Explanation:
Okay, so I need to find all the integer roots of the polynomial equation $x^4 + 5x^3 + 9x^2 - x - 14 = 0$. Hmm, let me think about how to approach this.

First, I remember that for polynomials with integer coefficients, the Rational Root Theorem can be helpful. The theorem states that any possible rational root, expressed in its lowest terms $\frac{p}{q}$, must have $p$ as a factor of the constant term and $q$ as a factor of the leading coefficient. Since we're looking for integer roots, $q$ must be 1 or -1, so the possible integer roots are just the factors of the constant term.

Looking at the given polynomial, the constant term is -14, and the leading coefficient is 1. Therefore, the possible integer roots are the factors of -14. Let me list those factors. The factors of -14 are: ±1, ±2, ±7, ±14. So those are the possible integer roots we need to test.

Alright, let me test each of these values one by one by substituting them into the polynomial and checking if the result is zero.

Starting with x = 1:
Plugging into the polynomial:
$1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14$. Let's calculate that step by step:
1 + 5 = 6, 6 + 9 = 15, 15 - 1 = 14, 14 - 14 = 0. Oh, that equals zero! So x = 1 is a root. Great, found one integer root.

Now let's check x = -1:
$(-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14 = 1 - 5 + 9 + 1 - 14$. Calculating step by step:
1 - 5 = -4, -4 + 9 = 5, 5 + 1 = 6, 6 - 14 = -8. That's not zero. So x = -1 is not a root.

Next, x = 2:
$2^4 + 5(2)^3 + 9(2)^2 - 2 - 14$. Calculating each term:
2^4 = 16, 5*(8) = 40, 9*(4) = 36, so 16 + 40 + 36 - 2 - 14. Let's add them up:
16 + 40 = 56, 56 + 36 = 92, 92 - 2 = 90, 90 - 14 = 76. Not zero. So x = 2 is not a root.

How about x = -2:
$(-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) - 14 = 16 + 5*(-8) + 9*(4) + 2 -14$. Calculating each term:
16 - 40 + 36 + 2 -14. Let me compute step by step:
16 - 40 = -24, -24 + 36 = 12, 12 + 2 = 14, 14 -14 = 0. Oh! That equals zero. So x = -2 is another root.

Alright, moving on to x = 7:
$7^4 + 5*(7)^3 + 9*(7)^2 -7 -14$. Let me compute this. Hmm, this might be a bit tedious. Let me calculate each term:

First, 7^4 is 7*7*7*7. 7*7 is 49, 49*7 is 343, 343*7 is 2401. So 2401.

Then 5*(7)^3: 7^3 is 343, so 5*343. Let me compute that. 343*5: 300*5=1500, 40*5=200, 3*5=15. Total is 1500 + 200 = 1700 + 15 = 1715.

Next, 9*(7)^2: 7^2 is 49, so 9*49. 49*9: 40*9=360, 9*9=81, so 360+81=441.

So putting it all together: 2401 + 1715 + 441 -7 -14.

Adding up the positive terms first: 2401 + 1715 = 4116, 4116 + 441 = 4557. Now subtracting the rest: 4557 -7 = 4550, 4550 -14 = 4536. That's a huge number, not zero. So x=7 isn't a root.

Now checking x = -7:

$(-7)^4 + 5*(-7)^3 + 9*(-7)^2 - (-7) -14$. Let's compute term by term:

(-7)^4 is positive 7^4, which is 2401.

5*(-7)^3: (-7)^3 is -343, so 5*(-343) = -1715.

9*(-7)^2: same as 9*(49) = 441.

Then -(-7) is +7. Then -14.

So altogether: 2401 -1715 + 441 +7 -14. Let me compute step by step:

2401 - 1715. Let me subtract: 2401 - 1700 = 701, then subtract 15 more: 701 - 15 = 686.

686 + 441 = 1127. Then 1127 +7 = 1134. Then 1134 -14 = 1120. Not zero. So x=-7 is not a root.

Now x = 14:

Let me check x=14. But given the previous results, this might be a huge number, but let me try:

$14^4 + 5*(14)^3 + 9*(14)^2 -14 -14$. Obviously, this will be a very large number. Let me see if I can estimate:

14^4 is 14*14*14*14. 14^2=196, so 14^4=196^2=38416. Then 5*(14)^3: 14^3=2744, so 5*2744=13720. 9*(14)^2=9*196=1764. Then subtract 14 and 14. So total is 38416 + 13720 + 1764 -14 -14. Adding up the first three terms: 38416 + 13720 = 52136, +1764 = 53900. Then subtract 28: 53900 -28=53872. Not zero. So x=14 is not a root.

Similarly, x=-14 would be even more massive. Let me check quickly:

(-14)^4 is same as 14^4=38416, 5*(-14)^3=5*(-2744)= -13720. 9*(-14)^2=9*196=1764. Then - (-14) is +14, and -14. So total: 38416 -13720 +1764 +14 -14. Let me compute:

38416 -13720. Let's see, 38416 - 13000 = 25416, then subtract 720 more: 25416 - 720 = 24696. Then +1764: 24696 + 1764 = 26460. Then +14 -14 cancels. So total is 26460. Not zero. So x=-14 is not a root.

So from the possible integer roots, we found x=1 and x=-2 are roots.

Now, since we have a quartic polynomial (degree 4), and we found two roots, we can factor the polynomial as (x - 1)(x + 2)(quadratic). Let me perform polynomial division or use synthetic division to factor out these roots and find the remaining quadratic, then check if it has integer roots.

Alternatively, since we know two roots, we can factor them out step by step. Let me try synthetic division first.

First, let's factor out (x - 1). Let me use synthetic division with root 1.

The coefficients of the polynomial are 1 (x^4), 5 (x^3), 9 (x^2), -1 (x), -14 (constant).

Set up synthetic division:

1 | 1   5   9   -1   -14
  |     1   6   15   14
  ------------------------
    1   6   15   14   0

Wait, so after dividing by (x - 1), the quotient polynomial is x^3 + 6x^2 + 15x + 14.

Now, let me check if x = -2 is a root of this cubic. Let me do synthetic division again with -2 on the cubic polynomial x^3 + 6x^2 + 15x + 14.

Set up synthetic division with root -2:

-2 | 1   6   15   14
    |     -2   -8   -14
    --------------------
      1   4   7   0

So, the quotient polynomial is x^2 + 4x + 7. Therefore, the original polynomial factors as (x - 1)(x + 2)(x^2 + 4x + 7).

Now, we need to check if the quadratic x^2 + 4x + 7 has any integer roots. To do this, let's compute its discriminant. The discriminant D = b² - 4ac = 16 - 28 = -12. Since the discriminant is negative, there are no real roots, let alone integer roots. Therefore, the only integer roots of the original equation are x = 1 and x = -2.

Wait, but let me just verify my synthetic division steps again to make sure I didn't make a mistake. Sometimes, when I do synthetic division quickly, I might mess up the signs or coefficients.

First, dividing by (x - 1):

Original coefficients: 1, 5, 9, -1, -14

Bring down the 1.

Multiply by 1: 1. Add to next coefficient: 5 + 1 = 6

Multiply by 1: 6. Add to next coefficient: 9 + 6 = 15

Multiply by 1: 15. Add to next coefficient: -1 + 15 = 14

Multiply by 1: 14. Add to last coefficient: -14 + 14 = 0. Correct. So that gives the cubic x³ + 6x² + 15x +14.

Then dividing this cubic by (x + 2), which is equivalent to root -2:

Coefficients: 1, 6, 15, 14

Bring down the 1.

Multiply by -2: -2. Add to next coefficient: 6 + (-2) = 4.

Multiply by -2: -8. Add to next coefficient: 15 + (-8) = 7.

Multiply by -2: -14. Add to last coefficient: 14 + (-14) = 0. Correct. So quotient polynomial is x² + 4x +7. Correct.

Discriminant: 16 - 28 = -12. So indeed, no real roots. Therefore, only integer roots are 1 and -2.

But just to be thorough, let me check whether there's a possibility that maybe I missed some roots or made an error in calculation earlier. For example, when I tested x= -2, I had:

Original polynomial evaluated at x=-2: 16 + 5*(-8) + 9*(4) + 2 -14. Let me recalculate this:

16 (which is (-2)^4) + 5*(-8) is -40, +9*(4) is +36, then -(-2) is +2, then -14. So 16 -40 is -24, plus 36 is +12, plus 2 is +14, minus 14 is 0. Correct. So x=-2 is indeed a root.

And for x=1, 1 + 5 + 9 -1 -14 is 0 as well. Correct.

So seems like all steps are correct. Therefore, the integer roots are 1 and -2.

But just to make sure, let me try factoring the original polynomial in another way, maybe grouping terms? Let me see.

Original polynomial: x⁴ + 5x³ + 9x² - x - 14.

Looking for grouping... Not sure if it's easy. But since we already factored it into (x -1)(x +2)(x² +4x +7), and that quadratic doesn't factor further over integers, so yes, those are the only integer roots.

Alternatively, perhaps trying to factor the original polynomial as a product of quadratics? Let me try. Suppose it factors as (x² + ax + b)(x² + cx + d). Then expanding:

x⁴ + (a + c)x³ + (ac + b + d)x² + (ad + bc)x + bd.

Comparing coefficients:

1. a + c = 5 (coefficient of x³)
2. ac + b + d = 9 (coefficient of x²)
3. ad + bc = -1 (coefficient of x)
4. bd = -14 (constant term)

We need integers a, b, c, d satisfying these equations. Let me see if this approach gives us anything.

Since bd = -14, possible integer pairs (b, d) are:

(1, -14), (-1, 14), (2, -7), (-2, 7), (7, -2), (-7, 2), (14, -1), (-14, 1).

Let me try these possibilities.

First, let's try (b, d) = (2, -7). Then bd = 2*(-7) = -14. Then:

From equation 1: a + c = 5.

From equation 3: ad + bc = a*(-7) + c*(2) = -7a + 2c = -1.

So we have:

a + c = 5

-7a + 2c = -1

Let me solve this system. From first equation, c = 5 - a. Substitute into second equation:

-7a + 2*(5 - a) = -1

-7a + 10 - 2a = -1

-9a + 10 = -1

-9a = -11

a = 11/9. Not integer. So discard this pair.

Next, try (b, d) = (-2, 7). Then:

bd = -2*7 = -14.

Equation 3: ad + bc = a*7 + c*(-2) = 7a - 2c = -1.

With a + c = 5.

So:

From equation 1: c = 5 - a.

Substitute into equation 3:

7a - 2*(5 - a) = -1

7a -10 + 2a = -1

9a -10 = -1

9a = 9 => a = 1. Then c = 5 -1 = 4.

Now check equation 2: ac + b + d = (1)(4) + (-2) +7 = 4 -2 +7 = 9. Which matches the coefficient of x² (which is 9). Perfect! So this works.

Therefore, the polynomial factors as (x² + ax + b)(x² + cx + d) = (x² + 1x -2)(x² + 4x +7). Let's check if this multiplication gives the original polynomial:

Multiply (x² + x -2)(x² +4x +7):

First, x²*(x² +4x +7) = x⁴ +4x³ +7x²

Then, x*(x² +4x +7) = x³ +4x² +7x

Then, -2*(x² +4x +7) = -2x² -8x -14

Adding them all together:

x⁴ +4x³ +7x² + x³ +4x² +7x -2x² -8x -14.

Combine like terms:

x⁴ + (4x³ +x³) + (7x² +4x² -2x²) + (7x -8x) -14

Which simplifies to:

x⁴ +5x³ +9x² -x -14. Perfect, matches the original polynomial!

Therefore, the factorization is correct. So the quadratic factors are (x² + x -2) and (x² +4x +7). Wait, but earlier when I did synthetic division, I had (x -1)(x +2)(x² +4x +7). Let me check if (x² + x -2) factors further.

Yes, x² + x -2 can be factored as (x + 2)(x - 1). Let me confirm:

(x + 2)(x -1) = x² -x + 2x -2 = x² +x -2. Correct.

Therefore, the full factorization of the original polynomial is (x -1)(x + 2)(x + 2)(x² +4x +7)? Wait, no:

Wait, original factorization from synthetic division was (x -1)(x + 2)(x² +4x +7). However, the grouping method gave (x² +x -2)(x² +4x +7) which is (x -1)(x +2)(x² +4x +7). So same thing. Hence, the roots are x=1, x=-2, and roots of x² +4x +7=0. Which we already determined are complex. Therefore, the integer roots are indeed 1 and -2.

Therefore, after verifying through multiple methods—Rational Root Theorem, synthetic division, factoring by grouping—it's confirmed that the only integer roots are 1 and -2.

Just to recap, the possible integer roots were ±1, ±2, ±7, ±14. Testing them, only 1 and -2 worked. Factoring the polynomial confirms that. So I can be confident these are the only integer roots.

I think that's thorough. I don't see any mistakes in my reasoning. So the final answer should be 1 and -2.

**Final Answer**
The integer roots are \boxed{-2} and \boxed{1}.

To find the integer roots of the polynomial
$$
x^4 + 5x^3 + 9x^2 - x - 14 = 0,
$$
we can use the **Rational Root Theorem**, which tells us that any possible rational root of the polynomial must be of the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. Since the leading coefficient is 1 and the constant term is $-14$, any rational root must be one of the factors of $-14$. These possible integer roots are:

$$
\pm 1, \pm 2, \pm 7, \pm 14.
$$

---

### Step 1: Test Possible Integer Roots

We will test each value of $x$ by substitution into the polynomial to see which values yield $0$.

- **For $x = 1$:**
  $$
  1^4 + 5 \cdot 1^3 + 9 \cdot 1^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14 = 0 \Rightarrow x = 1 \text{ is a root}.
  $$

- **For $x = -1$:**
  $$
  (-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14 = 1 - 5 + 9 + 1 - 14 = -8 \neq 0.
  $$

- **For $x = 2$:**
  $$
  2^4 + 5 \cdot 2^3 + 9 \cdot 2^2 - 2 - 14 = 16 + 40 + 36 - 2 - 14 = 76 \neq 0.
  $$

- **For $x = -2$:**
  $$
  (-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) - 14 = 16 - 40 + 36 + 2 - 14 = 0 \Rightarrow x = -2 \text{ is a root}.
  $$

- **For $x = 7$:**
  $$
  7^4 + 5 \cdot 7^3 + 9 \cdot 7^2 - 7 - 14 = 2401 + 1715 + 441 - 7 - 14 = \text{a large positive number} \neq 0.
  $$

- **For $x = -7$:**
  $$
  (-7)^4 + 5(-7)^3 + 9(-7)^2 - (-7) - 14 = 2401 - 1715 + 441 + 7 - 14 = 1120 \neq 0.
  $$

- **For $x = 14$** and $x = -14$:
  Similar large positive results are obtained, again yielding **non-zero** values.

So, the only **integer roots** are $x = 1$ and $x = -2$.

---

### Step 2: Factor the Polynomial

We now divide the original polynomial by $(x - 1)(x + 2)$, which are the factors corresponding to the roots we found. We'll do this using **synthetic division**.

#### First: Divide by $(x - 1)$

Coefficients: $1, 5, 9, -1, -14$

$$
\begin{array}{r|rrrrr}
1 & 1 & 5 & 9 & -1 & -14 \\
  &   & 1 & 6 & 15 & 14 \\
\hline
  & 1 & 6 & 15 & 14 & 0 \\
\end{array}
$$

The quotient is:
$$
x^3 + 6x^2 + 15x + 14
$$

---

#### Next: Divide by $(x + 2)$

Use synthetic division with $x = -2$ on the cubic:

$$
\begin{array}{r|rrrr}
-2 & 1 & 6 & 15 & 14 \\
   &   & -2 & -8 & -14 \\
\hline
   & 1 & 4 & 7 & 0 \\
\end{array}
$$

The quotient is:
$$
x^2 + 4x + 7
$$

Now, the original polynomial factors as:
$$
(x - 1)(x + 2)(x^2 + 4x + 7)
$$

---

### Step 3: Analyze the Remaining Quadratic

We examine the quadratic factor $x^2 + 4x + 7$. The discriminant of this quadratic is:

$$
\Delta = b^2 - 4ac = 4^2 - 4(1)(7) = 16 - 28 = -12
$$

Since the discriminant is negative, the quadratic has **no real roots** — only complex ones. Thus, there are **no more integer roots**.

---

### Final Conclusion

The only **integer roots** of the polynomial
$$
x^4 + 5x^3 + 9x^2 - x - 14 = 0
$$
are:

$$
\boxed{-2} \text{ and } \boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to find all the integer roots of the polynomial equation $x^4 + 5x^3 + 9x^2 - x - 14 = 0$. Hmm, let me think about how to approach this.

First, I remember that for polynomials with integer coefficients, the Rational Root Theorem can be helpful. The theorem states that any possible rational root, expressed in its lowest terms $\frac{p}{q}$, must have $p$ as a factor of the constant term and $q$ as a factor of the leading coefficient. Since we're looking for integer roots, $q$ must be 1 or -1, so the possible integer roots are just the factors of the constant term.

Looking at the given polynomial, the constant term is -14, and the leading coefficient is 1. Therefore, the possible integer roots are the factors of -14. Let me list those factors. The factors of -14 are: ±1, ±2, ±7, ±14. So those are the possible integer roots we need to test.

Alright, let me test each of these values one by one by substituting them into the polynomial and checking if the result is zero.

Starting with x = 1:
Plugging into the polynomial:
$1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14$. Let's calculate that step by step:
1 + 5 = 6, 6 + 9 = 15, 15 - 1 = 14, 14 - 14 = 0. Oh, that equals zero! So x = 1 is a root. Great, found one integer root.

Now let's check x = -1:
$(-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14 = 1 - 5 + 9 + 1 - 14$. Calculating step by step:
1 - 5 = -4, -4 + 9 = 5, 5 + 1 = 6, 6 - 14 = -8. That's not zero. So x = -1 is not a root.

Next, x = 2:
$2^4 + 5(2)^3 + 9(2)^2 - 2 - 14$. Calculating each term:
2^4 = 16, 5*(8) = 40, 9*(4) = 36, so 16 + 40 + 36 - 2 - 14. Let's add them up:
16 + 40 = 56, 56 + 36 = 92, 92 - 2 = 90, 90 - 14 = 76. Not zero. So x = 2 is not a root.

How about x = -2:
$(-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) - 14 = 16 + 5*(-8) + 9*(4) + 2 -14$. Calculating each term:
16 - 40 + 36 + 2 -14. Let me compute step by step:
16 - 40 = -24, -24 + 36 = 12, 12 + 2 = 14, 14 -14 = 0. Oh! That equals zero. So x = -2 is another root.

Alright, moving on to x = 7:
$7^4 + 5*(7)^3 + 9*(7)^2 -7 -14$. Let me compute this. Hmm, this might be a bit tedious. Let me calculate each term:

First, 7^4 is 7*7*7*7. 7*7 is 49, 49*7 is 343, 343*7 is 2401. So 2401.

Then 5*(7)^3: 7^3 is 343, so 5*343. Let me compute that. 343*5: 300*5=1500, 40*5=200, 3*5=15. Total is 1500 + 200 = 1700 + 15 = 1715.

Next, 9*(7)^2: 7^2 is 49, so 9*49. 49*9: 40*9=360, 9*9=81, so 360+81=441.

So putting it all together: 2401 + 1715 + 441 -7 -14.

Adding up the positive terms first: 2401 + 1715 = 4116, 4116 + 441 = 4557. Now subtracting the rest: 4557 -7 = 4550, 4550 -14 = 4536. That's a huge number, not zero. So x=7 isn't a root.

Now checking x = -7:

$(-7)^4 + 5*(-7)^3 + 9*(-7)^2 - (-7) -14$. Let's compute term by term:

(-7)^4 is positive 7^4, which is 2401.

5*(-7)^3: (-7)^3 is -343, so 5*(-343) = -1715.

9*(-7)^2: same as 9*(49) = 441.

Then -(-7) is +7. Then -14.

So altogether: 2401 -1715 + 441 +7 -14. Let me compute step by step:

2401 - 1715. Let me subtract: 2401 - 1700 = 701, then subtract 15 more: 701 - 15 = 686.

686 + 441 = 1127. Then 1127 +7 = 1134. Then 1134 -14 = 1120. Not zero. So x=-7 is not a root.

Now x = 14:

Let me check x=14. But given the previous results, this might be a huge number, but let me try:

$14^4 + 5*(14)^3 + 9*(14)^2 -14 -14$. Obviously, this will be a very large number. Let me see if I can estimate:

14^4 is 14*14*14*14. 14^2=196, so 14^4=196^2=38416. Then 5*(14)^3: 14^3=2744, so 5*2744=13720. 9*(14)^2=9*196=1764. Then subtract 14 and 14. So total is 38416 + 13720 + 1764 -14 -14. Adding up the first three terms: 38416 + 13720 = 52136, +1764 = 53900. Then subtract 28: 53900 -28=53872. Not zero. So x=14 is not a root.

Similarly, x=-14 would be even more massive. Let me check quickly:

(-14)^4 is same as 14^4=38416, 5*(-14)^3=5*(-2744)= -13720. 9*(-14)^2=9*196=1764. Then - (-14) is +14, and -14. So total: 38416 -13720 +1764 +14 -14. Let me compute:

38416 -13720. Let's see, 38416 - 13000 = 25416, then subtract 720 more: 25416 - 720 = 24696. Then +1764: 24696 + 1764 = 26460. Then +14 -14 cancels. So total is 26460. Not zero. So x=-14 is not a root.

So from the possible integer roots, we found x=1 and x=-2 are roots.

Now, since we have a quartic polynomial (degree 4), and we found two roots, we can factor the polynomial as (x - 1)(x + 2)(quadratic). Let me perform polynomial division or use synthetic division to factor out these roots and find the remaining quadratic, then check if it has integer roots.

Alternatively, since we know two roots, we can factor them out step by step. Let me try synthetic division first.

First, let's factor out (x - 1). Let me use synthetic division with root 1.

The coefficients of the polynomial are 1 (x^4), 5 (x^3), 9 (x^2), -1 (x), -14 (constant).

Set up synthetic division:

1 | 1   5   9   -1   -14
  |     1   6   15   14
  ------------------------
    1   6   15   14   0

Wait, so after dividing by (x - 1), the quotient polynomial is x^3 + 6x^2 + 15x + 14.

Now, let me check if x = -2 is a root of this cubic. Let me do synthetic division again with -2 on the cubic polynomial x^3 + 6x^2 + 15x + 14.

Set up synthetic division with root -2:

-2 | 1   6   15   14
    |     -2   -8   -14
    --------------------
      1   4   7   0

So, the quotient polynomial is x^2 + 4x + 7. Therefore, the original polynomial factors as (x - 1)(x + 2)(x^2 + 4x + 7).

Now, we need to check if the quadratic x^2 + 4x + 7 has any integer roots. To do this, let's compute its discriminant. The discriminant D = b² - 4ac = 16 - 28 = -12. Since the discriminant is negative, there are no real roots, let alone integer roots. Therefore, the only integer roots of the original equation are x = 1 and x = -2.

Wait, but let me just verify my synthetic division steps again to make sure I didn't make a mistake. Sometimes, when I do synthetic division quickly, I might mess up the signs or coefficients.

First, dividing by (x - 1):

Original coefficients: 1, 5, 9, -1, -14

Bring down the 1.

Multiply by 1: 1. Add to next coefficient: 5 + 1 = 6

Multiply by 1: 6. Add to next coefficient: 9 + 6 = 15

Multiply by 1: 15. Add to next coefficient: -1 + 15 = 14

Multiply by 1: 14. Add to last coefficient: -14 + 14 = 0. Correct. So that gives the cubic x³ + 6x² + 15x +14.

Then dividing this cubic by (x + 2), which is equivalent to root -2:

Coefficients: 1, 6, 15, 14

Bring down the 1.

Multiply by -2: -2. Add to next coefficient: 6 + (-2) = 4.

Multiply by -2: -8. Add to next coefficient: 15 + (-8) = 7.

Multiply by -2: -14. Add to last coefficient: 14 + (-14) = 0. Correct. So quotient polynomial is x² + 4x +7. Correct.

Discriminant: 16 - 28 = -12. So indeed, no real roots. Therefore, only integer roots are 1 and -2.

But just to be thorough, let me check whether there's a possibility that maybe I missed some roots or made an error in calculation earlier. For example, when I tested x= -2, I had:

Original polynomial evaluated at x=-2: 16 + 5*(-8) + 9*(4) + 2 -14. Let me recalculate this:

16 (which is (-2)^4) + 5*(-8) is -40, +9*(4) is +36, then -(-2) is +2, then -14. So 16 -40 is -24, plus 36 is +12, plus 2 is +14, minus 14 is 0. Correct. So x=-2 is indeed a root.

And for x=1, 1 + 5 + 9 -1 -14 is 0 as well. Correct.

So seems like all steps are correct. Therefore, the integer roots are 1 and -2.

But just to make sure, let me try factoring the original polynomial in another way, maybe grouping terms? Let me see.

Original polynomial: x⁴ + 5x³ + 9x² - x - 14.

Looking for grouping... Not sure if it's easy. But since we already factored it into (x -1)(x +2)(x² +4x +7), and that quadratic doesn't factor further over integers, so yes, those are the only integer roots.

Alternatively, perhaps trying to factor the original polynomial as a product of quadratics? Let me try. Suppose it factors as (x² + ax + b)(x² + cx + d). Then expanding:

x⁴ + (a + c)x³ + (ac + b + d)x² + (ad + bc)x + bd.

Comparing coefficients:

1. a + c = 5 (coefficient of x³)
2. ac + b + d = 9 (coefficient of x²)
3. ad + bc = -1 (coefficient of x)
4. bd = -14 (constant term)

We need integers a, b, c, d satisfying these equations. Let me see if this approach gives us anything.

Since bd = -14, possible integer pairs (b, d) are:

(1, -14), (-1, 14), (2, -7), (-2, 7), (7, -2), (-7, 2), (14, -1), (-14, 1).

Let me try these possibilities.

First, let's try (b, d) = (2, -7). Then bd = 2*(-7) = -14. Then:

From equation 1: a + c = 5.

From equation 3: ad + bc = a*(-7) + c*(2) = -7a + 2c = -1.

So we have:

a + c = 5

-7a + 2c = -1

Let me solve this system. From first equation, c = 5 - a. Substitute into second equation:

-7a + 2*(5 - a) = -1

-7a + 10 - 2a = -1

-9a + 10 = -1

-9a = -11

a = 11/9. Not integer. So discard this pair.

Next, try (b, d) = (-2, 7). Then:

bd = -2*7 = -14.

Equation 3: ad + bc = a*7 + c*(-2) = 7a - 2c = -1.

With a + c = 5.

So:

From equation 1: c = 5 - a.

Substitute into equation 3:

7a - 2*(5 - a) = -1

7a -10 + 2a = -1

9a -10 = -1

9a = 9 => a = 1. Then c = 5 -1 = 4.

Now check equation 2: ac + b + d = (1)(4) + (-2) +7 = 4 -2 +7 = 9. Which matches the coefficient of x² (which is 9). Perfect! So this works.

Therefore, the polynomial factors as (x² + ax + b)(x² + cx + d) = (x² + 1x -2)(x² + 4x +7). Let's check if this multiplication gives the original polynomial:

Multiply (x² + x -2)(x² +4x +7):

First, x²*(x² +4x +7) = x⁴ +4x³ +7x²

Then, x*(x² +4x +7) = x³ +4x² +7x

Then, -2*(x² +4x +7) = -2x² -8x -14

Adding them all together:

x⁴ +4x³ +7x² + x³ +4x² +7x -2x² -8x -14.

Combine like terms:

x⁴ + (4x³ +x³) + (7x² +4x² -2x²) + (7x -8x) -14

Which simplifies to:

x⁴ +5x³ +9x² -x -14. Perfect, matches the original polynomial!

Therefore, the factorization is correct. So the quadratic factors are (x² + x -2) and (x² +4x +7). Wait, but earlier when I did synthetic division, I had (x -1)(x +2)(x² +4x +7). Let me check if (x² + x -2) factors further.

Yes, x² + x -2 can be factored as (x + 2)(x - 1). Let me confirm:

(x + 2)(x -1) = x² -x + 2x -2 = x² +x -2. Correct.

Therefore, the full factorization of the original polynomial is (x -1)(x + 2)(x + 2)(x² +4x +7)? Wait, no:

Wait, original factorization from synthetic division was (x -1)(x + 2)(x² +4x +7). However, the grouping method gave (x² +x -2)(x² +4x +7) which is (x -1)(x +2)(x² +4x +7). So same thing. Hence, the roots are x=1, x=-2, and roots of x² +4x +7=0. Which we already determined are complex. Therefore, the integer roots are indeed 1 and -2.

Therefore, after verifying through multiple methods—Rational Root Theorem, synthetic division, factoring by grouping—it's confirmed that the only integer roots are 1 and -2.

Just to recap, the possible integer roots were ±1, ±2, ±7, ±14. Testing them, only 1 and -2 worked. Factoring the polynomial confirms that. So I can be confident these are the only integer roots.

I think that's thorough. I don't see any mistakes in my reasoning. So the final answer should be 1 and -2.

**Final Answer**
The integer roots are \boxed{-2} and \boxed{1}.

To find the integer roots of the polynomial
$$
x^4 + 5x^3 + 9x^2 - x - 14 = 0,
$$
we can use the **Rational Root Theorem**, which tells us that any possible rational root of the polynomial must be of the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. Since the leading coefficient is 1 and the constant term is $-14$, any rational root must be one of the factors of $-14$. These possible integer roots are:

$$
\pm 1, \pm 2, \pm 7, \pm 14.
$$

---

### Step 1: Test Possible Integer Roots

We will test each value of $x$ by substitution into the polynomial to see which values yield $0$.

- **For $x = 1$:**
  $$
  1^4 + 5 \cdot 1^3 + 9 \cdot 1^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14 = 0 \Rightarrow x = 1 \text{ is a root}.
  $$

- **For $x = -1$:**
  $$
  (-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14 = 1 - 5 + 9 + 1 - 14 = -8 \neq 0.
  $$

- **For $x = 2$:**
  $$
  2^4 + 5 \cdot 2^3 + 9 \cdot 2^2 - 2 - 14 = 16 + 40 + 36 - 2 - 14 = 76 \neq 0.
  $$

- **For $x = -2$:**
  $$
  (-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) - 14 = 16 - 40 + 36 + 2 - 14 = 0 \Rightarrow x = -2 \text{ is a root}.
  $$

- **For $x = 7$:**
  $$
  7^4 + 5 \cdot 7^3 + 9 \cdot 7^2 - 7 - 14 = 2401 + 1715 + 441 - 7 - 14 = \text{a large positive number} \neq 0.
  $$

- **For $x = -7$:**
  $$
  (-7)^4 + 5(-7)^3 + 9(-7)^2 - (-7) - 14 = 2401 - 1715 + 441 + 7 - 14 = 1120 \neq 0.
  $$

- **For $x = 14$** and $x = -14$:
  Similar large positive results are obtained, again yielding **non-zero** values.

So, the only **integer roots** are $x = 1$ and $x = -2$.

---

### Step 2: Factor the Polynomial

We now divide the original polynomial by $(x - 1)(x + 2)$, which are the factors corresponding to the roots we found. We'll do this using **synthetic division**.

#### First: Divide by $(x - 1)$

Coefficients: $1, 5, 9, -1, -14$

$$
\begin{array}{r|rrrrr}
1 & 1 & 5 & 9 & -1 & -14 \\
  &   & 1 & 6 & 15 & 14 \\
\hline
  & 1 & 6 & 15 & 14 & 0 \\
\end{array}
$$

The quotient is:
$$
x^3 + 6x^2 + 15x + 14
$$

---

#### Next: Divide by $(x + 2)$

Use synthetic division with $x = -2$ on the cubic:

$$
\begin{array}{r|rrrr}
-2 & 1 & 6 & 15 & 14 \\
   &   & -2 & -8 & -14 \\
\hline
   & 1 & 4 & 7 & 0 \\
\end{array}
$$

The quotient is:
$$
x^2 + 4x + 7
$$

Now, the original polynomial factors as:
$$
(x - 1)(x + 2)(x^2 + 4x + 7)
$$

---

### Step 3: Analyze the Remaining Quadratic

We examine the quadratic factor $x^2 + 4x + 7$. The discriminant of this quadratic is:

$$
\Delta = b^2 - 4ac = 4^2 - 4(1)(7) = 16 - 28 = -12
$$

Since the discriminant is negative, the quadratic has **no real roots** — only complex ones. Thus, there are **no more integer roots**.

---

### Final Conclusion

The only **integer roots** of the polynomial
$$
x^4 + 5x^3 + 9x^2 - x - 14 = 0
$$
are:

$$
\boxed{-2} \text{ and } \boxed{1}
$$

Solution: By the Integer Root Theorem, the possible integer roots are all the divisors of 14 (including negative divisors), which are $-14,$ $-7,$ $-2,$ $-1,$ $1,$ $2,$ $7,$ and $14.$  Checking, we find that the only integer roots are $\boxed{-2,1}.$


Explanation:
Okay, so I need to figure out the remainder when 54 is divided by 6. Let me start by recalling what modulo means. From what I remember, when we say "a mod b," we're looking for the remainder after dividing a by b. So in this case, 54 divided by 6, and then find the remainder.

First, let me do the division. How many times does 6 go into 54? Well, 6 times 9 is 54, right? Let me check that: 6 multiplied by 9. 6*9 is 54. So if 6 goes into 54 exactly 9 times with no remainder, then the remainder should be zero. Wait, is that correct?

Let me verify. If I take 6 times 9, that equals 54. So 54 divided by 6 is exactly 9, which means there's no leftover. Therefore, the remainder is zero. So 54 mod 6 should be 0.

But maybe I should approach this another way to be sure. Let's think about division with remainder. The formula is: dividend = divisor * quotient + remainder. So here, dividend is 54, divisor is 6. So 54 = 6 * q + r, where r is the remainder, and 0 ≤ r < 6.

If I solve for q and r, then since 6*9 is 54, then q is 9 and r is 0. So that checks out. Therefore, the remainder is zero.

Alternatively, maybe I can think in terms of smaller steps. Let me try dividing 54 by 6 step by step.

Starting with 6 into 54.

First, 6 into 5 is 0, so we consider 54.

6 into 54: how many times does 6 go into 54? Well, 6 times 1 is 6, 6 times 2 is 12, 6 times 3 is 18, 6 times 4 is 24, 6 times 5 is 30, 6 times 6 is 36, 6 times 7 is 42, 6 times 8 is 48, 6 times 9 is 54. So exactly 9 times. Therefore, 6*9 is 54, so there's no remainder.

So again, that gives me remainder zero.

Wait, maybe I can think about multiples of 6. Let me list some multiples of 6 and see where 54 falls.

Starting from 0: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54... So the multiples of 6 are 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, etc. So 54 is actually a multiple of 6, which means when you divide 54 by 6, there's no remainder.

Alternatively, if I think about the number line, 54 is exactly at the point where 6 lands. So if you divide 54 by 6, you reach exactly 9, so nothing left over.

Is there another way to check this? Maybe using subtraction. Let's subtract multiples of 6 from 54 until we get a number less than 6.

Starting with 54. Subtract 6: 54 - 6 = 48. Still larger than 6. Subtract 6 again: 48 - 6 = 42. Still larger. Continue this process: 42-6=36, 36-6=30, 30-6=24, 24-6=18, 18-6=12, 12-6=6, 6-6=0. So how many times did I subtract 6? Let's count: 1, 2, 3, 4, 5, 6, 7, 8, 9 times. So 9 subtractions of 6 bring me down to 0. Therefore, the remainder is 0.

Alternatively, if I had stopped when the result was less than 6, but in this case, it ended up at 0, which is less than 6.

Another method: using the modulo operation properties. For example, 54 divided by 6. Since 6 is a factor of 54, as 54 is 6*9, so it's divisible. Therefore, remainder is zero.

Wait, maybe I can think about smaller numbers. For example, 6 mod 6 is 0. 12 mod 6 is 0. So 6*9 is 54, so same thing.

Alternatively, think about 54 divided by 6. If I do 54 divided by 6 on a calculator, it's exactly 9. So decimal-wise, it's a whole number, so the remainder is zero.

But just to be thorough, let me think if there's any chance that I made a mistake here. For example, if the number was 55, then 55 divided by 6 is 9 with remainder 1, because 6*9=54, 55-54=1. But here it's 54, so 54-54=0.

Alternatively, if I have 54 divided by 6, and if I think about division algorithm, which states that for any integers a and b (with b > 0), there exist unique integers q and r such that a = bq + r and 0 ≤ r < b. Here, a is 54, b is 6, so 54 = 6*q + r. As we saw, q is 9 and r is 0. So that satisfies the condition because 0 ≤ 0 < 6.

Another way to check: using negative numbers? Wait, not sure if that's helpful here. But maybe if I consider 54 divided by 6, if there was a remainder, it would have to be between 0 and 5. So if I think about 54 divided by 6, if there was a remainder, it would be something like 54 = 6*9 + r, so r = 54 - 6*9 = 54 - 54 = 0.

Alternatively, maybe using modular arithmetic properties. For example, 54 mod 6. Since 54 is 6*9, then 54 mod 6 is 0.

Alternatively, breaking down 54 into smaller parts. For example, 54 can be written as 50 + 4. Then 50 mod 6 and 4 mod 6. Let's see. 50 divided by 6: 6*8=48, so 50-48=2, so 50 mod 6 is 2. Then 4 mod 6 is 4. Then adding them together: 2 + 4 = 6. Then 6 mod 6 is 0. So that gives the same result.

Alternatively, if I use the property that (a + b) mod m = [(a mod m) + (b mod m)] mod m. So 50 + 4 mod 6 is (50 mod 6 + 4 mod 6) mod 6. Which is (2 + 4) mod 6 = 6 mod 6 = 0.

Another example: breaking down 54 as 60 - 6. 60 mod 6 is 0, and 6 mod 6 is 0. So 0 - 0 = 0 mod 6.

Alternatively, maybe using division with remainder. If I have 54 divided by 6, the quotient is 9, remainder 0.

Alternatively, think in terms of clocks. If you have a clock with 6 hours, then 54 hours later, how many hours past the clock? Well, since the clock cycles every 6 hours, 54 divided by 6 is 9 full cycles, so the remainder is 0.

Alternatively, if you have 54 apples and you want to distribute them into baskets with 6 apples each, how many apples are left? Since 54 divided by 6 is exactly 9, there are no apples left, so remainder 0.

Wait, maybe I can check with another modulus. For example, 54 mod 6 is equivalent to 54 - 6*floor(54/6). So floor(54/6) is 9, so 54 - 6*9 = 54 - 54 = 0.

Alternatively, think of the remainder as the value that when added to the divisor times quotient gives the original number. Since 6*9=54, adding remainder 0 gives 54, which is the original number.

I think all these methods lead me to the conclusion that the remainder is 0.

But just to make sure, let me think if there's any case where this might not hold. For example, if the number was negative? But in this case, the number is positive. Also, if I had a non-integer division, but here it's exact.

Another thing: sometimes people get confused between remainder and modulo in programming contexts, but in mathematics, the remainder is the same as the modulo operation when dealing with positive integers.

Let me also recall that 6 is a factor of 54, so 54 is a multiple of 6. Therefore, when divided by 6, there's no remainder.

Therefore, after checking all these different methods—direct division, division algorithm, breaking down numbers, modular arithmetic properties, real-world examples—it's consistent that the remainder is 0.

So I think I can confidently say that 54 mod 6 is 0.

**Final Answer**
The remainder of 54 divided by 6 is \boxed{0}.

To find the remainder of $ 54 \mod 6 $, we are essentially looking for the **remainder** when 54 is divided by 6.

---

### Step 1: Understand the Division

We begin by performing the division:

$$
54 \div 6 = 9 \text{ with no remainder}
$$

This means that 6 divides into 54 exactly 9 times, without any leftover value.

---

### Step 2: Verify Using the Division Algorithm

The division algorithm states:

$$
a = b \cdot q + r \quad \text{where} \quad 0 \leq r < b
$$

For $ a = 54 $, $ b = 6 $, we find that:

$$
54 = 6 \cdot 9 + 0
$$

Here, the **remainder $ r $** is 0, and since $ 0 < 6 $, the condition is satisfied.

---

### Step 3: Use Modular Arithmetic Properties

From modular arithmetic, we know:

$$
a \mod b = a - b \cdot \left\lfloor \frac{a}{b} \right\rfloor
$$

Calculating:

$$
54 \mod 6 = 54 - 6 \cdot \left\lfloor \frac{54}{6} \right\rfloor = 54 - 6 \cdot 9 = 54 - 54 = 0
$$

---

### Step 4: Interpret in Terms of Multiples

Since 54 is exactly divisible by 6 (i.e., 54 is a multiple of 6), the remainder is 0.

---

### Final Answer

$$
\boxed{0}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the remainder when 54 is divided by 6. Let me start by recalling what modulo means. From what I remember, when we say "a mod b," we're looking for the remainder after dividing a by b. So in this case, 54 divided by 6, and then find the remainder.

First, let me do the division. How many times does 6 go into 54? Well, 6 times 9 is 54, right? Let me check that: 6 multiplied by 9. 6*9 is 54. So if 6 goes into 54 exactly 9 times with no remainder, then the remainder should be zero. Wait, is that correct?

Let me verify. If I take 6 times 9, that equals 54. So 54 divided by 6 is exactly 9, which means there's no leftover. Therefore, the remainder is zero. So 54 mod 6 should be 0.

But maybe I should approach this another way to be sure. Let's think about division with remainder. The formula is: dividend = divisor * quotient + remainder. So here, dividend is 54, divisor is 6. So 54 = 6 * q + r, where r is the remainder, and 0 ≤ r < 6.

If I solve for q and r, then since 6*9 is 54, then q is 9 and r is 0. So that checks out. Therefore, the remainder is zero.

Alternatively, maybe I can think in terms of smaller steps. Let me try dividing 54 by 6 step by step.

Starting with 6 into 54.

First, 6 into 5 is 0, so we consider 54.

6 into 54: how many times does 6 go into 54? Well, 6 times 1 is 6, 6 times 2 is 12, 6 times 3 is 18, 6 times 4 is 24, 6 times 5 is 30, 6 times 6 is 36, 6 times 7 is 42, 6 times 8 is 48, 6 times 9 is 54. So exactly 9 times. Therefore, 6*9 is 54, so there's no remainder.

So again, that gives me remainder zero.

Wait, maybe I can think about multiples of 6. Let me list some multiples of 6 and see where 54 falls.

Starting from 0: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54... So the multiples of 6 are 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, etc. So 54 is actually a multiple of 6, which means when you divide 54 by 6, there's no remainder.

Alternatively, if I think about the number line, 54 is exactly at the point where 6 lands. So if you divide 54 by 6, you reach exactly 9, so nothing left over.

Is there another way to check this? Maybe using subtraction. Let's subtract multiples of 6 from 54 until we get a number less than 6.

Starting with 54. Subtract 6: 54 - 6 = 48. Still larger than 6. Subtract 6 again: 48 - 6 = 42. Still larger. Continue this process: 42-6=36, 36-6=30, 30-6=24, 24-6=18, 18-6=12, 12-6=6, 6-6=0. So how many times did I subtract 6? Let's count: 1, 2, 3, 4, 5, 6, 7, 8, 9 times. So 9 subtractions of 6 bring me down to 0. Therefore, the remainder is 0.

Alternatively, if I had stopped when the result was less than 6, but in this case, it ended up at 0, which is less than 6.

Another method: using the modulo operation properties. For example, 54 divided by 6. Since 6 is a factor of 54, as 54 is 6*9, so it's divisible. Therefore, remainder is zero.

Wait, maybe I can think about smaller numbers. For example, 6 mod 6 is 0. 12 mod 6 is 0. So 6*9 is 54, so same thing.

Alternatively, think about 54 divided by 6. If I do 54 divided by 6 on a calculator, it's exactly 9. So decimal-wise, it's a whole number, so the remainder is zero.

But just to be thorough, let me think if there's any chance that I made a mistake here. For example, if the number was 55, then 55 divided by 6 is 9 with remainder 1, because 6*9=54, 55-54=1. But here it's 54, so 54-54=0.

Alternatively, if I have 54 divided by 6, and if I think about division algorithm, which states that for any integers a and b (with b > 0), there exist unique integers q and r such that a = bq + r and 0 ≤ r < b. Here, a is 54, b is 6, so 54 = 6*q + r. As we saw, q is 9 and r is 0. So that satisfies the condition because 0 ≤ 0 < 6.

Another way to check: using negative numbers? Wait, not sure if that's helpful here. But maybe if I consider 54 divided by 6, if there was a remainder, it would have to be between 0 and 5. So if I think about 54 divided by 6, if there was a remainder, it would be something like 54 = 6*9 + r, so r = 54 - 6*9 = 54 - 54 = 0.

Alternatively, maybe using modular arithmetic properties. For example, 54 mod 6. Since 54 is 6*9, then 54 mod 6 is 0.

Alternatively, breaking down 54 into smaller parts. For example, 54 can be written as 50 + 4. Then 50 mod 6 and 4 mod 6. Let's see. 50 divided by 6: 6*8=48, so 50-48=2, so 50 mod 6 is 2. Then 4 mod 6 is 4. Then adding them together: 2 + 4 = 6. Then 6 mod 6 is 0. So that gives the same result.

Alternatively, if I use the property that (a + b) mod m = [(a mod m) + (b mod m)] mod m. So 50 + 4 mod 6 is (50 mod 6 + 4 mod 6) mod 6. Which is (2 + 4) mod 6 = 6 mod 6 = 0.

Another example: breaking down 54 as 60 - 6. 60 mod 6 is 0, and 6 mod 6 is 0. So 0 - 0 = 0 mod 6.

Alternatively, maybe using division with remainder. If I have 54 divided by 6, the quotient is 9, remainder 0.

Alternatively, think in terms of clocks. If you have a clock with 6 hours, then 54 hours later, how many hours past the clock? Well, since the clock cycles every 6 hours, 54 divided by 6 is 9 full cycles, so the remainder is 0.

Alternatively, if you have 54 apples and you want to distribute them into baskets with 6 apples each, how many apples are left? Since 54 divided by 6 is exactly 9, there are no apples left, so remainder 0.

Wait, maybe I can check with another modulus. For example, 54 mod 6 is equivalent to 54 - 6*floor(54/6). So floor(54/6) is 9, so 54 - 6*9 = 54 - 54 = 0.

Alternatively, think of the remainder as the value that when added to the divisor times quotient gives the original number. Since 6*9=54, adding remainder 0 gives 54, which is the original number.

I think all these methods lead me to the conclusion that the remainder is 0.

But just to make sure, let me think if there's any case where this might not hold. For example, if the number was negative? But in this case, the number is positive. Also, if I had a non-integer division, but here it's exact.

Another thing: sometimes people get confused between remainder and modulo in programming contexts, but in mathematics, the remainder is the same as the modulo operation when dealing with positive integers.

Let me also recall that 6 is a factor of 54, so 54 is a multiple of 6. Therefore, when divided by 6, there's no remainder.

Therefore, after checking all these different methods—direct division, division algorithm, breaking down numbers, modular arithmetic properties, real-world examples—it's consistent that the remainder is 0.

So I think I can confidently say that 54 mod 6 is 0.

**Final Answer**
The remainder of 54 divided by 6 is \boxed{0}.

To find the remainder of $ 54 \mod 6 $, we are essentially looking for the **remainder** when 54 is divided by 6.

---

### Step 1: Understand the Division

We begin by performing the division:

$$
54 \div 6 = 9 \text{ with no remainder}
$$

This means that 6 divides into 54 exactly 9 times, without any leftover value.

---

### Step 2: Verify Using the Division Algorithm

The division algorithm states:

$$
a = b \cdot q + r \quad \text{where} \quad 0 \leq r < b
$$

For $ a = 54 $, $ b = 6 $, we find that:

$$
54 = 6 \cdot 9 + 0
$$

Here, the **remainder $ r $** is 0, and since $ 0 < 6 $, the condition is satisfied.

---

### Step 3: Use Modular Arithmetic Properties

From modular arithmetic, we know:

$$
a \mod b = a - b \cdot \left\lfloor \frac{a}{b} \right\rfloor
$$

Calculating:

$$
54 \mod 6 = 54 - 6 \cdot \left\lfloor \frac{54}{6} \right\rfloor = 54 - 6 \cdot 9 = 54 - 54 = 0
$$

---

### Step 4: Interpret in Terms of Multiples

Since 54 is exactly divisible by 6 (i.e., 54 is a multiple of 6), the remainder is 0.

---

### Final Answer

$$
\boxed{0}
$$

Solution: $54 = 9 \cdot 6 + 0 \Rightarrow 54 \equiv \boxed{0} \pmod{6}$.


Explanation:
Okay, so I need to figure out how far a car would travel if it's going 50 miles per hour for 2 and 3/4 hours. Hmm, let me think. Alright, distance equals speed multiplied by time, right? So the formula is distance = speed × time. Got that part down.

First, let me make sure I understand the problem. The speed is given as 50 miles per hour, and the time is 2 and 3/4 hours. I need to calculate the distance. Since the time is a mixed number, maybe I should convert that to an improper fraction or a decimal to make the multiplication easier? Let me try both ways and see which one works better.

Starting with converting the mixed number to an improper fraction. 2 and 3/4 hours... To convert that, I multiply the whole number part (2) by the denominator (4), which gives me 8, then add the numerator (3), so 8 + 3 = 11. Therefore, 2 and 3/4 hours is the same as 11/4 hours. Let me check that again. Yes, because 2 is 8/4, and adding 3/4 gives 11/4. Okay, that seems right.

Alternatively, converting to a decimal. 3/4 is 0.75, so 2 and 3/4 hours is 2.75 hours. Either way, both 11/4 and 2.75 represent the same amount of time. I can use either one for the calculation. Let me pick fractions since the answer needs to be a mixed number, and maybe fractions will be more precise.

So, distance = 50 mph × 11/4 hours. Let me do the multiplication. 50 times 11 divided by 4. Let me calculate 50 × 11 first. 50 × 10 is 500, and 50 × 1 is 50, so adding those together gives 550. Then divide by 4. So 550 divided by 4. Let me compute that. 550 ÷ 4. Well, 4 × 137 is 548, right? Because 4 × 100 is 400, 4 × 30 is 120, 4 × 7 is 28. So 400 + 120 is 520, plus 28 is 548. So 4 × 137 = 548. Therefore, 550 divided by 4 is 137 with a remainder of 2. So that's 137 and 2/4. Simplifying 2/4 to 1/2, so the distance is 137 and 1/2 miles. Let me check that again.

Wait, so 50 multiplied by 11/4. Alternatively, I could think of 50 divided by 4 times 11. 50 divided by 4 is 12.5, and then 12.5 multiplied by 11. Let me do that. 12.5 × 10 is 125, and 12.5 × 1 is 12.5. Adding them together gives 125 + 12.5 = 137.5. Which is the same as 137 and 1/2. So that's consistent. So whether I do fractions or decimals, I end up with 137.5 miles, which is 137 and 1/2 as a mixed number.

Let me just confirm once more to make sure I didn't make a mistake. If the car is going 50 mph, in 1 hour it would go 50 miles. In 2 hours, that's 100 miles. Then, for the remaining 3/4 of an hour. So 3/4 of an hour at 50 mph. So 50 × 3/4. Let me calculate that. 50 divided by 4 is 12.5, times 3 is 37.5. So adding that to the 100 miles from the first 2 hours, gives 137.5 miles. Yep, same result. So that seems right.

Alternatively, maybe breaking down the time. 2 and 3/4 hours is 2 hours plus 45 minutes. Wait, 3/4 of an hour is 45 minutes. So if I calculate the distance for 2 hours and then for 45 minutes. 2 hours is 100 miles as before. 45 minutes is 0.75 hours. So 50 × 0.75 is 37.5. So 100 + 37.5 is 137.5. Yep, same answer.

So regardless of the method, it seems like the answer is 137.5 miles, which as a mixed number is 137 1/2. Wait, 0.5 is 1/2. So yes, 137 and 1/2. Let me confirm that there are no calculation errors here. 50 times 2 is 100, 50 times 0.75 is 37.5, sum is 137.5. Yes. So converting that to a mixed number, the decimal part is 0.5, which is 1/2. Therefore, 137 1/2 miles.

Alternatively, using fractions: 50 is equal to 50/1. Multiply by 11/4. So (50 × 11) / (1 × 4) = 550/4. Dividing numerator and denominator by 2 gives 275/2. Which is equal to 137.5. And as a mixed number, that's 137 and 1/2. So that checks out.

Therefore, I think the answer is 137 and 1/2 miles. Let me just make sure that the question didn't have any tricks or anything. It says "Express your answer as a mixed number," so 137 1/2 is correct. There's no need to simplify further because 1/2 is already in simplest terms.

So yeah, I think that's the answer. All methods lead to 137.5, which is 137 1/2 as a mixed number. I don't see any mistakes in the reasoning steps. So I feel confident about this answer.

**Final Answer**
The car would travel \boxed{137\frac{1}{2}} miles.

To determine how far a car would travel at a constant speed of **50 miles per hour** over a time interval of **$2\frac{3}{4}$ hours**, we use the fundamental formula for distance:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

---

### Step 1: Convert the mixed number to an improper fraction

We start by converting the time, $2\frac{3}{4}$ hours, into an improper fraction:

$$
2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{11}{4}
$$

---

### Step 2: Multiply the speed by the time

Now, we multiply the speed (50 mph) by the time ($\frac{11}{4}$ hours):

$$
\text{Distance} = 50 \times \frac{11}{4} = \frac{50 \times 11}{4} = \frac{550}{4}
$$

---

### Step 3: Simplify the resulting fraction

Let's divide the numerator 550 by 4 to simplify:

$$
\frac{550}{4} = 137 \text{ remainder } 2 = 137 + \frac{2}{4} = 137\frac{1}{2}
$$

---

### Final Answer

After simplifying, the car would travel:

$$
\boxed{137\frac{1}{2}}
$$

miles.

[Assistant]
Explanation:
Okay, so I need to figure out how far a car would travel if it's going 50 miles per hour for 2 and 3/4 hours. Hmm, let me think. Alright, distance equals speed multiplied by time, right? So the formula is distance = speed × time. Got that part down.

First, let me make sure I understand the problem. The speed is given as 50 miles per hour, and the time is 2 and 3/4 hours. I need to calculate the distance. Since the time is a mixed number, maybe I should convert that to an improper fraction or a decimal to make the multiplication easier? Let me try both ways and see which one works better.

Starting with converting the mixed number to an improper fraction. 2 and 3/4 hours... To convert that, I multiply the whole number part (2) by the denominator (4), which gives me 8, then add the numerator (3), so 8 + 3 = 11. Therefore, 2 and 3/4 hours is the same as 11/4 hours. Let me check that again. Yes, because 2 is 8/4, and adding 3/4 gives 11/4. Okay, that seems right.

Alternatively, converting to a decimal. 3/4 is 0.75, so 2 and 3/4 hours is 2.75 hours. Either way, both 11/4 and 2.75 represent the same amount of time. I can use either one for the calculation. Let me pick fractions since the answer needs to be a mixed number, and maybe fractions will be more precise.

So, distance = 50 mph × 11/4 hours. Let me do the multiplication. 50 times 11 divided by 4. Let me calculate 50 × 11 first. 50 × 10 is 500, and 50 × 1 is 50, so adding those together gives 550. Then divide by 4. So 550 divided by 4. Let me compute that. 550 ÷ 4. Well, 4 × 137 is 548, right? Because 4 × 100 is 400, 4 × 30 is 120, 4 × 7 is 28. So 400 + 120 is 520, plus 28 is 548. So 4 × 137 = 548. Therefore, 550 divided by 4 is 137 with a remainder of 2. So that's 137 and 2/4. Simplifying 2/4 to 1/2, so the distance is 137 and 1/2 miles. Let me check that again.

Wait, so 50 multiplied by 11/4. Alternatively, I could think of 50 divided by 4 times 11. 50 divided by 4 is 12.5, and then 12.5 multiplied by 11. Let me do that. 12.5 × 10 is 125, and 12.5 × 1 is 12.5. Adding them together gives 125 + 12.5 = 137.5. Which is the same as 137 and 1/2. So that's consistent. So whether I do fractions or decimals, I end up with 137.5 miles, which is 137 and 1/2 as a mixed number.

Let me just confirm once more to make sure I didn't make a mistake. If the car is going 50 mph, in 1 hour it would go 50 miles. In 2 hours, that's 100 miles. Then, for the remaining 3/4 of an hour. So 3/4 of an hour at 50 mph. So 50 × 3/4. Let me calculate that. 50 divided by 4 is 12.5, times 3 is 37.5. So adding that to the 100 miles from the first 2 hours, gives 137.5 miles. Yep, same result. So that seems right.

Alternatively, maybe breaking down the time. 2 and 3/4 hours is 2 hours plus 45 minutes. Wait, 3/4 of an hour is 45 minutes. So if I calculate the distance for 2 hours and then for 45 minutes. 2 hours is 100 miles as before. 45 minutes is 0.75 hours. So 50 × 0.75 is 37.5. So 100 + 37.5 is 137.5. Yep, same answer.

So regardless of the method, it seems like the answer is 137.5 miles, which as a mixed number is 137 1/2. Wait, 0.5 is 1/2. So yes, 137 and 1/2. Let me confirm that there are no calculation errors here. 50 times 2 is 100, 50 times 0.75 is 37.5, sum is 137.5. Yes. So converting that to a mixed number, the decimal part is 0.5, which is 1/2. Therefore, 137 1/2 miles.

Alternatively, using fractions: 50 is equal to 50/1. Multiply by 11/4. So (50 × 11) / (1 × 4) = 550/4. Dividing numerator and denominator by 2 gives 275/2. Which is equal to 137.5. And as a mixed number, that's 137 and 1/2. So that checks out.

Therefore, I think the answer is 137 and 1/2 miles. Let me just make sure that the question didn't have any tricks or anything. It says "Express your answer as a mixed number," so 137 1/2 is correct. There's no need to simplify further because 1/2 is already in simplest terms.

So yeah, I think that's the answer. All methods lead to 137.5, which is 137 1/2 as a mixed number. I don't see any mistakes in the reasoning steps. So I feel confident about this answer.

**Final Answer**
The car would travel \boxed{137\frac{1}{2}} miles.

To determine how far a car would travel at a constant speed of **50 miles per hour** over a time interval of **$2\frac{3}{4}$ hours**, we use the fundamental formula for distance:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

---

### Step 1: Convert the mixed number to an improper fraction

We start by converting the time, $2\frac{3}{4}$ hours, into an improper fraction:

$$
2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{11}{4}
$$

---

### Step 2: Multiply the speed by the time

Now, we multiply the speed (50 mph) by the time ($\frac{11}{4}$ hours):

$$
\text{Distance} = 50 \times \frac{11}{4} = \frac{50 \times 11}{4} = \frac{550}{4}
$$

---

### Step 3: Simplify the resulting fraction

Let's divide the numerator 550 by 4 to simplify:

$$
\frac{550}{4} = 137 \text{ remainder } 2 = 137 + \frac{2}{4} = 137\frac{1}{2}
$$

---

### Final Answer

After simplifying, the car would travel:

$$
\boxed{137\frac{1}{2}}
$$

miles.

Solution: In two hours, a car traveling at $50$ mph would travel $50$ mph $\times 2$ hours $= 100$ miles. Now we find how far a car can travel in $3/4$ of an hour which is $50$ mph $\times \frac{3}{4}$ hours $ = \frac{150}{4} = 37 \frac{1}{2}$ miles. Thus, the car would travel a total of $100 + 37 \frac{1}{2}= \boxed{137 \frac{1}{2}}$ miles.


Explanation:
Okay, so I need to find the value of this product of logarithms: log base 2 of 3 times log base 3 of 4 times log base 4 of 5, and so on up to log base 7 of 8. Hmm, let me write that down to make sure I have it right.

It's log₂3 · log₃4 · log₄5 · log₅6 · log₆7 · log₇8. Alright, so each logarithm is in the form logₙ(n+1), starting from n=2 up to n=7. So there are six terms here.

I remember that there's a property of logarithms called the change of base formula. Is that useful here? Let me recall: log_b a = (ln a)/(ln b). Maybe if I convert each term to natural logs, something will cancel out? Let me try that.

So, let me rewrite each term:

log₂3 = ln3 / ln2

log₃4 = ln4 / ln3

log₄5 = ln5 / ln4

log₅6 = ln6 / ln5

log₆7 = ln7 / ln6

log₇8 = ln8 / ln7

So if I multiply all these together, the numerator of each term is the denominator of the previous term. Let me check:

First term: ln3 / ln2

Second term: ln4 / ln3 → the ln3 cancels with the denominator of the first term.

Third term: ln5 / ln4 → ln4 cancels with the denominator of the second term.

Same with the rest: ln6 cancels with the denominator of the third term, ln7 cancels with the denominator of the fourth term, and ln8 remains in the numerator. The denominators left are ln2 from the first term, and the numerators go up to ln8.

Wait, so when you multiply them all, the intermediates cancel out, and you get ln8 / ln2.

So that simplifies to ln8 divided by ln2. But ln8 is ln(2^3) which is 3 ln2. Therefore, 3 ln2 / ln2 = 3.

So is the answer 3? Wait, that seems too straightforward. Let me verify.

Alternatively, I remember there's a property that log_b a * log_c b = log_c a. Wait, maybe that's the chain rule for logarithms? Let me think. If I have log_b a * log_c b, then that would be (ln a / ln b) * (ln b / ln c) = ln a / ln c = log_c a. So that's correct.

So in this case, if I have a product of logs where each subsequent log has the previous base, then they telescope. So starting with log₂3 * log₃4 * log₄5 * ... * log₇8. So each log's base is the previous number, so the bases and arguments are consecutive numbers. So using the property above, each pair cancels.

For example, log₂3 * log₃4 = log₂4. Then log₂4 * log₄5 = log₂5. Then log₂5 * log₅6 = log₂6. Then log₂6 * log₆7 = log₂7. Then log₂7 * log₇8 = log₂8.

So after all the multiplications, it's log base 2 of 8. Which is log₂8. Since 2^3 = 8, that is 3. So the answer is 3.

So both methods give me 3. That seems to confirm it.

But just to be thorough, let me compute each term numerically and multiply them to check.

First, compute each logarithm:

log₂3: log base 2 of 3. Since 2^1 = 2, 2^1.58496 ≈ 3. So approximately 1.58496.

log₃4: log base 3 of 4. 3^1 = 3, 3^1.26186 ≈ 4. So approximately 1.26186.

log₄5: log base 4 of 5. 4^1 = 4, 4^1.16096 ≈ 5. Approximately 1.16096.

log₅6: log base 5 of 6. 5^1 = 5, 5^1.11328 ≈ 6. Approximately 1.11328.

log₆7: log base 6 of 7. 6^1 = 6, 6^1.0860 ≈ 7. Approximately 1.0860.

log₇8: log base 7 of 8. 7^1 = 7, 7^1.0686 ≈ 8. Approximately 1.0686.

Now let me multiply these approximate values:

First, 1.58496 * 1.26186 ≈ Let me compute that:

1.58496 * 1.26186. Let me approximate:

1.58 * 1.26 ≈ 1.58 * 1.26. 1.58*1 = 1.58, 1.58*0.26 ≈ 0.4108, total ≈ 1.9908. So roughly 2.

Then multiply by 1.16096: 2 * 1.16096 ≈ 2.32192

Then multiply by 1.11328: 2.32192 * 1.11328 ≈ Let me approximate 2.32 * 1.11 ≈ 2.32*1 + 2.32*0.11 = 2.32 + 0.2552 ≈ 2.5752

Then multiply by 1.0860: 2.5752 * 1.086 ≈ 2.5752 * 1 + 2.5752 * 0.086 ≈ 2.5752 + 0.2215 ≈ 2.7967

Then multiply by 1.0686: 2.7967 * 1.0686 ≈ 2.7967 + 2.7967*0.0686 ≈ 2.7967 + 0.1916 ≈ 2.9883

So approximately 2.9883, which is very close to 3. Considering that my approximated values for each log had some rounding errors, this seems to confirm that the exact value is 3.

Therefore, I can be confident that the answer is 3.

Alternatively, maybe even more straightforward with exponents. Since log₂3 · log₃4 · ... · log₇8 = log₂8 as per telescoping. Since log₂8 is 3. So that's even simpler.

So the key idea is recognizing that the product telescopes, with each denominator canceling the numerator of the previous term. Therefore, the entire product simplifies to ln8 / ln2 = log₂8 = 3.

So, yeah, the answer is 3.

Just to recap, in case I missed something. Each log term is of the form log_{n}(n+1). So when you multiply them together, using the change of base formula, they all cancel out except for the first denominator and the last numerator. So ln3/ln2 * ln4/ln3 * ln5/ln4 * ... * ln8/ln7 = ln8/ln2 = log₂8 = 3.

Yes, that makes sense. So I think that's solid. Therefore, the value is 3.

**Final Answer**
\boxed{3}

To evaluate the expression:

$$
\log_2{3} \cdot \log_3{4} \cdot \log_4{5} \cdot \log_5{6} \cdot \log_6{7} \cdot \log_7{8}
$$

we begin by recognizing a key property of logarithms: the **change of base formula**, which allows us to rewrite any logarithm in terms of a common base. This property is:

$$
\log_b{a} = \frac{\log_c{a}}{\log_c{b}}
$$

Using this, each logarithmic term in the product can be expressed as a fraction of natural logarithms (for example):

$$
\log_2{3} = \frac{\ln 3}{\ln 2}, \quad \log_3{4} = \frac{\ln 4}{\ln 3}, \quad \log_4{5} = \frac{\ln 5}{\ln 4}, \quad \dots, \quad \log_7{8} = \frac{\ln 8}{\ln 7}
$$

Now, multiplying all these together:

$$
\left( \frac{\ln 3}{\ln 2} \right) \cdot \left( \frac{\ln 4}{\ln 3} \right) \cdot \left( \frac{\ln 5}{\ln 4} \right) \cdot \dots \cdot \left( \frac{\ln 8}{\ln 7} \right)
$$

we observe that the intermediate terms **telescope** — that is, each denominator cancels with the numerator of the preceding term. The overall product simplifies to:

$$
\frac{\ln 8}{\ln 2}
$$

This expression is recognized as:

$$
\log_2{8}
$$

Because $8 = 2^3$, we have:

$$
\log_2{8} = 3
$$

---

**Final Answer:**

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of this product of logarithms: log base 2 of 3 times log base 3 of 4 times log base 4 of 5, and so on up to log base 7 of 8. Hmm, let me write that down to make sure I have it right.

It's log₂3 · log₃4 · log₄5 · log₅6 · log₆7 · log₇8. Alright, so each logarithm is in the form logₙ(n+1), starting from n=2 up to n=7. So there are six terms here.

I remember that there's a property of logarithms called the change of base formula. Is that useful here? Let me recall: log_b a = (ln a)/(ln b). Maybe if I convert each term to natural logs, something will cancel out? Let me try that.

So, let me rewrite each term:

log₂3 = ln3 / ln2

log₃4 = ln4 / ln3

log₄5 = ln5 / ln4

log₅6 = ln6 / ln5

log₆7 = ln7 / ln6

log₇8 = ln8 / ln7

So if I multiply all these together, the numerator of each term is the denominator of the previous term. Let me check:

First term: ln3 / ln2

Second term: ln4 / ln3 → the ln3 cancels with the denominator of the first term.

Third term: ln5 / ln4 → ln4 cancels with the denominator of the second term.

Same with the rest: ln6 cancels with the denominator of the third term, ln7 cancels with the denominator of the fourth term, and ln8 remains in the numerator. The denominators left are ln2 from the first term, and the numerators go up to ln8.

Wait, so when you multiply them all, the intermediates cancel out, and you get ln8 / ln2.

So that simplifies to ln8 divided by ln2. But ln8 is ln(2^3) which is 3 ln2. Therefore, 3 ln2 / ln2 = 3.

So is the answer 3? Wait, that seems too straightforward. Let me verify.

Alternatively, I remember there's a property that log_b a * log_c b = log_c a. Wait, maybe that's the chain rule for logarithms? Let me think. If I have log_b a * log_c b, then that would be (ln a / ln b) * (ln b / ln c) = ln a / ln c = log_c a. So that's correct.

So in this case, if I have a product of logs where each subsequent log has the previous base, then they telescope. So starting with log₂3 * log₃4 * log₄5 * ... * log₇8. So each log's base is the previous number, so the bases and arguments are consecutive numbers. So using the property above, each pair cancels.

For example, log₂3 * log₃4 = log₂4. Then log₂4 * log₄5 = log₂5. Then log₂5 * log₅6 = log₂6. Then log₂6 * log₆7 = log₂7. Then log₂7 * log₇8 = log₂8.

So after all the multiplications, it's log base 2 of 8. Which is log₂8. Since 2^3 = 8, that is 3. So the answer is 3.

So both methods give me 3. That seems to confirm it.

But just to be thorough, let me compute each term numerically and multiply them to check.

First, compute each logarithm:

log₂3: log base 2 of 3. Since 2^1 = 2, 2^1.58496 ≈ 3. So approximately 1.58496.

log₃4: log base 3 of 4. 3^1 = 3, 3^1.26186 ≈ 4. So approximately 1.26186.

log₄5: log base 4 of 5. 4^1 = 4, 4^1.16096 ≈ 5. Approximately 1.16096.

log₅6: log base 5 of 6. 5^1 = 5, 5^1.11328 ≈ 6. Approximately 1.11328.

log₆7: log base 6 of 7. 6^1 = 6, 6^1.0860 ≈ 7. Approximately 1.0860.

log₇8: log base 7 of 8. 7^1 = 7, 7^1.0686 ≈ 8. Approximately 1.0686.

Now let me multiply these approximate values:

First, 1.58496 * 1.26186 ≈ Let me compute that:

1.58496 * 1.26186. Let me approximate:

1.58 * 1.26 ≈ 1.58 * 1.26. 1.58*1 = 1.58, 1.58*0.26 ≈ 0.4108, total ≈ 1.9908. So roughly 2.

Then multiply by 1.16096: 2 * 1.16096 ≈ 2.32192

Then multiply by 1.11328: 2.32192 * 1.11328 ≈ Let me approximate 2.32 * 1.11 ≈ 2.32*1 + 2.32*0.11 = 2.32 + 0.2552 ≈ 2.5752

Then multiply by 1.0860: 2.5752 * 1.086 ≈ 2.5752 * 1 + 2.5752 * 0.086 ≈ 2.5752 + 0.2215 ≈ 2.7967

Then multiply by 1.0686: 2.7967 * 1.0686 ≈ 2.7967 + 2.7967*0.0686 ≈ 2.7967 + 0.1916 ≈ 2.9883

So approximately 2.9883, which is very close to 3. Considering that my approximated values for each log had some rounding errors, this seems to confirm that the exact value is 3.

Therefore, I can be confident that the answer is 3.

Alternatively, maybe even more straightforward with exponents. Since log₂3 · log₃4 · ... · log₇8 = log₂8 as per telescoping. Since log₂8 is 3. So that's even simpler.

So the key idea is recognizing that the product telescopes, with each denominator canceling the numerator of the previous term. Therefore, the entire product simplifies to ln8 / ln2 = log₂8 = 3.

So, yeah, the answer is 3.

Just to recap, in case I missed something. Each log term is of the form log_{n}(n+1). So when you multiply them together, using the change of base formula, they all cancel out except for the first denominator and the last numerator. So ln3/ln2 * ln4/ln3 * ln5/ln4 * ... * ln8/ln7 = ln8/ln2 = log₂8 = 3.

Yes, that makes sense. So I think that's solid. Therefore, the value is 3.

**Final Answer**
\boxed{3}

To evaluate the expression:

$$
\log_2{3} \cdot \log_3{4} \cdot \log_4{5} \cdot \log_5{6} \cdot \log_6{7} \cdot \log_7{8}
$$

we begin by recognizing a key property of logarithms: the **change of base formula**, which allows us to rewrite any logarithm in terms of a common base. This property is:

$$
\log_b{a} = \frac{\log_c{a}}{\log_c{b}}
$$

Using this, each logarithmic term in the product can be expressed as a fraction of natural logarithms (for example):

$$
\log_2{3} = \frac{\ln 3}{\ln 2}, \quad \log_3{4} = \frac{\ln 4}{\ln 3}, \quad \log_4{5} = \frac{\ln 5}{\ln 4}, \quad \dots, \quad \log_7{8} = \frac{\ln 8}{\ln 7}
$$

Now, multiplying all these together:

$$
\left( \frac{\ln 3}{\ln 2} \right) \cdot \left( \frac{\ln 4}{\ln 3} \right) \cdot \left( \frac{\ln 5}{\ln 4} \right) \cdot \dots \cdot \left( \frac{\ln 8}{\ln 7} \right)
$$

we observe that the intermediate terms **telescope** — that is, each denominator cancels with the numerator of the preceding term. The overall product simplifies to:

$$
\frac{\ln 8}{\ln 2}
$$

This expression is recognized as:

$$
\log_2{8}
$$

Because $8 = 2^3$, we have:

$$
\log_2{8} = 3
$$

---

**Final Answer:**

$$
\boxed{3}
$$

Solution:
We use the change-of-base identity $\log_a{b}=\frac{\log{b}}{\log{a}}$ to find $$\log_2{3} \cdot \log_3{4} \cdot \log_4{5} \cdot \log_5{6} \cdot \log_6{7} \cdot \log_7{8}=
\frac{\log3}{\log2} \cdot \frac{\log4}{\log3} \cdot \frac{\log5}{\log4} \cdot \frac{\log6}{\log5} \cdot \frac{\log7}{\log6} \cdot \frac{\log8}{\log7}.$$Simplifying, we get $\frac{\log8}{\log2}=\log_2{8}=\boxed{3}$.


Explanation:
Okay, so I need to figure out the remainder when the sum of numbers from 1 to 12 is divided by 4. Let me think about how to approach this.

First, maybe I should just calculate the total sum and then divide by 4 to find the remainder. But adding all those numbers might take some time. Wait, there's a formula for the sum of the first n numbers, right? It's n(n + 1)/2. Let me recall... Yeah, I think that's correct. So if I use that formula, maybe it will be faster.

Let me check with a small example. If n is 3, then the sum is 1+2+3=6. Using the formula: 3(3+1)/2 = 3*4/2 = 12/2 = 6. Yep, that works. So for n=12, the sum should be 12*13/2. Let me calculate that. 12 divided by 2 is 6, so 6*13. 6*10 is 60, 6*3 is 18, so total is 60+18=78. So the total sum is 78. Now, I need to find 78 modulo 4.

But maybe there's a smarter way than calculating the entire sum? Maybe I can find each term modulo 4 and then add them up? Let me try both methods and see if they match.

First, the total sum is 78. Dividing 78 by 4: 4*19 is 76, so 78 - 76 = 2. Therefore, the remainder is 2. So the answer should be 2.

But let me verify using the other method. Let's take each number from 1 to 12, find their remainders modulo 4, then add those up and take modulo 4 again. That might help me check.

Let's list them:

1 mod 4 = 1

2 mod 4 = 2

3 mod 4 = 3

4 mod 4 = 0

5 mod 4 = 1 (since 5 - 4 =1)

6 mod 4 = 2 (6 - 4 = 2)

7 mod 4 = 3 (7 - 4*1=3)

8 mod 4 = 0

9 mod 4 = 1 (9 - 8=1)

10 mod 4 = 2 (10 - 8=2)

11 mod 4 = 3 (11 - 8=3)

12 mod 4 = 0 (12 - 12=0)

So the remainders are: 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0.

Now let's add them up step by step:

Starting with 0:

0 + 1 = 1

1 + 2 = 3

3 + 3 = 6

6 + 0 = 6

6 + 1 = 7

7 + 2 = 9

9 + 3 = 12

12 + 0 = 12

12 + 1 = 13

13 + 2 = 15

15 + 3 = 18

18 + 0 = 18

So the total sum of remainders is 18. Now, 18 divided by 4 is 4*4=16, so 18-16=2. Therefore, 18 mod 4 is 2. Which matches the previous result. So that's good.

Alternatively, maybe I can group the numbers to make it easier. Let me think. Since modulo 4, every 4 numbers will have a pattern. Let me see:

Numbers 1-4: 1,2,3,0. Sum is 1+2+3+0=6. 6 mod 4 is 2.

Numbers 5-8: same pattern, 1,2,3,0. Sum is 6 mod 4 is 2.

Numbers 9-12: same pattern, 1,2,3,0. Sum is 6 mod 4 is 2.

So total sum is 2 + 2 + 2 = 6. Then 6 mod 4 is 2. Again, same answer. So that's another way to think about it.

So all methods are leading me to the same answer, 2. Therefore, I think the remainder when the sum is divided by 4 is 2.

But just to make sure I didn't make any calculation errors... Let me check the sum again using another approach. Let me pair numbers:

1 + 12 = 13

2 + 11 = 13

3 + 10 = 13

4 + 9 = 13

5 + 8 = 13

6 + 7 = 13

So that's 6 pairs, each summing to 13. So total sum is 6*13=78. Which is the same as before. So 78 mod 4.

Dividing 78 by 4: 4*19=76, 78-76=2. Yep. So remainder is 2.

Therefore, all methods confirm that the answer is 2. I think that's solid.

Alternatively, maybe using properties of modular arithmetic. Since I need the sum mod 4, and the sum is 1+2+...+12.

But since sum is n(n+1)/2, so 12*13/2 = 78. So 78 mod 4.

But 78 divided by 4: 4*10=40, 4*20=80. So 78 is 80-2. Therefore, 78 ≡ -2 mod 4. But -2 mod 4 is equivalent to 2 mod 4. So that's another way to see it.

So yeah, multiple approaches confirm that the remainder is 2.

Hence, I can be confident that the answer is 2.

**Final Answer**
The modulo 4 remainder of the sum is \boxed{2}.

To determine the **modulo 4 remainder** of the sum:

$$
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12,
$$

we can approach this using a combination of **summing the sequence**, **modulo arithmetic**, and **modular patterns**.

---

### Step 1: Use the Summation Formula

The sum of the first $ n $ positive integers is given by the formula:

$$
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}.
$$

In this case, $ n = 12 $, so the total sum is:

$$
\frac{12 \cdot 13}{2} = 6 \cdot 13 = 78.
$$

Now, we want to find the modulo 4 remainder of this total, that is:

$$
78 \mod 4.
$$

---

### Step 2: Compute the Modulo

Divide 78 by 4:

$$
78 \div 4 = 19 \text{ remainder } 2.
$$

So,

$$
78 \equiv 2 \pmod{4}.
$$

This is our answer, but let’s verify using an alternative method for confirmation.

---

### Step 3: Confirmation by Modulo of Each Term

We can compute the modulo 4 of each term individually and sum the results, then take the modulo 4 again.

$$
\begin{align*}
1 &\equiv 1 \pmod{4} \\
2 &\equiv 2 \pmod{4} \\
3 &\equiv 3 \pmod{4} \\
4 &\equiv 0 \pmod{4} \\
5 &\equiv 1 \pmod{4} \\
6 &\equiv 2 \pmod{4} \\
7 &\equiv 3 \pmod{4} \\
8 &\equiv 0 \pmod{4} \\
9 &\equiv 1 \pmod{4} \\
10 &\equiv 2 \pmod{4} \\
11 &\equiv 3 \pmod{4} \\
12 &\equiv 0 \pmod{4} \\
\end{align*}
$$

Adding these remainders:

$$
1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 = 18.
$$

Now compute:

$$
18 \mod 4 = 2.
$$

This confirms that the remainder is $ \boxed{2} $.

---

### Step 4: Modular Pattern Confirmation

Note that the sequence of numbers $ 1 $ to $ 12 $ contains exactly 3 full cycles of numbers modulo 4:

- $ 1, 2, 3, 0 $
- $ 1, 2, 3, 0 $
- $ 1, 2, 3, 0 $

Each cycle sums to $ 6 $, so the total sum modulo 4 is:

$$
6 + 6 + 6 = 18 \Rightarrow 18 \mod 4 = 2.
$$

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the remainder when the sum of numbers from 1 to 12 is divided by 4. Let me think about how to approach this.

First, maybe I should just calculate the total sum and then divide by 4 to find the remainder. But adding all those numbers might take some time. Wait, there's a formula for the sum of the first n numbers, right? It's n(n + 1)/2. Let me recall... Yeah, I think that's correct. So if I use that formula, maybe it will be faster.

Let me check with a small example. If n is 3, then the sum is 1+2+3=6. Using the formula: 3(3+1)/2 = 3*4/2 = 12/2 = 6. Yep, that works. So for n=12, the sum should be 12*13/2. Let me calculate that. 12 divided by 2 is 6, so 6*13. 6*10 is 60, 6*3 is 18, so total is 60+18=78. So the total sum is 78. Now, I need to find 78 modulo 4.

But maybe there's a smarter way than calculating the entire sum? Maybe I can find each term modulo 4 and then add them up? Let me try both methods and see if they match.

First, the total sum is 78. Dividing 78 by 4: 4*19 is 76, so 78 - 76 = 2. Therefore, the remainder is 2. So the answer should be 2.

But let me verify using the other method. Let's take each number from 1 to 12, find their remainders modulo 4, then add those up and take modulo 4 again. That might help me check.

Let's list them:

1 mod 4 = 1

2 mod 4 = 2

3 mod 4 = 3

4 mod 4 = 0

5 mod 4 = 1 (since 5 - 4 =1)

6 mod 4 = 2 (6 - 4 = 2)

7 mod 4 = 3 (7 - 4*1=3)

8 mod 4 = 0

9 mod 4 = 1 (9 - 8=1)

10 mod 4 = 2 (10 - 8=2)

11 mod 4 = 3 (11 - 8=3)

12 mod 4 = 0 (12 - 12=0)

So the remainders are: 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0.

Now let's add them up step by step:

Starting with 0:

0 + 1 = 1

1 + 2 = 3

3 + 3 = 6

6 + 0 = 6

6 + 1 = 7

7 + 2 = 9

9 + 3 = 12

12 + 0 = 12

12 + 1 = 13

13 + 2 = 15

15 + 3 = 18

18 + 0 = 18

So the total sum of remainders is 18. Now, 18 divided by 4 is 4*4=16, so 18-16=2. Therefore, 18 mod 4 is 2. Which matches the previous result. So that's good.

Alternatively, maybe I can group the numbers to make it easier. Let me think. Since modulo 4, every 4 numbers will have a pattern. Let me see:

Numbers 1-4: 1,2,3,0. Sum is 1+2+3+0=6. 6 mod 4 is 2.

Numbers 5-8: same pattern, 1,2,3,0. Sum is 6 mod 4 is 2.

Numbers 9-12: same pattern, 1,2,3,0. Sum is 6 mod 4 is 2.

So total sum is 2 + 2 + 2 = 6. Then 6 mod 4 is 2. Again, same answer. So that's another way to think about it.

So all methods are leading me to the same answer, 2. Therefore, I think the remainder when the sum is divided by 4 is 2.

But just to make sure I didn't make any calculation errors... Let me check the sum again using another approach. Let me pair numbers:

1 + 12 = 13

2 + 11 = 13

3 + 10 = 13

4 + 9 = 13

5 + 8 = 13

6 + 7 = 13

So that's 6 pairs, each summing to 13. So total sum is 6*13=78. Which is the same as before. So 78 mod 4.

Dividing 78 by 4: 4*19=76, 78-76=2. Yep. So remainder is 2.

Therefore, all methods confirm that the answer is 2. I think that's solid.

Alternatively, maybe using properties of modular arithmetic. Since I need the sum mod 4, and the sum is 1+2+...+12.

But since sum is n(n+1)/2, so 12*13/2 = 78. So 78 mod 4.

But 78 divided by 4: 4*10=40, 4*20=80. So 78 is 80-2. Therefore, 78 ≡ -2 mod 4. But -2 mod 4 is equivalent to 2 mod 4. So that's another way to see it.

So yeah, multiple approaches confirm that the remainder is 2.

Hence, I can be confident that the answer is 2.

**Final Answer**
The modulo 4 remainder of the sum is \boxed{2}.

To determine the **modulo 4 remainder** of the sum:

$$
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12,
$$

we can approach this using a combination of **summing the sequence**, **modulo arithmetic**, and **modular patterns**.

---

### Step 1: Use the Summation Formula

The sum of the first $ n $ positive integers is given by the formula:

$$
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}.
$$

In this case, $ n = 12 $, so the total sum is:

$$
\frac{12 \cdot 13}{2} = 6 \cdot 13 = 78.
$$

Now, we want to find the modulo 4 remainder of this total, that is:

$$
78 \mod 4.
$$

---

### Step 2: Compute the Modulo

Divide 78 by 4:

$$
78 \div 4 = 19 \text{ remainder } 2.
$$

So,

$$
78 \equiv 2 \pmod{4}.
$$

This is our answer, but let’s verify using an alternative method for confirmation.

---

### Step 3: Confirmation by Modulo of Each Term

We can compute the modulo 4 of each term individually and sum the results, then take the modulo 4 again.

$$
\begin{align*}
1 &\equiv 1 \pmod{4} \\
2 &\equiv 2 \pmod{4} \\
3 &\equiv 3 \pmod{4} \\
4 &\equiv 0 \pmod{4} \\
5 &\equiv 1 \pmod{4} \\
6 &\equiv 2 \pmod{4} \\
7 &\equiv 3 \pmod{4} \\
8 &\equiv 0 \pmod{4} \\
9 &\equiv 1 \pmod{4} \\
10 &\equiv 2 \pmod{4} \\
11 &\equiv 3 \pmod{4} \\
12 &\equiv 0 \pmod{4} \\
\end{align*}
$$

Adding these remainders:

$$
1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 = 18.
$$

Now compute:

$$
18 \mod 4 = 2.
$$

This confirms that the remainder is $ \boxed{2} $.

---

### Step 4: Modular Pattern Confirmation

Note that the sequence of numbers $ 1 $ to $ 12 $ contains exactly 3 full cycles of numbers modulo 4:

- $ 1, 2, 3, 0 $
- $ 1, 2, 3, 0 $
- $ 1, 2, 3, 0 $

Each cycle sums to $ 6 $, so the total sum modulo 4 is:

$$
6 + 6 + 6 = 18 \Rightarrow 18 \mod 4 = 2.
$$

---

### Final Answer

$$
\boxed{2}
$$

Solution:
Grouping residues helps make some series computations easier:  \begin{align*}
1 + 2 + 3 + 0 + 1 + 2& + 3 + 0 + 1 + 2 + 3 + 0\\&\equiv 3(1 + 2 + 3 + 0) \\
&\equiv 18\\
& \equiv \boxed{2} \pmod{4}.
\end{align*}