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 them out: ±1, ±2, ±7, ±14.

Alright, so I need to test each of these values in the polynomial to see if they satisfy the equation. Let me start with the smaller numbers first.

Let me test x = 1. Plugging into the polynomial:

$1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14$. Let me compute 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, that's one integer root.

Now let me check x = -1. Plugging in:

$(-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14$. Let me compute each term:

(-1)^4 = 1; 5(-1)^3 = 5*(-1) = -5; 9(-1)^2 = 9*1 = 9; -(-1) = +1; and then -14. So adding them up:

1 - 5 + 9 + 1 - 14. Let me compute step by step:

1 - 5 = -4; -4 + 9 = 5; 5 + 1 = 6; 6 - 14 = -8. Not zero. So x = -1 is not a root.

Next, let's try x = 2. Plugging in:

$2^4 + 5(2)^3 + 9(2)^2 - 2 - 14$. Calculating each term:

2^4 = 16; 5*(8) = 40; 9*(4) = 36; then -2 and -14. So total:

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? Let's compute:

(-2)^4 + 5*(-2)^3 + 9*(-2)^2 - (-2) -14.

Calculating each term:

(-2)^4 = 16; 5*(-8) = -40; 9*(4) = 36; -(-2) = +2; and -14.

Adding them up: 16 - 40 = -24; -24 + 36 = 12; 12 + 2 = 14; 14 -14 = 0. Oh! That equals zero. So x = -2 is also a root. Nice, so now we have two integer roots: 1 and -2.

Moving on to x = 7. Let me check that.

7^4 + 5*(7)^3 + 9*(7)^2 - 7 -14.

Calculating each term:

7^4 is 2401; 5*(343) = 1715; 9*(49) = 441; then -7 and -14.

Adding them up: 2401 + 1715 = 4116; 4116 + 441 = 4557; 4557 - 7 = 4550; 4550 -14 = 4536. Definitely not zero. So x = 7 is not a root.

How about x = -7? Let me try that.

(-7)^4 + 5*(-7)^3 + 9*(-7)^2 - (-7) -14.

Calculating each term:

(-7)^4 = 2401; 5*(-343) = -1715; 9*(49) = 441; -(-7) = +7; and -14.

Adding them up: 2401 - 1715 = 686; 686 + 441 = 1127; 1127 + 7 = 1134; 1134 -14 = 1120. Not zero. So x = -7 is not a root.

Next, x = 14. Let me check that.

14^4 + 5*(14)^3 + 9*(14)^2 -14 -14.

This seems like a huge number. Let me compute step by step:

14^2 = 196; 14^3 = 14*196 = 2744; 14^4 = 14*2744 = 38416.

So:

38416 + 5*(2744) + 9*(196) -14 -14.

Calculating each term:

5*2744 = 13720; 9*196 = 1764.

Adding up: 38416 + 13720 = 52136; 52136 + 1764 = 53900; 53900 -14 = 53886; 53886 -14 = 53872. Definitely not zero. So x = 14 is not a root.

Similarly, x = -14 would be even larger in magnitude, so I can safely say it's not a root.

So far, we have found two integer roots: 1 and -2. But since the polynomial is of degree 4, there could be more roots. However, since we are only looking for integer roots, and we have already tested all possible candidates from the Rational Root Theorem, maybe there are no more integer roots? Wait, but maybe there are other integer roots that are not factors of -14? Wait, no, the Rational Root Theorem says that any rational root must be a factor of the constant term over a factor of the leading coefficient. Since the leading coefficient is 1, the only possible rational roots are the factors of -14, which we have already checked. Therefore, there are no other possible integer roots.

But just to be thorough, let me check if maybe there are multiple roots or something. Wait, but we found two roots already. Let me try to factor the polynomial using these roots.

If x = 1 and x = -2 are roots, then (x - 1) and (x + 2) are factors of the polynomial. Let me perform polynomial division or use synthetic division to factor them out.

First, let me divide the original polynomial by (x - 1). Wait, but synthetic division is easier for that. Let me try 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 with root 1:

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

So the remainder is 0, as expected. The quotient polynomial is x^3 + 6x^2 + 15x + 14.

Now, let me divide this cubic polynomial by (x + 2), since x = -2 is another root. Let me use synthetic division again, this time with root -2.

Coefficients of the cubic: 1 (x^3), 6 (x^2), 15 (x), 14 (constant).

Set up synthetic division with -2:

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

So the remainder is 0, as expected. The quotient polynomial is x^2 + 4x + 7.

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

Now, to find all integer roots, we need to check if the quadratic factor x^2 + 4x + 7 has any integer roots. Let me compute its discriminant: D = 16 - 4*1*7 = 16 - 28 = -12. Since the discriminant is negative, there are no real roots, let alone integer roots. Therefore, the only integer roots are x = 1 and x = -2.

Wait, but let me double-check my factoring. Let me multiply out (x - 1)(x + 2)(x^2 + 4x + 7) to ensure that it equals the original polynomial.

First, multiply (x - 1)(x + 2):

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

Now multiply this by (x^2 + 4x + 7):

Multiply x^2 + x - 2 by x^2 + 4x + 7.

Let me do term by term:

First, x^2*(x^2 + 4x + 7) = x^4 + 4x^3 + 7x^2.

Then, x*(x^2 + 4x + 7) = x^3 + 4x^2 + 7x.

Then, -2*(x^2 + 4x + 7) = -2x^2 -8x -14.

Now, add all these together:

x^4 + 4x^3 + 7x^2 + x^3 + 4x^2 + 7x -2x^2 -8x -14.

Combine like terms:

x^4 + (4x^3 + x^3) + (7x^2 + 4x^2 - 2x^2) + (7x -8x) -14.

Which is:

x^4 + 5x^3 + 9x^2 - x -14.

Yes! That matches the original polynomial. So the factoring is correct. Therefore, the only integer roots are indeed 1 and -2.

But just to make sure I didn't miss any other possible roots, let me check if maybe there are multiple roots or something. Wait, the original polynomial is degree 4, so there are four roots in total (real or complex). We found two real roots, and the other two are complex. Since the quadratic factor has discriminant -12, they are complex conjugates. Therefore, there are no other real roots, so no other integer roots.

Therefore, the integer roots are 1 and -2.

But let me just check once again if I made a mistake in testing the possible roots. For example, maybe I miscalculated for x = -2? Let me recheck that.

Original equation: x^4 +5x^3 +9x^2 -x -14.

For x = -2:

(-2)^4 = 16; 5*(-2)^3 = 5*(-8) = -40; 9*(-2)^2 = 9*4 = 36; -(-2) = +2; -14.

So total: 16 -40 +36 +2 -14.

Calculating step by step:

16 -40 = -24; -24 +36 = 12; 12 +2 =14; 14 -14 = 0. Correct.

And for x = 1: 1 +5 +9 -1 -14 = 0. Correct.

So no mistakes there.

Therefore, I think I can confidently say that the integer roots are 1 and -2.

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

To find all the **integer roots** of the polynomial

$$
x^4 + 5x^3 + 9x^2 - x - 14 = 0,
$$

we begin by applying the **Rational Root Theorem**. This theorem tells us that any rational root of a polynomial with integer coefficients must be of the form $\frac{p}{q}$, where:

- $p$ is a factor of the **constant term** (here, $-14$),
- $q$ is a factor of the **leading coefficient** (here, $1$).

Since the leading coefficient is 1, the only possible rational roots are the **integer factors** of $-14$, which are:

$$
\pm1, \pm2, \pm7, \pm14.
$$

---

### Step 1: Test Each Possible Integer Root

We substitute each of these values into the polynomial to check if they yield zero.

- **$x = 1$:**
  $$
  1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14 = 0
  $$
  ✅ **Root: $x = 1$**

- **$x = -1$:**
  $$
  (-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14 = 1 - 5 + 9 + 1 - 14 = -8
  $$
  ❌ Not a root

- **$x = 2$:**
  $$
  2^4 + 5(2)^3 + 9(2)^2 - 2 - 14 = 16 + 40 + 36 - 2 - 14 = 76
  $$
  ❌ Not a root

- **$x = -2$:**
  $$
  (-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) - 14 = 16 - 40 + 36 + 2 - 14 = 0
  $$
  ✅ **Root: $x = -2$**

- **$x = 7$:**
  $$
  7^4 + 5(7)^3 + 9(7)^2 - 7 - 14 = 2401 + 1715 + 441 - 7 - 14 = 4536
  $$
  ❌ Not a root

- **$x = -7$:**
  $$
  (-7)^4 + 5(-7)^3 + 9(-7)^2 - (-7) - 14 = 2401 - 1715 + 441 + 7 - 14 = 1120
  $$
  ❌ Not a root

- **$x = 14$:**
  $$
  14^4 + 5(14)^3 + 9(14)^2 - 14 - 14 = 38416 + 13720 + 1764 - 14 - 14 = 53872
  $$
  ❌ Not a root

- **$x = -14$:**
  $$
  (-14)^4 + 5(-14)^3 + 9(-14)^2 - (-14) - 14 = 38416 - 13720 + 1764 + 14 - 14 = 26460
  $$
  ❌ Not a root

---

### Step 2: Factor the Polynomial

We have found two integer roots: $x = 1$ and $x = -2$. This means that the polynomial is divisible by $(x - 1)(x + 2)$. We can now perform **polynomial division** or **synthetic division** to factor the original polynomial.

Using synthetic division:

- Divide the original polynomial by $(x - 1)$, then divide the result by $(x + 2)$, we get:

$$
x^4 + 5x^3 + 9x^2 - x - 14 = (x - 1)(x + 2)(x^2 + 4x + 7)
$$

---

### Step 3: Analyze the Quadratic Factor

Now, we examine the quadratic factor:

$$
x^2 + 4x + 7
$$

Compute its **discriminant**:

$$
D = b^2 - 4ac = 16 - 28 = -12
$$

Since the discriminant is **negative**, the quadratic has **no real roots**, and thus **no integer roots**.

---

### Final Conclusion

The only **integer roots** of the polynomial are:

$$
\boxed{-2}, \boxed{1}
$$

[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 them out: ±1, ±2, ±7, ±14.

Alright, so I need to test each of these values in the polynomial to see if they satisfy the equation. Let me start with the smaller numbers first.

Let me test x = 1. Plugging into the polynomial:

$1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14$. Let me compute 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, that's one integer root.

Now let me check x = -1. Plugging in:

$(-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14$. Let me compute each term:

(-1)^4 = 1; 5(-1)^3 = 5*(-1) = -5; 9(-1)^2 = 9*1 = 9; -(-1) = +1; and then -14. So adding them up:

1 - 5 + 9 + 1 - 14. Let me compute step by step:

1 - 5 = -4; -4 + 9 = 5; 5 + 1 = 6; 6 - 14 = -8. Not zero. So x = -1 is not a root.

Next, let's try x = 2. Plugging in:

$2^4 + 5(2)^3 + 9(2)^2 - 2 - 14$. Calculating each term:

2^4 = 16; 5*(8) = 40; 9*(4) = 36; then -2 and -14. So total:

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? Let's compute:

(-2)^4 + 5*(-2)^3 + 9*(-2)^2 - (-2) -14.

Calculating each term:

(-2)^4 = 16; 5*(-8) = -40; 9*(4) = 36; -(-2) = +2; and -14.

Adding them up: 16 - 40 = -24; -24 + 36 = 12; 12 + 2 = 14; 14 -14 = 0. Oh! That equals zero. So x = -2 is also a root. Nice, so now we have two integer roots: 1 and -2.

Moving on to x = 7. Let me check that.

7^4 + 5*(7)^3 + 9*(7)^2 - 7 -14.

Calculating each term:

7^4 is 2401; 5*(343) = 1715; 9*(49) = 441; then -7 and -14.

Adding them up: 2401 + 1715 = 4116; 4116 + 441 = 4557; 4557 - 7 = 4550; 4550 -14 = 4536. Definitely not zero. So x = 7 is not a root.

How about x = -7? Let me try that.

(-7)^4 + 5*(-7)^3 + 9*(-7)^2 - (-7) -14.

Calculating each term:

(-7)^4 = 2401; 5*(-343) = -1715; 9*(49) = 441; -(-7) = +7; and -14.

Adding them up: 2401 - 1715 = 686; 686 + 441 = 1127; 1127 + 7 = 1134; 1134 -14 = 1120. Not zero. So x = -7 is not a root.

Next, x = 14. Let me check that.

14^4 + 5*(14)^3 + 9*(14)^2 -14 -14.

This seems like a huge number. Let me compute step by step:

14^2 = 196; 14^3 = 14*196 = 2744; 14^4 = 14*2744 = 38416.

So:

38416 + 5*(2744) + 9*(196) -14 -14.

Calculating each term:

5*2744 = 13720; 9*196 = 1764.

Adding up: 38416 + 13720 = 52136; 52136 + 1764 = 53900; 53900 -14 = 53886; 53886 -14 = 53872. Definitely not zero. So x = 14 is not a root.

Similarly, x = -14 would be even larger in magnitude, so I can safely say it's not a root.

So far, we have found two integer roots: 1 and -2. But since the polynomial is of degree 4, there could be more roots. However, since we are only looking for integer roots, and we have already tested all possible candidates from the Rational Root Theorem, maybe there are no more integer roots? Wait, but maybe there are other integer roots that are not factors of -14? Wait, no, the Rational Root Theorem says that any rational root must be a factor of the constant term over a factor of the leading coefficient. Since the leading coefficient is 1, the only possible rational roots are the factors of -14, which we have already checked. Therefore, there are no other possible integer roots.

But just to be thorough, let me check if maybe there are multiple roots or something. Wait, but we found two roots already. Let me try to factor the polynomial using these roots.

If x = 1 and x = -2 are roots, then (x - 1) and (x + 2) are factors of the polynomial. Let me perform polynomial division or use synthetic division to factor them out.

First, let me divide the original polynomial by (x - 1). Wait, but synthetic division is easier for that. Let me try 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 with root 1:

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

So the remainder is 0, as expected. The quotient polynomial is x^3 + 6x^2 + 15x + 14.

Now, let me divide this cubic polynomial by (x + 2), since x = -2 is another root. Let me use synthetic division again, this time with root -2.

Coefficients of the cubic: 1 (x^3), 6 (x^2), 15 (x), 14 (constant).

Set up synthetic division with -2:

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

So the remainder is 0, as expected. The quotient polynomial is x^2 + 4x + 7.

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

Now, to find all integer roots, we need to check if the quadratic factor x^2 + 4x + 7 has any integer roots. Let me compute its discriminant: D = 16 - 4*1*7 = 16 - 28 = -12. Since the discriminant is negative, there are no real roots, let alone integer roots. Therefore, the only integer roots are x = 1 and x = -2.

Wait, but let me double-check my factoring. Let me multiply out (x - 1)(x + 2)(x^2 + 4x + 7) to ensure that it equals the original polynomial.

First, multiply (x - 1)(x + 2):

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

Now multiply this by (x^2 + 4x + 7):

Multiply x^2 + x - 2 by x^2 + 4x + 7.

Let me do term by term:

First, x^2*(x^2 + 4x + 7) = x^4 + 4x^3 + 7x^2.

Then, x*(x^2 + 4x + 7) = x^3 + 4x^2 + 7x.

Then, -2*(x^2 + 4x + 7) = -2x^2 -8x -14.

Now, add all these together:

x^4 + 4x^3 + 7x^2 + x^3 + 4x^2 + 7x -2x^2 -8x -14.

Combine like terms:

x^4 + (4x^3 + x^3) + (7x^2 + 4x^2 - 2x^2) + (7x -8x) -14.

Which is:

x^4 + 5x^3 + 9x^2 - x -14.

Yes! That matches the original polynomial. So the factoring is correct. Therefore, the only integer roots are indeed 1 and -2.

But just to make sure I didn't miss any other possible roots, let me check if maybe there are multiple roots or something. Wait, the original polynomial is degree 4, so there are four roots in total (real or complex). We found two real roots, and the other two are complex. Since the quadratic factor has discriminant -12, they are complex conjugates. Therefore, there are no other real roots, so no other integer roots.

Therefore, the integer roots are 1 and -2.

But let me just check once again if I made a mistake in testing the possible roots. For example, maybe I miscalculated for x = -2? Let me recheck that.

Original equation: x^4 +5x^3 +9x^2 -x -14.

For x = -2:

(-2)^4 = 16; 5*(-2)^3 = 5*(-8) = -40; 9*(-2)^2 = 9*4 = 36; -(-2) = +2; -14.

So total: 16 -40 +36 +2 -14.

Calculating step by step:

16 -40 = -24; -24 +36 = 12; 12 +2 =14; 14 -14 = 0. Correct.

And for x = 1: 1 +5 +9 -1 -14 = 0. Correct.

So no mistakes there.

Therefore, I think I can confidently say that the integer roots are 1 and -2.

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

To find all the **integer roots** of the polynomial

$$
x^4 + 5x^3 + 9x^2 - x - 14 = 0,
$$

we begin by applying the **Rational Root Theorem**. This theorem tells us that any rational root of a polynomial with integer coefficients must be of the form $\frac{p}{q}$, where:

- $p$ is a factor of the **constant term** (here, $-14$),
- $q$ is a factor of the **leading coefficient** (here, $1$).

Since the leading coefficient is 1, the only possible rational roots are the **integer factors** of $-14$, which are:

$$
\pm1, \pm2, \pm7, \pm14.
$$

---

### Step 1: Test Each Possible Integer Root

We substitute each of these values into the polynomial to check if they yield zero.

- **$x = 1$:**
  $$
  1^4 + 5(1)^3 + 9(1)^2 - 1 - 14 = 1 + 5 + 9 - 1 - 14 = 0
  $$
  ✅ **Root: $x = 1$**

- **$x = -1$:**
  $$
  (-1)^4 + 5(-1)^3 + 9(-1)^2 - (-1) - 14 = 1 - 5 + 9 + 1 - 14 = -8
  $$
  ❌ Not a root

- **$x = 2$:**
  $$
  2^4 + 5(2)^3 + 9(2)^2 - 2 - 14 = 16 + 40 + 36 - 2 - 14 = 76
  $$
  ❌ Not a root

- **$x = -2$:**
  $$
  (-2)^4 + 5(-2)^3 + 9(-2)^2 - (-2) - 14 = 16 - 40 + 36 + 2 - 14 = 0
  $$
  ✅ **Root: $x = -2$**

- **$x = 7$:**
  $$
  7^4 + 5(7)^3 + 9(7)^2 - 7 - 14 = 2401 + 1715 + 441 - 7 - 14 = 4536
  $$
  ❌ Not a root

- **$x = -7$:**
  $$
  (-7)^4 + 5(-7)^3 + 9(-7)^2 - (-7) - 14 = 2401 - 1715 + 441 + 7 - 14 = 1120
  $$
  ❌ Not a root

- **$x = 14$:**
  $$
  14^4 + 5(14)^3 + 9(14)^2 - 14 - 14 = 38416 + 13720 + 1764 - 14 - 14 = 53872
  $$
  ❌ Not a root

- **$x = -14$:**
  $$
  (-14)^4 + 5(-14)^3 + 9(-14)^2 - (-14) - 14 = 38416 - 13720 + 1764 + 14 - 14 = 26460
  $$
  ❌ Not a root

---

### Step 2: Factor the Polynomial

We have found two integer roots: $x = 1$ and $x = -2$. This means that the polynomial is divisible by $(x - 1)(x + 2)$. We can now perform **polynomial division** or **synthetic division** to factor the original polynomial.

Using synthetic division:

- Divide the original polynomial by $(x - 1)$, then divide the result by $(x + 2)$, we get:

$$
x^4 + 5x^3 + 9x^2 - x - 14 = (x - 1)(x + 2)(x^2 + 4x + 7)
$$

---

### Step 3: Analyze the Quadratic Factor

Now, we examine the quadratic factor:

$$
x^2 + 4x + 7
$$

Compute its **discriminant**:

$$
D = b^2 - 4ac = 16 - 28 = -12
$$

Since the discriminant is **negative**, the quadratic has **no real roots**, and thus **no integer roots**.

---

### Final Conclusion

The only **integer roots** of the polynomial are:

$$
\boxed{-2}, \boxed{1}
$$

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 think. Hmm, modulus operations can sometimes be tricky, but I remember that the remainder is what's left after dividing the number by the modulus. So, basically, I need to divide 54 by 6 and find out what's left over.

Wait, let me recall the definition. The remainder when a number 'a' is divided by 'n' is the integer 'r' such that 0 ≤ r < n and a = qn + r, where q is the quotient. So in this case, a is 54 and n is 6. So I need to find q and r such that 54 = 6q + r, with r being between 0 and 5, since 6 is the modulus.

Let me try dividing 54 by 6. Well, 6 times 9 is 54, right? 6*9=54. So if I divide 54 by 6, the quotient q is 9, and the remainder r would be 0 because 6*9 is exactly 54. So there's no remainder left. Therefore, the remainder should be 0.

But wait, let me check that again. Maybe I made a mistake. Let me do the division step by step. If I take 54 divided by 6, how many times does 6 go into 54? Let me count. 6*1=6, 6*2=12, 6*3=18, 6*4=24, 6*5=30, 6*6=36, 6*7=42, 6*8=48, 6*9=54. Oh, so 6 goes into 54 exactly 9 times with no remainder. So yes, the remainder is 0.

Alternatively, maybe I can think of it in terms of multiples. Since 6 is a factor of 54, then 54 is a multiple of 6, which means when you divide them, there's no remainder. So the remainder is 0.

But just to make sure I'm not missing something, let me try another approach. Let's subtract multiples of 6 from 54 until I get a number less than 6. Starting with 54, subtract 6: 54 - 6 = 48. Then subtract 6 again: 48 - 6 = 42. Keep going: 42 - 6 = 36, 36 - 6 = 30, 30 - 6 = 24, 24 - 6 = 18, 18 - 6 = 12, 12 - 6 = 6, 6 - 6 = 0. So after subtracting 6 nine times, I get to 0. That means 54 divided by 6 is 9 with a remainder of 0.

Another way to think about it is using the modulo operation properties. For example, 54 divided by 6 is 9, so 54 mod 6 is 0. But maybe I can use some other properties? Like, if I know that 6*9=54, then 54 mod 6 is 0.

Wait, maybe I can use division with remainder. The division algorithm 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. So here, a is 54, b is 6. So 54 = 6q + r, and 0 ≤ r < 6. We need to find q and r.

Since 6*9=54, then q=9 and r=0. Therefore, the remainder is 0.

Alternatively, if I didn't know that 6*9=54, maybe I could do the division step by step. Let me try long division. Dividing 54 by 6.

6 goes into 5 (the first digit) 0 times. So we consider 54. 6 goes into 54 how many times? Well, 6*9=54, so 9 times. Multiply 9 by 6, which is 54. Subtract 54 from 54, which gives 0. So the remainder is 0.

Alternatively, maybe I can use modular arithmetic properties. For example, 54 divided by 6. Since 6 is 6, and 54 is 6*9, so 54 is congruent to 0 mod 6.

Wait, but maybe I can think of smaller numbers. Let me check with smaller numbers to see if my reasoning is correct. For example, 6 mod 6 is 0, 12 mod 6 is 0, 18 mod 6 is 0, so 54, which is 6*9, should also be 0 mod 6.

Alternatively, maybe I can use the fact that 54 divided by 6 is 9, so the remainder is 0.

Is there any chance that the remainder could be something else? For example, if the division wasn't exact, then the remainder would be the leftover. But since 54 is exactly divisible by 6, there's nothing left.

Wait, let me check with another example. Let's say 55 divided by 6. Then 6*9=54, so 55-54=1, so remainder is 1. Similarly, 54 divided by 6 is 9*6, so remainder is 0.

Another way: think of the number line. 6*9=54, so 54 is at the multiple of 6, so the remainder is 0.

Alternatively, if I use the formula: remainder = dividend - divisor * quotient. So here, remainder = 54 - 6*9 = 54 - 54 = 0.

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

But just to make sure, let me check with a calculator. If I divide 54 by 6, the result is 9.0, which is an integer, so there's no remainder. Therefore, the remainder is 0.

Alternatively, if I use a calculator to compute 54 mod 6, it should give me 0.

Wait, but maybe I can think of it in terms of negative numbers? For example, if I have a negative number, the remainder can be different, but in this case, 54 is positive, and 6 is positive. So the remainder should be between 0 and 5.

Alternatively, if I use the division algorithm, since 54 is positive and 6 is positive, the quotient is positive, and the remainder is non-negative.

So, all these different approaches confirm that the remainder is 0.

Therefore, I think the answer is 0.

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

To determine the remainder of $ 54 \mod 6 $, we apply the **division algorithm**, which states that for any integers $ a $ and $ b $ (with $ b > 0 $), there exist unique integers $ q $ (the quotient) and $ r $ (the remainder) such that:

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

---

### Step-by-Step Breakdown:

- Here, $ a = 54 $ and $ b = 6 $.
- We want to find $ r $, the remainder when 54 is divided by 6.

#### Step 1: Perform the Division
We divide 54 by 6:

$$
54 \div 6 = 9
$$

This means that 6 fits into 54 exactly 9 times with no remainder.

#### Step 2: Apply the Division Algorithm
Using the formula:

$$
54 = 6 \times 9 + r
$$

$$
54 = 54 + r \Rightarrow r = 0
$$

So, the remainder $ r $ is 0.

---

### Alternative Verification:

- Since $ 6 \times 9 = 54 $, the number 54 is a **multiple** of 6.
- Therefore, when 54 is divided by 6, the remainder is **zero**.

---

### Final Answer:

$$
\boxed{0}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the remainder when 54 is divided by 6. Let me think. Hmm, modulus operations can sometimes be tricky, but I remember that the remainder is what's left after dividing the number by the modulus. So, basically, I need to divide 54 by 6 and find out what's left over.

Wait, let me recall the definition. The remainder when a number 'a' is divided by 'n' is the integer 'r' such that 0 ≤ r < n and a = qn + r, where q is the quotient. So in this case, a is 54 and n is 6. So I need to find q and r such that 54 = 6q + r, with r being between 0 and 5, since 6 is the modulus.

Let me try dividing 54 by 6. Well, 6 times 9 is 54, right? 6*9=54. So if I divide 54 by 6, the quotient q is 9, and the remainder r would be 0 because 6*9 is exactly 54. So there's no remainder left. Therefore, the remainder should be 0.

But wait, let me check that again. Maybe I made a mistake. Let me do the division step by step. If I take 54 divided by 6, how many times does 6 go into 54? Let me count. 6*1=6, 6*2=12, 6*3=18, 6*4=24, 6*5=30, 6*6=36, 6*7=42, 6*8=48, 6*9=54. Oh, so 6 goes into 54 exactly 9 times with no remainder. So yes, the remainder is 0.

Alternatively, maybe I can think of it in terms of multiples. Since 6 is a factor of 54, then 54 is a multiple of 6, which means when you divide them, there's no remainder. So the remainder is 0.

But just to make sure I'm not missing something, let me try another approach. Let's subtract multiples of 6 from 54 until I get a number less than 6. Starting with 54, subtract 6: 54 - 6 = 48. Then subtract 6 again: 48 - 6 = 42. Keep going: 42 - 6 = 36, 36 - 6 = 30, 30 - 6 = 24, 24 - 6 = 18, 18 - 6 = 12, 12 - 6 = 6, 6 - 6 = 0. So after subtracting 6 nine times, I get to 0. That means 54 divided by 6 is 9 with a remainder of 0.

Another way to think about it is using the modulo operation properties. For example, 54 divided by 6 is 9, so 54 mod 6 is 0. But maybe I can use some other properties? Like, if I know that 6*9=54, then 54 mod 6 is 0.

Wait, maybe I can use division with remainder. The division algorithm 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. So here, a is 54, b is 6. So 54 = 6q + r, and 0 ≤ r < 6. We need to find q and r.

Since 6*9=54, then q=9 and r=0. Therefore, the remainder is 0.

Alternatively, if I didn't know that 6*9=54, maybe I could do the division step by step. Let me try long division. Dividing 54 by 6.

6 goes into 5 (the first digit) 0 times. So we consider 54. 6 goes into 54 how many times? Well, 6*9=54, so 9 times. Multiply 9 by 6, which is 54. Subtract 54 from 54, which gives 0. So the remainder is 0.

Alternatively, maybe I can use modular arithmetic properties. For example, 54 divided by 6. Since 6 is 6, and 54 is 6*9, so 54 is congruent to 0 mod 6.

Wait, but maybe I can think of smaller numbers. Let me check with smaller numbers to see if my reasoning is correct. For example, 6 mod 6 is 0, 12 mod 6 is 0, 18 mod 6 is 0, so 54, which is 6*9, should also be 0 mod 6.

Alternatively, maybe I can use the fact that 54 divided by 6 is 9, so the remainder is 0.

Is there any chance that the remainder could be something else? For example, if the division wasn't exact, then the remainder would be the leftover. But since 54 is exactly divisible by 6, there's nothing left.

Wait, let me check with another example. Let's say 55 divided by 6. Then 6*9=54, so 55-54=1, so remainder is 1. Similarly, 54 divided by 6 is 9*6, so remainder is 0.

Another way: think of the number line. 6*9=54, so 54 is at the multiple of 6, so the remainder is 0.

Alternatively, if I use the formula: remainder = dividend - divisor * quotient. So here, remainder = 54 - 6*9 = 54 - 54 = 0.

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

But just to make sure, let me check with a calculator. If I divide 54 by 6, the result is 9.0, which is an integer, so there's no remainder. Therefore, the remainder is 0.

Alternatively, if I use a calculator to compute 54 mod 6, it should give me 0.

Wait, but maybe I can think of it in terms of negative numbers? For example, if I have a negative number, the remainder can be different, but in this case, 54 is positive, and 6 is positive. So the remainder should be between 0 and 5.

Alternatively, if I use the division algorithm, since 54 is positive and 6 is positive, the quotient is positive, and the remainder is non-negative.

So, all these different approaches confirm that the remainder is 0.

Therefore, I think the answer is 0.

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

To determine the remainder of $ 54 \mod 6 $, we apply the **division algorithm**, which states that for any integers $ a $ and $ b $ (with $ b > 0 $), there exist unique integers $ q $ (the quotient) and $ r $ (the remainder) such that:

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

---

### Step-by-Step Breakdown:

- Here, $ a = 54 $ and $ b = 6 $.
- We want to find $ r $, the remainder when 54 is divided by 6.

#### Step 1: Perform the Division
We divide 54 by 6:

$$
54 \div 6 = 9
$$

This means that 6 fits into 54 exactly 9 times with no remainder.

#### Step 2: Apply the Division Algorithm
Using the formula:

$$
54 = 6 \times 9 + r
$$

$$
54 = 54 + r \Rightarrow r = 0
$$

So, the remainder $ r $ is 0.

---

### Alternative Verification:

- Since $ 6 \times 9 = 54 $, the number 54 is a **multiple** of 6.
- Therefore, when 54 is divided by 6, the remainder is **zero**.

---

### 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. I remember that distance equals speed multiplied by time. So the formula is distance = speed × time. Right?

Alright, so the speed here is 50 miles per hour, and the time is 2 and 3/4 hours. Wait, 2 and 3/4 hours 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 is simpler.

First, converting 2 3/4 hours to an improper fraction. The whole number part is 2, which is 2/1. The fractional part is 3/4. To add them together, I need a common denominator. So 2 is 8/4, right? Because 2 times 4 is 8. Then adding 8/4 and 3/4 gives me 11/4. So 2 3/4 hours is equal to 11/4 hours. Okay, that seems right.

Alternatively, if I convert it to a decimal, 3/4 is 0.75, so 2.75 hours. Either way, I can use that in the calculation. Let me try both methods to check my answer.

Starting with the improper fraction: distance = 50 mph × 11/4 hours. Let me compute that. Multiplying 50 by 11 first. 50 times 11 is... 550. Then divide by 4. So 550 divided by 4. Let me do that division. 4 goes into 550 how many times?

Well, 4 × 137 is 548, because 4 × 100 is 400, 4 × 30 is 120, 4 × 7 is 28. So 400 + 120 is 520, plus 28 is 548. Then 550 minus 548 is 2. So that would be 137 and 2/4, which simplifies to 137 and 1/2. So 137.5 miles? Wait, but the question says to express the answer as a mixed number. So 137 1/2 miles. Let me check that again.

Alternatively, using decimals: 50 mph × 2.75 hours. Let me compute that. 50 × 2 is 100, and 50 × 0.75 is 37.5. Adding them together, 100 + 37.5 is 137.5. Which is the same as 137 1/2. So that matches. So both methods give me the same answer. That seems good.

Wait, let me just make sure I didn't make any calculation errors. Let me verify the improper fraction method again. 50 multiplied by 11 is indeed 550. Divided by 4. 550 divided by 4. Let me divide 550 by 2 first, which is 275, then divide by 2 again, which is 137.5. Yep, same result. So that's correct.

Alternatively, if I think about breaking down the time. 2 and 3/4 hours is 2 hours plus 45 minutes. Because 3/4 of an hour is 45 minutes. So maybe I can calculate the distance for 2 hours and then for 45 minutes and add them together.

For 2 hours, at 50 mph, that's 50 × 2 = 100 miles. Then for 45 minutes, which is 0.75 hours, that's 50 × 0.75. Let me compute that. 50 × 0.75. Well, 50 × 0.7 is 35, and 50 × 0.05 is 2.5, so adding those together gives 35 + 2.5 = 37.5 miles. Then adding that to the 100 miles gives 137.5 miles. Again, same answer. So that's consistent.

So all methods lead me to 137.5 miles, which as a mixed number is 137 1/2. Therefore, the answer should be 137 1/2 miles. Let me just check once more if there's any chance I messed up the conversion from the mixed number to the improper fraction.

Original time: 2 3/4 hours. To convert to improper fraction: 2 × 4 = 8, plus 3 is 11. So 11/4. Yes, that's correct. So 50 × 11/4. 50 × 11 is 550. Divided by 4. 550 divided by 4. 4 × 100 is 400, 550 - 400 is 150. 4 × 37 is 148, so 100 + 37 is 137, with a remainder of 2. So 137 and 2/4, which reduces to 137 1/2. Yep, that's correct.

Alternatively, if I think about fractions. 50 is a whole number, so multiplying by 11/4. 50 can be written as 50/1. So multiplying 50/1 × 11/4 = (50 × 11)/(1 × 4) = 550/4. Simplify that fraction. Divide numerator and denominator by 2: 275/2. Which is equal to 137 1/2. Yep, same result. So that's correct.

So all methods confirm that the distance is 137 1/2 miles. Therefore, the answer is 137 and 1/2, which as a mixed number is already expressed. So I think that's the 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 period 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

The time given is $ 2\frac{3}{4} $ hours. To convert this to an improper fraction:

$$
2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4}
$$

---

### Step 2: Multiply Speed by Time

Now, multiply the speed (50 miles per hour) 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 Fraction

Divide the numerator by the denominator:

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

Simplify the fractional part:

$$
\frac{2}{4} = \frac{1}{2}
$$

So, the final distance is:

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

---

### Final Answer

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

[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. I remember that distance equals speed multiplied by time. So the formula is distance = speed × time. Right?

Alright, so the speed here is 50 miles per hour, and the time is 2 and 3/4 hours. Wait, 2 and 3/4 hours 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 is simpler.

First, converting 2 3/4 hours to an improper fraction. The whole number part is 2, which is 2/1. The fractional part is 3/4. To add them together, I need a common denominator. So 2 is 8/4, right? Because 2 times 4 is 8. Then adding 8/4 and 3/4 gives me 11/4. So 2 3/4 hours is equal to 11/4 hours. Okay, that seems right.

Alternatively, if I convert it to a decimal, 3/4 is 0.75, so 2.75 hours. Either way, I can use that in the calculation. Let me try both methods to check my answer.

Starting with the improper fraction: distance = 50 mph × 11/4 hours. Let me compute that. Multiplying 50 by 11 first. 50 times 11 is... 550. Then divide by 4. So 550 divided by 4. Let me do that division. 4 goes into 550 how many times?

Well, 4 × 137 is 548, because 4 × 100 is 400, 4 × 30 is 120, 4 × 7 is 28. So 400 + 120 is 520, plus 28 is 548. Then 550 minus 548 is 2. So that would be 137 and 2/4, which simplifies to 137 and 1/2. So 137.5 miles? Wait, but the question says to express the answer as a mixed number. So 137 1/2 miles. Let me check that again.

Alternatively, using decimals: 50 mph × 2.75 hours. Let me compute that. 50 × 2 is 100, and 50 × 0.75 is 37.5. Adding them together, 100 + 37.5 is 137.5. Which is the same as 137 1/2. So that matches. So both methods give me the same answer. That seems good.

Wait, let me just make sure I didn't make any calculation errors. Let me verify the improper fraction method again. 50 multiplied by 11 is indeed 550. Divided by 4. 550 divided by 4. Let me divide 550 by 2 first, which is 275, then divide by 2 again, which is 137.5. Yep, same result. So that's correct.

Alternatively, if I think about breaking down the time. 2 and 3/4 hours is 2 hours plus 45 minutes. Because 3/4 of an hour is 45 minutes. So maybe I can calculate the distance for 2 hours and then for 45 minutes and add them together.

For 2 hours, at 50 mph, that's 50 × 2 = 100 miles. Then for 45 minutes, which is 0.75 hours, that's 50 × 0.75. Let me compute that. 50 × 0.75. Well, 50 × 0.7 is 35, and 50 × 0.05 is 2.5, so adding those together gives 35 + 2.5 = 37.5 miles. Then adding that to the 100 miles gives 137.5 miles. Again, same answer. So that's consistent.

So all methods lead me to 137.5 miles, which as a mixed number is 137 1/2. Therefore, the answer should be 137 1/2 miles. Let me just check once more if there's any chance I messed up the conversion from the mixed number to the improper fraction.

Original time: 2 3/4 hours. To convert to improper fraction: 2 × 4 = 8, plus 3 is 11. So 11/4. Yes, that's correct. So 50 × 11/4. 50 × 11 is 550. Divided by 4. 550 divided by 4. 4 × 100 is 400, 550 - 400 is 150. 4 × 37 is 148, so 100 + 37 is 137, with a remainder of 2. So 137 and 2/4, which reduces to 137 1/2. Yep, that's correct.

Alternatively, if I think about fractions. 50 is a whole number, so multiplying by 11/4. 50 can be written as 50/1. So multiplying 50/1 × 11/4 = (50 × 11)/(1 × 4) = 550/4. Simplify that fraction. Divide numerator and denominator by 2: 275/2. Which is equal to 137 1/2. Yep, same result. So that's correct.

So all methods confirm that the distance is 137 1/2 miles. Therefore, the answer is 137 and 1/2, which as a mixed number is already expressed. So I think that's the 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 period 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

The time given is $ 2\frac{3}{4} $ hours. To convert this to an improper fraction:

$$
2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4}
$$

---

### Step 2: Multiply Speed by Time

Now, multiply the speed (50 miles per hour) 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 Fraction

Divide the numerator by the denominator:

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

Simplify the fractional part:

$$
\frac{2}{4} = \frac{1}{2}
$$

So, the final distance is:

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

---

### Final Answer

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

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. Let me write that down to visualize it better:

log₂3 × log₃4 × log₄5 × log₅6 × log₆7 × log₇8

Hmm, I remember that there's a property of logarithms that can help simplify products of logs with consecutive bases. Let me recall... Oh yeah! The change of base formula. Wait, but isn't there something called the chain rule for logarithms? Or maybe it's called the telescoping product?

Let me think. If I have log_a b × log_b c, that's equal to log_a c, right? Wait, is that true? Let me check with numbers. Let's take log₂4 × log₄8. Log base 2 of 4 is 2, log base 4 of 8 is log₄(4×2) = log₄4 + log₄2 = 1 + 0.5 = 1.5. Multiplying them gives 2 × 1.5 = 3. On the other hand, log₂8 is 3. So yes! It works. So log_a b × log_b c = log_a c. That seems to be the case. So this is a property where the product of logs with consecutive bases simplifies to a single logarithm.

So applying that here, if I have log₂3 × log₃4 × log₄5 × ... × log₇8, then each time the base of the next log is the result of the previous log's argument. So this should telescope down to log₂8. Wait, let me verify with a smaller example. Let's take log₂3 × log₃4. According to the property, that should be log₂4, which is 2. Let me compute it directly: log₂3 is approximately 1.58496, log₃4 is approximately 1.26186, multiplying them gives roughly 1.58496 × 1.26186 ≈ 2, which is indeed log₂4. So that works. So if I have more terms, like log₂3 × log₃4 × log₄5, that should be log₂5. Let me check with numbers. log₂3 ≈ 1.58496, log₃4 ≈ 1.26186, log₄5 ≈ 1.16096. Multiplying them: 1.58496 × 1.26186 ≈ 2, then × 1.16096 ≈ 2.3219, which is log₂5 ≈ 2.3219. Yep, that works. So the pattern is that each time, the product telescopes to log base 2 of the last argument, which in this case is 8. Therefore, the entire product should be log₂8.

But log₂8 is 3, since 2^3 = 8. Therefore, the value of the product is 3. Wait, is that really all? Let me check with another example. Suppose I have log₂3 × log₃4 × log₄5 × log₅6 × log₆7 × log₇8. If I apply the property step by step:

First, log₂3 × log₃4 = log₂4

Then, log₂4 × log₄5 = log₂5

Then, log₂5 × log₅6 = log₂6

Then, log₂6 × log₆7 = log₂7

Then, log₂7 × log₇8 = log₂8

Yes! So each multiplication cancels out the previous base and argument, leaving the next one. So after all the multiplications, it's log₂8, which is 3. Therefore, the answer is 3. That seems straightforward, but let me make sure I didn't miss anything.

Alternatively, maybe I can use the change of base formula to convert all logs to the same base and see if that gives the same result. Let me try that approach as a verification.

The change of base formula says that log_a b = (ln b)/(ln a). So applying that to each term:

log₂3 = ln3 / ln2

log₃4 = ln4 / ln3

log₄5 = ln5 / ln4

log₅6 = ln6 / ln5

log₆7 = ln7 / ln6

log₇8 = ln8 / ln7

Multiplying all these together:

(ln3 / ln2) × (ln4 / ln3) × (ln5 / ln4) × (ln6 / ln5) × (ln7 / ln6) × (ln8 / ln7)

Now, let's see, the numerator of each fraction cancels with the denominator of the next fraction. For example, ln3 in the numerator of the first term cancels with ln3 in the denominator of the second term. Similarly, ln4 cancels between the second numerator and third denominator, and so on. So all the intermediate terms cancel out, leaving:

ln8 / ln2

Which is log₂8, as before. And log₂8 is 3. So that confirms the same result. Therefore, regardless of the method, the answer is 3.

Wait, but let me check if there's any chance that I misapplied the property. For example, is the property log_a b × log_b c = log_a c always true? Let me think. Let me take another example where I can compute both sides. Let me take log_2 8 × log_8 16. According to the property, that should be log_2 16 = 4. Let me compute it directly. log_2 8 is 3, log_8 16 is log_8 (8 × 2) = log_8 8 + log_8 2 = 1 + (1/3) ≈ 1.333... Multiplying 3 × 1.333... ≈ 4, which is indeed log_2 16. So that works. Therefore, the property holds.

Therefore, applying this property repeatedly, the entire product telescopes to log_2 8 = 3. Therefore, the answer is 3.

Alternatively, maybe I can think of it in terms of exponents. Let me consider each logarithm as an exponent. For example, log₂3 is the exponent to which we need to raise 2 to get 3. Similarly, log₃4 is the exponent to which we need to raise 3 to get 4, and so on. So if I multiply all these exponents together, what does that represent?

Let me denote:

Let’s say log₂3 = a, so 2^a = 3

log₃4 = b, so 3^b = 4

log₄5 = c, so 4^c = 5

log₅6 = d, so 5^d = 6

log₆7 = e, so 6^e = 7

log₇8 = f, so 7^f = 8

Then the product is a × b × c × d × e × f.

If I substitute each variable:

a = log₂3, so 2^a = 3

Then 3^b = 4, so substituting 3 from above: (2^a)^b = 2^{ab} = 4

Similarly, 4^c = 5, so substituting 4 from above: (2^{ab})^c = 2^{abc} = 5

Continuing this way:

5^d = 6 => (2^{abc})^d = 2^{abcd} = 6

6^e = 7 => (2^{abcd})^e = 2^{abcde} = 7

7^f = 8 => (2^{abcde})^f = 2^{abcdef} = 8

Therefore, 2^{abcdef} = 8 = 2^3, so abcdef = 3. Therefore, the product a × b × c × d × e × f = 3. Therefore, the value is 3.

This method also leads to the same conclusion. So regardless of the approach—whether using the telescoping property, converting to natural logs and seeing cancellation, or expressing each log as an exponent and substituting—the result is 3.

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

But just to make sure, let me compute the numerical value step by step using approximate decimal values for each logarithm and multiply them. Let me try that.

First, compute each log:

log₂3: log base 2 of 3. Since 2^1 = 2 and 2^2 = 4, so it's between 1 and 2. Using calculator, log₂3 ≈ 1.58496

log₃4: log base 3 of 4. Similarly, between 1 and 2. Since 3^1 = 3, 3^1.26186 ≈ 4. Let me check with calculator: log(4)/log(3) ≈ 1.26186

log₄5: log base 4 of 5. 4^1 = 4, 4^1.16096 ≈ 5. Let me compute log(5)/log(4) ≈ 1.16096

log₅6: log base 5 of 6. 5^1 = 5, 5^1.11328 ≈ 6. Let me compute log(6)/log(5) ≈ 1.11328

log₆7: log base 6 of 7. 6^1 = 6, 6^1.0860 ≈ 7. Let me compute log(7)/log(6) ≈ 1.0860

log₇8: log base 7 of 8. 7^1 = 7, 7^1.0686 ≈ 8. Let me compute log(8)/log(7) ≈ 1.0686

Now multiply all these approximate values:

Start with 1.58496 × 1.26186 ≈ Let me compute that first.

1.58496 × 1.26186. Let me approximate:

First, 1.5 × 1.26 = 1.89. But more accurately:

1.58496 × 1.26186 ≈ Let me use calculator steps:

Multiply 1.58496 × 1.26186:

Well, 1.58496 × 1 = 1.58496

1.58496 × 0.2 = 0.316992

1.58496 × 0.06 = 0.0950976

1.58496 × 0.00186 ≈ approx 0.00295

Adding them up: 1.58496 + 0.316992 = 1.901952 + 0.0950976 = 1.9970496 + 0.00295 ≈ 2.0. So approximately 2.0.

Then multiply by the next term, which is 1.16096:

2.0 × 1.16096 ≈ 2.32192

Next term is 1.11328:

2.32192 × 1.11328 ≈ Let me compute 2.32192 × 1.1 = 2.554112, plus 2.32192 × 0.01328 ≈ approx 0.0308. Total ≈ 2.554112 + 0.0308 ≈ 2.5849

Next term is 1.0860:

2.5849 × 1.0860 ≈ Let me compute 2.5849 × 1 = 2.5849, plus 2.5849 × 0.08 = 0.20679, plus 2.5849 × 0.006 ≈ 0.0155. Total ≈ 2.5849 + 0.20679 = 2.79169 + 0.0155 ≈ 2.8072

Next term is 1.0686:

2.8072 × 1.0686 ≈ Let me compute 2.8072 × 1 = 2.8072, plus 2.8072 × 0.06 = 0.16843, plus 2.8072 × 0.0086 ≈ approx 0.0241. Total ≈ 2.8072 + 0.16843 = 2.9756 + 0.0241 ≈ 3.0.

Wow, so after multiplying all the approximate decimal values, I end up with approximately 3.0. That's very close to 3, which confirms the exact value is indeed 3.

Therefore, all methods lead to the conclusion that the product is 3.

I think that's thorough enough. I considered multiple approaches: telescoping product, change of base formula, expressing logs as exponents, and numerical approximation. All of them confirm that the value is 3. So I can be confident that the answer 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 can utilize a key property of logarithms known as the **telescoping property**. This property states that:

$$
\log_a{b} \cdot \log_b{c} = \log_a{c}
$$

This is a direct consequence of the **change of base formula**:

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

Using this, we can rewrite each logarithm in terms of a common base (say, base 2), and then observe that the intermediate terms cancel out.

---

### Step-by-Step Simplification

Let’s apply the change of base formula to each term:

$$
\log_2{3} = \frac{\ln{3}}{\ln{2}}, \quad
\log_3{4} = \frac{\ln{4}}{\ln{3}}, \quad
\log_4{5} = \frac{\ln{5}}{\ln{4}}, \quad
\log_5{6} = \frac{\ln{6}}{\ln{5}}, \quad
\log_6{7} = \frac{\ln{7}}{\ln{6}}, \quad
\log_7{8} = \frac{\ln{8}}{\ln{7}}
$$

Now, multiplying all these together:

$$
\left( \frac{\ln{3}}{\ln{2}} \right) \cdot \left( \frac{\ln{4}}{\ln{3}} \right) \cdot \left( \frac{\ln{5}}{\ln{4}} \right) \cdot \left( \frac{\ln{6}}{\ln{5}} \right) \cdot \left( \frac{\ln{7}}{\ln{6}} \right) \cdot \left( \frac{\ln{8}}{\ln{7}} \right)
$$

Notice that the numerator of each fraction cancels with the denominator of the next one:

- $\ln{3}$ cancels with $\ln{3}$
- $\ln{4}$ cancels with $\ln{4}$
- $\ln{5}$ cancels with $\ln{5}$
- $\ln{6}$ cancels with $\ln{6}$
- $\ln{7}$ cancels with $\ln{7}$

This leaves:

$$
\frac{\ln{8}}{\ln{2}} = \log_2{8}
$$

---

### Final Evaluation

We now compute:

$$
\log_2{8} = \frac{\ln{8}}{\ln{2}} = \frac{\ln{2^3}}{\ln{2}} = \frac{3\ln{2}}{\ln{2}} = 3
$$

---

### Conclusion

Thus, the value of the original expression is:

$$
\boxed{3}
$$

[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. Let me write that down to visualize it better:

log₂3 × log₃4 × log₄5 × log₅6 × log₆7 × log₇8

Hmm, I remember that there's a property of logarithms that can help simplify products of logs with consecutive bases. Let me recall... Oh yeah! The change of base formula. Wait, but isn't there something called the chain rule for logarithms? Or maybe it's called the telescoping product?

Let me think. If I have log_a b × log_b c, that's equal to log_a c, right? Wait, is that true? Let me check with numbers. Let's take log₂4 × log₄8. Log base 2 of 4 is 2, log base 4 of 8 is log₄(4×2) = log₄4 + log₄2 = 1 + 0.5 = 1.5. Multiplying them gives 2 × 1.5 = 3. On the other hand, log₂8 is 3. So yes! It works. So log_a b × log_b c = log_a c. That seems to be the case. So this is a property where the product of logs with consecutive bases simplifies to a single logarithm.

So applying that here, if I have log₂3 × log₃4 × log₄5 × ... × log₇8, then each time the base of the next log is the result of the previous log's argument. So this should telescope down to log₂8. Wait, let me verify with a smaller example. Let's take log₂3 × log₃4. According to the property, that should be log₂4, which is 2. Let me compute it directly: log₂3 is approximately 1.58496, log₃4 is approximately 1.26186, multiplying them gives roughly 1.58496 × 1.26186 ≈ 2, which is indeed log₂4. So that works. So if I have more terms, like log₂3 × log₃4 × log₄5, that should be log₂5. Let me check with numbers. log₂3 ≈ 1.58496, log₃4 ≈ 1.26186, log₄5 ≈ 1.16096. Multiplying them: 1.58496 × 1.26186 ≈ 2, then × 1.16096 ≈ 2.3219, which is log₂5 ≈ 2.3219. Yep, that works. So the pattern is that each time, the product telescopes to log base 2 of the last argument, which in this case is 8. Therefore, the entire product should be log₂8.

But log₂8 is 3, since 2^3 = 8. Therefore, the value of the product is 3. Wait, is that really all? Let me check with another example. Suppose I have log₂3 × log₃4 × log₄5 × log₅6 × log₆7 × log₇8. If I apply the property step by step:

First, log₂3 × log₃4 = log₂4

Then, log₂4 × log₄5 = log₂5

Then, log₂5 × log₅6 = log₂6

Then, log₂6 × log₆7 = log₂7

Then, log₂7 × log₇8 = log₂8

Yes! So each multiplication cancels out the previous base and argument, leaving the next one. So after all the multiplications, it's log₂8, which is 3. Therefore, the answer is 3. That seems straightforward, but let me make sure I didn't miss anything.

Alternatively, maybe I can use the change of base formula to convert all logs to the same base and see if that gives the same result. Let me try that approach as a verification.

The change of base formula says that log_a b = (ln b)/(ln a). So applying that to each term:

log₂3 = ln3 / ln2

log₃4 = ln4 / ln3

log₄5 = ln5 / ln4

log₅6 = ln6 / ln5

log₆7 = ln7 / ln6

log₇8 = ln8 / ln7

Multiplying all these together:

(ln3 / ln2) × (ln4 / ln3) × (ln5 / ln4) × (ln6 / ln5) × (ln7 / ln6) × (ln8 / ln7)

Now, let's see, the numerator of each fraction cancels with the denominator of the next fraction. For example, ln3 in the numerator of the first term cancels with ln3 in the denominator of the second term. Similarly, ln4 cancels between the second numerator and third denominator, and so on. So all the intermediate terms cancel out, leaving:

ln8 / ln2

Which is log₂8, as before. And log₂8 is 3. So that confirms the same result. Therefore, regardless of the method, the answer is 3.

Wait, but let me check if there's any chance that I misapplied the property. For example, is the property log_a b × log_b c = log_a c always true? Let me think. Let me take another example where I can compute both sides. Let me take log_2 8 × log_8 16. According to the property, that should be log_2 16 = 4. Let me compute it directly. log_2 8 is 3, log_8 16 is log_8 (8 × 2) = log_8 8 + log_8 2 = 1 + (1/3) ≈ 1.333... Multiplying 3 × 1.333... ≈ 4, which is indeed log_2 16. So that works. Therefore, the property holds.

Therefore, applying this property repeatedly, the entire product telescopes to log_2 8 = 3. Therefore, the answer is 3.

Alternatively, maybe I can think of it in terms of exponents. Let me consider each logarithm as an exponent. For example, log₂3 is the exponent to which we need to raise 2 to get 3. Similarly, log₃4 is the exponent to which we need to raise 3 to get 4, and so on. So if I multiply all these exponents together, what does that represent?

Let me denote:

Let’s say log₂3 = a, so 2^a = 3

log₃4 = b, so 3^b = 4

log₄5 = c, so 4^c = 5

log₅6 = d, so 5^d = 6

log₆7 = e, so 6^e = 7

log₇8 = f, so 7^f = 8

Then the product is a × b × c × d × e × f.

If I substitute each variable:

a = log₂3, so 2^a = 3

Then 3^b = 4, so substituting 3 from above: (2^a)^b = 2^{ab} = 4

Similarly, 4^c = 5, so substituting 4 from above: (2^{ab})^c = 2^{abc} = 5

Continuing this way:

5^d = 6 => (2^{abc})^d = 2^{abcd} = 6

6^e = 7 => (2^{abcd})^e = 2^{abcde} = 7

7^f = 8 => (2^{abcde})^f = 2^{abcdef} = 8

Therefore, 2^{abcdef} = 8 = 2^3, so abcdef = 3. Therefore, the product a × b × c × d × e × f = 3. Therefore, the value is 3.

This method also leads to the same conclusion. So regardless of the approach—whether using the telescoping property, converting to natural logs and seeing cancellation, or expressing each log as an exponent and substituting—the result is 3.

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

But just to make sure, let me compute the numerical value step by step using approximate decimal values for each logarithm and multiply them. Let me try that.

First, compute each log:

log₂3: log base 2 of 3. Since 2^1 = 2 and 2^2 = 4, so it's between 1 and 2. Using calculator, log₂3 ≈ 1.58496

log₃4: log base 3 of 4. Similarly, between 1 and 2. Since 3^1 = 3, 3^1.26186 ≈ 4. Let me check with calculator: log(4)/log(3) ≈ 1.26186

log₄5: log base 4 of 5. 4^1 = 4, 4^1.16096 ≈ 5. Let me compute log(5)/log(4) ≈ 1.16096

log₅6: log base 5 of 6. 5^1 = 5, 5^1.11328 ≈ 6. Let me compute log(6)/log(5) ≈ 1.11328

log₆7: log base 6 of 7. 6^1 = 6, 6^1.0860 ≈ 7. Let me compute log(7)/log(6) ≈ 1.0860

log₇8: log base 7 of 8. 7^1 = 7, 7^1.0686 ≈ 8. Let me compute log(8)/log(7) ≈ 1.0686

Now multiply all these approximate values:

Start with 1.58496 × 1.26186 ≈ Let me compute that first.

1.58496 × 1.26186. Let me approximate:

First, 1.5 × 1.26 = 1.89. But more accurately:

1.58496 × 1.26186 ≈ Let me use calculator steps:

Multiply 1.58496 × 1.26186:

Well, 1.58496 × 1 = 1.58496

1.58496 × 0.2 = 0.316992

1.58496 × 0.06 = 0.0950976

1.58496 × 0.00186 ≈ approx 0.00295

Adding them up: 1.58496 + 0.316992 = 1.901952 + 0.0950976 = 1.9970496 + 0.00295 ≈ 2.0. So approximately 2.0.

Then multiply by the next term, which is 1.16096:

2.0 × 1.16096 ≈ 2.32192

Next term is 1.11328:

2.32192 × 1.11328 ≈ Let me compute 2.32192 × 1.1 = 2.554112, plus 2.32192 × 0.01328 ≈ approx 0.0308. Total ≈ 2.554112 + 0.0308 ≈ 2.5849

Next term is 1.0860:

2.5849 × 1.0860 ≈ Let me compute 2.5849 × 1 = 2.5849, plus 2.5849 × 0.08 = 0.20679, plus 2.5849 × 0.006 ≈ 0.0155. Total ≈ 2.5849 + 0.20679 = 2.79169 + 0.0155 ≈ 2.8072

Next term is 1.0686:

2.8072 × 1.0686 ≈ Let me compute 2.8072 × 1 = 2.8072, plus 2.8072 × 0.06 = 0.16843, plus 2.8072 × 0.0086 ≈ approx 0.0241. Total ≈ 2.8072 + 0.16843 = 2.9756 + 0.0241 ≈ 3.0.

Wow, so after multiplying all the approximate decimal values, I end up with approximately 3.0. That's very close to 3, which confirms the exact value is indeed 3.

Therefore, all methods lead to the conclusion that the product is 3.

I think that's thorough enough. I considered multiple approaches: telescoping product, change of base formula, expressing logs as exponents, and numerical approximation. All of them confirm that the value is 3. So I can be confident that the answer 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 can utilize a key property of logarithms known as the **telescoping property**. This property states that:

$$
\log_a{b} \cdot \log_b{c} = \log_a{c}
$$

This is a direct consequence of the **change of base formula**:

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

Using this, we can rewrite each logarithm in terms of a common base (say, base 2), and then observe that the intermediate terms cancel out.

---

### Step-by-Step Simplification

Let’s apply the change of base formula to each term:

$$
\log_2{3} = \frac{\ln{3}}{\ln{2}}, \quad
\log_3{4} = \frac{\ln{4}}{\ln{3}}, \quad
\log_4{5} = \frac{\ln{5}}{\ln{4}}, \quad
\log_5{6} = \frac{\ln{6}}{\ln{5}}, \quad
\log_6{7} = \frac{\ln{7}}{\ln{6}}, \quad
\log_7{8} = \frac{\ln{8}}{\ln{7}}
$$

Now, multiplying all these together:

$$
\left( \frac{\ln{3}}{\ln{2}} \right) \cdot \left( \frac{\ln{4}}{\ln{3}} \right) \cdot \left( \frac{\ln{5}}{\ln{4}} \right) \cdot \left( \frac{\ln{6}}{\ln{5}} \right) \cdot \left( \frac{\ln{7}}{\ln{6}} \right) \cdot \left( \frac{\ln{8}}{\ln{7}} \right)
$$

Notice that the numerator of each fraction cancels with the denominator of the next one:

- $\ln{3}$ cancels with $\ln{3}$
- $\ln{4}$ cancels with $\ln{4}$
- $\ln{5}$ cancels with $\ln{5}$
- $\ln{6}$ cancels with $\ln{6}$
- $\ln{7}$ cancels with $\ln{7}$

This leaves:

$$
\frac{\ln{8}}{\ln{2}} = \log_2{8}
$$

---

### Final Evaluation

We now compute:

$$
\log_2{8} = \frac{\ln{8}}{\ln{2}} = \frac{\ln{2^3}}{\ln{2}} = \frac{3\ln{2}}{\ln{2}} = 3
$$

---

### Conclusion

Thus, the value of the original expression is:

$$
\boxed{3}
$$

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 find 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 check if I remember that correctly. Yeah, I think that's right. So if I use that formula, I can find the sum quickly.

Let me apply the formula here. The numbers go from 1 to 12, so n is 12. Plugging into the formula: 12 times (12 + 1) divided by 2. Let me compute that. 12 + 1 is 13, so 12 times 13 is... 12*10 is 120, 12*3 is 36, so 120 + 36 is 156. Then divide by 2: 156 divided by 2 is 78. So the total sum is 78.

Now, I need to find 78 modulo 4. That means I need to divide 78 by 4 and find the remainder. Let me do that division. 4 times 19 is 76, right? Because 4*20 is 80, which is too big. So 4*19 is 76. Subtract 76 from 78, and the remainder is 2. So the remainder is 2.

Wait, but maybe there's another way to do this without calculating the entire sum? Maybe by taking each number modulo 4 first and then adding them up? Let me try that method to verify.

Let me list the numbers from 1 to 12 and find each modulo 4:

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 (8 is divisible by 4)

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 is divisible by 4)

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

Now let me add these up:

Starting from the beginning:

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 the remainders is 18. Now, I need to find 18 modulo 4. Let me divide 18 by 4. 4*4 is 16, so 18 - 16 = 2. Therefore, the remainder is 2.

Hmm, same result as before. So that's good, both methods give me 2. So the answer should be 2.

But just to be thorough, let me check another way. Maybe grouping numbers. Since modulo 4, every 4 numbers will have a certain pattern. Let me see.

From 1 to 12, there are 12 numbers. Let me group them into sets of 4:

First group: 1, 2, 3, 4. Their sum is 1+2+3+4=10. 10 mod 4 is 2.

Second group: 5, 6, 7, 8. Their sum is 5+6+7+8=26. 26 mod 4: 4*6=24, 26-24=2. So remainder 2.

Third group: 9, 10, 11, 12. Their sum is 9+10+11+12=42. 42 mod 4: 4*10=40, 42-40=2. Remainder 2.

So each group of 4 numbers contributes a remainder of 2. There are 3 groups, so total remainder would be 2 + 2 + 2 = 6. Then 6 mod 4 is 2. Again, same answer.

So all three methods give me 2. That seems pretty solid.

Alternatively, maybe I can use properties of modular arithmetic. For example, since the sum from 1 to n is n(n+1)/2, then modulo 4, we can compute that expression modulo 4. Let me try that.

So n is 12. So the sum is 12*13/2 = 78. So 78 mod 4. As before, 78 divided by 4 is 19*4=76, remainder 2.

Alternatively, maybe compute 12*13/2 mod 4. Let me see if I can compute this without calculating the entire number.

First, note that division by 2 in modular arithmetic can be tricky, but since 2 and 4 are not coprime, we have to be careful. However, since 12 is even, 12*13 is divisible by 2, so 12*13/2 is an integer. Let me compute 12*13 first modulo 8? Wait, maybe not. Let me think.

Alternatively, compute 12 mod 4 is 0, 13 mod 4 is 1. So 12*13 mod 4 is 0*1 = 0. Then divide by 2? Wait, but division isn't straightforward here.

Wait, maybe I need to compute the entire expression modulo 4. Let me think.

The formula is (n(n+1))/2. So for n=12, that's (12*13)/2. Let me compute this modulo 4.

First, note that 12 is divisible by 4, so 12 ≡ 0 mod 4. Then 12*13 ≡ 0*13 ≡ 0 mod 4. Then divide by 2. But division by 2 in mod 4? Hmm.

Wait, if I have 0 divided by 2 mod 4, that's 0/2 mod 4. But 0 divided by 2 is 0, so 0 mod 4. But that contradicts our previous result of 2. So this approach must be wrong.

Wait, why? Because when we do modular arithmetic, division isn't always straightforward. Let me think again.

Actually, the formula (n(n+1))/2 is an integer, but when we take modulo 4, we need to ensure that we handle the division properly.

Alternatively, maybe compute the numerator first modulo 8? Because when you divide by 2, if you have a number that's even, then modulo 4, you can compute it as (numerator mod 8)/2. Let me try that.

So compute 12*13 mod 8.

12 mod 8 is 4, 13 mod 8 is 5. So 4*5 = 20. 20 mod 8 is 4. Then divide by 2: 4/2 = 2. So 2 mod 4 is 2. Which matches our previous result.

Ah, so that works. So the key is that when dividing by 2 in mod 4, if the numerator is even, we can compute it modulo 8 first, then divide by 2.

So that's another way to see it.

So, in summary, all methods lead to the conclusion that the remainder is 2.

But just to make sure, let me actually add the numbers step by step and take modulo 4 at each step. Maybe that's the most straightforward way.

Starting with 0.

Add 1: 0 + 1 = 1 mod 4 is 1.

Add 2: 1 + 2 = 3 mod 4 is 3.

Add 3: 3 + 3 = 6 mod 4 is 2 (since 6 - 4 = 2).

Add 4: 2 + 4 = 6 mod 4 is 2 (same as before).

Add 5: 2 + 5 = 7 mod 4 is 3 (7 - 4 = 3).

Add 6: 3 + 6 = 9 mod 4 is 1 (9 - 8 = 1).

Add 7: 1 + 7 = 8 mod 4 is 0 (8 is divisible by 4).

Add 8: 0 + 8 = 8 mod 4 is 0.

Add 9: 0 + 9 = 9 mod 4 is 1.

Add 10: 1 + 10 = 11 mod 4 is 3 (11 - 8 = 3).

Add 11: 3 + 11 = 14 mod 4 is 2 (14 - 12 = 2).

Add 12: 2 + 12 = 14 mod 4 is 2 (same as before).

So after adding all numbers, the remainder is 2.

Yep, same answer again.

So regardless of the method I use—direct calculation, using the formula, breaking into groups, or step-by-step addition with modulo at each step—the result is 2.

Therefore, I can be confident that the modulo 4 remainder of the sum from 1 to 12 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 the problem in multiple ways, all of which lead to the same conclusion.

---

### **Method 1: Using the Sum Formula**

The sum of the first $ n $ natural numbers is given by the formula:

$$
\text{Sum} = \frac{n(n+1)}{2}
$$

For $ n = 12 $:

$$
\text{Sum} = \frac{12 \times 13}{2} = \frac{156}{2} = 78
$$

Now, compute $ 78 \mod 4 $:

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

So, the remainder is:

$$
\boxed{2}
$$

---

### **Method 2: Modulo 4 of Each Term**

We can compute the modulo 4 of each number from 1 to 12 and then sum the results:

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

Now sum the remainders:

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

Now compute $ 18 \mod 4 $:

$$
18 \div 4 = 4 \text{ remainder } 2
$$

So, the remainder is again:

$$
\boxed{2}
$$

---

### **Method 3: Grouping by Modulo 4 Cycles**

Since the modulo 4 pattern repeats every 4 numbers, we can group the numbers into sets of 4:

- Group 1: $ 1 + 2 + 3 + 4 = 10 \Rightarrow 10 \mod 4 = 2 $
- Group 2: $ 5 + 6 + 7 + 8 = 26 \Rightarrow 26 \mod 4 = 2 $
- Group 3: $ 9 + 10 + 11 + 12 = 42 \Rightarrow 42 \mod 4 = 2 $

Each group contributes a remainder of 2, and there are 3 such groups:

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

Again, the remainder is:

$$
\boxed{2}
$$

---

### **Conclusion**

All methods consistently yield the same result. Therefore, the modulo 4 remainder of the sum $ 1 + 2 + 3 + \dots + 12 $ is:

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to find 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 check if I remember that correctly. Yeah, I think that's right. So if I use that formula, I can find the sum quickly.

Let me apply the formula here. The numbers go from 1 to 12, so n is 12. Plugging into the formula: 12 times (12 + 1) divided by 2. Let me compute that. 12 + 1 is 13, so 12 times 13 is... 12*10 is 120, 12*3 is 36, so 120 + 36 is 156. Then divide by 2: 156 divided by 2 is 78. So the total sum is 78.

Now, I need to find 78 modulo 4. That means I need to divide 78 by 4 and find the remainder. Let me do that division. 4 times 19 is 76, right? Because 4*20 is 80, which is too big. So 4*19 is 76. Subtract 76 from 78, and the remainder is 2. So the remainder is 2.

Wait, but maybe there's another way to do this without calculating the entire sum? Maybe by taking each number modulo 4 first and then adding them up? Let me try that method to verify.

Let me list the numbers from 1 to 12 and find each modulo 4:

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 (8 is divisible by 4)

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 is divisible by 4)

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

Now let me add these up:

Starting from the beginning:

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 the remainders is 18. Now, I need to find 18 modulo 4. Let me divide 18 by 4. 4*4 is 16, so 18 - 16 = 2. Therefore, the remainder is 2.

Hmm, same result as before. So that's good, both methods give me 2. So the answer should be 2.

But just to be thorough, let me check another way. Maybe grouping numbers. Since modulo 4, every 4 numbers will have a certain pattern. Let me see.

From 1 to 12, there are 12 numbers. Let me group them into sets of 4:

First group: 1, 2, 3, 4. Their sum is 1+2+3+4=10. 10 mod 4 is 2.

Second group: 5, 6, 7, 8. Their sum is 5+6+7+8=26. 26 mod 4: 4*6=24, 26-24=2. So remainder 2.

Third group: 9, 10, 11, 12. Their sum is 9+10+11+12=42. 42 mod 4: 4*10=40, 42-40=2. Remainder 2.

So each group of 4 numbers contributes a remainder of 2. There are 3 groups, so total remainder would be 2 + 2 + 2 = 6. Then 6 mod 4 is 2. Again, same answer.

So all three methods give me 2. That seems pretty solid.

Alternatively, maybe I can use properties of modular arithmetic. For example, since the sum from 1 to n is n(n+1)/2, then modulo 4, we can compute that expression modulo 4. Let me try that.

So n is 12. So the sum is 12*13/2 = 78. So 78 mod 4. As before, 78 divided by 4 is 19*4=76, remainder 2.

Alternatively, maybe compute 12*13/2 mod 4. Let me see if I can compute this without calculating the entire number.

First, note that division by 2 in modular arithmetic can be tricky, but since 2 and 4 are not coprime, we have to be careful. However, since 12 is even, 12*13 is divisible by 2, so 12*13/2 is an integer. Let me compute 12*13 first modulo 8? Wait, maybe not. Let me think.

Alternatively, compute 12 mod 4 is 0, 13 mod 4 is 1. So 12*13 mod 4 is 0*1 = 0. Then divide by 2? Wait, but division isn't straightforward here.

Wait, maybe I need to compute the entire expression modulo 4. Let me think.

The formula is (n(n+1))/2. So for n=12, that's (12*13)/2. Let me compute this modulo 4.

First, note that 12 is divisible by 4, so 12 ≡ 0 mod 4. Then 12*13 ≡ 0*13 ≡ 0 mod 4. Then divide by 2. But division by 2 in mod 4? Hmm.

Wait, if I have 0 divided by 2 mod 4, that's 0/2 mod 4. But 0 divided by 2 is 0, so 0 mod 4. But that contradicts our previous result of 2. So this approach must be wrong.

Wait, why? Because when we do modular arithmetic, division isn't always straightforward. Let me think again.

Actually, the formula (n(n+1))/2 is an integer, but when we take modulo 4, we need to ensure that we handle the division properly.

Alternatively, maybe compute the numerator first modulo 8? Because when you divide by 2, if you have a number that's even, then modulo 4, you can compute it as (numerator mod 8)/2. Let me try that.

So compute 12*13 mod 8.

12 mod 8 is 4, 13 mod 8 is 5. So 4*5 = 20. 20 mod 8 is 4. Then divide by 2: 4/2 = 2. So 2 mod 4 is 2. Which matches our previous result.

Ah, so that works. So the key is that when dividing by 2 in mod 4, if the numerator is even, we can compute it modulo 8 first, then divide by 2.

So that's another way to see it.

So, in summary, all methods lead to the conclusion that the remainder is 2.

But just to make sure, let me actually add the numbers step by step and take modulo 4 at each step. Maybe that's the most straightforward way.

Starting with 0.

Add 1: 0 + 1 = 1 mod 4 is 1.

Add 2: 1 + 2 = 3 mod 4 is 3.

Add 3: 3 + 3 = 6 mod 4 is 2 (since 6 - 4 = 2).

Add 4: 2 + 4 = 6 mod 4 is 2 (same as before).

Add 5: 2 + 5 = 7 mod 4 is 3 (7 - 4 = 3).

Add 6: 3 + 6 = 9 mod 4 is 1 (9 - 8 = 1).

Add 7: 1 + 7 = 8 mod 4 is 0 (8 is divisible by 4).

Add 8: 0 + 8 = 8 mod 4 is 0.

Add 9: 0 + 9 = 9 mod 4 is 1.

Add 10: 1 + 10 = 11 mod 4 is 3 (11 - 8 = 3).

Add 11: 3 + 11 = 14 mod 4 is 2 (14 - 12 = 2).

Add 12: 2 + 12 = 14 mod 4 is 2 (same as before).

So after adding all numbers, the remainder is 2.

Yep, same answer again.

So regardless of the method I use—direct calculation, using the formula, breaking into groups, or step-by-step addition with modulo at each step—the result is 2.

Therefore, I can be confident that the modulo 4 remainder of the sum from 1 to 12 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 the problem in multiple ways, all of which lead to the same conclusion.

---

### **Method 1: Using the Sum Formula**

The sum of the first $ n $ natural numbers is given by the formula:

$$
\text{Sum} = \frac{n(n+1)}{2}
$$

For $ n = 12 $:

$$
\text{Sum} = \frac{12 \times 13}{2} = \frac{156}{2} = 78
$$

Now, compute $ 78 \mod 4 $:

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

So, the remainder is:

$$
\boxed{2}
$$

---

### **Method 2: Modulo 4 of Each Term**

We can compute the modulo 4 of each number from 1 to 12 and then sum the results:

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

Now sum the remainders:

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

Now compute $ 18 \mod 4 $:

$$
18 \div 4 = 4 \text{ remainder } 2
$$

So, the remainder is again:

$$
\boxed{2}
$$

---

### **Method 3: Grouping by Modulo 4 Cycles**

Since the modulo 4 pattern repeats every 4 numbers, we can group the numbers into sets of 4:

- Group 1: $ 1 + 2 + 3 + 4 = 10 \Rightarrow 10 \mod 4 = 2 $
- Group 2: $ 5 + 6 + 7 + 8 = 26 \Rightarrow 26 \mod 4 = 2 $
- Group 3: $ 9 + 10 + 11 + 12 = 42 \Rightarrow 42 \mod 4 = 2 $

Each group contributes a remainder of 2, and there are 3 such groups:

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

Again, the remainder is:

$$
\boxed{2}
$$

---

### **Conclusion**

All methods consistently yield the same result. Therefore, the modulo 4 remainder of the sum $ 1 + 2 + 3 + \dots + 12 $ is:

$$
\boxed{2}
$$

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*}