Explanation:
Okay, so I need to find the largest x-value where the graphs of f(x) and g(x) intersect. Let me first write down both functions to see what they look like.

The functions are:

f(x) = e^{3x² - |⌊x⌋|!} + C(22 + 735235|⌊x⌋|, 2356) + φ(|⌊x⌋| + 1) + 72x⁴ + 3x³ - 6x² + 2x + 1

g(x) = e^{3x² - |⌊x⌋|!} + C(22 + 735235|⌊x⌋|, 2356) + φ(|⌊x⌋| + 1) + 72x⁴ + 4x³ - 11x² - 6x + 13

Wait, so both functions have a lot of similar terms. Let me check if I can subtract them to find where they are equal.

Let me compute f(x) - g(x):

f(x) - g(x) = [e^{3x² - |⌊x⌋|!} + C(...) + φ(...) + 72x⁴ + 3x³ - 6x² + 2x + 1] - [e^{3x² - |⌊x⌋|!} + C(...) + φ(...) + 72x⁴ + 4x³ - 11x² - 6x + 13]

So, the exponential terms cancel out, the binomial coefficients cancel out, the φ terms cancel out, and the 72x⁴ terms cancel out. Then we have:

(3x³ - 6x² + 2x + 1) - (4x³ - 11x² - 6x + 13) = 3x³ - 6x² + 2x + 1 - 4x³ + 11x² + 6x - 13

Combine like terms:

(3x³ - 4x³) + (-6x² + 11x²) + (2x + 6x) + (1 - 13) = (-x³) + (5x²) + (8x) - 12

So f(x) - g(x) simplifies to -x³ + 5x² + 8x - 12.

Therefore, the equation f(x) = g(x) is equivalent to -x³ + 5x² + 8x - 12 = 0, or equivalently x³ - 5x² - 8x + 12 = 0.

Wait, so solving x³ - 5x² - 8x + 12 = 0 will give me the x-values where f(x) and g(x) intersect. But hold on, is this true? Because the original functions have terms involving floor(x). So, does the simplification hold for all x?

Wait, the floor function is inside the exponent, binomial coefficient, and φ function. However, when I subtracted f(x) - g(x), the terms with floor(x) were the same in both functions, so they canceled out. Therefore, the difference f(x) - g(x) only depends on the polynomial parts? Wait, but the floor function is inside the exponent, binomial coefficient, and φ function. However, since those terms are the same in both f(x) and g(x), they cancel out. Therefore, the difference f(x) - g(x) is indeed equal to the difference of the polynomial parts. Therefore, solving f(x) = g(x) reduces to solving -x³ + 5x² + 8x - 12 = 0, which is equivalent to x³ - 5x² - 8x + 12 = 0.

But wait, hold on. Is that correct? Let me check again. The floor function is inside the exponent, binomial coefficient, and φ function. However, the floor function is a function of x, so if x is in an interval [n, n+1), where n is integer, then floor(x) = n. Therefore, for each interval [n, n+1), the terms involving floor(x) are constants. Therefore, in each interval [n, n+1), the functions f(x) and g(x) are equal to:

For f(x):

e^{3x² - n!} + C(22 + 735235n, 2356) + φ(n + 1) + 72x⁴ + 3x³ - 6x² + 2x + 1

Similarly for g(x):

e^{3x² - n!} + C(22 + 735235n, 2356) + φ(n + 1) + 72x⁴ + 4x³ - 11x² - 6x + 13

Therefore, when we subtract them, the terms with floor(x) cancel out, and the rest is the polynomial difference. Therefore, in each interval [n, n+1), the equation f(x) = g(x) reduces to the same polynomial equation: -x³ + 5x² + 8x - 12 = 0. Therefore, the solutions to this equation in each interval [n, n+1) would be the points of intersection. However, we need to check if the solutions lie within the interval [n, n+1). Therefore, the problem reduces to solving x³ - 5x² - 8x + 12 = 0 and then checking which roots lie in the intervals [n, n+1) for integer n, and then finding the largest such x.

But first, let me solve the equation x³ - 5x² - 8x + 12 = 0.

To solve this cubic equation, let me try rational roots. The possible rational roots are factors of 12 over factors of 1, so ±1, ±2, ±3, ±4, ±6, ±12.

Let me test x=1: 1 - 5 - 8 + 12 = 0. 1 -5 is -4, -4 -8 is -12, -12 +12 is 0. So x=1 is a root.

Therefore, we can factor (x - 1) out of the cubic. Let's perform polynomial division or use synthetic division.

Dividing x³ -5x² -8x +12 by (x - 1):

Using synthetic division:

Coefficients: 1 | -5 | -8 | 12

Bring down the 1.

Multiply by 1: 1.

Add to next coefficient: -5 +1 = -4.

Multiply by 1: -4.

Add to next coefficient: -8 + (-4) = -12.

Multiply by 1: -12.

Add to last coefficient: 12 + (-12) = 0. Perfect.

So the cubic factors as (x - 1)(x² - 4x - 12).

Now, solving x² - 4x - 12 = 0.

Using quadratic formula: x = [4 ± sqrt(16 + 48)] / 2 = [4 ± sqrt(64)] / 2 = [4 ± 8]/2.

Therefore, x = (4 + 8)/2 = 12/2 = 6, and x = (4 - 8)/2 = -4/2 = -2.

Therefore, the roots are x = 1, x = -2, and x = 6.

So the equation f(x) = g(x) reduces to x = -2, 1, 6. However, we need to check if these roots lie in the intervals where the floor function is constant. Wait, but earlier reasoning said that in each interval [n, n+1), the equation reduces to the same polynomial equation. Therefore, the roots x=-2, 1, 6 must lie in some interval [n, n+1). However, we need to check if these roots are valid. Wait, but for example, x=1 is in [1, 2), x=-2 is in [-2, -1), and x=6 is in [6,7). However, we need to check if these roots are actually solutions to the original equation f(x)=g(x). Because even though the polynomial equation is satisfied, we need to ensure that the floor function terms are correctly evaluated. Wait, but since we already canceled out the floor function terms, the equation f(x) = g(x) is equivalent to the polynomial equation in each interval. Therefore, if x is in [n, n+1), then floor(x) = n, so the terms involving floor(x) are constants. Therefore, the equation f(x) = g(x) is equivalent to the polynomial equation in each interval. Therefore, the roots x=-2, 1, 6 are solutions. However, we need to check if they are in the correct intervals. For example, x=-2 is in [-2, -1), x=1 is in [1, 2), and x=6 is in [6,7). Therefore, these are valid solutions. Therefore, the solutions are x=-2, 1, 6. Therefore, the largest x-value is 6. But wait, the problem says "the largest x-value at which the graphs intersect". So is 6 the answer? But wait, maybe there are other solutions? Because the polynomial equation is x³ -5x² -8x +12=0, which only has roots at -2, 1, 6. Therefore, only these three roots. Therefore, the largest x is 6. However, I need to check if there are other solutions where floor(x) is different? Wait, but the equation f(x) = g(x) reduces to the polynomial equation in each interval. Therefore, if there are other roots in other intervals, but since the polynomial equation only has three roots, then only those three. Therefore, the answer is 6. However, let me check if x=6 is indeed in the interval [6,7). Yes, x=6 is in [6,7). Therefore, floor(x) = 6. Therefore, the terms involving floor(x) are evaluated at 6. Therefore, the original functions f(x) and g(x) would have those terms. However, since we already canceled them out, the equation f(x)=g(x) reduces to the polynomial equation. Therefore, x=6 is a valid solution. Therefore, the answer is 6. But wait, let me check if maybe there are other roots in other intervals? For example, suppose that in some interval [n, n+1), the equation f(x)=g(x) could have solutions other than the roots of the polynomial? But no, because in each interval, the equation reduces to the same polynomial equation. Therefore, the roots of the polynomial equation are the only solutions. Therefore, the answer is 6. However, let me check if maybe the polynomial equation has more roots? But we already factored it into (x - 1)(x - 6)(x + 2). So only three roots. Therefore, the answer is 6. However, let me check if x=6 is actually a solution. Let me plug x=6 into f(x) and g(x). Wait, but since we already derived that f(x) - g(x) = -x³ +5x² +8x -12, which is zero at x=6. Therefore, f(6) - g(6) = 0, so they intersect at x=6. Therefore, the answer is 6. However, the problem says "the largest x-value". But maybe there are other solutions? For example, suppose that in some interval [n, n+1), the equation f(x)=g(x) could have solutions outside of the roots of the polynomial? But no, because in each interval, the equation reduces to the same polynomial equation. Therefore, the only solutions are the roots of that polynomial. Therefore, the answer is 6. However, let me check if maybe the polynomial equation has other roots? For example, maybe there are multiple roots? But the cubic equation only has three roots. Therefore, the answer is 6. Therefore, the largest x is 6. However, let me check if there is a possibility that for some x, even though the polynomial equation is satisfied, the floor function terms might not be consistent? Wait, for example, suppose that x is exactly an integer. Let me check x=1. At x=1, floor(x) = 1. So, the terms involving floor(x) are evaluated at 1. However, if x is exactly an integer, then it's included in the interval [n, n+1) where n is the integer. For example, x=1 is in [1, 2). Therefore, floor(x) = 1. Therefore, the terms are evaluated correctly. Similarly, x=6 is in [6,7), floor(x)=6. Therefore, the terms are evaluated correctly. Therefore, the solutions x=-2, 1, 6 are valid. Therefore, the answer is 6. However, let me check if maybe there are other solutions where the floor function changes? For example, suppose that between two integers, the equation f(x)=g(x) could have solutions? But since in each interval [n, n+1), the equation reduces to the same polynomial equation, which only has roots at -2, 1, 6. Therefore, if those roots lie in the interval [n, n+1), then they are solutions. For example, x=-2 is in [-2, -1), x=1 is in [1, 2), x=6 is in [6,7). Therefore, those are the only solutions. Therefore, the answer is 6. Therefore, the largest x is 6. However, let me check if maybe there are other roots in other intervals. For example, suppose that the polynomial equation x³ -5x² -8x +12=0 has roots outside of the intervals where floor(x) is constant? But no, the roots are -2, 1, 6. So, for example, if there was a root between, say, 2 and 3, but the polynomial equation doesn't have such roots. Therefore, the answer is 6. Therefore, I think the answer is 6. However, let me check if maybe the problem is more complicated. Wait, the original functions have terms like e^{3x² - |floor(x)|!}. Let me check if for x=6, floor(x)=6, so |floor(x)|! = 6! = 720. Therefore, the exponent is 3x² - 720. However, for other x in [6,7), floor(x)=6, so the exponent is same for all x in [6,7). Therefore, the term e^{3x² - 6!} is same for all x in [6,7). Similarly, the binomial coefficient term is C(22 + 735235*6, 2356). Which is a constant for all x in [6,7). Similarly, φ(|floor(x)| +1) = φ(6 +1) = φ(7). Which is 6, since 7 is prime. Therefore, those terms are constants in [6,7). Therefore, the difference between f(x) and g(x) is only the polynomial part. Therefore, solving the polynomial equation gives the solutions. Therefore, x=6 is a solution. Therefore, the answer is 6. Therefore, I think the answer is 6. However, let me check if maybe there's a mistake in my calculation. Let me check the polynomial difference again. Original f(x) - g(x):

The polynomial part of f(x) is 72x⁴ + 3x³ -6x² + 2x +1

The polynomial part of g(x) is 72x⁴ + 4x³ -11x² -6x +13

Subtracting, we get:

(72x⁴ -72x⁴) + (3x³ -4x³) + (-6x² - (-11x²)) + (2x - (-6x)) + (1 -13)

Which is 0 -x³ +5x² +8x -12. So yes, that's correct. Therefore, f(x) - g(x) = -x³ +5x² +8x -12. Therefore, setting equal to zero gives x³ -5x² -8x +12=0. Which factors as (x -1)(x -6)(x +2). Therefore, roots at 1, 6, -2. Therefore, the answer is 6. Therefore, I think the answer is 6. However, let me check if maybe there's a mistake in the problem statement? For example, maybe the floor function is inside the exponent? Wait, the problem says:

f(x)=e^{3x² - |⌊x⌋|!} + ... So yes, the exponent is 3x² minus the absolute value of floor(x) factorial. But since floor(x) is an integer, and absolute value of floor(x) is just floor(x) if floor(x) is non-negative, which it is for x ≥ 0. For x negative, floor(x) is negative, so |floor(x)| is positive. However, in the exponent, it's 3x² - |floor(x)|! So for example, if x is in [-2, -1), floor(x) = -2, so |floor(x)| = 2, so |floor(x)|! = 2! = 2. Therefore, the exponent is 3x² - 2. However, for x in [n, n+1), floor(x) = n, so |floor(x)| = |n|. Therefore, the exponent is 3x² - |n|! So for example, if n is negative, |n| is positive, so |n|! is factorial of positive integer. However, in the case of x=-2, floor(x)=-2, so |floor(x)|=2, so exponent is 3x² - 2! = 3x² - 2. However, when we subtract f(x) - g(x), the exponent terms cancel out. Therefore, even though the exponent terms depend on floor(x), they are same in both f(x) and g(x), so they cancel. Therefore, the difference is only the polynomial part. Therefore, the roots are indeed -2, 1, 6. Therefore, the answer is 6. Therefore, I think the answer is 6. However, let me check if maybe there's a mistake in the problem's phrasing? For example, maybe the floor function is inside the exponent? Wait, the problem says:

"e^{3x² - |\lfloor x \rfloor|!} + ...". So yes, the exponent is 3x² minus the absolute value of floor(x) factorial. So that is correct. Therefore, the previous reasoning holds. Therefore, the answer is 6. Therefore, I think the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To determine the largest $ x $-value at which the graphs of the functions

$$
f(x) = e^{3x^2 - |\lfloor x \rfloor|!} + \binom{22 + 735235|\lfloor x \rfloor|}{2356} + \phi(|\lfloor x \rfloor| + 1) + 72x^4 + 3x^3 - 6x^2 + 2x + 1
$$
and
$$
g(x) = e^{3x^2 - |\lfloor x \rfloor|!} + \binom{22 + 735235|\lfloor x \rfloor|}{2356} + \phi(|\lfloor x \rfloor| + 1) + 72x^4 + 4x^3 - 11x^2 - 6x + 13
$$
intersect, we begin by analyzing the structure of both functions.

---

### Step 1: Simplify the Difference

We compute the difference $ f(x) - g(x) $:

$$
f(x) - g(x) = \left[ \text{common terms} \right] + \left( 72x^4 + 3x^3 - 6x^2 + 2x + 1 \right) - \left( 72x^4 + 4x^3 - 11x^2 - 6x + 13 \right)
$$

The exponential, binomial coefficient, and Euler's totient function terms are identical in both functions, so they cancel out. The remaining polynomial terms give:

$$
f(x) - g(x) = (3x^3 - 6x^2 + 2x + 1) - (4x^3 - 11x^2 - 6x + 13)
$$

Simplifying:

$$
f(x) - g(x) = -x^3 + 5x^2 + 8x - 12
$$

Thus, the equation $ f(x) = g(x) $ reduces to:

$$
-x^3 + 5x^2 + 8x - 12 = 0 \quad \Rightarrow \quad x^3 - 5x^2 - 8x + 12 = 0
$$

---

### Step 2: Solve the Cubic Equation

We solve the cubic equation:

$$
x^3 - 5x^2 - 8x + 12 = 0
$$

Using the Rational Root Theorem, we test possible rational roots. Trying $ x = 1 $:

$$
1^3 - 5(1)^2 - 8(1) + 12 = 1 - 5 - 8 + 12 = 0
$$

So $ x = 1 $ is a root. We factor the cubic as:

$$
(x - 1)(x^2 - 4x - 12) = 0
$$

Now solve the quadratic:

$$
x^2 - 4x - 12 = 0 \Rightarrow x = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2}
$$

$$
x = 6 \quad \text{or} \quad x = -2
$$

So the roots of the cubic are:

$$
x = -2, \quad x = 1, \quad x = 6
$$

---

### Step 3: Determine Valid Solutions

We now check which of these roots lie in intervals where the floor function is constant. The floor function $ \lfloor x \rfloor $ is constant on intervals of the form $ [n, n+1) $, where $ n \in \mathbb{Z} $.

- $ x = -2 $ lies in $ [-2, -1) $, so $ \lfloor x \rfloor = -2 $
- $ x = 1 $ lies in $ [1, 2) $, so $ \lfloor x \rfloor = 1 $
- $ x = 6 $ lies in $ [6, 7) $, so $ \lfloor x \rfloor = 6 $

Since the floor function is constant in each of these intervals, and the difference $ f(x) - g(x) $ only depends on the polynomial part (which we already simplified), all three roots are valid solutions to the original equation $ f(x) = g(x) $.

---

### Step 4: Identify the Largest $ x $-Value

Among the valid solutions $ x = -2, 1, 6 $, the largest is:

$$
\boxed{6}
$$

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the largest x-value where the graphs of f(x) and g(x) intersect. Let me first write down both functions to see what they look like.

The functions are:

f(x) = e^{3x² - |⌊x⌋|!} + C(22 + 735235|⌊x⌋|, 2356) + φ(|⌊x⌋| + 1) + 72x⁴ + 3x³ - 6x² + 2x + 1

g(x) = e^{3x² - |⌊x⌋|!} + C(22 + 735235|⌊x⌋|, 2356) + φ(|⌊x⌋| + 1) + 72x⁴ + 4x³ - 11x² - 6x + 13

Wait, so both functions have a lot of similar terms. Let me check if I can subtract them to find where they are equal.

Let me compute f(x) - g(x):

f(x) - g(x) = [e^{3x² - |⌊x⌋|!} + C(...) + φ(...) + 72x⁴ + 3x³ - 6x² + 2x + 1] - [e^{3x² - |⌊x⌋|!} + C(...) + φ(...) + 72x⁴ + 4x³ - 11x² - 6x + 13]

So, the exponential terms cancel out, the binomial coefficients cancel out, the φ terms cancel out, and the 72x⁴ terms cancel out. Then we have:

(3x³ - 6x² + 2x + 1) - (4x³ - 11x² - 6x + 13) = 3x³ - 6x² + 2x + 1 - 4x³ + 11x² + 6x - 13

Combine like terms:

(3x³ - 4x³) + (-6x² + 11x²) + (2x + 6x) + (1 - 13) = (-x³) + (5x²) + (8x) - 12

So f(x) - g(x) simplifies to -x³ + 5x² + 8x - 12.

Therefore, the equation f(x) = g(x) is equivalent to -x³ + 5x² + 8x - 12 = 0, or equivalently x³ - 5x² - 8x + 12 = 0.

Wait, so solving x³ - 5x² - 8x + 12 = 0 will give me the x-values where f(x) and g(x) intersect. But hold on, is this true? Because the original functions have terms involving floor(x). So, does the simplification hold for all x?

Wait, the floor function is inside the exponent, binomial coefficient, and φ function. However, when I subtracted f(x) - g(x), the terms with floor(x) were the same in both functions, so they canceled out. Therefore, the difference f(x) - g(x) only depends on the polynomial parts? Wait, but the floor function is inside the exponent, binomial coefficient, and φ function. However, since those terms are the same in both f(x) and g(x), they cancel out. Therefore, the difference f(x) - g(x) is indeed equal to the difference of the polynomial parts. Therefore, solving f(x) = g(x) reduces to solving -x³ + 5x² + 8x - 12 = 0, which is equivalent to x³ - 5x² - 8x + 12 = 0.

But wait, hold on. Is that correct? Let me check again. The floor function is inside the exponent, binomial coefficient, and φ function. However, the floor function is a function of x, so if x is in an interval [n, n+1), where n is integer, then floor(x) = n. Therefore, for each interval [n, n+1), the terms involving floor(x) are constants. Therefore, in each interval [n, n+1), the functions f(x) and g(x) are equal to:

For f(x):

e^{3x² - n!} + C(22 + 735235n, 2356) + φ(n + 1) + 72x⁴ + 3x³ - 6x² + 2x + 1

Similarly for g(x):

e^{3x² - n!} + C(22 + 735235n, 2356) + φ(n + 1) + 72x⁴ + 4x³ - 11x² - 6x + 13

Therefore, when we subtract them, the terms with floor(x) cancel out, and the rest is the polynomial difference. Therefore, in each interval [n, n+1), the equation f(x) = g(x) reduces to the same polynomial equation: -x³ + 5x² + 8x - 12 = 0. Therefore, the solutions to this equation in each interval [n, n+1) would be the points of intersection. However, we need to check if the solutions lie within the interval [n, n+1). Therefore, the problem reduces to solving x³ - 5x² - 8x + 12 = 0 and then checking which roots lie in the intervals [n, n+1) for integer n, and then finding the largest such x.

But first, let me solve the equation x³ - 5x² - 8x + 12 = 0.

To solve this cubic equation, let me try rational roots. The possible rational roots are factors of 12 over factors of 1, so ±1, ±2, ±3, ±4, ±6, ±12.

Let me test x=1: 1 - 5 - 8 + 12 = 0. 1 -5 is -4, -4 -8 is -12, -12 +12 is 0. So x=1 is a root.

Therefore, we can factor (x - 1) out of the cubic. Let's perform polynomial division or use synthetic division.

Dividing x³ -5x² -8x +12 by (x - 1):

Using synthetic division:

Coefficients: 1 | -5 | -8 | 12

Bring down the 1.

Multiply by 1: 1.

Add to next coefficient: -5 +1 = -4.

Multiply by 1: -4.

Add to next coefficient: -8 + (-4) = -12.

Multiply by 1: -12.

Add to last coefficient: 12 + (-12) = 0. Perfect.

So the cubic factors as (x - 1)(x² - 4x - 12).

Now, solving x² - 4x - 12 = 0.

Using quadratic formula: x = [4 ± sqrt(16 + 48)] / 2 = [4 ± sqrt(64)] / 2 = [4 ± 8]/2.

Therefore, x = (4 + 8)/2 = 12/2 = 6, and x = (4 - 8)/2 = -4/2 = -2.

Therefore, the roots are x = 1, x = -2, and x = 6.

So the equation f(x) = g(x) reduces to x = -2, 1, 6. However, we need to check if these roots lie in the intervals where the floor function is constant. Wait, but earlier reasoning said that in each interval [n, n+1), the equation reduces to the same polynomial equation. Therefore, the roots x=-2, 1, 6 must lie in some interval [n, n+1). However, we need to check if these roots are valid. Wait, but for example, x=1 is in [1, 2), x=-2 is in [-2, -1), and x=6 is in [6,7). However, we need to check if these roots are actually solutions to the original equation f(x)=g(x). Because even though the polynomial equation is satisfied, we need to ensure that the floor function terms are correctly evaluated. Wait, but since we already canceled out the floor function terms, the equation f(x) = g(x) is equivalent to the polynomial equation in each interval. Therefore, if x is in [n, n+1), then floor(x) = n, so the terms involving floor(x) are constants. Therefore, the equation f(x) = g(x) is equivalent to the polynomial equation in each interval. Therefore, the roots x=-2, 1, 6 are solutions. However, we need to check if they are in the correct intervals. For example, x=-2 is in [-2, -1), x=1 is in [1, 2), and x=6 is in [6,7). Therefore, these are valid solutions. Therefore, the solutions are x=-2, 1, 6. Therefore, the largest x-value is 6. But wait, the problem says "the largest x-value at which the graphs intersect". So is 6 the answer? But wait, maybe there are other solutions? Because the polynomial equation is x³ -5x² -8x +12=0, which only has roots at -2, 1, 6. Therefore, only these three roots. Therefore, the largest x is 6. However, I need to check if there are other solutions where floor(x) is different? Wait, but the equation f(x) = g(x) reduces to the polynomial equation in each interval. Therefore, if there are other roots in other intervals, but since the polynomial equation only has three roots, then only those three. Therefore, the answer is 6. However, let me check if x=6 is indeed in the interval [6,7). Yes, x=6 is in [6,7). Therefore, floor(x) = 6. Therefore, the terms involving floor(x) are evaluated at 6. Therefore, the original functions f(x) and g(x) would have those terms. However, since we already canceled them out, the equation f(x)=g(x) reduces to the polynomial equation. Therefore, x=6 is a valid solution. Therefore, the answer is 6. But wait, let me check if maybe there are other roots in other intervals? For example, suppose that in some interval [n, n+1), the equation f(x)=g(x) could have solutions other than the roots of the polynomial? But no, because in each interval, the equation reduces to the same polynomial equation. Therefore, the roots of the polynomial equation are the only solutions. Therefore, the answer is 6. However, let me check if maybe the polynomial equation has more roots? But we already factored it into (x - 1)(x - 6)(x + 2). So only three roots. Therefore, the answer is 6. However, let me check if x=6 is actually a solution. Let me plug x=6 into f(x) and g(x). Wait, but since we already derived that f(x) - g(x) = -x³ +5x² +8x -12, which is zero at x=6. Therefore, f(6) - g(6) = 0, so they intersect at x=6. Therefore, the answer is 6. However, the problem says "the largest x-value". But maybe there are other solutions? For example, suppose that in some interval [n, n+1), the equation f(x)=g(x) could have solutions outside of the roots of the polynomial? But no, because in each interval, the equation reduces to the same polynomial equation. Therefore, the only solutions are the roots of that polynomial. Therefore, the answer is 6. However, let me check if maybe the polynomial equation has other roots? For example, maybe there are multiple roots? But the cubic equation only has three roots. Therefore, the answer is 6. Therefore, the largest x is 6. However, let me check if there is a possibility that for some x, even though the polynomial equation is satisfied, the floor function terms might not be consistent? Wait, for example, suppose that x is exactly an integer. Let me check x=1. At x=1, floor(x) = 1. So, the terms involving floor(x) are evaluated at 1. However, if x is exactly an integer, then it's included in the interval [n, n+1) where n is the integer. For example, x=1 is in [1, 2). Therefore, floor(x) = 1. Therefore, the terms are evaluated correctly. Similarly, x=6 is in [6,7), floor(x)=6. Therefore, the terms are evaluated correctly. Therefore, the solutions x=-2, 1, 6 are valid. Therefore, the answer is 6. However, let me check if maybe there are other solutions where the floor function changes? For example, suppose that between two integers, the equation f(x)=g(x) could have solutions? But since in each interval [n, n+1), the equation reduces to the same polynomial equation, which only has roots at -2, 1, 6. Therefore, if those roots lie in the interval [n, n+1), then they are solutions. For example, x=-2 is in [-2, -1), x=1 is in [1, 2), x=6 is in [6,7). Therefore, those are the only solutions. Therefore, the answer is 6. Therefore, the largest x is 6. However, let me check if maybe there are other roots in other intervals. For example, suppose that the polynomial equation x³ -5x² -8x +12=0 has roots outside of the intervals where floor(x) is constant? But no, the roots are -2, 1, 6. So, for example, if there was a root between, say, 2 and 3, but the polynomial equation doesn't have such roots. Therefore, the answer is 6. Therefore, I think the answer is 6. However, let me check if maybe the problem is more complicated. Wait, the original functions have terms like e^{3x² - |floor(x)|!}. Let me check if for x=6, floor(x)=6, so |floor(x)|! = 6! = 720. Therefore, the exponent is 3x² - 720. However, for other x in [6,7), floor(x)=6, so the exponent is same for all x in [6,7). Therefore, the term e^{3x² - 6!} is same for all x in [6,7). Similarly, the binomial coefficient term is C(22 + 735235*6, 2356). Which is a constant for all x in [6,7). Similarly, φ(|floor(x)| +1) = φ(6 +1) = φ(7). Which is 6, since 7 is prime. Therefore, those terms are constants in [6,7). Therefore, the difference between f(x) and g(x) is only the polynomial part. Therefore, solving the polynomial equation gives the solutions. Therefore, x=6 is a solution. Therefore, the answer is 6. Therefore, I think the answer is 6. However, let me check if maybe there's a mistake in my calculation. Let me check the polynomial difference again. Original f(x) - g(x):

The polynomial part of f(x) is 72x⁴ + 3x³ -6x² + 2x +1

The polynomial part of g(x) is 72x⁴ + 4x³ -11x² -6x +13

Subtracting, we get:

(72x⁴ -72x⁴) + (3x³ -4x³) + (-6x² - (-11x²)) + (2x - (-6x)) + (1 -13)

Which is 0 -x³ +5x² +8x -12. So yes, that's correct. Therefore, f(x) - g(x) = -x³ +5x² +8x -12. Therefore, setting equal to zero gives x³ -5x² -8x +12=0. Which factors as (x -1)(x -6)(x +2). Therefore, roots at 1, 6, -2. Therefore, the answer is 6. Therefore, I think the answer is 6. However, let me check if maybe there's a mistake in the problem statement? For example, maybe the floor function is inside the exponent? Wait, the problem says:

f(x)=e^{3x² - |⌊x⌋|!} + ... So yes, the exponent is 3x² minus the absolute value of floor(x) factorial. But since floor(x) is an integer, and absolute value of floor(x) is just floor(x) if floor(x) is non-negative, which it is for x ≥ 0. For x negative, floor(x) is negative, so |floor(x)| is positive. However, in the exponent, it's 3x² - |floor(x)|! So for example, if x is in [-2, -1), floor(x) = -2, so |floor(x)| = 2, so |floor(x)|! = 2! = 2. Therefore, the exponent is 3x² - 2. However, for x in [n, n+1), floor(x) = n, so |floor(x)| = |n|. Therefore, the exponent is 3x² - |n|! So for example, if n is negative, |n| is positive, so |n|! is factorial of positive integer. However, in the case of x=-2, floor(x)=-2, so |floor(x)|=2, so exponent is 3x² - 2! = 3x² - 2. However, when we subtract f(x) - g(x), the exponent terms cancel out. Therefore, even though the exponent terms depend on floor(x), they are same in both f(x) and g(x), so they cancel. Therefore, the difference is only the polynomial part. Therefore, the roots are indeed -2, 1, 6. Therefore, the answer is 6. Therefore, I think the answer is 6. However, let me check if maybe there's a mistake in the problem's phrasing? For example, maybe the floor function is inside the exponent? Wait, the problem says:

"e^{3x² - |\lfloor x \rfloor|!} + ...". So yes, the exponent is 3x² minus the absolute value of floor(x) factorial. So that is correct. Therefore, the previous reasoning holds. Therefore, the answer is 6. Therefore, I think the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To determine the largest $ x $-value at which the graphs of the functions

$$
f(x) = e^{3x^2 - |\lfloor x \rfloor|!} + \binom{22 + 735235|\lfloor x \rfloor|}{2356} + \phi(|\lfloor x \rfloor| + 1) + 72x^4 + 3x^3 - 6x^2 + 2x + 1
$$
and
$$
g(x) = e^{3x^2 - |\lfloor x \rfloor|!} + \binom{22 + 735235|\lfloor x \rfloor|}{2356} + \phi(|\lfloor x \rfloor| + 1) + 72x^4 + 4x^3 - 11x^2 - 6x + 13
$$
intersect, we begin by analyzing the structure of both functions.

---

### Step 1: Simplify the Difference

We compute the difference $ f(x) - g(x) $:

$$
f(x) - g(x) = \left[ \text{common terms} \right] + \left( 72x^4 + 3x^3 - 6x^2 + 2x + 1 \right) - \left( 72x^4 + 4x^3 - 11x^2 - 6x + 13 \right)
$$

The exponential, binomial coefficient, and Euler's totient function terms are identical in both functions, so they cancel out. The remaining polynomial terms give:

$$
f(x) - g(x) = (3x^3 - 6x^2 + 2x + 1) - (4x^3 - 11x^2 - 6x + 13)
$$

Simplifying:

$$
f(x) - g(x) = -x^3 + 5x^2 + 8x - 12
$$

Thus, the equation $ f(x) = g(x) $ reduces to:

$$
-x^3 + 5x^2 + 8x - 12 = 0 \quad \Rightarrow \quad x^3 - 5x^2 - 8x + 12 = 0
$$

---

### Step 2: Solve the Cubic Equation

We solve the cubic equation:

$$
x^3 - 5x^2 - 8x + 12 = 0
$$

Using the Rational Root Theorem, we test possible rational roots. Trying $ x = 1 $:

$$
1^3 - 5(1)^2 - 8(1) + 12 = 1 - 5 - 8 + 12 = 0
$$

So $ x = 1 $ is a root. We factor the cubic as:

$$
(x - 1)(x^2 - 4x - 12) = 0
$$

Now solve the quadratic:

$$
x^2 - 4x - 12 = 0 \Rightarrow x = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2}
$$

$$
x = 6 \quad \text{or} \quad x = -2
$$

So the roots of the cubic are:

$$
x = -2, \quad x = 1, \quad x = 6
$$

---

### Step 3: Determine Valid Solutions

We now check which of these roots lie in intervals where the floor function is constant. The floor function $ \lfloor x \rfloor $ is constant on intervals of the form $ [n, n+1) $, where $ n \in \mathbb{Z} $.

- $ x = -2 $ lies in $ [-2, -1) $, so $ \lfloor x \rfloor = -2 $
- $ x = 1 $ lies in $ [1, 2) $, so $ \lfloor x \rfloor = 1 $
- $ x = 6 $ lies in $ [6, 7) $, so $ \lfloor x \rfloor = 6 $

Since the floor function is constant in each of these intervals, and the difference $ f(x) - g(x) $ only depends on the polynomial part (which we already simplified), all three roots are valid solutions to the original equation $ f(x) = g(x) $.

---

### Step 4: Identify the Largest $ x $-Value

Among the valid solutions $ x = -2, 1, 6 $, the largest is:

$$
\boxed{6}
$$

---

### Final Answer

$$
\boxed{6}
$$

Solution: The messy parts of the functions are irrelevant. All that matters for intersection is whether $f(x)-g(x)=0$. As $g(x)-f(x)=x^3-5x^2-8x+12=(x-6)(x+2)(x-1)$, the largest value of $x$ at which the graphs intersect is $x=\boxed{6}$.


Explanation:
Okay, so I need to figure out the greatest length of rope that can be saved by replacing ropes HC and HD with a single rope HP, where P is on the line between C and D. The balloon is still at H above O, and the original ropes HC and HD have lengths 150 m and 130 m respectively. The distance between C and D is 140 m.

First, let me visualize the setup. There's a point O, with A north, B west, C south, D east. So, if I imagine a coordinate system with O at the origin, then A would be along the positive y-axis, B along the negative x-axis, C along the negative y-axis, and D along the positive x-axis. The points C and D are 140 meters apart. Since they are directly south and east of O, the distance between them should be the hypotenuse of a right triangle with legs equal to the distances from O to C and O to D. Wait, but the problem says the distance between C and D is 140 m. So if I let the distance from O to C be, say, y, and from O to D be x, then the distance between C and D would be sqrt(x² + y²) = 140 m.

But maybe I need to assign coordinates to all points. Let me try that. Let me set O at (0, 0). Then, since A is directly north of O, let me assume A is at (0, a) for some a. Similarly, B is directly west of O, so B is at (-b, 0) for some b. C is directly south of O, so C is at (0, -c) for some c. D is directly east of O, so D is at (d, 0) for some d.

But the distance between C and D is 140 m. So the coordinates of C are (0, -c) and D are (d, 0). The distance between them is sqrt[(d - 0)^2 + (0 - (-c))^2] = sqrt(d² + c²) = 140. So that gives me c² + d² = 140².

Now, the balloon is at H directly above O, so H has coordinates (0, h) for some h. The ropes HA, HB, HC, HD are from H to each of the points. The lengths of HC and HD are given as 150 m and 130 m. Let me write equations for those.

HC is the distance from H(0, h) to C(0, -c). That distance is sqrt[(0 - 0)^2 + (h - (-c))^2] = sqrt[(h + c)^2] = h + c. Wait, that's interesting. Since H is above O, and C is below O, the vertical distance between H and C is h + c. So HC = h + c = 150 m. Similarly, HD is the distance from H(0, h) to D(d, 0). That distance is sqrt[(d - 0)^2 + (0 - h)^2] = sqrt(d² + h²) = 130 m.

So we have two equations:

1. h + c = 150
2. sqrt(d² + h²) = 130

Also, from the distance between C and D, we have:

3. c² + d² = 140² = 19600

So, let me note that equation 2 is sqrt(d² + h²) = 130, so squaring both sides gives d² + h² = 16900.

But from equation 3, c² + d² = 19600. If I subtract equation 2 squared from equation 3, I get:

(c² + d²) - (d² + h²) = 19600 - 16900 => c² - h² = 2700.

But from equation 1, h + c = 150. Let me write that as c = 150 - h. Then, substitute into c² - h² = 2700.

So, c² - h² = (c - h)(c + h) = (c - h)(150) = 2700. Therefore, (c - h) = 2700 / 150 = 18.

So c - h = 18. But we also have c + h = 150. So we can solve for c and h.

Adding the two equations:

(c - h) + (c + h) = 18 + 150 => 2c = 168 => c = 84. Then h = 150 - c = 150 - 84 = 66.

So c = 84 m, h = 66 m. Then, from equation 2, d² + h² = 16900. So d² = 16900 - h² = 16900 - 66². Let me compute 66²: 66*66 = 4356. Then 16900 - 4356 = 12544. Then d = sqrt(12544). Let me check sqrt(12544). Well, 112² = 12544 because 100²=10000, 110²=12100, 112²= (110 + 2)^2 = 110² + 4*110 + 4 = 12100 + 440 + 4 = 12544. So d = 112 m.

So, coordinates:

O is (0, 0)

A is (0, a). Wait, but we haven't determined a yet. Wait, but maybe we don't need a? Because the problem is about replacing HC and HD with HP where P is on CD. But maybe we need coordinates for P?

Wait, but maybe we can figure out coordinates for all points. Let me check.

We have:

- C is (0, -c) = (0, -84)

- D is (d, 0) = (112, 0)

So the line between C and D goes from (0, -84) to (112, 0). So parametric equations for CD can be written as x = 112t, y = -84 + 84t, where t ranges from 0 to 1. Wait, when t=0, x=0, y=-84 (point C); when t=1, x=112, y=0 (point D). Alternatively, maybe parameterizing differently. Let me think.

Alternatively, the line CD can be parametrized as starting at C(0, -84) and moving towards D(112, 0). So direction vector is (112 - 0, 0 - (-84)) = (112, 84). So parametric equations can be written as:

x = 0 + 112s

y = -84 + 84s

where s ∈ [0, 1]. So any point P on CD can be represented as (112s, -84 + 84s) for some s between 0 and 1.

Alternatively, since the problem says P is on the straight line between C and D, so we can parametrize it as such.

Now, the original ropes HC and HD have lengths 150 m and 130 m. The total length of these two ropes is 150 + 130 = 280 m. If we replace them with a single rope HP, then the length of HP will be less than or equal to the sum of HC and HD? Wait, no. Wait, actually, the problem says "to reduce the total length of rope used, rope HC and rope HD are to be replaced by a single rope HP". So the total length saved would be (HC + HD) - HP. Therefore, to maximize the length saved, we need to minimize HP.

Therefore, the problem reduces to finding the minimum possible length of HP where P lies on CD. Then, the maximum length saved is (150 + 130) - HP_min = 280 - HP_min.

Therefore, the key is to find the minimal HP where P is on CD.

So, H is at (0, h) = (0, 66). So coordinates of H are (0, 66).

So, we need to find the point P on CD such that the distance from H(0, 66) to P is minimized.

So, this is a classic problem of finding the shortest distance from a point to a line segment. However, since CD is a straight line, and H is a point not on CD, the minimal distance from H to CD is the perpendicular distance from H to the line CD. However, we need to check if the foot of the perpendicular lies on the segment CD. If it does, then that's the minimal distance. If not, then the minimal distance is the distance to the nearest endpoint.

Therefore, first, let me find the equation of line CD.

Points C(0, -84) and D(112, 0). Let me compute the slope of CD.

Slope m = (0 - (-84)) / (112 - 0) = 84 / 112 = 3/4.

So the equation of line CD can be written as y - (-84) = (3/4)(x - 0), so y + 84 = (3/4)x => y = (3/4)x - 84.

Alternatively, in standard form: 3x - 4y - 336 = 0. Let me check:

Starting from y = (3/4)x - 84, multiply both sides by 4: 4y = 3x - 336 => 3x - 4y - 336 = 0. Yes.

So the line CD is 3x - 4y - 336 = 0.

Now, the distance from H(0, 66) to line CD is given by the formula:

Distance = |Ax + By + C| / sqrt(A² + B²), where the line is Ax + By + C = 0.

But our line is 3x - 4y - 336 = 0, so A = 3, B = -4, C = -336.

Wait, the formula is |Ax + By + C| / sqrt(A² + B²). Wait, but in standard form, it's usually written as Ax + By + C = 0. So here, 3x - 4y - 336 = 0, so A = 3, B = -4, C = -336.

So plugging in H(0, 66):

Distance = |3*0 + (-4)*66 + (-336)| / sqrt(3² + (-4)^2) = |0 - 264 - 336| / 5 = |-600| / 5 = 600 / 5 = 120.

So the minimal distance from H to line CD is 120 meters. However, we need to check if the foot of the perpendicular from H to CD lies on the segment CD. If it does, then the minimal HP is 120 m. If not, then the minimal distance would be to one of the endpoints.

So let me find the coordinates of the foot of the perpendicular from H(0, 66) to line CD.

The formula for the foot of the perpendicular from a point (x0, y0) to the line ax + by + c = 0 is:

x = x0 - a*(ax0 + by0 + c)/(a² + b²)

y = y0 - b*(ax0 + by0 + c)/(a² + b²)

Wait, but let me recall the formula. Alternatively, since we have the line CD: y = (3/4)x - 84.

The slope of CD is 3/4, so the slope of the perpendicular is -4/3.

So the line perpendicular to CD passing through H(0, 66) has equation y - 66 = (-4/3)(x - 0), so y = (-4/3)x + 66.

Find the intersection point between this line and CD.

Set equations equal:

(3/4)x - 84 = (-4/3)x + 66

Multiply both sides by 12 to eliminate denominators:

12*(3/4)x - 12*84 = 12*(-4/3)x + 12*66

Which simplifies to:

9x - 1008 = -16x + 792

Bring all terms to left side:

9x + 16x - 1008 - 792 = 0 => 25x - 1800 = 0 => 25x = 1800 => x = 1800 / 25 = 72.

Then, substitute x = 72 into the equation of CD: y = (3/4)(72) - 84 = 54 - 84 = -30.

So the foot of the perpendicular is at (72, -30).

Now, check if this point lies on segment CD.

Point C is (0, -84), D is (112, 0). The x-coordinate of the foot is 72, which is between 0 and 112. The y-coordinate is -30, which is between -84 and 0. Therefore, yes, the foot of the perpendicular lies on segment CD. Therefore, the minimal HP is 120 meters.

Therefore, the maximum length saved is (150 + 130) - 120 = 280 - 120 = 160 meters.

Wait, but hold on. Let me confirm. The original total length of HC and HD is 150 + 130 = 280. If we replace them with HP, which is 120, then the length saved is 280 - 120 = 160. Therefore, the answer is 160 meters.

But let me check if there's a mistake here. Because sometimes when you replace two ropes with one, you have to make sure that the single rope is sufficient to hold the balloon. But the problem says "the balloon remains at the same position H above O as described above." So the position of H is fixed, and P is on CD. So the rope HP is just a single rope from H to P. So the length of HP is the distance from H to P, and since P is on CD, the minimal HP is 120, so replacing HC and HD with HP saves 280 - 120 = 160.

But let me check if my calculation of the foot of the perpendicular is correct.

Earlier, I found that the foot is at (72, -30). Let me check if that point is on CD.

Parametrize CD as from C(0, -84) to D(112, 0). Let me check if (72, -30) is between them.

The vector from C to D is (112, 84). The vector from C to the foot is (72 - 0, -30 - (-84)) = (72, 54).

Check if this vector is a scalar multiple of (112, 84).

Let me see: 72 / 112 = 9/14, and 54 / 84 = 9/14. Yes, same ratio. Therefore, the foot is indeed on CD.

Therefore, the minimal HP is 120 m. Therefore, the maximum length saved is 160 m.

But let me check if there's another way to approach this problem, maybe using coordinates.

Alternatively, since P is on CD, we can parametrize P as (112s, -84 + 84s) where s ∈ [0, 1]. Then, compute HP^2 as a function of s, find its minimum, then take the square root.

Let me try that.

Coordinates of H: (0, 66). Coordinates of P: (112s, -84 + 84s).

Then, HP^2 = (112s - 0)^2 + (-84 + 84s - 66)^2 = (112s)^2 + (-150 + 84s)^2.

Let me compute that:

HP² = (112s)² + (-150 + 84s)²

Let me expand this:

= 12544s² + (84s - 150)²

Compute (84s - 150)^2:

= (84s)^2 - 2*84s*150 + 150²

= 7056s² - 25200s + 22500

Therefore, HP² = 12544s² + 7056s² - 25200s + 22500

Combine like terms:

(12544 + 7056)s² - 25200s + 22500

Calculate 12544 + 7056: 12544 + 7056 = 19600

So HP² = 19600s² - 25200s + 22500

To find the minimum of this quadratic function in s, we can take derivative or use vertex formula.

Since it's a quadratic in s, the minimum occurs at s = -b/(2a), where a = 19600, b = -25200.

So s = -(-25200)/(2*19600) = 25200 / 39200 = 252 / 392 = divide numerator and denominator by 28: 9 / 14 ≈ 0.642857.

Which is between 0 and 1, so valid.

Therefore, the minimal HP² is at s = 9/14.

Compute HP² at s = 9/14:

HP² = 19600*(9/14)^2 - 25200*(9/14) + 22500

First compute 19600*(81/196): since (9/14)^2 = 81/196.

Note that 19600 / 196 = 100. Therefore, 19600*(81/196) = 100*81 = 8100.

Then, -25200*(9/14) = - (25200 / 14)*9 = -1800*9 = -16200

Then, +22500.

So total HP² = 8100 - 16200 + 22500 = (8100 + 22500) - 16200 = 30600 - 16200 = 14400

Therefore, HP = sqrt(14400) = 120 m.

So same result as before. Therefore, the minimal HP is indeed 120 m. Therefore, the maximum length saved is 280 - 120 = 160 m.

But let me check if there's a possibility that the minimal HP is not achieved at the foot of the perpendicular? But according to both methods, the minimal HP is 120 m.

Alternatively, maybe I made a mistake in assuming that replacing HC and HD with HP would save (HC + HD) - HP. Let me think.

Originally, we have two ropes: HC and HD. Their total length is 150 + 130 = 280. If we replace them with a single rope HP, then the total length is HP. Therefore, the amount saved is 280 - HP. Therefore, to maximize the saved length, we need to minimize HP.

Yes, that seems correct. Therefore, the answer is 160 m.

But let me check with coordinates. For example, if P is at C, then HP is HC = 150 m. If P is at D, then HP is HD = 130 m. The minimal HP is 120 m, which is less than 130 m. Therefore, replacing with HP at the foot of the perpendicular gives a shorter rope, saving 280 - 120 = 160 m.

Therefore, the answer should be 160 meters.

But let me check if there's any other constraints or if I misread the problem.

Wait, the problem says "rope HC and rope HD are to be replaced by a single rope HP where P is a point on the straight line between C and D." So P is on CD, which we considered.

Also, the Asymptote code shows some other points, but maybe that's just for drawing.

Alternatively, maybe there's a mistake in assuming that the minimal HP is 120. Let me check with coordinates.

The foot of the perpendicular is at (72, -30). Let me compute the distance from H(0, 66) to (72, -30).

Distance squared is (72 - 0)^2 + (-30 - 66)^2 = 72² + (-96)^2.

Compute 72²: 5184, 96²: 9216. Total: 5184 + 9216 = 14400. sqrt(14400) = 120. Correct.

Therefore, yes, that's correct.

Therefore, the maximum length saved is 160 meters.

But let me check if the problem says "the greatest length of rope that can be saved". So, yes, 160 m.

But let me check if there's a different interpretation. For example, maybe the problem is not about replacing HC and HD with HP, but replacing HC and HD with a single rope that is connected to H and P, but maybe there are other constraints?

But according to the problem statement: "rope HC and rope HD are to be replaced by a single rope HP where P is a point on the straight line between C and D." So, the original ropes are HA, HB, HC, HD. Now, they are replacing HC and HD with HP. So, the total length of ropes is HA + HB + HP. Previously, it was HA + HB + HC + HD. Therefore, the difference is (HC + HD) - HP. Therefore, the amount saved is (HC + HD) - HP. Therefore, to maximize the saved length, minimize HP.

Therefore, the answer is 160.

But let me check if the problem says "the greatest length of rope that can be saved". So, if HP is minimized, then the saved length is maximized. Therefore, 160 m.

Alternatively, maybe there's a mistake in the calculation of HC and HD? Let me check again.

Earlier, we had:

From HC = h + c = 150, and HD = sqrt(d² + h²) = 130. Also, c² + d² = 140² = 19600.

Then, we found c = 84, h = 66, d = 112.

Check HC: h + c = 66 + 84 = 150. Correct.

Check HD: sqrt(d² + h²) = sqrt(112² + 66²). Let me compute 112² = 12544, 66² = 4356. Sum is 12544 + 4356 = 16900. sqrt(16900) = 130. Correct.

Distance between C and D: sqrt(112² + 84²) = sqrt(12544 + 7056) = sqrt(19600) = 140. Correct.

Therefore, all the values are consistent.

Therefore, the answer is 160 meters.

But just to be thorough, let me check if there's another way that HP could be shorter? For example, if the foot of the perpendicular is not on CD? But we checked that it is.

Alternatively, maybe the problem is in three dimensions? But the problem says "a flat field", so all points are on a plane, and H is directly above O, so it's a vertical line. Therefore, the coordinates are in 2D with H at (0, h) in the plane.

Therefore, the calculation is correct.

Therefore, I think the answer is 160 meters.

**Final Answer**
\boxed{160}

To determine the **greatest length of rope that can be saved**, we analyze the scenario where two ropes, $ HC $ and $ HD $, are replaced by a single rope $ HP $, with point $ P $ lying on the straight line segment $ CD $. The goal is to **minimize the length of $ HP $** to **maximize the length saved**, which is:

$$
\text{Length saved} = (HC + HD) - HP
$$

---

### Step 1: Coordinate Setup

Let’s place the points in a coordinate system:

- Let $ O = (0, 0) $
- Since $ A $ is directly north of $ O $, let $ A = (0, a) $
- $ B $ is directly west of $ O $, so $ B = (-b, 0) $
- $ C $ is directly south of $ O $, so $ C = (0, -c) $
- $ D $ is directly east of $ O $, so $ D = (d, 0) $

Given:
- $ HC = 150 $ m
- $ HD = 130 $ m
- $ CD = 140 $ m

From the coordinates:
- $ HC = \text{distance from } H \text{ to } C = h + c = 150 $
- $ HD = \text{distance from } H \text{ to } D = \sqrt{d^2 + h^2} = 130 $
- $ CD = \sqrt{d^2 + c^2} = 140 $

---

### Step 2: Solve for Coordinates

From the equations:

1. $ h + c = 150 $
2. $ \sqrt{d^2 + h^2} = 130 \Rightarrow d^2 + h^2 = 16900 $
3. $ \sqrt{d^2 + c^2} = 140 \Rightarrow d^2 + c^2 = 19600 $

Subtracting (2) from (3):

$$
(c^2 + d^2) - (d^2 + h^2) = 19600 - 16900 \Rightarrow c^2 - h^2 = 2700
$$

Now, from (1): $ c = 150 - h $

Substitute into $ c^2 - h^2 = 2700 $:

$$
(150 - h)^2 - h^2 = 2700 \Rightarrow 22500 - 300h + h^2 - h^2 = 2700 \Rightarrow 22500 - 300h = 2700
$$

$$
300h = 19800 \Rightarrow h = 66
$$

Then $ c = 150 - 66 = 84 $

Now, from (2): $ d^2 + h^2 = 16900 \Rightarrow d^2 = 16900 - 66^2 = 16900 - 4356 = 12544 \Rightarrow d = 112 $

So the coordinates are:

- $ C = (0, -84) $
- $ D = (112, 0) $
- $ H = (0, 66) $

---

### Step 3: Find the Minimum Length of $ HP $

We are to find the **minimum** length of $ HP $, where $ P $ lies on the line segment $ CD $.

The line $ CD $ passes through $ C = (0, -84) $ and $ D = (112, 0) $. Its slope is:

$$
m = \frac{0 - (-84)}{112 - 0} = \frac{84}{112} = \frac{3}{4}
$$

So the equation of line $ CD $ is:

$$
y = \frac{3}{4}x - 84
$$

The point $ H = (0, 66) $ lies above this line. The **shortest distance** from $ H $ to the line $ CD $ is the **perpendicular distance** from $ H $ to the line.

Using the point-to-line distance formula:

$$
\text{Distance} = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}
$$

Rewriting the line equation in standard form:

$$
3x - 4y - 336 = 0
$$

So $ A = 3 $, $ B = -4 $, $ C = -336 $

Plug in $ H = (0, 66) $:

$$
\text{Distance} = \frac{|3(0) - 4(66) - 336|}{\sqrt{3^2 + (-4)^2}} = \frac{|0 - 264 - 336|}{5} = \frac{600}{5} = 120
$$

Now, we check whether the **foot of the perpendicular** lies on the segment $ CD $.

The line perpendicular to $ CD $ through $ H $ has slope $ -\frac{4}{3} $, so its equation is:

$$
y - 66 = -\frac{4}{3}x \Rightarrow y = -\frac{4}{3}x + 66
$$

Solving the system:

$$
\frac{3}{4}x - 84 = -\frac{4}{3}x + 66
$$

Multiply both sides by 12:

$$
9x - 1008 = -16x + 792 \Rightarrow 25x = 1800 \Rightarrow x = 72
$$

Then $ y = \frac{3}{4}(72) - 84 = 54 - 84 = -30 $

So the foot of the perpendicular is at $ (72, -30) $, which lies between $ C = (0, -84) $ and $ D = (112, 0) $, so it is on the segment $ CD $.

Thus, the **minimum length of $ HP $** is:

$$
HP = 120 \text{ m}
$$

---

### Step 4: Compute the Maximum Length Saved

Original total length of ropes $ HC $ and $ HD $:

$$
HC + HD = 150 + 130 = 280 \text{ m}
$$

New total length with rope $ HP $:

$$
HP = 120 \text{ m}
$$

So the **maximum length saved** is:

$$
280 - 120 = \boxed{160}
$$

---

### Final Answer

$$
\boxed{160}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the greatest length of rope that can be saved by replacing ropes HC and HD with a single rope HP, where P is on the line between C and D. The balloon is still at H above O, and the original ropes HC and HD have lengths 150 m and 130 m respectively. The distance between C and D is 140 m.

First, let me visualize the setup. There's a point O, with A north, B west, C south, D east. So, if I imagine a coordinate system with O at the origin, then A would be along the positive y-axis, B along the negative x-axis, C along the negative y-axis, and D along the positive x-axis. The points C and D are 140 meters apart. Since they are directly south and east of O, the distance between them should be the hypotenuse of a right triangle with legs equal to the distances from O to C and O to D. Wait, but the problem says the distance between C and D is 140 m. So if I let the distance from O to C be, say, y, and from O to D be x, then the distance between C and D would be sqrt(x² + y²) = 140 m.

But maybe I need to assign coordinates to all points. Let me try that. Let me set O at (0, 0). Then, since A is directly north of O, let me assume A is at (0, a) for some a. Similarly, B is directly west of O, so B is at (-b, 0) for some b. C is directly south of O, so C is at (0, -c) for some c. D is directly east of O, so D is at (d, 0) for some d.

But the distance between C and D is 140 m. So the coordinates of C are (0, -c) and D are (d, 0). The distance between them is sqrt[(d - 0)^2 + (0 - (-c))^2] = sqrt(d² + c²) = 140. So that gives me c² + d² = 140².

Now, the balloon is at H directly above O, so H has coordinates (0, h) for some h. The ropes HA, HB, HC, HD are from H to each of the points. The lengths of HC and HD are given as 150 m and 130 m. Let me write equations for those.

HC is the distance from H(0, h) to C(0, -c). That distance is sqrt[(0 - 0)^2 + (h - (-c))^2] = sqrt[(h + c)^2] = h + c. Wait, that's interesting. Since H is above O, and C is below O, the vertical distance between H and C is h + c. So HC = h + c = 150 m. Similarly, HD is the distance from H(0, h) to D(d, 0). That distance is sqrt[(d - 0)^2 + (0 - h)^2] = sqrt(d² + h²) = 130 m.

So we have two equations:

1. h + c = 150
2. sqrt(d² + h²) = 130

Also, from the distance between C and D, we have:

3. c² + d² = 140² = 19600

So, let me note that equation 2 is sqrt(d² + h²) = 130, so squaring both sides gives d² + h² = 16900.

But from equation 3, c² + d² = 19600. If I subtract equation 2 squared from equation 3, I get:

(c² + d²) - (d² + h²) = 19600 - 16900 => c² - h² = 2700.

But from equation 1, h + c = 150. Let me write that as c = 150 - h. Then, substitute into c² - h² = 2700.

So, c² - h² = (c - h)(c + h) = (c - h)(150) = 2700. Therefore, (c - h) = 2700 / 150 = 18.

So c - h = 18. But we also have c + h = 150. So we can solve for c and h.

Adding the two equations:

(c - h) + (c + h) = 18 + 150 => 2c = 168 => c = 84. Then h = 150 - c = 150 - 84 = 66.

So c = 84 m, h = 66 m. Then, from equation 2, d² + h² = 16900. So d² = 16900 - h² = 16900 - 66². Let me compute 66²: 66*66 = 4356. Then 16900 - 4356 = 12544. Then d = sqrt(12544). Let me check sqrt(12544). Well, 112² = 12544 because 100²=10000, 110²=12100, 112²= (110 + 2)^2 = 110² + 4*110 + 4 = 12100 + 440 + 4 = 12544. So d = 112 m.

So, coordinates:

O is (0, 0)

A is (0, a). Wait, but we haven't determined a yet. Wait, but maybe we don't need a? Because the problem is about replacing HC and HD with HP where P is on CD. But maybe we need coordinates for P?

Wait, but maybe we can figure out coordinates for all points. Let me check.

We have:

- C is (0, -c) = (0, -84)

- D is (d, 0) = (112, 0)

So the line between C and D goes from (0, -84) to (112, 0). So parametric equations for CD can be written as x = 112t, y = -84 + 84t, where t ranges from 0 to 1. Wait, when t=0, x=0, y=-84 (point C); when t=1, x=112, y=0 (point D). Alternatively, maybe parameterizing differently. Let me think.

Alternatively, the line CD can be parametrized as starting at C(0, -84) and moving towards D(112, 0). So direction vector is (112 - 0, 0 - (-84)) = (112, 84). So parametric equations can be written as:

x = 0 + 112s

y = -84 + 84s

where s ∈ [0, 1]. So any point P on CD can be represented as (112s, -84 + 84s) for some s between 0 and 1.

Alternatively, since the problem says P is on the straight line between C and D, so we can parametrize it as such.

Now, the original ropes HC and HD have lengths 150 m and 130 m. The total length of these two ropes is 150 + 130 = 280 m. If we replace them with a single rope HP, then the length of HP will be less than or equal to the sum of HC and HD? Wait, no. Wait, actually, the problem says "to reduce the total length of rope used, rope HC and rope HD are to be replaced by a single rope HP". So the total length saved would be (HC + HD) - HP. Therefore, to maximize the length saved, we need to minimize HP.

Therefore, the problem reduces to finding the minimum possible length of HP where P lies on CD. Then, the maximum length saved is (150 + 130) - HP_min = 280 - HP_min.

Therefore, the key is to find the minimal HP where P is on CD.

So, H is at (0, h) = (0, 66). So coordinates of H are (0, 66).

So, we need to find the point P on CD such that the distance from H(0, 66) to P is minimized.

So, this is a classic problem of finding the shortest distance from a point to a line segment. However, since CD is a straight line, and H is a point not on CD, the minimal distance from H to CD is the perpendicular distance from H to the line CD. However, we need to check if the foot of the perpendicular lies on the segment CD. If it does, then that's the minimal distance. If not, then the minimal distance is the distance to the nearest endpoint.

Therefore, first, let me find the equation of line CD.

Points C(0, -84) and D(112, 0). Let me compute the slope of CD.

Slope m = (0 - (-84)) / (112 - 0) = 84 / 112 = 3/4.

So the equation of line CD can be written as y - (-84) = (3/4)(x - 0), so y + 84 = (3/4)x => y = (3/4)x - 84.

Alternatively, in standard form: 3x - 4y - 336 = 0. Let me check:

Starting from y = (3/4)x - 84, multiply both sides by 4: 4y = 3x - 336 => 3x - 4y - 336 = 0. Yes.

So the line CD is 3x - 4y - 336 = 0.

Now, the distance from H(0, 66) to line CD is given by the formula:

Distance = |Ax + By + C| / sqrt(A² + B²), where the line is Ax + By + C = 0.

But our line is 3x - 4y - 336 = 0, so A = 3, B = -4, C = -336.

Wait, the formula is |Ax + By + C| / sqrt(A² + B²). Wait, but in standard form, it's usually written as Ax + By + C = 0. So here, 3x - 4y - 336 = 0, so A = 3, B = -4, C = -336.

So plugging in H(0, 66):

Distance = |3*0 + (-4)*66 + (-336)| / sqrt(3² + (-4)^2) = |0 - 264 - 336| / 5 = |-600| / 5 = 600 / 5 = 120.

So the minimal distance from H to line CD is 120 meters. However, we need to check if the foot of the perpendicular from H to CD lies on the segment CD. If it does, then the minimal HP is 120 m. If not, then the minimal distance would be to one of the endpoints.

So let me find the coordinates of the foot of the perpendicular from H(0, 66) to line CD.

The formula for the foot of the perpendicular from a point (x0, y0) to the line ax + by + c = 0 is:

x = x0 - a*(ax0 + by0 + c)/(a² + b²)

y = y0 - b*(ax0 + by0 + c)/(a² + b²)

Wait, but let me recall the formula. Alternatively, since we have the line CD: y = (3/4)x - 84.

The slope of CD is 3/4, so the slope of the perpendicular is -4/3.

So the line perpendicular to CD passing through H(0, 66) has equation y - 66 = (-4/3)(x - 0), so y = (-4/3)x + 66.

Find the intersection point between this line and CD.

Set equations equal:

(3/4)x - 84 = (-4/3)x + 66

Multiply both sides by 12 to eliminate denominators:

12*(3/4)x - 12*84 = 12*(-4/3)x + 12*66

Which simplifies to:

9x - 1008 = -16x + 792

Bring all terms to left side:

9x + 16x - 1008 - 792 = 0 => 25x - 1800 = 0 => 25x = 1800 => x = 1800 / 25 = 72.

Then, substitute x = 72 into the equation of CD: y = (3/4)(72) - 84 = 54 - 84 = -30.

So the foot of the perpendicular is at (72, -30).

Now, check if this point lies on segment CD.

Point C is (0, -84), D is (112, 0). The x-coordinate of the foot is 72, which is between 0 and 112. The y-coordinate is -30, which is between -84 and 0. Therefore, yes, the foot of the perpendicular lies on segment CD. Therefore, the minimal HP is 120 meters.

Therefore, the maximum length saved is (150 + 130) - 120 = 280 - 120 = 160 meters.

Wait, but hold on. Let me confirm. The original total length of HC and HD is 150 + 130 = 280. If we replace them with HP, which is 120, then the length saved is 280 - 120 = 160. Therefore, the answer is 160 meters.

But let me check if there's a mistake here. Because sometimes when you replace two ropes with one, you have to make sure that the single rope is sufficient to hold the balloon. But the problem says "the balloon remains at the same position H above O as described above." So the position of H is fixed, and P is on CD. So the rope HP is just a single rope from H to P. So the length of HP is the distance from H to P, and since P is on CD, the minimal HP is 120, so replacing HC and HD with HP saves 280 - 120 = 160.

But let me check if my calculation of the foot of the perpendicular is correct.

Earlier, I found that the foot is at (72, -30). Let me check if that point is on CD.

Parametrize CD as from C(0, -84) to D(112, 0). Let me check if (72, -30) is between them.

The vector from C to D is (112, 84). The vector from C to the foot is (72 - 0, -30 - (-84)) = (72, 54).

Check if this vector is a scalar multiple of (112, 84).

Let me see: 72 / 112 = 9/14, and 54 / 84 = 9/14. Yes, same ratio. Therefore, the foot is indeed on CD.

Therefore, the minimal HP is 120 m. Therefore, the maximum length saved is 160 m.

But let me check if there's another way to approach this problem, maybe using coordinates.

Alternatively, since P is on CD, we can parametrize P as (112s, -84 + 84s) where s ∈ [0, 1]. Then, compute HP^2 as a function of s, find its minimum, then take the square root.

Let me try that.

Coordinates of H: (0, 66). Coordinates of P: (112s, -84 + 84s).

Then, HP^2 = (112s - 0)^2 + (-84 + 84s - 66)^2 = (112s)^2 + (-150 + 84s)^2.

Let me compute that:

HP² = (112s)² + (-150 + 84s)²

Let me expand this:

= 12544s² + (84s - 150)²

Compute (84s - 150)^2:

= (84s)^2 - 2*84s*150 + 150²

= 7056s² - 25200s + 22500

Therefore, HP² = 12544s² + 7056s² - 25200s + 22500

Combine like terms:

(12544 + 7056)s² - 25200s + 22500

Calculate 12544 + 7056: 12544 + 7056 = 19600

So HP² = 19600s² - 25200s + 22500

To find the minimum of this quadratic function in s, we can take derivative or use vertex formula.

Since it's a quadratic in s, the minimum occurs at s = -b/(2a), where a = 19600, b = -25200.

So s = -(-25200)/(2*19600) = 25200 / 39200 = 252 / 392 = divide numerator and denominator by 28: 9 / 14 ≈ 0.642857.

Which is between 0 and 1, so valid.

Therefore, the minimal HP² is at s = 9/14.

Compute HP² at s = 9/14:

HP² = 19600*(9/14)^2 - 25200*(9/14) + 22500

First compute 19600*(81/196): since (9/14)^2 = 81/196.

Note that 19600 / 196 = 100. Therefore, 19600*(81/196) = 100*81 = 8100.

Then, -25200*(9/14) = - (25200 / 14)*9 = -1800*9 = -16200

Then, +22500.

So total HP² = 8100 - 16200 + 22500 = (8100 + 22500) - 16200 = 30600 - 16200 = 14400

Therefore, HP = sqrt(14400) = 120 m.

So same result as before. Therefore, the minimal HP is indeed 120 m. Therefore, the maximum length saved is 280 - 120 = 160 m.

But let me check if there's a possibility that the minimal HP is not achieved at the foot of the perpendicular? But according to both methods, the minimal HP is 120 m.

Alternatively, maybe I made a mistake in assuming that replacing HC and HD with HP would save (HC + HD) - HP. Let me think.

Originally, we have two ropes: HC and HD. Their total length is 150 + 130 = 280. If we replace them with a single rope HP, then the total length is HP. Therefore, the amount saved is 280 - HP. Therefore, to maximize the saved length, we need to minimize HP.

Yes, that seems correct. Therefore, the answer is 160 m.

But let me check with coordinates. For example, if P is at C, then HP is HC = 150 m. If P is at D, then HP is HD = 130 m. The minimal HP is 120 m, which is less than 130 m. Therefore, replacing with HP at the foot of the perpendicular gives a shorter rope, saving 280 - 120 = 160 m.

Therefore, the answer should be 160 meters.

But let me check if there's any other constraints or if I misread the problem.

Wait, the problem says "rope HC and rope HD are to be replaced by a single rope HP where P is a point on the straight line between C and D." So P is on CD, which we considered.

Also, the Asymptote code shows some other points, but maybe that's just for drawing.

Alternatively, maybe there's a mistake in assuming that the minimal HP is 120. Let me check with coordinates.

The foot of the perpendicular is at (72, -30). Let me compute the distance from H(0, 66) to (72, -30).

Distance squared is (72 - 0)^2 + (-30 - 66)^2 = 72² + (-96)^2.

Compute 72²: 5184, 96²: 9216. Total: 5184 + 9216 = 14400. sqrt(14400) = 120. Correct.

Therefore, yes, that's correct.

Therefore, the maximum length saved is 160 meters.

But let me check if the problem says "the greatest length of rope that can be saved". So, yes, 160 m.

But let me check if there's a different interpretation. For example, maybe the problem is not about replacing HC and HD with HP, but replacing HC and HD with a single rope that is connected to H and P, but maybe there are other constraints?

But according to the problem statement: "rope HC and rope HD are to be replaced by a single rope HP where P is a point on the straight line between C and D." So, the original ropes are HA, HB, HC, HD. Now, they are replacing HC and HD with HP. So, the total length of ropes is HA + HB + HP. Previously, it was HA + HB + HC + HD. Therefore, the difference is (HC + HD) - HP. Therefore, the amount saved is (HC + HD) - HP. Therefore, to maximize the saved length, minimize HP.

Therefore, the answer is 160.

But let me check if the problem says "the greatest length of rope that can be saved". So, if HP is minimized, then the saved length is maximized. Therefore, 160 m.

Alternatively, maybe there's a mistake in the calculation of HC and HD? Let me check again.

Earlier, we had:

From HC = h + c = 150, and HD = sqrt(d² + h²) = 130. Also, c² + d² = 140² = 19600.

Then, we found c = 84, h = 66, d = 112.

Check HC: h + c = 66 + 84 = 150. Correct.

Check HD: sqrt(d² + h²) = sqrt(112² + 66²). Let me compute 112² = 12544, 66² = 4356. Sum is 12544 + 4356 = 16900. sqrt(16900) = 130. Correct.

Distance between C and D: sqrt(112² + 84²) = sqrt(12544 + 7056) = sqrt(19600) = 140. Correct.

Therefore, all the values are consistent.

Therefore, the answer is 160 meters.

But just to be thorough, let me check if there's another way that HP could be shorter? For example, if the foot of the perpendicular is not on CD? But we checked that it is.

Alternatively, maybe the problem is in three dimensions? But the problem says "a flat field", so all points are on a plane, and H is directly above O, so it's a vertical line. Therefore, the coordinates are in 2D with H at (0, h) in the plane.

Therefore, the calculation is correct.

Therefore, I think the answer is 160 meters.

**Final Answer**
\boxed{160}

To determine the **greatest length of rope that can be saved**, we analyze the scenario where two ropes, $ HC $ and $ HD $, are replaced by a single rope $ HP $, with point $ P $ lying on the straight line segment $ CD $. The goal is to **minimize the length of $ HP $** to **maximize the length saved**, which is:

$$
\text{Length saved} = (HC + HD) - HP
$$

---

### Step 1: Coordinate Setup

Let’s place the points in a coordinate system:

- Let $ O = (0, 0) $
- Since $ A $ is directly north of $ O $, let $ A = (0, a) $
- $ B $ is directly west of $ O $, so $ B = (-b, 0) $
- $ C $ is directly south of $ O $, so $ C = (0, -c) $
- $ D $ is directly east of $ O $, so $ D = (d, 0) $

Given:
- $ HC = 150 $ m
- $ HD = 130 $ m
- $ CD = 140 $ m

From the coordinates:
- $ HC = \text{distance from } H \text{ to } C = h + c = 150 $
- $ HD = \text{distance from } H \text{ to } D = \sqrt{d^2 + h^2} = 130 $
- $ CD = \sqrt{d^2 + c^2} = 140 $

---

### Step 2: Solve for Coordinates

From the equations:

1. $ h + c = 150 $
2. $ \sqrt{d^2 + h^2} = 130 \Rightarrow d^2 + h^2 = 16900 $
3. $ \sqrt{d^2 + c^2} = 140 \Rightarrow d^2 + c^2 = 19600 $

Subtracting (2) from (3):

$$
(c^2 + d^2) - (d^2 + h^2) = 19600 - 16900 \Rightarrow c^2 - h^2 = 2700
$$

Now, from (1): $ c = 150 - h $

Substitute into $ c^2 - h^2 = 2700 $:

$$
(150 - h)^2 - h^2 = 2700 \Rightarrow 22500 - 300h + h^2 - h^2 = 2700 \Rightarrow 22500 - 300h = 2700
$$

$$
300h = 19800 \Rightarrow h = 66
$$

Then $ c = 150 - 66 = 84 $

Now, from (2): $ d^2 + h^2 = 16900 \Rightarrow d^2 = 16900 - 66^2 = 16900 - 4356 = 12544 \Rightarrow d = 112 $

So the coordinates are:

- $ C = (0, -84) $
- $ D = (112, 0) $
- $ H = (0, 66) $

---

### Step 3: Find the Minimum Length of $ HP $

We are to find the **minimum** length of $ HP $, where $ P $ lies on the line segment $ CD $.

The line $ CD $ passes through $ C = (0, -84) $ and $ D = (112, 0) $. Its slope is:

$$
m = \frac{0 - (-84)}{112 - 0} = \frac{84}{112} = \frac{3}{4}
$$

So the equation of line $ CD $ is:

$$
y = \frac{3}{4}x - 84
$$

The point $ H = (0, 66) $ lies above this line. The **shortest distance** from $ H $ to the line $ CD $ is the **perpendicular distance** from $ H $ to the line.

Using the point-to-line distance formula:

$$
\text{Distance} = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}
$$

Rewriting the line equation in standard form:

$$
3x - 4y - 336 = 0
$$

So $ A = 3 $, $ B = -4 $, $ C = -336 $

Plug in $ H = (0, 66) $:

$$
\text{Distance} = \frac{|3(0) - 4(66) - 336|}{\sqrt{3^2 + (-4)^2}} = \frac{|0 - 264 - 336|}{5} = \frac{600}{5} = 120
$$

Now, we check whether the **foot of the perpendicular** lies on the segment $ CD $.

The line perpendicular to $ CD $ through $ H $ has slope $ -\frac{4}{3} $, so its equation is:

$$
y - 66 = -\frac{4}{3}x \Rightarrow y = -\frac{4}{3}x + 66
$$

Solving the system:

$$
\frac{3}{4}x - 84 = -\frac{4}{3}x + 66
$$

Multiply both sides by 12:

$$
9x - 1008 = -16x + 792 \Rightarrow 25x = 1800 \Rightarrow x = 72
$$

Then $ y = \frac{3}{4}(72) - 84 = 54 - 84 = -30 $

So the foot of the perpendicular is at $ (72, -30) $, which lies between $ C = (0, -84) $ and $ D = (112, 0) $, so it is on the segment $ CD $.

Thus, the **minimum length of $ HP $** is:

$$
HP = 120 \text{ m}
$$

---

### Step 4: Compute the Maximum Length Saved

Original total length of ropes $ HC $ and $ HD $:

$$
HC + HD = 150 + 130 = 280 \text{ m}
$$

New total length with rope $ HP $:

$$
HP = 120 \text{ m}
$$

So the **maximum length saved** is:

$$
280 - 120 = \boxed{160}
$$

---

### Final Answer

$$
\boxed{160}
$$

Solution:
To save the most rope, we must have $HP$ having minimum length.
For $HP$ to have minimum length, $HP$ must be perpendicular to $CD$. [asy]
pair C, D, H, P;
H=(90,120);
C=(0,0);
D=(140,0);
P=(90,0);
draw(H--C--D--H--P);
label("H", H, N);
label("C", C, SW);
label("D", D, SE);
label("P", P, S);
label("150", (C+H)/2, NW);
label("130", (D+H)/2, NE);
[/asy] (Among other things, we can see from this diagram that sliding $P$ away from the perpendicular position does make $HP$ longer.)
In the diagram, $HC=150$, $HD=130$ and $CD=140$.
Let $HP=x$ and $PD=a$.  Then $CP=140-a$.
By the Pythagorean Theorem in $\triangle HPC$, $x^2 + (140-a)^2 = 150^2$.
By the Pythagorean Theorem in $\triangle HPD$, $x^2+a^2 = 130^2$.
Subtracting the second equation from the first, we obtain \begin{align*}
(140-a)^2 - a^2 & = 150^2 - 130^2 \\
(19600 - 280a+a^2)-a^2 & = 5600 \\
19600 -280a & = 5600 \\
280a & = 14000 \\
a & = 50
\end{align*} Therefore, $x^2 + 90^2 = 150^2$ or $x^2 = 150^2 - 90^2 = 22500 - 8100 = 14400$ so $x =120$.
So the shortest possible rope that we can use is 120 m, which saves $130+150-120 = \boxed{160}$ m of rope.


Explanation:
Okay, so Paul and Jesse each pick a number at random from the first six primes. I need to find the probability that the sum of their numbers is even. Let me think about how to approach this.

First, let me recall the first six prime numbers. Primes are numbers greater than 1 that have no divisors other than 1 and themselves. The first six primes are 2, 3, 5, 7, 11, and 13. Let me confirm that: 2 is prime, then 3, 5, 7, 11, 13. Yep, those are the first six. So each of them has these numbers to choose from.

Now, the question is about the probability that the sum of their numbers is even. So, I need to figure out in how many cases the sum is even, divided by the total number of possible outcomes.

First, let me figure out the total number of possible outcomes. Since both Paul and Jesse are choosing a number independently, each has 6 choices. So the total number of possible pairs is 6 * 6 = 36. So the denominator of my probability will be 36.

Now, the numerator is the number of pairs where the sum is even. So I need to find how many such pairs there are.

To do this, I should recall that the sum of two numbers is even if both numbers are even or both are odd. Because even + even = even, odd + odd = even, and even + odd = odd. So, the sum is even if both numbers are even or both are odd.

Therefore, I need to find the number of pairs where both numbers are even or both are odd. So, I can calculate the number of ways both are even, plus the number of ways both are odd.

So, first, let me figure out how many even numbers are in the first six primes. The primes are 2, 3, 5, 7, 11, 13. Let me check which are even. Well, 2 is the only even prime number. All other primes are odd because any other even number would be divisible by 2. So, in the first six primes, there is only one even number: 2. The rest are odd: 3, 5, 7, 11, 13. So that's 5 odd numbers and 1 even number.

Therefore, the number of ways both Paul and Jesse choose even numbers: since there's only one even number, 2. So Paul can choose 2, and Jesse can choose 2. So that's 1 * 1 = 1 way.

Wait, but hold on. Wait, if there is only one even number, then the number of ways both choose even numbers is 1 * 1 = 1? But actually, each of them can choose any of the numbers, so if there is only one even number, then the number of ways both choose even numbers is 1 (for Paul) * 1 (for Jesse) = 1. But wait, is that correct?

Wait, no. Wait, the number of even numbers is 1. So, the number of ways Paul can choose an even number is 1, and the same for Jesse. So the number of pairs where both are even is 1 * 1 = 1. But then, the number of ways both are odd is... Well, there are 5 odd numbers. So Paul can choose any of the 5 odd numbers, and Jesse can choose any of the 5 odd numbers. So that's 5 * 5 = 25. Therefore, total number of favorable outcomes is 1 + 25 = 26. Therefore, the probability is 26/36, which simplifies to 13/18. Is that correct?

Wait, let me check again. Let me think again. So total number of possible outcomes is 6*6=36. Number of favorable outcomes: both even or both odd. Since there is only one even number, the number of ways both are even is 1*1=1. The number of ways both are odd is 5*5=25. So total favorable is 1 + 25 = 26. Therefore, probability is 26/36 = 13/18. So that would be the probability. But let me check if there's another way to think about this.

Alternatively, maybe I can think about the probability that both numbers are even or both are odd. So, the probability that both are even is (number of even numbers / total numbers)^2. Similarly for both odd. So, since there is 1 even number, probability both even is (1/6)*(1/6) = 1/36. Probability both odd is (5/6)*(5/6) = 25/36. So total probability is 1/36 + 25/36 = 26/36 = 13/18. So same result. So that seems to check out.

But let me think again. Wait, is there a possibility that I made a mistake in counting the number of even numbers? Let me check the first six primes again. First six primes: 2, 3, 5, 7, 11, 13. Yes, only 2 is even. So that's correct. So only one even number. Therefore, the rest are odd. So that seems correct.

Alternatively, maybe I can think about parity. Since the sum is even if both numbers are even or both are odd. Since there is only one even number, the chance that both are even is very low. But since there are five odd numbers, the chance that both are odd is high. So 25/36 + 1/36 = 26/36. So that seems correct.

But let me think of another approach. Let me think about all possible pairs. Let me list them. But that might take time, but maybe for verification.

First, the first six primes: 2, 3, 5, 7, 11, 13. Let me think of all possible pairs. But since there are 36 pairs, that's a lot, but maybe I can count the number of pairs where the sum is even.

Alternatively, think of the parity of each number. Since only 2 is even, the rest are odd. So, if Paul picks 2, then Jesse needs to pick 2 to get an even sum. If Paul picks an odd number, then Jesse needs to pick an odd number to get an even sum.

So, let's compute the number of favorable cases:

Case 1: Paul picks 2. Then Jesse must pick 2. There is 1 choice for Paul (2) and 1 choice for Jesse (2). So 1*1=1.

Case 2: Paul picks an odd number. There are 5 odd numbers. Then Jesse must pick an odd number. There are 5 choices for Jesse. So 5*5=25.

Total favorable cases: 1 + 25 = 26. So same as before. Therefore, probability is 26/36 = 13/18. So that seems consistent.

Alternatively, maybe I can think of the probability that the sum is even as follows: the sum is even if both numbers are even or both are odd. Since there is only one even number, the probability that both are even is (1/6)(1/6) = 1/36. The probability that both are odd is (5/6)(5/6) = 25/36. So total probability is 1/36 + 25/36 = 26/36 = 13/18. Same answer.

Alternatively, maybe I can think of the probability that the sum is even as equal to the probability that both are even plus the probability that both are odd. Which is the same as above.

Alternatively, maybe I can think of the probability that the sum is even as 1 minus the probability that the sum is odd. Let me check that. The sum is odd if one is even and the other is odd. So, the probability that the sum is odd is (probability Paul even and Jesse odd) + (probability Paul odd and Jesse even). Which is (1/6 * 5/6) + (5/6 * 1/6) = 5/36 + 5/36 = 10/36. Therefore, the probability that the sum is even is 1 - 10/36 = 26/36 = 13/18. Same answer again. So that seems to confirm it.

Therefore, regardless of the approach, I get 13/18. Therefore, the probability is 13/18.

But let me check once more with actual enumeration. Let me think of the possible pairs. Since there are only 6 numbers, maybe I can think of how many even and odd numbers there are.

Number of even numbers: 1 (which is 2). Number of odd numbers: 5.

So, when Paul and Jesse pick numbers, the sum is even in the following cases:

- Both pick even: only possible if both pick 2. So that's 1 case.

- Both pick odd: there are 5 odd numbers, so for Paul, 5 choices, for Jesse, 5 choices, so 5*5=25 cases.

Total: 1 + 25 = 26. Total possible cases: 6*6=36. So 26/36=13/18. Yep, same result.

Alternatively, maybe I can think of the possible sums. Let me think of the numbers:

Paul's numbers: 2, 3, 5, 7, 11, 13.

Jesse's numbers: same.

So, for example, if Paul picks 2 (even), then Jesse needs to pick 2 (even) to get even sum. If Paul picks 3 (odd), Jesse needs to pick odd (3,5,7,11,13). Similarly for other odd numbers.

So, for each of Paul's choices:

If Paul picks 2 (even): Jesse has 1 choice (2) to make the sum even.

If Paul picks 3 (odd): Jesse has 5 choices (all odd numbers) to make the sum even.

Same for Paul picking 5, 7, 11, 13. Each of these 5 odd numbers for Paul would result in 5 choices for Jesse. So total number of favorable cases is 1*1 + 5*5 = 1 + 25 = 26. Yep, same as before.

Therefore, I think this is solid. So the probability is 13/18.

But just to be thorough, let me check if there's any other way this could be approached, or if I made a mistake in considering the parity.

Wait, another thought: sometimes when dealing with probabilities, people make mistakes with dependent vs independent events. But in this case, since Paul and Jesse are choosing independently, the choices are independent. So the probability calculations are straightforward.

Alternatively, maybe I can think of the possible parities. Since only one of the numbers is even, the rest are odd. So, the probability that Paul's number is even is 1/6, odd is 5/6. Similarly for Jesse.

So, the probability that both are even: (1/6)(1/6) = 1/36.

Probability both are odd: (5/6)(5/6) = 25/36.

Total probability: 1/36 + 25/36 = 26/36 = 13/18. Yep.

Alternatively, if there were more even numbers, the calculation would be different, but since there's only one, it's straightforward.

Wait, but let me think again: is there a case where even + even is even? Yes, 2 + 2 = 4, which is even. And odd + odd is even, like 3 + 5 = 8, which is even. So that's correct.

But if there were more even numbers, say, for example, if there were two even numbers, then the number of ways both are even would be 2*2=4, and so on. But in this case, only one even number.

Therefore, I think my answer is correct. So the probability is 13/18.

But just to make sure, let me think of another angle. Let me think of the total number of pairs where the sum is even. Let me think of all possible pairs:

First, list all possible pairs where the sum is even. Since there are only 6 numbers, maybe I can think of how many pairs satisfy the condition.

But since there are 36 pairs, it's tedious, but maybe I can think of it in terms of the numbers.

First, note that the only even number is 2. So, if one person picks 2, the other person must also pick 2 to get an even sum. If neither picks 2, then both have to pick odd numbers, which will add up to even.

So, how many pairs have both picking 2? Only one pair: (2,2).

How many pairs have neither picking 2? That is, both picking from the 5 odd numbers. So, 5*5=25 pairs. So total pairs with even sum: 1 + 25 = 26. So same as before. Therefore, 26/36=13/18. Yep.

Alternatively, if I think of the number of pairs where the sum is even, it's equal to the number of pairs where both are even or both are odd. Since there is only one even number, the number of pairs where both are even is 1, and the number of pairs where both are odd is 5*5=25. So again, 26.

Therefore, I think this is correct. Therefore, the probability is 13/18.

I don't see any mistakes in my reasoning. All approaches lead to the same answer. Therefore, I think the probability is 13/18.

**Final Answer**
The probability that the sum of the numbers they choose is even is \boxed{\dfrac{13}{18}}.

To determine the probability that the sum of the numbers chosen by Paul and Jesse is even, we begin by analyzing the set of numbers they are choosing from.

---

### Step 1: Identify the First Six Prime Numbers

The first six prime numbers are:

$$
\{2, 3, 5, 7, 11, 13\}
$$

These are the numbers from which both Paul and Jesse independently choose one number at random.

---

### Step 2: Total Number of Outcomes

Since each person chooses one number from the 6 primes, and their choices are independent, the total number of possible outcomes is:

$$
6 \times 6 = 36
$$

---

### Step 3: Determine When the Sum is Even

The sum of two numbers is **even** if:

- **Both numbers are even**, or
- **Both numbers are odd**

Let’s analyze the parity (even or odd) of the numbers in the set:

- **Even number**: Only **2** is even.
- **Odd numbers**: **3, 5, 7, 11, 13** — that’s 5 odd numbers.

So, there is **1 even number** and **5 odd numbers**.

---

### Step 4: Count Favorable Outcomes

We now count the number of pairs where the sum is even.

#### Case 1: Both choose even numbers

There is only **1 even number** (2), so the only pair where both choose even numbers is:

$$
(2, 2)
$$

So, **1 favorable outcome**.

#### Case 2: Both choose odd numbers

There are **5 odd numbers**, so the number of pairs where both choose odd numbers is:

$$
5 \times 5 = 25
$$

So, **25 favorable outcomes**.

---

### Step 5: Total Favorable Outcomes

Adding both cases:

$$
1 + 25 = 26
$$

---

### Step 6: Compute the Probability

The probability that the sum is even is the ratio of favorable outcomes to total outcomes:

$$
\frac{26}{36} = \frac{13}{18}
$$

---

### Final Answer

$$
\boxed{\dfrac{13}{18}}
$$

[Assistant]
Explanation:
Okay, so Paul and Jesse each pick a number at random from the first six primes. I need to find the probability that the sum of their numbers is even. Let me think about how to approach this.

First, let me recall the first six prime numbers. Primes are numbers greater than 1 that have no divisors other than 1 and themselves. The first six primes are 2, 3, 5, 7, 11, and 13. Let me confirm that: 2 is prime, then 3, 5, 7, 11, 13. Yep, those are the first six. So each of them has these numbers to choose from.

Now, the question is about the probability that the sum of their numbers is even. So, I need to figure out in how many cases the sum is even, divided by the total number of possible outcomes.

First, let me figure out the total number of possible outcomes. Since both Paul and Jesse are choosing a number independently, each has 6 choices. So the total number of possible pairs is 6 * 6 = 36. So the denominator of my probability will be 36.

Now, the numerator is the number of pairs where the sum is even. So I need to find how many such pairs there are.

To do this, I should recall that the sum of two numbers is even if both numbers are even or both are odd. Because even + even = even, odd + odd = even, and even + odd = odd. So, the sum is even if both numbers are even or both are odd.

Therefore, I need to find the number of pairs where both numbers are even or both are odd. So, I can calculate the number of ways both are even, plus the number of ways both are odd.

So, first, let me figure out how many even numbers are in the first six primes. The primes are 2, 3, 5, 7, 11, 13. Let me check which are even. Well, 2 is the only even prime number. All other primes are odd because any other even number would be divisible by 2. So, in the first six primes, there is only one even number: 2. The rest are odd: 3, 5, 7, 11, 13. So that's 5 odd numbers and 1 even number.

Therefore, the number of ways both Paul and Jesse choose even numbers: since there's only one even number, 2. So Paul can choose 2, and Jesse can choose 2. So that's 1 * 1 = 1 way.

Wait, but hold on. Wait, if there is only one even number, then the number of ways both choose even numbers is 1 * 1 = 1? But actually, each of them can choose any of the numbers, so if there is only one even number, then the number of ways both choose even numbers is 1 (for Paul) * 1 (for Jesse) = 1. But wait, is that correct?

Wait, no. Wait, the number of even numbers is 1. So, the number of ways Paul can choose an even number is 1, and the same for Jesse. So the number of pairs where both are even is 1 * 1 = 1. But then, the number of ways both are odd is... Well, there are 5 odd numbers. So Paul can choose any of the 5 odd numbers, and Jesse can choose any of the 5 odd numbers. So that's 5 * 5 = 25. Therefore, total number of favorable outcomes is 1 + 25 = 26. Therefore, the probability is 26/36, which simplifies to 13/18. Is that correct?

Wait, let me check again. Let me think again. So total number of possible outcomes is 6*6=36. Number of favorable outcomes: both even or both odd. Since there is only one even number, the number of ways both are even is 1*1=1. The number of ways both are odd is 5*5=25. So total favorable is 1 + 25 = 26. Therefore, probability is 26/36 = 13/18. So that would be the probability. But let me check if there's another way to think about this.

Alternatively, maybe I can think about the probability that both numbers are even or both are odd. So, the probability that both are even is (number of even numbers / total numbers)^2. Similarly for both odd. So, since there is 1 even number, probability both even is (1/6)*(1/6) = 1/36. Probability both odd is (5/6)*(5/6) = 25/36. So total probability is 1/36 + 25/36 = 26/36 = 13/18. So same result. So that seems to check out.

But let me think again. Wait, is there a possibility that I made a mistake in counting the number of even numbers? Let me check the first six primes again. First six primes: 2, 3, 5, 7, 11, 13. Yes, only 2 is even. So that's correct. So only one even number. Therefore, the rest are odd. So that seems correct.

Alternatively, maybe I can think about parity. Since the sum is even if both numbers are even or both are odd. Since there is only one even number, the chance that both are even is very low. But since there are five odd numbers, the chance that both are odd is high. So 25/36 + 1/36 = 26/36. So that seems correct.

But let me think of another approach. Let me think about all possible pairs. Let me list them. But that might take time, but maybe for verification.

First, the first six primes: 2, 3, 5, 7, 11, 13. Let me think of all possible pairs. But since there are 36 pairs, that's a lot, but maybe I can count the number of pairs where the sum is even.

Alternatively, think of the parity of each number. Since only 2 is even, the rest are odd. So, if Paul picks 2, then Jesse needs to pick 2 to get an even sum. If Paul picks an odd number, then Jesse needs to pick an odd number to get an even sum.

So, let's compute the number of favorable cases:

Case 1: Paul picks 2. Then Jesse must pick 2. There is 1 choice for Paul (2) and 1 choice for Jesse (2). So 1*1=1.

Case 2: Paul picks an odd number. There are 5 odd numbers. Then Jesse must pick an odd number. There are 5 choices for Jesse. So 5*5=25.

Total favorable cases: 1 + 25 = 26. So same as before. Therefore, probability is 26/36 = 13/18. So that seems consistent.

Alternatively, maybe I can think of the probability that the sum is even as follows: the sum is even if both numbers are even or both are odd. Since there is only one even number, the probability that both are even is (1/6)(1/6) = 1/36. The probability that both are odd is (5/6)(5/6) = 25/36. So total probability is 1/36 + 25/36 = 26/36 = 13/18. Same answer.

Alternatively, maybe I can think of the probability that the sum is even as equal to the probability that both are even plus the probability that both are odd. Which is the same as above.

Alternatively, maybe I can think of the probability that the sum is even as 1 minus the probability that the sum is odd. Let me check that. The sum is odd if one is even and the other is odd. So, the probability that the sum is odd is (probability Paul even and Jesse odd) + (probability Paul odd and Jesse even). Which is (1/6 * 5/6) + (5/6 * 1/6) = 5/36 + 5/36 = 10/36. Therefore, the probability that the sum is even is 1 - 10/36 = 26/36 = 13/18. Same answer again. So that seems to confirm it.

Therefore, regardless of the approach, I get 13/18. Therefore, the probability is 13/18.

But let me check once more with actual enumeration. Let me think of the possible pairs. Since there are only 6 numbers, maybe I can think of how many even and odd numbers there are.

Number of even numbers: 1 (which is 2). Number of odd numbers: 5.

So, when Paul and Jesse pick numbers, the sum is even in the following cases:

- Both pick even: only possible if both pick 2. So that's 1 case.

- Both pick odd: there are 5 odd numbers, so for Paul, 5 choices, for Jesse, 5 choices, so 5*5=25 cases.

Total: 1 + 25 = 26. Total possible cases: 6*6=36. So 26/36=13/18. Yep, same result.

Alternatively, maybe I can think of the possible sums. Let me think of the numbers:

Paul's numbers: 2, 3, 5, 7, 11, 13.

Jesse's numbers: same.

So, for example, if Paul picks 2 (even), then Jesse needs to pick 2 (even) to get even sum. If Paul picks 3 (odd), Jesse needs to pick odd (3,5,7,11,13). Similarly for other odd numbers.

So, for each of Paul's choices:

If Paul picks 2 (even): Jesse has 1 choice (2) to make the sum even.

If Paul picks 3 (odd): Jesse has 5 choices (all odd numbers) to make the sum even.

Same for Paul picking 5, 7, 11, 13. Each of these 5 odd numbers for Paul would result in 5 choices for Jesse. So total number of favorable cases is 1*1 + 5*5 = 1 + 25 = 26. Yep, same as before.

Therefore, I think this is solid. So the probability is 13/18.

But just to be thorough, let me check if there's any other way this could be approached, or if I made a mistake in considering the parity.

Wait, another thought: sometimes when dealing with probabilities, people make mistakes with dependent vs independent events. But in this case, since Paul and Jesse are choosing independently, the choices are independent. So the probability calculations are straightforward.

Alternatively, maybe I can think of the possible parities. Since only one of the numbers is even, the rest are odd. So, the probability that Paul's number is even is 1/6, odd is 5/6. Similarly for Jesse.

So, the probability that both are even: (1/6)(1/6) = 1/36.

Probability both are odd: (5/6)(5/6) = 25/36.

Total probability: 1/36 + 25/36 = 26/36 = 13/18. Yep.

Alternatively, if there were more even numbers, the calculation would be different, but since there's only one, it's straightforward.

Wait, but let me think again: is there a case where even + even is even? Yes, 2 + 2 = 4, which is even. And odd + odd is even, like 3 + 5 = 8, which is even. So that's correct.

But if there were more even numbers, say, for example, if there were two even numbers, then the number of ways both are even would be 2*2=4, and so on. But in this case, only one even number.

Therefore, I think my answer is correct. So the probability is 13/18.

But just to make sure, let me think of another angle. Let me think of the total number of pairs where the sum is even. Let me think of all possible pairs:

First, list all possible pairs where the sum is even. Since there are only 6 numbers, maybe I can think of how many pairs satisfy the condition.

But since there are 36 pairs, it's tedious, but maybe I can think of it in terms of the numbers.

First, note that the only even number is 2. So, if one person picks 2, the other person must also pick 2 to get an even sum. If neither picks 2, then both have to pick odd numbers, which will add up to even.

So, how many pairs have both picking 2? Only one pair: (2,2).

How many pairs have neither picking 2? That is, both picking from the 5 odd numbers. So, 5*5=25 pairs. So total pairs with even sum: 1 + 25 = 26. So same as before. Therefore, 26/36=13/18. Yep.

Alternatively, if I think of the number of pairs where the sum is even, it's equal to the number of pairs where both are even or both are odd. Since there is only one even number, the number of pairs where both are even is 1, and the number of pairs where both are odd is 5*5=25. So again, 26.

Therefore, I think this is correct. Therefore, the probability is 13/18.

I don't see any mistakes in my reasoning. All approaches lead to the same answer. Therefore, I think the probability is 13/18.

**Final Answer**
The probability that the sum of the numbers they choose is even is \boxed{\dfrac{13}{18}}.

To determine the probability that the sum of the numbers chosen by Paul and Jesse is even, we begin by analyzing the set of numbers they are choosing from.

---

### Step 1: Identify the First Six Prime Numbers

The first six prime numbers are:

$$
\{2, 3, 5, 7, 11, 13\}
$$

These are the numbers from which both Paul and Jesse independently choose one number at random.

---

### Step 2: Total Number of Outcomes

Since each person chooses one number from the 6 primes, and their choices are independent, the total number of possible outcomes is:

$$
6 \times 6 = 36
$$

---

### Step 3: Determine When the Sum is Even

The sum of two numbers is **even** if:

- **Both numbers are even**, or
- **Both numbers are odd**

Let’s analyze the parity (even or odd) of the numbers in the set:

- **Even number**: Only **2** is even.
- **Odd numbers**: **3, 5, 7, 11, 13** — that’s 5 odd numbers.

So, there is **1 even number** and **5 odd numbers**.

---

### Step 4: Count Favorable Outcomes

We now count the number of pairs where the sum is even.

#### Case 1: Both choose even numbers

There is only **1 even number** (2), so the only pair where both choose even numbers is:

$$
(2, 2)
$$

So, **1 favorable outcome**.

#### Case 2: Both choose odd numbers

There are **5 odd numbers**, so the number of pairs where both choose odd numbers is:

$$
5 \times 5 = 25
$$

So, **25 favorable outcomes**.

---

### Step 5: Total Favorable Outcomes

Adding both cases:

$$
1 + 25 = 26
$$

---

### Step 6: Compute the Probability

The probability that the sum is even is the ratio of favorable outcomes to total outcomes:

$$
\frac{26}{36} = \frac{13}{18}
$$

---

### Final Answer

$$
\boxed{\dfrac{13}{18}}
$$

Solution: The only way for the sum of the numbers Paul and Jesse choose to be odd is if one of them chooses 2 and the other chooses an odd prime. There are five ways for Paul to choose 2 and Jesse to choose an odd prime, and there are five ways for Jesse to choose 2 and Paul to choose an odd prime. Since there are $6\cdot 6=36$ total possible ways for Paul and Jesse to choose their numbers, the probability that the sum of the numbers Paul and Jesse choose is NOT even is $\frac{10}{36}=\frac{5}{18}$. Therefore, the probability that the sum of the numbers Paul and Jesse choose IS even is $1-\frac{5}{18}=\boxed{\frac{13}{18}}$.