's start by factoring 196. Let me try dividing by small primes. First, check if it's even. 196 ends with a 6, which is even, so it's divisible by 2. Let me divide 196 by 2. 196 divided by 2 is 98. Okay, so 2 is a prime factor. Now, 98 is also even, so divide by 2 again. 98 divided by 2 is 49. So now we have 2^2 times 49. Now, 49... I know that's 7 squared. Let me confirm: 7 times 7 is 49. So, 49 is 7^2. Therefore, putting it all together, the prime factorization of 196 is 2 squared times 7 squared. So, $ 196 = 2^2 \times 7^2 $.

Now, applying the formula for the number of divisors. The exponents here are 2 and 2. So, adding one to each exponent gives 2 + 1 = 3 and 2 + 1 = 3. Then multiplying those together: 3 times 3 is 9. Therefore, 196 should have 9 positive divisors. Wait, let me make sure I didn't skip any steps or make a mistake in the prime factorization.

Let me verify the prime factorization again. Starting with 196. Divided by 2 is 98, divided by 2 is 49. Then 49 divided by 7 is 7, and divided by 7 is 1. So, yes, that's 2^2 * 7^2. So that seems correct. So exponents are 2 and 2. Then (2+1)(2+1) = 9. So, 9 divisors. Let me try listing them out to check if that's accurate.

Starting with 1, since 1 divides every number. Then 2, because 196 divided by 2 is 98. Then 4, since 196 divided by 4 is 49. Then 7, because 196 divided by 7 is 28. Then 14, since 196 divided by 14 is 14. Then 28, which is 196 divided by 7. Wait, but I already have 7 and 14 and 28. Let me think. Maybe I need to list them systematically.

Alternatively, since the prime factors are 2^2 and 7^2, the divisors can be formed by taking exponents of 2 from 0 to 2 and exponents of 7 from 0 to 2. So, for each combination, we can have:

- 2^0 * 7^0 = 1
- 2^1 * 7^0 = 2
- 2^2 * 7^0 = 4
- 2^0 * 7^1 = 7
- 2^1 * 7^1 = 14
- 2^2 * 7^1 = 28
- 2^0 * 7^2 = 49
- 2^1 * 7^2 = 98
- 2^2 * 7^2 = 196

So that's 9 divisors: 1, 2, 4, 7, 14, 28, 49, 98, 196. Let me count them: 1, 2, 3, 4, 5, 6, 7, 8, 9. Yep, that's nine divisors. So that matches the formula result. Therefore, the number of positive whole-number divisors of 196 is 9.

But just to be thorough, let me check if there are any other divisors that I might have missed. Let me try dividing 196 by numbers from 1 upwards and see which ones divide evenly.

Starting with 1: 196 / 1 = 196, which is an integer. So 1 is a divisor.

2: 196 / 2 = 98, integer. So 2 is a divisor.

3: 196 divided by 3 is approximately 65.333... Not an integer. So 3 is not a divisor.

4: 196 / 4 = 49, which is integer. So 4 is a divisor.

5: 196 ends with 6, so not divisible by 5. 196 / 5 = 39.2, not integer.

6: Since it's divisible by 2 but not by 3, so not divisible by 6.

7: 196 / 7 = 28, integer. So 7 is a divisor.

8: 196 divided by 8 is 24.5, not integer.

9: 196 divided by 9 is about 21.777... Not integer.

10: Ends with 6, not divisible by 10.

11: Let me check. 11 times 17 is 187, 11 times 18 is 198, which is more than 196. So no.

12: Not divisible by 3, so no.

13: 13*15 is 195, 13*15 +1 = 196, so no.

14: 196 /14 = 14, which is integer. So 14 is a divisor.

15: Not divisible by 3 or 5.

16: 16*12 is 192, 16*12 +4 = 196, so no.

17: 17*11 is 187, 17*12 is 204, so no.

18: Not divisible by 3? Wait, 18 is 2*9, so since 196 isn't divisible by 9, no.

19: 19*10 is 190, 19*10 +6=196, so no.

20: Ends with 6, not divisible by 5.

21: 21*9 is 189, 21*9 +7=196, so no.

22: 22*8 is 176, 22*9 is 198, so no.

23: 23*8 is 184, 23*8 +12=196, nope.

24: Not divisible by 3?

Wait, 24 is 8*3. Since 196 isn't divisible by 3, no.

25: Ends with 6, not divisible by 5.

26: 26*7 is 182, 26*7 +14=196, so no.

27: Not divisible by 3? Wait, 27 is 3^3. Since 196 isn't divisible by 3, no.

28: 196 /28 = 7, which is integer. So 28 is a divisor.

29: 29*6 is 174, 29*7 is 203, so nope.

30: Not divisible by 5 or 3.

...and so on. So up to 28, we have already found all the divisors. The next ones would be 49, 98, 196. Let me check 49: 196 /49 = 4, which is integer. 98: 196 /98 = 2, integer. 196: 196 /196 = 1, integer. So those are the remaining ones. So total divisors are 1, 2, 4, 7, 14, 28, 49, 98, 196. That's nine numbers. So that's consistent with the earlier count.

Therefore, I think my answer is correct. The number of positive whole-number divisors of 196 is 9. So I can confidently say the answer is 9.

**Final Answer**
The number of positive whole-number divisors of 196 is \boxed{9}.

To determine how many **positive whole-number divisors** the number **196** has, we can follow a systematic approach based on **prime factorization** and a known formula for counting divisors.

---

### Step 1: Prime Factorization of 196

We begin by breaking down 196 into its prime factors:

- 196 is even, so divide by 2:
  $ 196 \div 2 = 98 $

- 98 is also even:
  $ 98 \div 2 = 49 $

- 49 is not divisible by 2, but it is divisible by 7:
  $ 49 \div 7 = 7 $

- 7 is a prime number.

So, the prime factorization of 196 is:

$$
196 = 2^2 \times 7^2
$$

---

### Step 2: Use the Divisor Formula

If a number $ N $ has a prime factorization of the form:

$$
N = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k}
$$

Then the total number of **positive divisors** of $ N $ is given by:

$$
(a_1 + 1)(a_2 + 1)\dots(a_k + 1)
$$

In our case:

- $ a_1 = 2 $ (for the prime factor 2)
- $ a_2 = 2 $ (for the prime factor 7)

So the number of divisors is:

$$
(2 + 1)(2 + 1) = 3 \times 3 = 9
$$

---

### Step 3: Verify by Listing the Divisors

To double-check, we can list all the divisors of 196 by considering all combinations of the prime factors $ 2^a \times 7^b $, where $ a = 0, 1, 2 $ and $ b = 0, 1, 2 $:

- $ 2^0 \times 7^0 = 1 $
- $ 2^1 \times 7^0 = 2 $
- $ 2^2 \times 7^0 = 4 $
- $ 2^0 \times 7^1 = 7 $
- $ 2^1 \times 7^1 = 14 $
- $ 2^2 \times 7^1 = 28 $
- $ 2^0 \times 7^2 = 49 $
- $ 2^1 \times 7^2 = 98 $
- $ 2^2 \times 7^2 = 196 $

These are the 9 positive divisors:
$$
1, 2, 4, 7, 14, 28, 49, 98, 196
$$

---

### Final Answer

$$
\boxed{9}
$$

Solution: First prime factorize $196=2^2\cdot7^2$.  The prime factorization of any divisor of 196 cannot include any primes other than 2 and 7.  We are free to choose either 0, 1, or 2 as the exponent of 2 in the prime factorization of a divisor of 196.  Similarly, we may choose 0, 1, or 2 as the exponent of 7.  In total, there are $3\times 3=9$ possibilities for the prime factorization of a divisor of 196.  Distinct prime factorizations correspond to distinct integers, so there are $\boxed{9}$ divisors of 196.


Explanation:
Okay, so I need to find the smallest positive perfect cube that can be written as the sum of three consecutive integers. Hmm, let me think about how to approach this.

First, let me recall what a perfect cube is. A perfect cube is a number that can be expressed as $ n^3 $ where $ n $ is an integer. So, examples are 1, 8, 27, 64, 125, etc. The question is asking for the smallest such number that can also be written as the sum of three consecutive integers.

Alright, so I need to find three consecutive integers whose sum is a perfect cube. Let me denote the three consecutive integers. Let me think, if I let the middle number be $ k $, then the three numbers would be $ k - 1 $, $ k $, and $ k + 1 $. Then their sum would be $ (k - 1) + k + (k + 1) $. Let me compute that:

$ (k - 1) + k + (k + 1) = k - 1 + k + k + 1 = 3k $.

Oh! So the sum of three consecutive integers is just three times the middle number. That simplifies things. Therefore, the problem reduces to finding the smallest positive perfect cube that is divisible by 3, right? Because the sum is $ 3k $, so $ 3k = n^3 $. Therefore, $ n^3 $ must be divisible by 3, which implies that $ n $ must be divisible by 3 as well.

So, if $ n $ is divisible by 3, let me write $ n = 3m $, where $ m $ is a positive integer. Then $ n^3 = (3m)^3 = 27m^3 $. Therefore, the sum $ 3k = 27m^3 $, so $ k = 9m^3 $.

Therefore, the sum of three consecutive integers is $ 27m^3 $, which is a perfect cube. Therefore, the smallest positive perfect cube would be when $ m = 1 $, which gives $ 27 \times 1^3 = 27 $.

Wait, so is 27 the answer? Let me check. Let me verify if 27 can be written as the sum of three consecutive integers.

If $ k = 9m^3 $, then when $ m = 1 $, $ k = 9 $. Therefore, the three consecutive integers would be 8, 9, 10. Let me add them up: 8 + 9 + 10 = 27. Yes, that works.

But wait, hold on. Let me check if there is a smaller perfect cube that can be expressed as the sum of three consecutive integers. The perfect cubes before 27 are 1, 8. Let me check if 1 or 8 can be written as the sum of three consecutive integers.

Starting with 1: Let me try to find three consecutive integers that add up to 1. Let me denote them as $ a $, $ a + 1 $, $ a + 2 $. Their sum is $ 3a + 3 = 3(a + 1) $. So, 3(a + 1) = 1. Then $ a + 1 = 1/3 $, so $ a = -2/3 $. But that's not an integer. Therefore, there are no three consecutive integers that add up to 1.

How about 8? Let me check. Again, the sum is $ 3(a + 1) = 8 $. Then $ a + 1 = 8/3 $, so $ a = 8/3 - 1 = 5/3 $. Again, not an integer. Therefore, 8 cannot be expressed as the sum of three consecutive integers.

Therefore, the next cube is 27, which works. Therefore, 27 is the answer? Wait, but let me think again. Is there a possibility that there is a smaller cube if we consider negative integers? Wait, the problem says "three consecutive integers," but doesn't specify they have to be positive. However, the question asks for the smallest positive perfect cube. So even if there are negative integers, the cube itself has to be positive.

But let me check if there's a smaller positive cube. For example, maybe there's a cube between 1 and 27 that can be expressed as the sum of three consecutive integers. But we saw that 1 and 8 can't. What about 64? Well, 64 is bigger than 27, so 27 is smaller.

Wait, but maybe there's another way to express the sum of three consecutive integers? Let me think. Suppose I don't take the middle number as $ k $, but instead take the first number as $ a $. Then the three numbers are $ a $, $ a + 1 $, $ a + 2 $. Their sum is $ 3a + 3 $, which is the same as before. So regardless of how I denote them, the sum is 3 times something. Therefore, the sum must be divisible by 3. Therefore, the cube must be divisible by 3, so the cube must be a multiple of 3. Therefore, the cube must be divisible by 3, so the cube root must be divisible by 3. Therefore, the smallest such cube is $ 3^3 = 27 $.

Therefore, 27 is the answer. But let me check again with another approach. Maybe there's a different way to get a smaller cube?

Alternatively, suppose I don't assume that the middle number is an integer. Wait, but the three consecutive integers must be integers, so their sum is 3 times the middle number, which must be an integer. Therefore, the sum must be divisible by 3. Therefore, the cube must be divisible by 3, so the cube must be 27, 216, etc. Therefore, 27 is the smallest.

But let me check with actual numbers. Let me try to find three consecutive integers that add up to 27. As before, 8 + 9 + 10 = 27. That works.

Alternatively, maybe there's another set of three consecutive integers that add up to a smaller cube? But since 1 and 8 don't work, and 27 is the next, then 27 is the answer.

Wait, but hold on. Let me think again. What if the three consecutive integers are not all positive? For example, could there be negative numbers that add up to a smaller positive cube? For example, let's say the sum is 1. But as we saw, that requires non-integer numbers. How about sum of 8? Also requires non-integer. How about sum of 27? That works with positive numbers.

Alternatively, maybe there's a cube that is smaller than 27, but achieved by negative numbers? For example, if we take three consecutive integers that include negative numbers, maybe their sum is a positive cube. But the cube itself has to be positive. Let me check.

For example, let me try to find three consecutive integers that add up to 1. As before, impossible. How about 8? Also impossible. How about 27? Possible. What about 64? Well, that's bigger.

Alternatively, maybe there's a cube between 1 and 27 that can be expressed as the sum of three consecutive integers. Let me check 0? But 0 is not positive. The problem says "smallest positive perfect cube," so 0 is excluded.

Alternatively, maybe there's a cube that is not divisible by 3, but somehow the sum of three consecutive integers equals it? Wait, but earlier we saw that the sum of three consecutive integers is always divisible by 3. Therefore, the cube must be divisible by 3. Therefore, the cube must be a multiple of 3. Therefore, the cube must be 27, 216, etc. Therefore, 27 is the smallest.

Therefore, I think the answer is 27.

But let me check another way. Let me suppose that the three consecutive integers are $ a - 1 $, $ a $, $ a + 1 $. Then their sum is $ 3a $. Therefore, $ 3a = n^3 $. Therefore, $ a = n^3 / 3 $. Since $ a $ must be an integer, $ n^3 $ must be divisible by 3, so $ n $ must be divisible by 3. Let $ n = 3k $, then $ n^3 = 27k^3 $, so $ a = 27k^3 / 3 = 9k^3 $. Therefore, the three numbers are $ 9k^3 - 1 $, $ 9k^3 $, $ 9k^3 + 1 $. Therefore, the smallest positive cube is when $ k = 1 $, which gives $ n^3 = 27 $.

Therefore, this seems to confirm the previous result.

But let me check for $ k = 1 $, the numbers are 8, 9, 10. Their sum is 27. Correct.

Is there a case where $ k $ is not an integer? No, because $ a $ has to be integer. Therefore, $ n $ must be divisible by 3. Therefore, the minimal cube is 27.

Therefore, I think the answer is 27.

But just to be thorough, let me check if there's another way to represent the sum of three consecutive integers. For example, maybe if the three numbers are not centered around an integer? Wait, but three consecutive integers will always have a middle number. For example, if you have three numbers, they can be represented as $ a $, $ a + 1 $, $ a + 2 $, which sum to $ 3a + 3 $, which is 3(a + 1). So regardless of how you choose them, the sum is three times some integer. Therefore, the sum must be divisible by 3. Therefore, the cube must be divisible by 3, so the cube must be 27, 216, etc.

Therefore, 27 is the smallest.

Alternatively, maybe there's a case where the three numbers are not in the form $ a $, $ a + 1 $, $ a + 2 $? No, because consecutive integers are always like that.

Alternatively, maybe considering different starting points? For example, starting from a negative number. Let me try. Let me try to find three consecutive integers that add up to 27. For example, if I take numbers like 7, 8, 9: 7 + 8 + 9 = 24. 8 + 9 + 10 = 27. So that's the one. If I take numbers like 9, 10, 11: 30. So 27 is between 24 and 30.

Alternatively, if I take negative numbers. Let me try. Let me take -1, 0, 1: sum is 0. Not helpful. How about -2, -1, 0: sum is -3. Not helpful. How about -3, -2, -1: sum is -6. Still negative. How about -4, -3, -2: sum is -9. Still negative. So if I take numbers that are negative and then some positive numbers? For example, let me try to get a positive sum. Let me try -5, -4, -3: sum is -12. Still negative. How about -1, 0, 1: 0. -2, -1, 0: -3. -1, 0, 1: 0. -0, 1, 2: 3. Wait, but 0 is not positive? Wait, the problem says "three consecutive integers," so they can be negative, zero, or positive. But the sum has to be a positive perfect cube.

So, for example, if I take numbers like 0, 1, 2: sum is 3. Which is not a cube. 1, 2, 3: sum is 6. Not a cube. 2, 3, 4: 9. Not a cube. 3, 4, 5: 12. Not a cube. 4, 5, 6: 15. Not a cube. 5, 6, 7: 18. Not a cube. 6, 7, 8: 21. Not a cube. 7, 8, 9: 24. Not a cube. 8, 9, 10: 27. There we go. So that's the first one.

So, even if I check all the sums starting from negative numbers, the first positive cube I get is 27. Therefore, 27 is indeed the answer.

But let me check another angle. Suppose someone thinks that maybe the three numbers are not in the middle of the cube? Wait, but the logic is that the sum is 3 times the middle number. Therefore, the cube must be divisible by 3. Therefore, the cube must be 27, 216, etc. Therefore, 27 is the smallest.

Alternatively, maybe there's a different way to express the sum? For example, if the three numbers are not equally spaced? But they have to be consecutive integers, so they are equally spaced by 1.

Alternatively, maybe the problem is considering something else? Let me check the problem statement again: "the smallest positive perfect cube that can be written as the sum of three consecutive integers."

So, the problem is straightforward. The sum of three consecutive integers is 3 times the middle number. Therefore, the cube must be divisible by 3, so the cube is 27, which is 3^3. Therefore, the answer is 27.

But just to be 100% sure, let me check if there's a cube between 1 and 27 that can be expressed as the sum of three consecutive integers.

For example, let me check 8. If 8 is the sum of three consecutive integers, then 3a + 3 = 8 => 3a = 5 => a = 5/3. Not integer.

For 1: 3a + 3 = 1 => 3a = -2 => a = -2/3. Not integer.

For 0: 3a + 3 = 0 => a = -1. So the numbers would be -2, -1, 0. Their sum is -3, not 0. Wait, wait, if a = -1, then numbers are -1 -1 = -2, -1, -1 + 1 = 0. So sum is -2 + (-1) + 0 = -3. Wait, but according to the formula, 3a + 3 = 3*(-1) + 3 = 0. Wait, but the actual sum is -3. Wait, there's a contradiction here. Wait, what's wrong?

Wait, earlier I said that if the three numbers are $ a $, $ a + 1 $, $ a + 2 $, then their sum is $ 3a + 3 $. But if $ a = -1 $, then the numbers are -1, 0, 1. Their sum is -1 + 0 + 1 = 0. Which is equal to $ 3a + 3 = 3*(-1) + 3 = 0 $. So that works. Wait, earlier I thought the numbers were -2, -1, 0, but that's if $ a = -2 $. Let me check:

If $ a = -2 $, then numbers are -2, -1, 0. Their sum is -3. Which is equal to $ 3a + 3 = 3*(-2) + 3 = -6 + 3 = -3 $. So that works. So if I set $ a = -1 $, the numbers are -1, 0, 1, sum is 0. So if the problem allows the sum to be zero, but the question says "smallest positive perfect cube," so zero is not positive. Therefore, even though 0 can be expressed as the sum of three consecutive integers, it's not positive.

Therefore, the next cube is 1, which we saw can't be expressed as such. Then 8, which also can't. Then 27, which can. Therefore, 27 is the answer.

Therefore, I think I can be confident that the answer is 27.

But just to make sure, let me check if there's a different interpretation of the problem. For example, maybe the problem is referring to three consecutive integers in a different way? For example, maybe not starting from an integer? But no, integers are whole numbers.

Alternatively, maybe the problem is considering three consecutive numbers in some other sequence? But no, consecutive integers are numbers that follow each other in order without gaps.

Alternatively, maybe the problem is in another base? But the question doesn't mention anything about bases, so I think it's safe to assume base 10.

Alternatively, maybe there's a mistake in assuming that the sum is 3 times the middle number? Let me check with an example. Let me take numbers 1, 2, 3. Sum is 6. Which is 3*2. Correct. Numbers 2, 3, 4: sum is 9, which is 3*3. Correct. Numbers 4, 5, 6: sum is 15, which is 3*5. Correct. So yes, the formula holds. Therefore, the sum is always three times the middle number.

Therefore, the conclusion is that the smallest positive perfect cube that can be written as the sum of three consecutive integers is 27.

Therefore, the answer is \boxed{27}.

**Final Answer**
The smallest positive perfect cube that can be written as the sum of three consecutive integers is \boxed{27}.

To solve the problem of finding the **smallest positive perfect cube** that can be written as the **sum of three consecutive integers**, we begin by analyzing the structure of such a sum.

---

### Step 1: Express the Sum of Three Consecutive Integers

Let the three consecutive integers be:

$$
a - 1, \quad a, \quad a + 1
$$

Their sum is:

$$
(a - 1) + a + (a + 1) = 3a
$$

So, the sum of any three consecutive integers is always **three times the middle number**.

---

### Step 2: Set the Sum Equal to a Perfect Cube

We are looking for a **positive perfect cube** that can be written in this form. Let the perfect cube be $ n^3 $, and set:

$$
3a = n^3
$$

This implies that $ n^3 $ must be divisible by 3. Since 3 is a prime number, this means that $ n $ itself must be divisible by 3. Let:

$$
n = 3k \quad \text{for some integer } k
$$

Then:

$$
n^3 = (3k)^3 = 27k^3
$$

So, the smallest such cube occurs when $ k = 1 $, giving:

$$
n^3 = 27 \times 1^3 = 27
$$

---

### Step 3: Verify the Result

We now check if 27 can be written as the sum of three consecutive integers. Let’s find the middle number $ a $ such that:

$$
3a = 27 \Rightarrow a = 9
$$

So the three consecutive integers are:

$$
8, \quad 9, \quad 10
$$

Their sum is:

$$
8 + 9 + 10 = 27
$$

This confirms that 27 is indeed the sum of three consecutive integers.

---

### Step 4: Check for Smaller Perfect Cubes

The perfect cubes less than 27 are:

- $ 1^3 = 1 $
- $ 2^3 = 8 $

Let’s check if these can be written as the sum of three consecutive integers.

- For $ 1 $: $ 3a = 1 \Rightarrow a = \frac{1}{3} $ → not an integer
- For $ 8 $: $ 3a = 8 \Rightarrow a = \frac{8}{3} $ → not an integer

So, neither 1 nor 8 can be written as the sum of three consecutive integers.

---

### Final Conclusion

The smallest positive perfect cube that can be written as the sum of three consecutive integers is:

$$
\boxed{27}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest positive perfect cube that can be written as the sum of three consecutive integers. Hmm, let me think about how to approach this.

First, let me recall what a perfect cube is. A perfect cube is a number that can be expressed as $ n^3 $ where $ n $ is an integer. So, examples are 1, 8, 27, 64, 125, etc. The question is asking for the smallest such number that can also be written as the sum of three consecutive integers.

Alright, so I need to find three consecutive integers whose sum is a perfect cube. Let me denote the three consecutive integers. Let me think, if I let the middle number be $ k $, then the three numbers would be $ k - 1 $, $ k $, and $ k + 1 $. Then their sum would be $ (k - 1) + k + (k + 1) $. Let me compute that:

$ (k - 1) + k + (k + 1) = k - 1 + k + k + 1 = 3k $.

Oh! So the sum of three consecutive integers is just three times the middle number. That simplifies things. Therefore, the problem reduces to finding the smallest positive perfect cube that is divisible by 3, right? Because the sum is $ 3k $, so $ 3k = n^3 $. Therefore, $ n^3 $ must be divisible by 3, which implies that $ n $ must be divisible by 3 as well.

So, if $ n $ is divisible by 3, let me write $ n = 3m $, where $ m $ is a positive integer. Then $ n^3 = (3m)^3 = 27m^3 $. Therefore, the sum $ 3k = 27m^3 $, so $ k = 9m^3 $.

Therefore, the sum of three consecutive integers is $ 27m^3 $, which is a perfect cube. Therefore, the smallest positive perfect cube would be when $ m = 1 $, which gives $ 27 \times 1^3 = 27 $.

Wait, so is 27 the answer? Let me check. Let me verify if 27 can be written as the sum of three consecutive integers.

If $ k = 9m^3 $, then when $ m = 1 $, $ k = 9 $. Therefore, the three consecutive integers would be 8, 9, 10. Let me add them up: 8 + 9 + 10 = 27. Yes, that works.

But wait, hold on. Let me check if there is a smaller perfect cube that can be expressed as the sum of three consecutive integers. The perfect cubes before 27 are 1, 8. Let me check if 1 or 8 can be written as the sum of three consecutive integers.

Starting with 1: Let me try to find three consecutive integers that add up to 1. Let me denote them as $ a $, $ a + 1 $, $ a + 2 $. Their sum is $ 3a + 3 = 3(a + 1) $. So, 3(a + 1) = 1. Then $ a + 1 = 1/3 $, so $ a = -2/3 $. But that's not an integer. Therefore, there are no three consecutive integers that add up to 1.

How about 8? Let me check. Again, the sum is $ 3(a + 1) = 8 $. Then $ a + 1 = 8/3 $, so $ a = 8/3 - 1 = 5/3 $. Again, not an integer. Therefore, 8 cannot be expressed as the sum of three consecutive integers.

Therefore, the next cube is 27, which works. Therefore, 27 is the answer? Wait, but let me think again. Is there a possibility that there is a smaller cube if we consider negative integers? Wait, the problem says "three consecutive integers," but doesn't specify they have to be positive. However, the question asks for the smallest positive perfect cube. So even if there are negative integers, the cube itself has to be positive.

But let me check if there's a smaller positive cube. For example, maybe there's a cube between 1 and 27 that can be expressed as the sum of three consecutive integers. But we saw that 1 and 8 can't. What about 64? Well, 64 is bigger than 27, so 27 is smaller.

Wait, but maybe there's another way to express the sum of three consecutive integers? Let me think. Suppose I don't take the middle number as $ k $, but instead take the first number as $ a $. Then the three numbers are $ a $, $ a + 1 $, $ a + 2 $. Their sum is $ 3a + 3 $, which is the same as before. So regardless of how I denote them, the sum is 3 times something. Therefore, the sum must be divisible by 3. Therefore, the cube must be divisible by 3, so the cube must be a multiple of 3. Therefore, the cube must be divisible by 3, so the cube root must be divisible by 3. Therefore, the smallest such cube is $ 3^3 = 27 $.

Therefore, 27 is the answer. But let me check again with another approach. Maybe there's a different way to get a smaller cube?

Alternatively, suppose I don't assume that the middle number is an integer. Wait, but the three consecutive integers must be integers, so their sum is 3 times the middle number, which must be an integer. Therefore, the sum must be divisible by 3. Therefore, the cube must be divisible by 3, so the cube must be 27, 216, etc. Therefore, 27 is the smallest.

But let me check with actual numbers. Let me try to find three consecutive integers that add up to 27. As before, 8 + 9 + 10 = 27. That works.

Alternatively, maybe there's another set of three consecutive integers that add up to a smaller cube? But since 1 and 8 don't work, and 27 is the next, then 27 is the answer.

Wait, but hold on. Let me think again. What if the three consecutive integers are not all positive? For example, could there be negative numbers that add up to a smaller positive cube? For example, let's say the sum is 1. But as we saw, that requires non-integer numbers. How about sum of 8? Also requires non-integer. How about sum of 27? That works with positive numbers.

Alternatively, maybe there's a cube that is smaller than 27, but achieved by negative numbers? For example, if we take three consecutive integers that include negative numbers, maybe their sum is a positive cube. But the cube itself has to be positive. Let me check.

For example, let me try to find three consecutive integers that add up to 1. As before, impossible. How about 8? Also impossible. How about 27? Possible. What about 64? Well, that's bigger.

Alternatively, maybe there's a cube between 1 and 27 that can be expressed as the sum of three consecutive integers. Let me check 0? But 0 is not positive. The problem says "smallest positive perfect cube," so 0 is excluded.

Alternatively, maybe there's a cube that is not divisible by 3, but somehow the sum of three consecutive integers equals it? Wait, but earlier we saw that the sum of three consecutive integers is always divisible by 3. Therefore, the cube must be divisible by 3. Therefore, the cube must be a multiple of 3. Therefore, the cube must be 27, 216, etc. Therefore, 27 is the smallest.

Therefore, I think the answer is 27.

But let me check another way. Let me suppose that the three consecutive integers are $ a - 1 $, $ a $, $ a + 1 $. Then their sum is $ 3a $. Therefore, $ 3a = n^3 $. Therefore, $ a = n^3 / 3 $. Since $ a $ must be an integer, $ n^3 $ must be divisible by 3, so $ n $ must be divisible by 3. Let $ n = 3k $, then $ n^3 = 27k^3 $, so $ a = 27k^3 / 3 = 9k^3 $. Therefore, the three numbers are $ 9k^3 - 1 $, $ 9k^3 $, $ 9k^3 + 1 $. Therefore, the smallest positive cube is when $ k = 1 $, which gives $ n^3 = 27 $.

Therefore, this seems to confirm the previous result.

But let me check for $ k = 1 $, the numbers are 8, 9, 10. Their sum is 27. Correct.

Is there a case where $ k $ is not an integer? No, because $ a $ has to be integer. Therefore, $ n $ must be divisible by 3. Therefore, the minimal cube is 27.

Therefore, I think the answer is 27.

But just to be thorough, let me check if there's another way to represent the sum of three consecutive integers. For example, maybe if the three numbers are not centered around an integer? Wait, but three consecutive integers will always have a middle number. For example, if you have three numbers, they can be represented as $ a $, $ a + 1 $, $ a + 2 $, which sum to $ 3a + 3 $, which is 3(a + 1). So regardless of how you choose them, the sum is three times some integer. Therefore, the sum must be divisible by 3. Therefore, the cube must be divisible by 3, so the cube must be 27, 216, etc.

Therefore, 27 is the smallest.

Alternatively, maybe there's a case where the three numbers are not in the form $ a $, $ a + 1 $, $ a + 2 $? No, because consecutive integers are always like that.

Alternatively, maybe considering different starting points? For example, starting from a negative number. Let me try. Let me try to find three consecutive integers that add up to 27. For example, if I take numbers like 7, 8, 9: 7 + 8 + 9 = 24. 8 + 9 + 10 = 27. So that's the one. If I take numbers like 9, 10, 11: 30. So 27 is between 24 and 30.

Alternatively, if I take negative numbers. Let me try. Let me take -1, 0, 1: sum is 0. Not helpful. How about -2, -1, 0: sum is -3. Not helpful. How about -3, -2, -1: sum is -6. Still negative. How about -4, -3, -2: sum is -9. Still negative. So if I take numbers that are negative and then some positive numbers? For example, let me try to get a positive sum. Let me try -5, -4, -3: sum is -12. Still negative. How about -1, 0, 1: 0. -2, -1, 0: -3. -1, 0, 1: 0. -0, 1, 2: 3. Wait, but 0 is not positive? Wait, the problem says "three consecutive integers," so they can be negative, zero, or positive. But the sum has to be a positive perfect cube.

So, for example, if I take numbers like 0, 1, 2: sum is 3. Which is not a cube. 1, 2, 3: sum is 6. Not a cube. 2, 3, 4: 9. Not a cube. 3, 4, 5: 12. Not a cube. 4, 5, 6: 15. Not a cube. 5, 6, 7: 18. Not a cube. 6, 7, 8: 21. Not a cube. 7, 8, 9: 24. Not a cube. 8, 9, 10: 27. There we go. So that's the first one.

So, even if I check all the sums starting from negative numbers, the first positive cube I get is 27. Therefore, 27 is indeed the answer.

But let me check another angle. Suppose someone thinks that maybe the three numbers are not in the middle of the cube? Wait, but the logic is that the sum is 3 times the middle number. Therefore, the cube must be divisible by 3. Therefore, the cube must be 27, 216, etc. Therefore, 27 is the smallest.

Alternatively, maybe there's a different way to express the sum? For example, if the three numbers are not equally spaced? But they have to be consecutive integers, so they are equally spaced by 1.

Alternatively, maybe the problem is considering something else? Let me check the problem statement again: "the smallest positive perfect cube that can be written as the sum of three consecutive integers."

So, the problem is straightforward. The sum of three consecutive integers is 3 times the middle number. Therefore, the cube must be divisible by 3, so the cube is 27, which is 3^3. Therefore, the answer is 27.

But just to be 100% sure, let me check if there's a cube between 1 and 27 that can be expressed as the sum of three consecutive integers.

For example, let me check 8. If 8 is the sum of three consecutive integers, then 3a + 3 = 8 => 3a = 5 => a = 5/3. Not integer.

For 1: 3a + 3 = 1 => 3a = -2 => a = -2/3. Not integer.

For 0: 3a + 3 = 0 => a = -1. So the numbers would be -2, -1, 0. Their sum is -3, not 0. Wait, wait, if a = -1, then numbers are -1 -1 = -2, -1, -1 + 1 = 0. So sum is -2 + (-1) + 0 = -3. Wait, but according to the formula, 3a + 3 = 3*(-1) + 3 = 0. Wait, but the actual sum is -3. Wait, there's a contradiction here. Wait, what's wrong?

Wait, earlier I said that if the three numbers are $ a $, $ a + 1 $, $ a + 2 $, then their sum is $ 3a + 3 $. But if $ a = -1 $, then the numbers are -1, 0, 1. Their sum is -1 + 0 + 1 = 0. Which is equal to $ 3a + 3 = 3*(-1) + 3 = 0 $. So that works. Wait, earlier I thought the numbers were -2, -1, 0, but that's if $ a = -2 $. Let me check:

If $ a = -2 $, then numbers are -2, -1, 0. Their sum is -3. Which is equal to $ 3a + 3 = 3*(-2) + 3 = -6 + 3 = -3 $. So that works. So if I set $ a = -1 $, the numbers are -1, 0, 1, sum is 0. So if the problem allows the sum to be zero, but the question says "smallest positive perfect cube," so zero is not positive. Therefore, even though 0 can be expressed as the sum of three consecutive integers, it's not positive.

Therefore, the next cube is 1, which we saw can't be expressed as such. Then 8, which also can't. Then 27, which can. Therefore, 27 is the answer.

Therefore, I think I can be confident that the answer is 27.

But just to make sure, let me check if there's a different interpretation of the problem. For example, maybe the problem is referring to three consecutive integers in a different way? For example, maybe not starting from an integer? But no, integers are whole numbers.

Alternatively, maybe the problem is considering three consecutive numbers in some other sequence? But no, consecutive integers are numbers that follow each other in order without gaps.

Alternatively, maybe the problem is in another base? But the question doesn't mention anything about bases, so I think it's safe to assume base 10.

Alternatively, maybe there's a mistake in assuming that the sum is 3 times the middle number? Let me check with an example. Let me take numbers 1, 2, 3. Sum is 6. Which is 3*2. Correct. Numbers 2, 3, 4: sum is 9, which is 3*3. Correct. Numbers 4, 5, 6: sum is 15, which is 3*5. Correct. So yes, the formula holds. Therefore, the sum is always three times the middle number.

Therefore, the conclusion is that the smallest positive perfect cube that can be written as the sum of three consecutive integers is 27.

Therefore, the answer is \boxed{27}.

**Final Answer**
The smallest positive perfect cube that can be written as the sum of three consecutive integers is \boxed{27}.

To solve the problem of finding the **smallest positive perfect cube** that can be written as the **sum of three consecutive integers**, we begin by analyzing the structure of such a sum.

---

### Step 1: Express the Sum of Three Consecutive Integers

Let the three consecutive integers be:

$$
a - 1, \quad a, \quad a + 1
$$

Their sum is:

$$
(a - 1) + a + (a + 1) = 3a
$$

So, the sum of any three consecutive integers is always **three times the middle number**.

---

### Step 2: Set the Sum Equal to a Perfect Cube

We are looking for a **positive perfect cube** that can be written in this form. Let the perfect cube be $ n^3 $, and set:

$$
3a = n^3
$$

This implies that $ n^3 $ must be divisible by 3. Since 3 is a prime number, this means that $ n $ itself must be divisible by 3. Let:

$$
n = 3k \quad \text{for some integer } k
$$

Then:

$$
n^3 = (3k)^3 = 27k^3
$$

So, the smallest such cube occurs when $ k = 1 $, giving:

$$
n^3 = 27 \times 1^3 = 27
$$

---

### Step 3: Verify the Result

We now check if 27 can be written as the sum of three consecutive integers. Let’s find the middle number $ a $ such that:

$$
3a = 27 \Rightarrow a = 9
$$

So the three consecutive integers are:

$$
8, \quad 9, \quad 10
$$

Their sum is:

$$
8 + 9 + 10 = 27
$$

This confirms that 27 is indeed the sum of three consecutive integers.

---

### Step 4: Check for Smaller Perfect Cubes

The perfect cubes less than 27 are:

- $ 1^3 = 1 $
- $ 2^3 = 8 $

Let’s check if these can be written as the sum of three consecutive integers.

- For $ 1 $: $ 3a = 1 \Rightarrow a = \frac{1}{3} $ → not an integer
- For $ 8 $: $ 3a = 8 \Rightarrow a = \frac{8}{3} $ → not an integer

So, neither 1 nor 8 can be written as the sum of three consecutive integers.

---

### Final Conclusion

The smallest positive perfect cube that can be written as the sum of three consecutive integers is:

$$
\boxed{27}
$$

Solution: The sum of three consecutive integers takes the form $(k-1)+(k)+(k+1)=3k$ and hence is a multiple of 3.  Conversely, if a number $n$ is a multiple of 3, then $n/3-1$, $n/3$, and $n/3+1$ are three consecutive integers that sum to give $n$.  Therefore, a number is a sum of three consecutive integers if and only if it is a multiple of 3.  The smallest positive perfect cube that is a multiple of 3 is $3^3=\boxed{27}$.


Explanation:
Okay, so I need to find the distance between the two points (2, -6) and (-4, 3). Hmm, distance formula, right? Let me remember. I think the distance formula is derived from the Pythagorean theorem. Yeah, if you have two points in a plane, the distance between them is the square root of the sum of the squares of the differences in their coordinates. Let me write that down.

The formula is: distance = √[(x2 - x1)² + (y2 - y1)²]. Let me make sure I got that right. So, if I have two points, (x1, y1) and (x2, y2), then the distance between them is the square root of [(x2 - x1) squared plus (y2 - y1) squared]. Yeah, that sounds familiar from the Pythagorean theorem. So, if I imagine a right triangle where the line connecting the two points is the hypotenuse, then the legs are the differences in the x-coordinates and the y-coordinates. So, that makes sense.

Alright, so let me assign the points. Let me take (2, -6) as (x1, y1) and (-4, 3) as (x2, y2). Wait, actually, it doesn't matter which one is first or second because squaring will take care of the sign. But just to be consistent, let me stick with that.

So, x1 is 2, y1 is -6, x2 is -4, y2 is 3. Let me compute the differences first. The difference in x-coordinates is x2 - x1, which is (-4) - 2. Let me calculate that. -4 minus 2 is... -6. Okay, so that's -6. Then the difference in y-coordinates is y2 - y1, which is 3 - (-6). Hmm, subtracting a negative is adding a positive, so that's 3 + 6, which is 9. So, the differences are -6 and 9.

Now, I need to square both of these differences. Let me do that. (-6) squared is... 36. And 9 squared is 81. Then, add those two results together. 36 + 81. Let me add those. 36 + 80 is 116, plus 1 is 117. So, the sum is 117. Then, the distance is the square root of 117. Wait, but the question says to express the answer in simplest radical form. So, I need to simplify √117.

Let me think. 117 can be factored into... Let me see. 117 divided by 3 is 39. Then 39 divided by 3 is 13. So, 117 is 3 times 3 times 13, which is 9 times 13. So, √117 is √(9*13). And since √(ab) is √a * √b, that becomes √9 * √13. √9 is 3, so that simplifies to 3√13. So, the simplified radical form is 3√13. Let me check if that's correct.

Wait, let me verify my steps again. First, coordinates: (2, -6) and (-4, 3). Difference in x: -4 - 2 = -6. Difference in y: 3 - (-6) = 9. Squared differences: (-6)^2 = 36, 9^2 = 81. Sum: 36 + 81 = 117. Square root of 117. Factoring 117: 117 divided by 3 is 39, divided by 3 again is 13. So, 3*3*13. So, √(9*13) = 3√13. Yes, that seems right.

But let me check if I did the differences correctly. Sometimes I mix up the order. Let me try switching the points. Let me take (x1, y1) as (-4, 3) and (x2, y2) as (2, -6). Then, x2 - x1 is 2 - (-4) = 2 + 4 = 6. And y2 - y1 is -6 - 3 = -9. Then, squaring those: 6^2 = 36, (-9)^2 = 81. Same result. So, adding them gives 117. So, same answer. So, regardless of which point is first or second, the differences squared are the same. So, that's good. So, the distance is √117, which simplifies to 3√13.

Wait, just to make sure that 3√13 is indeed the simplest form. Let me check if 13 is a prime number. Yes, 13 is a prime number, so it doesn't have any square factors other than 1. Therefore, √13 cannot be simplified further. Therefore, 3√13 is the simplest radical form.

So, I think that's the answer. Let me just recap all steps once again to make sure I didn't skip anything or make a calculation error.

Starting with points (2, -6) and (-4, 3). Calculated the differences in x: -4 - 2 = -6. Difference in y: 3 - (-6) = 9. Squared both: (-6)^2 = 36, 9^2 = 81. Added them: 36 + 81 = 117. Square root of 117 simplifies to 3√13. All steps check out.

Alternatively, maybe I can check using another method. Let me try plugging into the distance formula again.

Distance = √[(x2 - x1)^2 + (y2 - y1)^2] = √[(-4 - 2)^2 + (3 - (-6))^2] = √[(-6)^2 + (9)^2] = √[36 + 81] = √117 = 3√13. Yep, same result.

Alternatively, maybe I can compute the distance numerically to check. Let me compute √117. Let me approximate √117. Well, 10^2 is 100, 11^2 is 121. So, √117 is between 10 and 11. Let me compute 10.8^2 = 116.64. 10.8^2 = (10 + 0.8)^2 = 10^2 + 2*10*0.8 + 0.8^2 = 100 + 16 + 0.64 = 116.64. Then, 10.8^2 = 116.64. So, √117 is approximately 10.8 + (117 - 116.64)/(2*10.8). Using linear approximation. The difference is 117 - 116.64 = 0.36. The derivative of √x is 1/(2√x). So, approximate √(116.64 + 0.36) ≈ √116.64 + 0.36/(2*√116.64). Which is 10.8 + 0.36/(2*10.8) = 10.8 + 0.36/21.6 = 10.8 + 0.016666... ≈ 10.816666...

So, approximately 10.8167. Let me check 3√13. √13 is approximately 3.6055. So, 3*3.6055 ≈ 10.8165. Which matches the previous approximation. So, that seems correct. Therefore, 3√13 is the exact value, and the approximate decimal is about 10.816.

Therefore, I think my answer is correct. So, the distance between the two points is 3√13 units.

**Final Answer**
The distance between the points is \boxed{3\sqrt{13}} units.

To find the distance between the points $(2, -6)$ and $(-4, 3)$, we use the **distance formula**, which is derived from the Pythagorean Theorem:

$$
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

---

### Step 1: Identify the coordinates

Let:
- $(x_1, y_1) = (2, -6)$
- $(x_2, y_2) = (-4, 3)$

---

### Step 2: Compute the differences

$$
x_2 - x_1 = -4 - 2 = -6
$$
$$
y_2 - y_1 = 3 - (-6) = 3 + 6 = 9
$$

---

### Step 3: Square the differences

$$
(-6)^2 = 36
$$
$$
9^2 = 81
$$

---

### Step 4: Add the squares

$$
36 + 81 = 117
$$

---

### Step 5: Take the square root

$$
\sqrt{117}
$$

Now, simplify the radical:

$$
117 = 9 \times 13
$$
$$
\sqrt{117} = \sqrt{9 \times 13} = \sqrt{9} \cdot \sqrt{13} = 3\sqrt{13}
$$

---

### Final Answer

$$
\boxed{3\sqrt{13}}
$$

[Assistant]
Explanation:
Okay, so I need to find the distance between the two points (2, -6) and (-4, 3). Hmm, distance formula, right? Let me remember. I think the distance formula is derived from the Pythagorean theorem. Yeah, if you have two points in a plane, the distance between them is the square root of the sum of the squares of the differences in their coordinates. Let me write that down.

The formula is: distance = √[(x2 - x1)² + (y2 - y1)²]. Let me make sure I got that right. So, if I have two points, (x1, y1) and (x2, y2), then the distance between them is the square root of [(x2 - x1) squared plus (y2 - y1) squared]. Yeah, that sounds familiar from the Pythagorean theorem. So, if I imagine a right triangle where the line connecting the two points is the hypotenuse, then the legs are the differences in the x-coordinates and the y-coordinates. So, that makes sense.

Alright, so let me assign the points. Let me take (2, -6) as (x1, y1) and (-4, 3) as (x2, y2). Wait, actually, it doesn't matter which one is first or second because squaring will take care of the sign. But just to be consistent, let me stick with that.

So, x1 is 2, y1 is -6, x2 is -4, y2 is 3. Let me compute the differences first. The difference in x-coordinates is x2 - x1, which is (-4) - 2. Let me calculate that. -4 minus 2 is... -6. Okay, so that's -6. Then the difference in y-coordinates is y2 - y1, which is 3 - (-6). Hmm, subtracting a negative is adding a positive, so that's 3 + 6, which is 9. So, the differences are -6 and 9.

Now, I need to square both of these differences. Let me do that. (-6) squared is... 36. And 9 squared is 81. Then, add those two results together. 36 + 81. Let me add those. 36 + 80 is 116, plus 1 is 117. So, the sum is 117. Then, the distance is the square root of 117. Wait, but the question says to express the answer in simplest radical form. So, I need to simplify √117.

Let me think. 117 can be factored into... Let me see. 117 divided by 3 is 39. Then 39 divided by 3 is 13. So, 117 is 3 times 3 times 13, which is 9 times 13. So, √117 is √(9*13). And since √(ab) is √a * √b, that becomes √9 * √13. √9 is 3, so that simplifies to 3√13. So, the simplified radical form is 3√13. Let me check if that's correct.

Wait, let me verify my steps again. First, coordinates: (2, -6) and (-4, 3). Difference in x: -4 - 2 = -6. Difference in y: 3 - (-6) = 9. Squared differences: (-6)^2 = 36, 9^2 = 81. Sum: 36 + 81 = 117. Square root of 117. Factoring 117: 117 divided by 3 is 39, divided by 3 again is 13. So, 3*3*13. So, √(9*13) = 3√13. Yes, that seems right.

But let me check if I did the differences correctly. Sometimes I mix up the order. Let me try switching the points. Let me take (x1, y1) as (-4, 3) and (x2, y2) as (2, -6). Then, x2 - x1 is 2 - (-4) = 2 + 4 = 6. And y2 - y1 is -6 - 3 = -9. Then, squaring those: 6^2 = 36, (-9)^2 = 81. Same result. So, adding them gives 117. So, same answer. So, regardless of which point is first or second, the differences squared are the same. So, that's good. So, the distance is √117, which simplifies to 3√13.

Wait, just to make sure that 3√13 is indeed the simplest form. Let me check if 13 is a prime number. Yes, 13 is a prime number, so it doesn't have any square factors other than 1. Therefore, √13 cannot be simplified further. Therefore, 3√13 is the simplest radical form.

So, I think that's the answer. Let me just recap all steps once again to make sure I didn't skip anything or make a calculation error.

Starting with points (2, -6) and (-4, 3). Calculated the differences in x: -4 - 2 = -6. Difference in y: 3 - (-6) = 9. Squared both: (-6)^2 = 36, 9^2 = 81. Added them: 36 + 81 = 117. Square root of 117 simplifies to 3√13. All steps check out.

Alternatively, maybe I can check using another method. Let me try plugging into the distance formula again.

Distance = √[(x2 - x1)^2 + (y2 - y1)^2] = √[(-4 - 2)^2 + (3 - (-6))^2] = √[(-6)^2 + (9)^2] = √[36 + 81] = √117 = 3√13. Yep, same result.

Alternatively, maybe I can compute the distance numerically to check. Let me compute √117. Let me approximate √117. Well, 10^2 is 100, 11^2 is 121. So, √117 is between 10 and 11. Let me compute 10.8^2 = 116.64. 10.8^2 = (10 + 0.8)^2 = 10^2 + 2*10*0.8 + 0.8^2 = 100 + 16 + 0.64 = 116.64. Then, 10.8^2 = 116.64. So, √117 is approximately 10.8 + (117 - 116.64)/(2*10.8). Using linear approximation. The difference is 117 - 116.64 = 0.36. The derivative of √x is 1/(2√x). So, approximate √(116.64 + 0.36) ≈ √116.64 + 0.36/(2*√116.64). Which is 10.8 + 0.36/(2*10.8) = 10.8 + 0.36/21.6 = 10.8 + 0.016666... ≈ 10.816666...

So, approximately 10.8167. Let me check 3√13. √13 is approximately 3.6055. So, 3*3.6055 ≈ 10.8165. Which matches the previous approximation. So, that seems correct. Therefore, 3√13 is the exact value, and the approximate decimal is about 10.816.

Therefore, I think my answer is correct. So, the distance between the two points is 3√13 units.

**Final Answer**
The distance between the points is \boxed{3\sqrt{13}} units.

To find the distance between the points $(2, -6)$ and $(-4, 3)$, we use the **distance formula**, which is derived from the Pythagorean Theorem:

$$
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

---

### Step 1: Identify the coordinates

Let:
- $(x_1, y_1) = (2, -6)$
- $(x_2, y_2) = (-4, 3)$

---

### Step 2: Compute the differences

$$
x_2 - x_1 = -4 - 2 = -6
$$
$$
y_2 - y_1 = 3 - (-6) = 3 + 6 = 9
$$

---

### Step 3: Square the differences

$$
(-6)^2 = 36
$$
$$
9^2 = 81
$$

---

### Step 4: Add the squares

$$
36 + 81 = 117
$$

---

### Step 5: Take the square root

$$
\sqrt{117}
$$

Now, simplify the radical:

$$
117 = 9 \times 13
$$
$$
\sqrt{117} = \sqrt{9 \times 13} = \sqrt{9} \cdot \sqrt{13} = 3\sqrt{13}
$$

---

### Final Answer

$$
\boxed{3\sqrt{13}}
$$

Solution:
We use the distance formula:  \begin{align*}
\sqrt{(2 - (-4))^2 + ((-6) - 3)^2} &= \sqrt{6^2 + (-9)^2}\\
& = \sqrt{36 + 81}\\
& = \sqrt{117} = \boxed{3\sqrt{13}}.
\end{align*}