Explanation:
Okay, so I need to solve the equation √(3x - 5) = 2. Hmm, let me think. First, I remember that when there's a square root involved, I should probably get rid of it by squaring both sides of the equation. That way, I can solve for x more easily. Let me try that.

So, if I square both sides, the left side becomes (√(3x - 5))², which should just be 3x - 5, right? Because squaring a square root cancels them out. And the right side is 2 squared, which is 4. So now the equation is 3x - 5 = 4.

Alright, now I need to solve for x. Let me add 5 to both sides to get rid of the -5. That would give me 3x = 4 + 5, which is 3x = 9. Then, dividing both sides by 3, I get x = 9/3, so x = 3.

Wait, but hold on. I remember that sometimes when you square both sides of an equation, you might introduce extraneous solutions. So I should check if this solution actually works in the original equation. Let me plug x = 3 back into the original equation to verify.

Original equation: √(3x - 5) = 2. Plugging in x = 3, we get √(3*3 - 5) = √(9 - 5) = √4 = 2. Yep, that's exactly equal to 2. So x = 3 is a valid solution.

But the question says "find all possible values of x." Is there a possibility of another solution? Let me think. The square root function only gives the principal (non-negative) root, so √(something) is always non-negative. In this case, the right side is 2, which is positive, so there's no issue there.

But maybe I should consider if there's another way to approach this? For example, could there be a negative solution? Let me see. If I had, say, √(something) = -2, that would have no solution because square roots are always non-negative. But in our case, the right side is positive, so we don't have to worry about negative solutions.

So, squaring both sides gave me x = 3, and checking it works. Therefore, the only solution is x = 3.

Wait, but maybe I should check if there are any restrictions on the domain of the original equation? The expression under the square root, 3x - 5, must be greater than or equal to zero. So 3x - 5 ≥ 0. Solving that, 3x ≥ 5, so x ≥ 5/3. Since 5/3 is approximately 1.666..., and our solution x = 3 is greater than that, so it's within the domain.

Therefore, there are no other solutions because squaring both sides didn't introduce any extraneous solutions here, and the domain restriction is satisfied.

Let me recap. Start with √(3x - 5) = 2. Square both sides: 3x - 5 = 4. Solve for x: 3x = 9, so x = 3. Check the solution in original equation: works. Check domain: x = 3 is allowed. Therefore, only solution is x = 3.

I think that's all. I don't see any other possible solutions. Maybe if I consider complex numbers? But the problem doesn't mention anything about complex numbers, so I think we're dealing with real numbers here. In real numbers, square roots are defined only for non-negative numbers, and the equation √(something) = 2 has only one solution.

Alternatively, if I think about the graph of y = √(3x - 5) and y = 2. The square root function is increasing, so it will cross y = 2 at exactly one point. Which is x = 3.

So yeah, I think that's solid. The only possible value of x is 3.

**Final Answer**
The only possible value of $ x $ is \boxed{3}.

To solve the equation:

$$
\sqrt{3x - 5} = 2
$$

we begin by eliminating the square root. The most straightforward method is to **square both sides** of the equation, which will remove the radical:

$$
(\sqrt{3x - 5})^2 = 2^2
$$

$$
3x - 5 = 4
$$

Now, solve for $ x $ by isolating it:

$$
3x = 4 + 5
$$

$$
3x = 9
$$

$$
x = \frac{9}{3} = 3
$$

---

### Step 2: Verify the Solution

We must check whether this solution satisfies the original equation. Substitute $ x = 3 $ into the left-hand side:

$$
\sqrt{3(3) - 5} = \sqrt{9 - 5} = \sqrt{4} = 2
$$

This matches the right-hand side of the original equation, so $ x = 3 $ is a valid solution.

---

### Step 3: Consider Domain Restrictions

The expression under the square root, $ 3x - 5 $, must be **non-negative** because the square root of a negative number is not defined in the real number system. So we require:

$$
3x - 5 \geq 0 \Rightarrow x \geq \frac{5}{3}
$$

Since $ x = 3 $ satisfies this inequality, it is within the domain of the original equation.

---

### Step 4: Check for Additional Solutions

The square root function is **one-to-one** and only yields **non-negative** outputs. Therefore, the equation $ \sqrt{3x - 5} = 2 $ has **only one solution** in the real numbers. There is no possibility of a negative solution or any other value that would satisfy the equation.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation √(3x - 5) = 2. Hmm, let me think. First, I remember that when there's a square root involved, I should probably get rid of it by squaring both sides of the equation. That way, I can solve for x more easily. Let me try that.

So, if I square both sides, the left side becomes (√(3x - 5))², which should just be 3x - 5, right? Because squaring a square root cancels them out. And the right side is 2 squared, which is 4. So now the equation is 3x - 5 = 4.

Alright, now I need to solve for x. Let me add 5 to both sides to get rid of the -5. That would give me 3x = 4 + 5, which is 3x = 9. Then, dividing both sides by 3, I get x = 9/3, so x = 3.

Wait, but hold on. I remember that sometimes when you square both sides of an equation, you might introduce extraneous solutions. So I should check if this solution actually works in the original equation. Let me plug x = 3 back into the original equation to verify.

Original equation: √(3x - 5) = 2. Plugging in x = 3, we get √(3*3 - 5) = √(9 - 5) = √4 = 2. Yep, that's exactly equal to 2. So x = 3 is a valid solution.

But the question says "find all possible values of x." Is there a possibility of another solution? Let me think. The square root function only gives the principal (non-negative) root, so √(something) is always non-negative. In this case, the right side is 2, which is positive, so there's no issue there.

But maybe I should consider if there's another way to approach this? For example, could there be a negative solution? Let me see. If I had, say, √(something) = -2, that would have no solution because square roots are always non-negative. But in our case, the right side is positive, so we don't have to worry about negative solutions.

So, squaring both sides gave me x = 3, and checking it works. Therefore, the only solution is x = 3.

Wait, but maybe I should check if there are any restrictions on the domain of the original equation? The expression under the square root, 3x - 5, must be greater than or equal to zero. So 3x - 5 ≥ 0. Solving that, 3x ≥ 5, so x ≥ 5/3. Since 5/3 is approximately 1.666..., and our solution x = 3 is greater than that, so it's within the domain.

Therefore, there are no other solutions because squaring both sides didn't introduce any extraneous solutions here, and the domain restriction is satisfied.

Let me recap. Start with √(3x - 5) = 2. Square both sides: 3x - 5 = 4. Solve for x: 3x = 9, so x = 3. Check the solution in original equation: works. Check domain: x = 3 is allowed. Therefore, only solution is x = 3.

I think that's all. I don't see any other possible solutions. Maybe if I consider complex numbers? But the problem doesn't mention anything about complex numbers, so I think we're dealing with real numbers here. In real numbers, square roots are defined only for non-negative numbers, and the equation √(something) = 2 has only one solution.

Alternatively, if I think about the graph of y = √(3x - 5) and y = 2. The square root function is increasing, so it will cross y = 2 at exactly one point. Which is x = 3.

So yeah, I think that's solid. The only possible value of x is 3.

**Final Answer**
The only possible value of $ x $ is \boxed{3}.

To solve the equation:

$$
\sqrt{3x - 5} = 2
$$

we begin by eliminating the square root. The most straightforward method is to **square both sides** of the equation, which will remove the radical:

$$
(\sqrt{3x - 5})^2 = 2^2
$$

$$
3x - 5 = 4
$$

Now, solve for $ x $ by isolating it:

$$
3x = 4 + 5
$$

$$
3x = 9
$$

$$
x = \frac{9}{3} = 3
$$

---

### Step 2: Verify the Solution

We must check whether this solution satisfies the original equation. Substitute $ x = 3 $ into the left-hand side:

$$
\sqrt{3(3) - 5} = \sqrt{9 - 5} = \sqrt{4} = 2
$$

This matches the right-hand side of the original equation, so $ x = 3 $ is a valid solution.

---

### Step 3: Consider Domain Restrictions

The expression under the square root, $ 3x - 5 $, must be **non-negative** because the square root of a negative number is not defined in the real number system. So we require:

$$
3x - 5 \geq 0 \Rightarrow x \geq \frac{5}{3}
$$

Since $ x = 3 $ satisfies this inequality, it is within the domain of the original equation.

---

### Step 4: Check for Additional Solutions

The square root function is **one-to-one** and only yields **non-negative** outputs. Therefore, the equation $ \sqrt{3x - 5} = 2 $ has **only one solution** in the real numbers. There is no possibility of a negative solution or any other value that would satisfy the equation.

---

### Final Answer

$$
\boxed{3}
$$

Solution:
First, we start by squaring both sides of the equation \begin{align*} (\sqrt{3x-5})^2& =(2)^2
\\ \Rightarrow\qquad 3x-5& =4
\\\Rightarrow\qquad 3x& =9
\\\Rightarrow\qquad x& =\boxed{3}.
\end{align*}Testing, we find that this value of $x$ does indeed satisfy the equation.


Explanation:
Okay, so I need to figure out whether the composition of these functions, f(f(g(f(g(f(x)))))) is even, odd, or neither. Let me start by recalling what even and odd functions are.

An even function satisfies the condition that g(-x) = g(x) for all x in its domain. On the other hand, an odd function satisfies h(-x) = -h(x) for all x. So, the problem says that f(x) is odd and g(x) is even. Now, I need to analyze the composition of these functions.

Let me write down the composition step by step. The function given is f(f(g(f(g(f(x)))))). Let me parse this. Starting from the inside, it's f(x), then g(f(x)), then f(g(f(x))), then g(f(g(f(x)))), then f(g(f(g(f(x))))), and finally f(f(g(f(g(f(x)))))). So, there are multiple layers here.

Since the composition is quite complex, maybe I can work from the inside out, applying the properties of even and odd functions step by step. Let me try to substitute -x into the entire expression and see what happens.

Let me denote the entire function as H(x) = f(f(g(f(g(f(x)))))). I need to find H(-x) and compare it to H(x) or -H(x).

So, let's compute H(-x):

H(-x) = f(f(g(f(g(f(-x))))))

Now, let me work from the inside out. Let me start with f(-x). Since f is odd, f(-x) = -f(x).

So, replacing the innermost f(-x) with -f(x):

H(-x) = f(f(g(f(g(-f(x))))))

Now, moving one layer out: the next function is g applied to something. Let me look at the inner part: g(-f(x)). Since g is even, g(-y) = g(y) for any y. Therefore, g(-f(x)) = g(f(x)).

So substituting that in:

H(-x) = f(f(g(f(g(f(x))))))

Wait, hold on. Let me check that again. Let me go step by step.

Original expression for H(-x):

f(f(g(f(g(f(-x))))))

First, f(-x) = -f(x). So replacing that:

f(f(g(f(g(-f(x))))))

Now, the next layer is g(-f(x)). Since g is even, g(-f(x)) = g(f(x)). Therefore, replacing that:

f(f(g(f(g(f(x))))))

Wait, so now we have f(f(g(f(g(f(x)))))) which is exactly H(x). Therefore, H(-x) = H(x). So that would mean H(x) is even? But wait, let me check again because this seems too straightforward. Let me verify each step carefully.

Starting again:

H(-x) = f(f(g(f(g(f(-x))))))

First, compute f(-x) = -f(x). Therefore, the innermost function f(-x) becomes -f(x). So:

H(-x) = f(f(g(f(g(-f(x))))))

Now, moving one layer out: the next function is g applied to something. Let me look at the argument of g: it's g(-f(x)). Since g is even, g(-y) = g(y), so replacing y with f(x), we get g(-f(x)) = g(f(x)). Therefore:

H(-x) = f(f(g(f(g(f(x))))))

But wait, the expression inside is g(f(g(f(x))))? Wait, let me check the layers again. Let me write down the entire expression step by step:

Original H(-x):

f(f(g(f(g(f(-x))))))

First substitution: f(-x) = -f(x):

f(f(g(f(g(-f(x))))))

Now, the next layer is g(-f(x)) which becomes g(f(x)):

f(f(g(f(g(f(x))))))

Wait, so now the expression is f(f(g(f(g(f(x)))))) which is H(x). Therefore, H(-x) = H(x). Therefore, H is even? But that seems to suggest that the entire composition is even. But let me check again because maybe I made a mistake in the layers.

Wait, let me count the layers again. Let me write down the original function:

H(x) = f(f(g(f(g(f(x))))))

So, starting from the inside:

Start with f(x). Then, g(f(x)). Then f(g(f(x))). Then g(f(g(f(x)))). Then f(g(f(g(f(x))))). Then f(f(g(f(g(f(x)))))).

So, H(x) is f composed with f composed with g composed with f composed with g composed with f(x).

So, when we compute H(-x), we have:

Start with f(-x) = -f(x). Then, the next layer is g(f(-x)) = g(-f(x)) = g(f(x)) because g is even. Then, the next layer is f(g(f(-x))) = f(g(f(x))). Then, the next layer is g(f(g(f(-x)))) = g(f(g(f(x)))). Then, the next layer is f(g(f(g(f(-x))))) = f(g(f(g(f(x))))). Then, finally, the outermost f is applied: f(f(g(f(g(f(-x)))))) = f(f(g(f(g(f(x)))))) which is H(x). Therefore, H(-x) = H(x). Therefore, H is even.

Wait, so according to this, H(-x) = H(x), so H is even. But let me check again because sometimes when there are multiple layers, it's easy to miscount. Let me try to write each step with substitution:

Let me denote:

Let me define intermediate functions step by step. Let me define:

Let’s start from the innermost function:

Let’s define:

Let’s let’s call the innermost function as A(x) = f(x)

Then, the next layer is B(x) = g(A(x)) = g(f(x))

Then, the next layer is C(x) = f(B(x)) = f(g(f(x)))

Then, the next layer is D(x) = g(C(x)) = g(f(g(f(x))))

Then, the next layer is E(x) = f(D(x)) = f(g(f(g(f(x)))))

Then, the outermost layer is H(x) = f(E(x)) = f(f(g(f(g(f(x))))))

So, H(x) is as given. Now, let's compute H(-x):

Starting with A(-x) = f(-x) = -f(x)

Then, B(-x) = g(A(-x)) = g(-f(x)) = g(f(x)) because g is even.

Then, C(-x) = f(B(-x)) = f(g(f(x))) = C(x)

Then, D(-x) = g(C(-x)) = g(C(x)) = D(x)

Then, E(-x) = f(D(-x)) = f(D(x)) = E(x)

Then, H(-x) = f(E(-x)) = f(E(x)) = H(x)

Therefore, H(-x) = H(x), so H is even.

Wait, so according to this, the entire composition is even. But let me check again with another approach. Maybe I can think about the number of f and g functions and how their properties affect the composition.

Since f is odd and g is even, let's think about how composing them affects the parity.

Let me recall that composing an even function with an odd function:

If you have an even function g and an odd function f, then g(f(x)) is even? Let me check. Let me compute g(f(-x)) = g(-f(x)) because f is odd. Since g is even, g(-f(x)) = g(f(x)). Therefore, g(f(-x)) = g(f(x)), so g(f(x)) is even.

Similarly, if I have f(g(x)), since g is even, g(-x) = g(x). So f(g(-x)) = f(g(x)). But since f is odd, f(g(-x)) = f(g(x)) which is equal to f(g(x)). Wait, but that would mean that f(g(x)) is even? Wait, no. Wait, if I have f(g(-x)) = f(g(x)) because g(-x) = g(x). Therefore, f(g(-x)) = f(g(x)), so f(g(x)) is even? Wait, no. Wait, if f(g(-x)) = f(g(x)), then that means that the function f(g(x)) is even. Because replacing x with -x gives the same value. Therefore, f(g(x)) is even.

So, composing an odd function with an even function gives an even function.

Similarly, composing an even function with an odd function: Let me think. Suppose we have h(x) = f(g(x)), which we just saw is even. Then, if we compose another function.

But in our case, we have multiple layers. Let me try to analyze the composition step by step.

Starting from the innermost function:

Start with f(x). Then, the next layer is g(f(x)), which is even as we saw. Then, the next layer is f(g(f(x))). Since g(f(x)) is even, then applying f to it. Since f is odd, then f(even function). Let me think: if we have an even function h(x), then f(h(x)) is odd? Let me check. Let me compute f(h(-x)) = f(h(x)) because h is even. But since f is odd, f(h(-x)) = -f(h(x))? Wait, no. Wait, f(h(-x)) = f(h(x)) because h is even. But since f is odd, f(h(-x)) = -f(-h(x))? Wait, no. Wait, let me think again.

Let me take h(x) as even. Then, f(h(x)) is a function. Let me check if it's odd or even. Let me compute f(h(-x)) = f(h(x)) because h is even. Therefore, f(h(-x)) = f(h(x)). Therefore, if we have a function k(x) = f(h(x)), then k(-x) = f(h(-x)) = f(h(x)) = k(x). Therefore, k(x) is even. Wait, so if h(x) is even, then f(h(x)) is even? Wait, but f is odd. Wait, but if h(x) is even, then h(-x) = h(x). Then, f(h(-x)) = f(h(x)). Therefore, k(-x) = k(x). Therefore, k(x) is even. Therefore, composing an odd function with an even function gives an even function.

Wait, so if h(x) is even, then f(h(x)) is even? Wait, but f is odd. Let me take an example. Let me take f(x) = -x (which is odd), and h(x) = x² (which is even). Then, f(h(x)) = -x², which is even. Because -(-x)² = -x². So yes, it's even. Therefore, composing an odd function with an even function gives an even function.

So, going back. So, starting from the innermost function:

1. f(x) is odd.

2. Then, g(f(x)) is even, as we saw.

3. Then, f(g(f(x))) is odd? Wait, no. Wait, if we have f applied to an even function. Wait, if we have f applied to an even function, then as above, it's even? Wait, no. Wait, let me think again.

Wait, if we have a function h(x) which is even, then f(h(x)) is even? Wait, in the example above, yes. But let me check with another example. Let me take f(x) = -x (odd), h(x) = x² (even). Then f(h(x)) = -x², which is even. So yes, even.

But suppose f(x) is odd, and h(x) is even. Then f(h(x)) is even. Therefore, composing an odd function with an even function gives an even function.

Therefore, step 3: f(g(f(x))) is even.

Then, step 4: g(f(g(f(x)))). Since f(g(f(x))) is even, then applying g to it. Since g is even, then g(even function) is even? Wait, if we have an even function k(x), then g(k(x)) is even? Let me check.

If k(x) is even, then g(k(-x)) = g(k(x)) because k is even, and since g is even, g(k(-x)) = g(k(x)). Therefore, g(k(x)) is even. Therefore, composing an even function with an even function gives an even function.

Therefore, step 4: g(f(g(f(x)))) is even.

Then, step 5: f(g(f(g(f(x))))). Since the previous function is even, then applying f to it. As before, if we have an even function m(x), then f(m(x)) is even? Wait, no. Wait, earlier we saw that if h(x) is even, then f(h(x)) is even? Wait, no. Wait, in the previous case, when h(x) is even, f(h(x)) is even? Wait, in the example, yes. Wait, but let me think again.

If h(x) is even, then h(-x) = h(x). Then, f(h(-x)) = f(h(x)). But since f is odd, f(-y) = -f(y). However, here, h(-x) = h(x), so f(h(-x)) = f(h(x)). Therefore, if we have a function k(x) = f(h(x)), then k(-x) = f(h(-x)) = f(h(x)) = k(x). Therefore, k(x) is even. Therefore, yes, if h(x) is even, then f(h(x)) is even.

Therefore, step 5: f(g(f(g(f(x))))) is even.

Then, step 6: f(f(g(f(g(f(x)))))). So, applying f to an even function. So, again, if we have an even function n(x), then f(n(x)) is even? Wait, same as before. If n(x) is even, then f(n(-x)) = f(n(x)) because n(-x) = n(x). But since f is odd, f(n(-x)) = f(n(x)) which is equal to f(n(x)). Therefore, the function f(n(x)) is even. Therefore, step 6: f(f(g(f(g(f(x)))))) is even.

Therefore, the entire composition is even.

But wait, this seems to contradict my initial thought that maybe there could be some oddness. But according to this step-by-step analysis, each time we apply f or g, the parity is preserved or transformed in a way that the final result is even.

But let me check with a concrete example. Let me take specific functions for f and g. Let me choose f(x) = x (which is odd) and g(x) = x² (which is even). Then compute H(x) = f(f(g(f(g(f(x)))))) and check if it's even or odd.

Let me compute H(x):

First, f(x) = x.

Then, g(f(x)) = g(x) = x².

Then, f(g(f(x))) = f(x²) = x².

Then, g(f(g(f(x)))) = g(x²) = (x²)² = x⁴.

Then, f(g(f(g(f(x))))) = f(x⁴) = x⁴.

Then, H(x) = f(f(g(f(g(f(x))))) = f(x⁴) = x⁴.

So, H(x) = x⁴, which is even. Therefore, in this case, H(x) is even.

Another example: Let me take f(x) = -x (odd) and g(x) = x² (even). Then compute H(x):

Start with f(x) = -x.

Then, g(f(x)) = g(-x) = (-x)² = x².

Then, f(g(f(x))) = f(x²) = -x².

Then, g(f(g(f(x)))) = g(-x²) = (-x²)² = x⁴.

Then, f(g(f(g(f(x))))) = f(x⁴) = -x⁴.

Then, H(x) = f(f(g(f(g(f(x))))) = f(-x⁴) = -(-x⁴) = x⁴.

So, H(x) = x⁴, which is even.

Another example: Let me take f(x) = x³ (odd) and g(x) = x² (even). Then compute H(x):

Start with f(x) = x³.

g(f(x)) = g(x³) = (x³)² = x⁶.

f(g(f(x))) = f(x⁶) = (x⁶)³ = x¹⁸.

g(f(g(f(x)))) = g(x¹⁸) = (x¹⁸)² = x³⁶.

f(g(f(g(f(x))))) = f(x³⁶) = (x³⁶)³ = x¹⁰⁸.

H(x) = f(f(g(f(g(f(x))))) = f(x¹⁰⁸) = (x¹⁰⁸)³ = x³²⁴.

Which is even, since (-x)^324 = x^324.

So, in all these examples, H(x) is even. Therefore, this seems to confirm that the composition is even.

But let me try another example where maybe the exponents are different. Let me take f(x) = -x (odd) and g(x) = x² (even). Wait, I did that already.

Alternatively, take f(x) = sin(x) (odd) and g(x) = cos(x) (even). Let me compute H(x):

H(x) = f(f(g(f(g(f(x))))))

Compute step by step:

Start with f(x) = sin(x).

g(f(x)) = cos(sin(x)).

f(g(f(x))) = sin(cos(sin(x))).

g(f(g(f(x)))) = cos(sin(cos(sin(x)))).

f(g(f(g(f(x))))) = sin(cos(sin(cos(sin(x))))).

H(x) = f(f(g(f(g(f(x))))) = sin(sin(cos(sin(cos(sin(x)))))).

Now, let me check H(-x):

H(-x) = sin(sin(cos(sin(cos(sin(-x)))))).

But sin(-x) = -sin(x). Therefore:

H(-x) = sin(sin(cos(sin(cos(-sin(x)))))).

Now, cos(-sin(x)) = cos(sin(x)) because cosine is even. Then:

H(-x) = sin(sin(cos(sin(cos(sin(x)))))) which is equal to H(x). Therefore, H(-x) = H(x). Therefore, H is even.

So, even with trigonometric functions, it's even.

Therefore, all these examples suggest that H(x) is even.

But let me think again about the general case. Let me try to think about the number of times f is applied. Since f is odd, and g is even. Let me think about the composition.

Each time we have an odd function, it flips the sign, and even function preserves the sign.

But in the composition, starting from the inside, each time we apply f or g, we can track the parity.

But since we have multiple layers, maybe we can think about the number of odd functions applied.

But since the composition is f(f(g(f(g(f(x)))))), let me count the number of f and g functions.

Starting from the inside:

1. f(x) - odd

2. g(f(x)) - even

3. f(g(f(x))) - odd (since even function inside odd function)

Wait, hold on. Wait, earlier analysis said that f applied to even function is even? But in the example with f(x) = -x and g(x) = x², we saw that f(g(f(x))) = -x², which is even. Wait, but if you have an even function inside an odd function, is the result even or odd?

Wait, let me think. Let me take a function h(x) which is even. Then, f(h(x)) is odd? Or even?

Wait, if h(x) is even, then h(-x) = h(x). Then, f(h(-x)) = f(h(x)). But since f is odd, f(-y) = -f(y). However, here, h(-x) = h(x), so f(h(-x)) = f(h(x)). Therefore, if we define k(x) = f(h(x)), then k(-x) = f(h(-x)) = f(h(x)) = k(x). Therefore, k(x) is even. Therefore, f applied to an even function is even.

Therefore, if you have an even function inside an odd function, the result is even.

Wait, so in the previous step, step 3: f(g(f(x))) is f applied to an even function (since g(f(x)) is even), so f(even) is even.

Then step 4: g applied to that even function, so g(even) is even.

Step 5: f applied to even function, which is even.

Step 6: f applied to even function, which is even.

Therefore, the entire composition is even.

Therefore, regardless of the number of layers, as long as we start with an odd function and apply even and odd functions in a way that the number of odd functions... Wait, but in this case, the number of odd functions is 3? Let me check:

Original composition: f(f(g(f(g(f(x))))))

Breaking down:

Starting from the inside:

1. f(x) - odd

2. g(f(x)) - even

3. f(g(f(x))) - odd applied to even, which is even? Wait, no. Wait, step 3 is f(g(f(x))). Since g(f(x)) is even, then f applied to even is even. Therefore, step 3 is even.

Then step 4: g applied to even, which is even.

Step 5: f applied to even, which is even.

Step 6: f applied to even, which is even.

Therefore, the entire composition is even.

Therefore, regardless of the number of layers, as long as the composition ends with an even function, then applying f (odd) to it would result in even.

Therefore, the conclusion is that H(x) is even.

But let me check another example where maybe the number of odd functions is different. Suppose I have f(f(f(x))). Since f is odd, then f(f(f(x))) is odd? Let me check.

Let me take f(x) = -x. Then f(f(f(x))) = f(f(-x)) = f(-(-x)) = f(x) = -x. Which is odd.

But in our case, the composition is more complex. However, in our case, the number of times we apply f is 3? Wait, let me count:

Original composition: f(f(g(f(g(f(x))))))

Breaking down:

- The innermost is f(x)

- Then, g(f(x))

- Then, f(g(f(x)))

- Then, g(f(g(f(x))))

- Then, f(g(f(g(f(x)))))

- Then, f(f(g(f(g(f(x))))))

So, total number of f's: Let's count:

Starting from the innermost:

1. f(x)

2. g(f(x))

3. f(g(f(x)))

4. g(f(g(f(x))))

5. f(g(f(g(f(x)))))

6. f(f(g(f(g(f(x))))))

So, the number of f's is 3? Wait, no. Let me count:

Each time there's an f in the composition. Let me see:

The entire expression is f composed with f composed with g composed with f composed with g composed with f. So, the order is:

Start with f(x), then g, then f, then g, then f, then f. Wait, no. Wait, the composition is f(f(g(f(g(f(x)))))). So, starting from the inside:

First, f(x). Then, g(f(x)). Then, f(g(f(x))). Then, g(f(g(f(x)))). Then, f(g(f(g(f(x))))). Then, f(f(g(f(g(f(x)))))).

So, the number of f's is 3? Wait, no. Let me count:

- The first f is the innermost.

- Then, the next f is in the third position: f(g(f(x))).

- Then, the next f is in the fifth position: f(g(f(g(f(x))))).

- Then, the outermost f is the sixth position: f(f(...)).

Wait, so total number of f's is 4? Wait, no. Let me think again.

The entire function is f(f(g(f(g(f(x)))))). So, the number of f's is 4? Let me parse:

- The outermost is f(...)

- Then another f(...)

- Then inside that, g(...)

- Then f(...)

- Then g(...)

- Then f(...)

So, total of four f's? Let me count:

1. f(...)

2. f(...)

3. f(...)

4. f(...)

Yes, four f's. Wait, but in the expression f(f(g(f(g(f(x)))))), the number of f's is 4? Let me check:

Start from the inside:

f(x) is one.

Then, g(f(x)) is two functions: g and f.

Then, f(g(f(x))) is three functions: f, g, f.

Then, g(f(g(f(x)))) is four functions: g, f, g, f.

Then, f(g(f(g(f(x))))) is five functions: f, g, f, g, f.

Then, f(f(g(f(g(f(x))))) is six functions: f, f, g, f, g, f. Wait, no. Wait, the entire expression is f composed with f composed with g composed with f composed with g composed with f. So, the number of functions is 6? But the number of f's is 4? Let me count:

- The first f is the innermost.

- Then, the next f is in the third position.

- Then, the next f is in the fifth position.

- Then, the outermost f is the second f. Wait, no. Let me think of the composition as:

H(x) = f( f( g( f( g( f(x) ) ) ) ) )

So, starting from the inside:

1. f(x)

2. g(f(x))

3. f(g(f(x)))

4. g(f(g(f(x))))

5. f(g(f(g(f(x)))))

6. f(f(g(f(g(f(x))))))

So, the number of f's is 1 (innermost), then 3, then 5, then 6. Wait, no. The total number of f's is 4? Let me count:

- The innermost is f(x): 1

- Then, in step 3: f(g(f(x))): 2

- Then, in step 5: f(g(f(g(f(x))))): 3

- Then, in step 6: f(f(...)): 4

Yes, four f's. So, four f's and two g's.

But regardless of the number, the key is that each time we apply f or g, we can track the parity.

But since we saw through multiple examples and through the step-by-step substitution that H(-x) = H(x), then H is even.

Therefore, the answer should be even.

But let me check once more with another example where maybe the functions are more complex. Let me take f(x) = x^3 (odd) and g(x) = x^2 (even). Then compute H(x):

H(x) = f(f(g(f(g(f(x))))))

Compute step by step:

Start with f(x) = x³.

g(f(x)) = g(x³) = (x³)² = x⁶.

f(g(f(x))) = f(x⁶) = (x⁶)³ = x¹⁸.

g(f(g(f(x)))) = g(x¹⁸) = (x¹⁸)² = x³⁶.

f(g(f(g(f(x))))) = f(x³⁶) = (x³⁶)³ = x¹⁰⁸.

H(x) = f(f(g(f(g(f(x))))) = f(x¹⁰⁸) = (x¹⁰⁸)³ = x³²⁴.

Which is even, since (-x)^324 = x^324.

Therefore, again, H(x) is even.

Therefore, I think I can be confident that the composition is even.

But just to make sure, let me think if there is any case where this might not hold. Suppose that somewhere in the composition, the number of odd functions is odd, leading to an odd result. But in our case, the composition ends with two f's. Let me think about the parity transformations.

Each time we apply an odd function, it flips the sign, and even function preserves it.

But since we have multiple layers, let me think of the entire composition as a series of operations.

Starting with x, then applying f (odd), then g (even), then f (odd), then g (even), then f (odd), then f (odd).

Wait, let me think of the entire composition as a sequence of operations:

Starting with x:

1. Apply f: odd, so sign flips.

2. Apply g: even, sign remains.

3. Apply f: odd, sign flips.

4. Apply g: even, sign remains.

5. Apply f: odd, sign flips.

6. Apply f: odd, sign flips.

So, starting with x, after step 1: -x

After step 2: -x (since g is even)

After step 3: -(-x) = x

After step 4: x (since g is even)

After step 5: -x

After step 6: -(-x) = x

Therefore, overall, H(-x) = x, which is same as H(x) if H(x) is x? Wait, but in this case, H(x) would be x? Wait, no. Wait, this is a very rough analysis, assuming that each operation is just flipping the sign or not. But in reality, the functions are not just sign flips, but more complex. However, this line of thinking might help.

Wait, if we think of the composition as a series of sign flips and sign preservation.

Starting with x:

- Apply f (odd): -x

- Apply g (even): -x (since g is even, it doesn't change the sign)

- Apply f (odd): -(-x) = x

- Apply g (even): x

- Apply f (odd): -x

- Apply f (odd): -(-x) = x

Therefore, H(-x) = x, which is same as H(x) if H(x) is x. But in reality, H(x) is some function. However, this suggests that H(-x) = H(x). Therefore, H is even.

Therefore, even with this sign-flipping analysis, it seems that H(-x) = H(x).

Therefore, regardless of the specific functions (as long as f is odd and g is even), the composition H(x) is even.

Therefore, the answer should be even.

**Final Answer**
\boxed{even}

To determine whether the function
$$
f(f(g(f(g(f(x))))))
$$
is **even**, **odd**, or **neither**, we begin by analyzing the properties of the individual functions and how they behave under composition.

---

### Step 1: Understand the Function Properties

- **Odd function**: A function $ f $ is **odd** if
  $$
  f(-x) = -f(x)
  $$
- **Even function**: A function $ g $ is **even** if
  $$
  g(-x) = g(x)
  $$

We are told:
- $ f(x) $ is **odd**
- $ g(x) $ is **even**

---

### Step 2: Analyze the Composition

We are given the composition:
$$
H(x) = f(f(g(f(g(f(x))))))
$$

Let’s evaluate $ H(-x) $ and compare it to $ H(x) $ to determine the parity of $ H(x) $.

Start from the innermost function and work outward:

1. **Innermost**: $ f(x) $ is odd → $ f(-x) = -f(x) $
2. **Next layer**: $ g(f(x)) $
   Since $ f(x) $ is odd, $ f(-x) = -f(x) $, so
   $$
   g(f(-x)) = g(-f(x)) = g(f(x)) \quad \text{(since $ g $ is even)}
   $$
   So, $ g(f(x)) $ is **even**.

3. **Next layer**: $ f(g(f(x))) $
   Since $ g(f(x)) $ is even, and $ f $ is odd,
   $$
   f(g(-x)) = f(g(x)) \quad \text{(since $ g $ is even)}
   $$
   So, $ f(g(f(x))) $ is **even**.

4. **Next layer**: $ g(f(g(f(x)))) $
   Since $ f(g(f(x))) $ is even, and $ g $ is even,
   $$
   g(f(g(-x))) = g(f(g(x))) \quad \text{(since $ g $ is even)}
   $$
   So, $ g(f(g(f(x)))) $ is **even**.

5. **Next layer**: $ f(g(f(g(f(x))))) $
   Since $ g(f(g(f(x)))) $ is even, and $ f $ is odd,
   $$
   f(g(f(g(-x)))) = f(g(f(g(x)))) \quad \text{(since $ g $ is even)}
   $$
   So, $ f(g(f(g(f(x))))) $ is **even**.

6. **Outermost layer**: $ f(f(g(f(g(f(x))))) $
   Since $ f(g(f(g(f(x))))) $ is even, and $ f $ is odd,
   $$
   f(f(g(f(g(-x))))) = f(f(g(f(g(x))))) \quad \text{(since $ g $ is even)}
   $$
   So, the entire expression is **even**.

---

### Step 3: Conclusion

We have shown that:
$$
H(-x) = H(x)
$$
This means that the function $ H(x) = f(f(g(f(g(f(x)))))) $ is **even**.

---

### Final Answer

$$
\boxed{even}
$$

[Assistant]
Explanation:
Okay, so I need to figure out whether the composition of these functions, f(f(g(f(g(f(x)))))) is even, odd, or neither. Let me start by recalling what even and odd functions are.

An even function satisfies the condition that g(-x) = g(x) for all x in its domain. On the other hand, an odd function satisfies h(-x) = -h(x) for all x. So, the problem says that f(x) is odd and g(x) is even. Now, I need to analyze the composition of these functions.

Let me write down the composition step by step. The function given is f(f(g(f(g(f(x)))))). Let me parse this. Starting from the inside, it's f(x), then g(f(x)), then f(g(f(x))), then g(f(g(f(x)))), then f(g(f(g(f(x))))), and finally f(f(g(f(g(f(x)))))). So, there are multiple layers here.

Since the composition is quite complex, maybe I can work from the inside out, applying the properties of even and odd functions step by step. Let me try to substitute -x into the entire expression and see what happens.

Let me denote the entire function as H(x) = f(f(g(f(g(f(x)))))). I need to find H(-x) and compare it to H(x) or -H(x).

So, let's compute H(-x):

H(-x) = f(f(g(f(g(f(-x))))))

Now, let me work from the inside out. Let me start with f(-x). Since f is odd, f(-x) = -f(x).

So, replacing the innermost f(-x) with -f(x):

H(-x) = f(f(g(f(g(-f(x))))))

Now, moving one layer out: the next function is g applied to something. Let me look at the inner part: g(-f(x)). Since g is even, g(-y) = g(y) for any y. Therefore, g(-f(x)) = g(f(x)).

So substituting that in:

H(-x) = f(f(g(f(g(f(x))))))

Wait, hold on. Let me check that again. Let me go step by step.

Original expression for H(-x):

f(f(g(f(g(f(-x))))))

First, f(-x) = -f(x). So replacing that:

f(f(g(f(g(-f(x))))))

Now, the next layer is g(-f(x)). Since g is even, g(-f(x)) = g(f(x)). Therefore, replacing that:

f(f(g(f(g(f(x))))))

Wait, so now we have f(f(g(f(g(f(x)))))) which is exactly H(x). Therefore, H(-x) = H(x). So that would mean H(x) is even? But wait, let me check again because this seems too straightforward. Let me verify each step carefully.

Starting again:

H(-x) = f(f(g(f(g(f(-x))))))

First, compute f(-x) = -f(x). Therefore, the innermost function f(-x) becomes -f(x). So:

H(-x) = f(f(g(f(g(-f(x))))))

Now, moving one layer out: the next function is g applied to something. Let me look at the argument of g: it's g(-f(x)). Since g is even, g(-y) = g(y), so replacing y with f(x), we get g(-f(x)) = g(f(x)). Therefore:

H(-x) = f(f(g(f(g(f(x))))))

But wait, the expression inside is g(f(g(f(x))))? Wait, let me check the layers again. Let me write down the entire expression step by step:

Original H(-x):

f(f(g(f(g(f(-x))))))

First substitution: f(-x) = -f(x):

f(f(g(f(g(-f(x))))))

Now, the next layer is g(-f(x)) which becomes g(f(x)):

f(f(g(f(g(f(x))))))

Wait, so now the expression is f(f(g(f(g(f(x)))))) which is H(x). Therefore, H(-x) = H(x). Therefore, H is even? But that seems to suggest that the entire composition is even. But let me check again because maybe I made a mistake in the layers.

Wait, let me count the layers again. Let me write down the original function:

H(x) = f(f(g(f(g(f(x))))))

So, starting from the inside:

Start with f(x). Then, g(f(x)). Then f(g(f(x))). Then g(f(g(f(x)))). Then f(g(f(g(f(x))))). Then f(f(g(f(g(f(x)))))).

So, H(x) is f composed with f composed with g composed with f composed with g composed with f(x).

So, when we compute H(-x), we have:

Start with f(-x) = -f(x). Then, the next layer is g(f(-x)) = g(-f(x)) = g(f(x)) because g is even. Then, the next layer is f(g(f(-x))) = f(g(f(x))). Then, the next layer is g(f(g(f(-x)))) = g(f(g(f(x)))). Then, the next layer is f(g(f(g(f(-x))))) = f(g(f(g(f(x))))). Then, finally, the outermost f is applied: f(f(g(f(g(f(-x)))))) = f(f(g(f(g(f(x)))))) which is H(x). Therefore, H(-x) = H(x). Therefore, H is even.

Wait, so according to this, H(-x) = H(x), so H is even. But let me check again because sometimes when there are multiple layers, it's easy to miscount. Let me try to write each step with substitution:

Let me denote:

Let me define intermediate functions step by step. Let me define:

Let’s start from the innermost function:

Let’s define:

Let’s let’s call the innermost function as A(x) = f(x)

Then, the next layer is B(x) = g(A(x)) = g(f(x))

Then, the next layer is C(x) = f(B(x)) = f(g(f(x)))

Then, the next layer is D(x) = g(C(x)) = g(f(g(f(x))))

Then, the next layer is E(x) = f(D(x)) = f(g(f(g(f(x)))))

Then, the outermost layer is H(x) = f(E(x)) = f(f(g(f(g(f(x))))))

So, H(x) is as given. Now, let's compute H(-x):

Starting with A(-x) = f(-x) = -f(x)

Then, B(-x) = g(A(-x)) = g(-f(x)) = g(f(x)) because g is even.

Then, C(-x) = f(B(-x)) = f(g(f(x))) = C(x)

Then, D(-x) = g(C(-x)) = g(C(x)) = D(x)

Then, E(-x) = f(D(-x)) = f(D(x)) = E(x)

Then, H(-x) = f(E(-x)) = f(E(x)) = H(x)

Therefore, H(-x) = H(x), so H is even.

Wait, so according to this, the entire composition is even. But let me check again with another approach. Maybe I can think about the number of f and g functions and how their properties affect the composition.

Since f is odd and g is even, let's think about how composing them affects the parity.

Let me recall that composing an even function with an odd function:

If you have an even function g and an odd function f, then g(f(x)) is even? Let me check. Let me compute g(f(-x)) = g(-f(x)) because f is odd. Since g is even, g(-f(x)) = g(f(x)). Therefore, g(f(-x)) = g(f(x)), so g(f(x)) is even.

Similarly, if I have f(g(x)), since g is even, g(-x) = g(x). So f(g(-x)) = f(g(x)). But since f is odd, f(g(-x)) = f(g(x)) which is equal to f(g(x)). Wait, but that would mean that f(g(x)) is even? Wait, no. Wait, if I have f(g(-x)) = f(g(x)) because g(-x) = g(x). Therefore, f(g(-x)) = f(g(x)), so f(g(x)) is even? Wait, no. Wait, if f(g(-x)) = f(g(x)), then that means that the function f(g(x)) is even. Because replacing x with -x gives the same value. Therefore, f(g(x)) is even.

So, composing an odd function with an even function gives an even function.

Similarly, composing an even function with an odd function: Let me think. Suppose we have h(x) = f(g(x)), which we just saw is even. Then, if we compose another function.

But in our case, we have multiple layers. Let me try to analyze the composition step by step.

Starting from the innermost function:

Start with f(x). Then, the next layer is g(f(x)), which is even as we saw. Then, the next layer is f(g(f(x))). Since g(f(x)) is even, then applying f to it. Since f is odd, then f(even function). Let me think: if we have an even function h(x), then f(h(x)) is odd? Let me check. Let me compute f(h(-x)) = f(h(x)) because h is even. But since f is odd, f(h(-x)) = -f(h(x))? Wait, no. Wait, f(h(-x)) = f(h(x)) because h is even. But since f is odd, f(h(-x)) = -f(-h(x))? Wait, no. Wait, let me think again.

Let me take h(x) as even. Then, f(h(x)) is a function. Let me check if it's odd or even. Let me compute f(h(-x)) = f(h(x)) because h is even. Therefore, f(h(-x)) = f(h(x)). Therefore, if we have a function k(x) = f(h(x)), then k(-x) = f(h(-x)) = f(h(x)) = k(x). Therefore, k(x) is even. Wait, so if h(x) is even, then f(h(x)) is even? Wait, but f is odd. Wait, but if h(x) is even, then h(-x) = h(x). Then, f(h(-x)) = f(h(x)). Therefore, k(-x) = k(x). Therefore, k(x) is even. Therefore, composing an odd function with an even function gives an even function.

Wait, so if h(x) is even, then f(h(x)) is even? Wait, but f is odd. Let me take an example. Let me take f(x) = -x (which is odd), and h(x) = x² (which is even). Then, f(h(x)) = -x², which is even. Because -(-x)² = -x². So yes, it's even. Therefore, composing an odd function with an even function gives an even function.

So, going back. So, starting from the innermost function:

1. f(x) is odd.

2. Then, g(f(x)) is even, as we saw.

3. Then, f(g(f(x))) is odd? Wait, no. Wait, if we have f applied to an even function. Wait, if we have f applied to an even function, then as above, it's even? Wait, no. Wait, let me think again.

Wait, if we have a function h(x) which is even, then f(h(x)) is even? Wait, in the example above, yes. But let me check with another example. Let me take f(x) = -x (odd), h(x) = x² (even). Then f(h(x)) = -x², which is even. So yes, even.

But suppose f(x) is odd, and h(x) is even. Then f(h(x)) is even. Therefore, composing an odd function with an even function gives an even function.

Therefore, step 3: f(g(f(x))) is even.

Then, step 4: g(f(g(f(x)))). Since f(g(f(x))) is even, then applying g to it. Since g is even, then g(even function) is even? Wait, if we have an even function k(x), then g(k(x)) is even? Let me check.

If k(x) is even, then g(k(-x)) = g(k(x)) because k is even, and since g is even, g(k(-x)) = g(k(x)). Therefore, g(k(x)) is even. Therefore, composing an even function with an even function gives an even function.

Therefore, step 4: g(f(g(f(x)))) is even.

Then, step 5: f(g(f(g(f(x))))). Since the previous function is even, then applying f to it. As before, if we have an even function m(x), then f(m(x)) is even? Wait, no. Wait, earlier we saw that if h(x) is even, then f(h(x)) is even? Wait, no. Wait, in the previous case, when h(x) is even, f(h(x)) is even? Wait, in the example, yes. Wait, but let me think again.

If h(x) is even, then h(-x) = h(x). Then, f(h(-x)) = f(h(x)). But since f is odd, f(-y) = -f(y). However, here, h(-x) = h(x), so f(h(-x)) = f(h(x)). Therefore, if we have a function k(x) = f(h(x)), then k(-x) = f(h(-x)) = f(h(x)) = k(x). Therefore, k(x) is even. Therefore, yes, if h(x) is even, then f(h(x)) is even.

Therefore, step 5: f(g(f(g(f(x))))) is even.

Then, step 6: f(f(g(f(g(f(x)))))). So, applying f to an even function. So, again, if we have an even function n(x), then f(n(x)) is even? Wait, same as before. If n(x) is even, then f(n(-x)) = f(n(x)) because n(-x) = n(x). But since f is odd, f(n(-x)) = f(n(x)) which is equal to f(n(x)). Therefore, the function f(n(x)) is even. Therefore, step 6: f(f(g(f(g(f(x)))))) is even.

Therefore, the entire composition is even.

But wait, this seems to contradict my initial thought that maybe there could be some oddness. But according to this step-by-step analysis, each time we apply f or g, the parity is preserved or transformed in a way that the final result is even.

But let me check with a concrete example. Let me take specific functions for f and g. Let me choose f(x) = x (which is odd) and g(x) = x² (which is even). Then compute H(x) = f(f(g(f(g(f(x)))))) and check if it's even or odd.

Let me compute H(x):

First, f(x) = x.

Then, g(f(x)) = g(x) = x².

Then, f(g(f(x))) = f(x²) = x².

Then, g(f(g(f(x)))) = g(x²) = (x²)² = x⁴.

Then, f(g(f(g(f(x))))) = f(x⁴) = x⁴.

Then, H(x) = f(f(g(f(g(f(x))))) = f(x⁴) = x⁴.

So, H(x) = x⁴, which is even. Therefore, in this case, H(x) is even.

Another example: Let me take f(x) = -x (odd) and g(x) = x² (even). Then compute H(x):

Start with f(x) = -x.

Then, g(f(x)) = g(-x) = (-x)² = x².

Then, f(g(f(x))) = f(x²) = -x².

Then, g(f(g(f(x)))) = g(-x²) = (-x²)² = x⁴.

Then, f(g(f(g(f(x))))) = f(x⁴) = -x⁴.

Then, H(x) = f(f(g(f(g(f(x))))) = f(-x⁴) = -(-x⁴) = x⁴.

So, H(x) = x⁴, which is even.

Another example: Let me take f(x) = x³ (odd) and g(x) = x² (even). Then compute H(x):

Start with f(x) = x³.

g(f(x)) = g(x³) = (x³)² = x⁶.

f(g(f(x))) = f(x⁶) = (x⁶)³ = x¹⁸.

g(f(g(f(x)))) = g(x¹⁸) = (x¹⁸)² = x³⁶.

f(g(f(g(f(x))))) = f(x³⁶) = (x³⁶)³ = x¹⁰⁸.

H(x) = f(f(g(f(g(f(x))))) = f(x¹⁰⁸) = (x¹⁰⁸)³ = x³²⁴.

Which is even, since (-x)^324 = x^324.

So, in all these examples, H(x) is even. Therefore, this seems to confirm that the composition is even.

But let me try another example where maybe the exponents are different. Let me take f(x) = -x (odd) and g(x) = x² (even). Wait, I did that already.

Alternatively, take f(x) = sin(x) (odd) and g(x) = cos(x) (even). Let me compute H(x):

H(x) = f(f(g(f(g(f(x))))))

Compute step by step:

Start with f(x) = sin(x).

g(f(x)) = cos(sin(x)).

f(g(f(x))) = sin(cos(sin(x))).

g(f(g(f(x)))) = cos(sin(cos(sin(x)))).

f(g(f(g(f(x))))) = sin(cos(sin(cos(sin(x))))).

H(x) = f(f(g(f(g(f(x))))) = sin(sin(cos(sin(cos(sin(x)))))).

Now, let me check H(-x):

H(-x) = sin(sin(cos(sin(cos(sin(-x)))))).

But sin(-x) = -sin(x). Therefore:

H(-x) = sin(sin(cos(sin(cos(-sin(x)))))).

Now, cos(-sin(x)) = cos(sin(x)) because cosine is even. Then:

H(-x) = sin(sin(cos(sin(cos(sin(x)))))) which is equal to H(x). Therefore, H(-x) = H(x). Therefore, H is even.

So, even with trigonometric functions, it's even.

Therefore, all these examples suggest that H(x) is even.

But let me think again about the general case. Let me try to think about the number of times f is applied. Since f is odd, and g is even. Let me think about the composition.

Each time we have an odd function, it flips the sign, and even function preserves the sign.

But in the composition, starting from the inside, each time we apply f or g, we can track the parity.

But since we have multiple layers, maybe we can think about the number of odd functions applied.

But since the composition is f(f(g(f(g(f(x)))))), let me count the number of f and g functions.

Starting from the inside:

1. f(x) - odd

2. g(f(x)) - even

3. f(g(f(x))) - odd (since even function inside odd function)

Wait, hold on. Wait, earlier analysis said that f applied to even function is even? But in the example with f(x) = -x and g(x) = x², we saw that f(g(f(x))) = -x², which is even. Wait, but if you have an even function inside an odd function, is the result even or odd?

Wait, let me think. Let me take a function h(x) which is even. Then, f(h(x)) is odd? Or even?

Wait, if h(x) is even, then h(-x) = h(x). Then, f(h(-x)) = f(h(x)). But since f is odd, f(-y) = -f(y). However, here, h(-x) = h(x), so f(h(-x)) = f(h(x)). Therefore, if we define k(x) = f(h(x)), then k(-x) = f(h(-x)) = f(h(x)) = k(x). Therefore, k(x) is even. Therefore, f applied to an even function is even.

Therefore, if you have an even function inside an odd function, the result is even.

Wait, so in the previous step, step 3: f(g(f(x))) is f applied to an even function (since g(f(x)) is even), so f(even) is even.

Then step 4: g applied to that even function, so g(even) is even.

Step 5: f applied to even function, which is even.

Step 6: f applied to even function, which is even.

Therefore, the entire composition is even.

Therefore, regardless of the number of layers, as long as we start with an odd function and apply even and odd functions in a way that the number of odd functions... Wait, but in this case, the number of odd functions is 3? Let me check:

Original composition: f(f(g(f(g(f(x))))))

Breaking down:

Starting from the inside:

1. f(x) - odd

2. g(f(x)) - even

3. f(g(f(x))) - odd applied to even, which is even? Wait, no. Wait, step 3 is f(g(f(x))). Since g(f(x)) is even, then f applied to even is even. Therefore, step 3 is even.

Then step 4: g applied to even, which is even.

Step 5: f applied to even, which is even.

Step 6: f applied to even, which is even.

Therefore, the entire composition is even.

Therefore, regardless of the number of layers, as long as the composition ends with an even function, then applying f (odd) to it would result in even.

Therefore, the conclusion is that H(x) is even.

But let me check another example where maybe the number of odd functions is different. Suppose I have f(f(f(x))). Since f is odd, then f(f(f(x))) is odd? Let me check.

Let me take f(x) = -x. Then f(f(f(x))) = f(f(-x)) = f(-(-x)) = f(x) = -x. Which is odd.

But in our case, the composition is more complex. However, in our case, the number of times we apply f is 3? Wait, let me count:

Original composition: f(f(g(f(g(f(x))))))

Breaking down:

- The innermost is f(x)

- Then, g(f(x))

- Then, f(g(f(x)))

- Then, g(f(g(f(x))))

- Then, f(g(f(g(f(x)))))

- Then, f(f(g(f(g(f(x))))))

So, total number of f's: Let's count:

Starting from the innermost:

1. f(x)

2. g(f(x))

3. f(g(f(x)))

4. g(f(g(f(x))))

5. f(g(f(g(f(x)))))

6. f(f(g(f(g(f(x))))))

So, the number of f's is 3? Wait, no. Let me count:

Each time there's an f in the composition. Let me see:

The entire expression is f composed with f composed with g composed with f composed with g composed with f. So, the order is:

Start with f(x), then g, then f, then g, then f, then f. Wait, no. Wait, the composition is f(f(g(f(g(f(x)))))). So, starting from the inside:

First, f(x). Then, g(f(x)). Then, f(g(f(x))). Then, g(f(g(f(x)))). Then, f(g(f(g(f(x))))). Then, f(f(g(f(g(f(x)))))).

So, the number of f's is 3? Wait, no. Let me count:

- The first f is the innermost.

- Then, the next f is in the third position: f(g(f(x))).

- Then, the next f is in the fifth position: f(g(f(g(f(x))))).

- Then, the outermost f is the sixth position: f(f(...)).

Wait, so total number of f's is 4? Wait, no. Let me think again.

The entire function is f(f(g(f(g(f(x)))))). So, the number of f's is 4? Let me parse:

- The outermost is f(...)

- Then another f(...)

- Then inside that, g(...)

- Then f(...)

- Then g(...)

- Then f(...)

So, total of four f's? Let me count:

1. f(...)

2. f(...)

3. f(...)

4. f(...)

Yes, four f's. Wait, but in the expression f(f(g(f(g(f(x)))))), the number of f's is 4? Let me check:

Start from the inside:

f(x) is one.

Then, g(f(x)) is two functions: g and f.

Then, f(g(f(x))) is three functions: f, g, f.

Then, g(f(g(f(x)))) is four functions: g, f, g, f.

Then, f(g(f(g(f(x))))) is five functions: f, g, f, g, f.

Then, f(f(g(f(g(f(x))))) is six functions: f, f, g, f, g, f. Wait, no. Wait, the entire expression is f composed with f composed with g composed with f composed with g composed with f. So, the number of functions is 6? But the number of f's is 4? Let me count:

- The first f is the innermost.

- Then, the next f is in the third position.

- Then, the next f is in the fifth position.

- Then, the outermost f is the second f. Wait, no. Let me think of the composition as:

H(x) = f( f( g( f( g( f(x) ) ) ) ) )

So, starting from the inside:

1. f(x)

2. g(f(x))

3. f(g(f(x)))

4. g(f(g(f(x))))

5. f(g(f(g(f(x)))))

6. f(f(g(f(g(f(x))))))

So, the number of f's is 1 (innermost), then 3, then 5, then 6. Wait, no. The total number of f's is 4? Let me count:

- The innermost is f(x): 1

- Then, in step 3: f(g(f(x))): 2

- Then, in step 5: f(g(f(g(f(x))))): 3

- Then, in step 6: f(f(...)): 4

Yes, four f's. So, four f's and two g's.

But regardless of the number, the key is that each time we apply f or g, we can track the parity.

But since we saw through multiple examples and through the step-by-step substitution that H(-x) = H(x), then H is even.

Therefore, the answer should be even.

But let me check once more with another example where maybe the functions are more complex. Let me take f(x) = x^3 (odd) and g(x) = x^2 (even). Then compute H(x):

H(x) = f(f(g(f(g(f(x))))))

Compute step by step:

Start with f(x) = x³.

g(f(x)) = g(x³) = (x³)² = x⁶.

f(g(f(x))) = f(x⁶) = (x⁶)³ = x¹⁸.

g(f(g(f(x)))) = g(x¹⁸) = (x¹⁸)² = x³⁶.

f(g(f(g(f(x))))) = f(x³⁶) = (x³⁶)³ = x¹⁰⁸.

H(x) = f(f(g(f(g(f(x))))) = f(x¹⁰⁸) = (x¹⁰⁸)³ = x³²⁴.

Which is even, since (-x)^324 = x^324.

Therefore, again, H(x) is even.

Therefore, I think I can be confident that the composition is even.

But just to make sure, let me think if there is any case where this might not hold. Suppose that somewhere in the composition, the number of odd functions is odd, leading to an odd result. But in our case, the composition ends with two f's. Let me think about the parity transformations.

Each time we apply an odd function, it flips the sign, and even function preserves it.

But since we have multiple layers, let me think of the entire composition as a series of operations.

Starting with x, then applying f (odd), then g (even), then f (odd), then g (even), then f (odd), then f (odd).

Wait, let me think of the entire composition as a sequence of operations:

Starting with x:

1. Apply f: odd, so sign flips.

2. Apply g: even, sign remains.

3. Apply f: odd, sign flips.

4. Apply g: even, sign remains.

5. Apply f: odd, sign flips.

6. Apply f: odd, sign flips.

So, starting with x, after step 1: -x

After step 2: -x (since g is even)

After step 3: -(-x) = x

After step 4: x (since g is even)

After step 5: -x

After step 6: -(-x) = x

Therefore, overall, H(-x) = x, which is same as H(x) if H(x) is x? Wait, but in this case, H(x) would be x? Wait, no. Wait, this is a very rough analysis, assuming that each operation is just flipping the sign or not. But in reality, the functions are not just sign flips, but more complex. However, this line of thinking might help.

Wait, if we think of the composition as a series of sign flips and sign preservation.

Starting with x:

- Apply f (odd): -x

- Apply g (even): -x (since g is even, it doesn't change the sign)

- Apply f (odd): -(-x) = x

- Apply g (even): x

- Apply f (odd): -x

- Apply f (odd): -(-x) = x

Therefore, H(-x) = x, which is same as H(x) if H(x) is x. But in reality, H(x) is some function. However, this suggests that H(-x) = H(x). Therefore, H is even.

Therefore, even with this sign-flipping analysis, it seems that H(-x) = H(x).

Therefore, regardless of the specific functions (as long as f is odd and g is even), the composition H(x) is even.

Therefore, the answer should be even.

**Final Answer**
\boxed{even}

To determine whether the function
$$
f(f(g(f(g(f(x))))))
$$
is **even**, **odd**, or **neither**, we begin by analyzing the properties of the individual functions and how they behave under composition.

---

### Step 1: Understand the Function Properties

- **Odd function**: A function $ f $ is **odd** if
  $$
  f(-x) = -f(x)
  $$
- **Even function**: A function $ g $ is **even** if
  $$
  g(-x) = g(x)
  $$

We are told:
- $ f(x) $ is **odd**
- $ g(x) $ is **even**

---

### Step 2: Analyze the Composition

We are given the composition:
$$
H(x) = f(f(g(f(g(f(x))))))
$$

Let’s evaluate $ H(-x) $ and compare it to $ H(x) $ to determine the parity of $ H(x) $.

Start from the innermost function and work outward:

1. **Innermost**: $ f(x) $ is odd → $ f(-x) = -f(x) $
2. **Next layer**: $ g(f(x)) $
   Since $ f(x) $ is odd, $ f(-x) = -f(x) $, so
   $$
   g(f(-x)) = g(-f(x)) = g(f(x)) \quad \text{(since $ g $ is even)}
   $$
   So, $ g(f(x)) $ is **even**.

3. **Next layer**: $ f(g(f(x))) $
   Since $ g(f(x)) $ is even, and $ f $ is odd,
   $$
   f(g(-x)) = f(g(x)) \quad \text{(since $ g $ is even)}
   $$
   So, $ f(g(f(x))) $ is **even**.

4. **Next layer**: $ g(f(g(f(x)))) $
   Since $ f(g(f(x))) $ is even, and $ g $ is even,
   $$
   g(f(g(-x))) = g(f(g(x))) \quad \text{(since $ g $ is even)}
   $$
   So, $ g(f(g(f(x)))) $ is **even**.

5. **Next layer**: $ f(g(f(g(f(x))))) $
   Since $ g(f(g(f(x)))) $ is even, and $ f $ is odd,
   $$
   f(g(f(g(-x)))) = f(g(f(g(x)))) \quad \text{(since $ g $ is even)}
   $$
   So, $ f(g(f(g(f(x))))) $ is **even**.

6. **Outermost layer**: $ f(f(g(f(g(f(x))))) $
   Since $ f(g(f(g(f(x))))) $ is even, and $ f $ is odd,
   $$
   f(f(g(f(g(-x))))) = f(f(g(f(g(x))))) \quad \text{(since $ g $ is even)}
   $$
   So, the entire expression is **even**.

---

### Step 3: Conclusion

We have shown that:
$$
H(-x) = H(x)
$$
This means that the function $ H(x) = f(f(g(f(g(f(x)))))) $ is **even**.

---

### Final Answer

$$
\boxed{even}
$$

Solution:
We have that
\[f(f(g(f(g(f(-x)))))) = f(f(g(f(g(-f(x)))))) = f(f(g(f(g(f(x)))))),\]so the function is $\boxed{\text{even}}.$

More generally, if we have a composition of functions, and at least one of the functions is even, then the whole composition of functions is even.


Explanation:
Okay, so I need to solve the equation (x/2) + (x/3) = 5. Hmm, let me think about how to approach this. I remember that when you have fractions in an equation, it's often helpful to eliminate them by finding a common denominator or multiplying both sides by the least common multiple of the denominators. Let me try that.

First, let me identify the denominators here. They are 2 and 3. The least common multiple of 2 and 3 is 6, right? So if I multiply both sides of the equation by 6, that should eliminate the fractions. Let me write that down:

6 * [(x/2) + (x/3)] = 6 * 5

Now, distributing the 6 to each term inside the brackets:

6*(x/2) + 6*(x/3) = 30

Calculating each term:

For the first term, 6 divided by 2 is 3, so 3x. For the second term, 6 divided by 3 is 2, so 2x. So the equation becomes:

3x + 2x = 30

Combining like terms:

3x + 2x is 5x, so:

5x = 30

Now, to solve for x, I just need to divide both sides by 5:

x = 30 / 5

Which simplifies to:

x = 6

Wait, let me check if that makes sense. Let me substitute x = 6 back into the original equation to verify.

Original equation: (6/2) + (6/3) = 5

Calculating each term:

6 divided by 2 is 3, and 6 divided by 3 is 2. Adding them together: 3 + 2 = 5. Which matches the right side of the equation. So that checks out.

But just to make sure I didn't make any mistakes in my steps, let me go through them again. Starting with multiplying both sides by 6. That's a valid step because 6 is not zero, so it's allowed. Then distributing the 6 correctly: 6*(x/2) is indeed 3x, and 6*(x/3) is 2x. Adding those gives 5x, which equals 30. Dividing by 5 gives x = 6. Yep, that all seems correct.

Alternatively, maybe I can solve it another way to confirm. Let me try combining the fractions first before multiplying. The original equation is (x/2) + (x/3) = 5. To combine the fractions, I need a common denominator, which is 6 again. So converting each term:

x/2 is equivalent to 3x/6, and x/3 is equivalent to 2x/6. Adding them together:

3x/6 + 2x/6 = (3x + 2x)/6 = 5x/6

So the equation becomes:

5x/6 = 5

Then, solving for x, multiply both sides by 6:

5x = 30

Divide by 5:

x = 6

Same result. So that confirms it again. So whether I eliminate the denominators first or combine the fractions first, I end up with x = 6.

Another way to think about it: maybe using decimal equivalents? Let me try that. Let's convert the fractions to decimals. 1/2 is 0.5 and 1/3 is approximately 0.333... So the equation becomes:

0.5x + 0.333...x = 5

Adding those decimals: 0.5 + 0.333... is approximately 0.833... So:

0.833...x = 5

Then, solving for x: x = 5 / 0.833...

Calculating that, 5 divided by 0.833... Well, 0.833... is 5/6, right? Because 5 divided by 6 is approximately 0.833... So 5 divided by (5/6) is 5 * (6/5) = 6. So again, x = 6.

So all methods lead to the same answer. Therefore, I can be confident that x = 6 is the correct solution.

I think that's thorough enough. I considered different methods to solve the equation and verified the answer by substitution. All steps check out, so I don't see any mistakes here.

**Final Answer**
The solution is \boxed{6}.

To solve the equation:

$$
\frac{x}{2} + \frac{x}{3} = 5
$$

we aim to isolate $ x $ by eliminating the fractions and simplifying the expression.

---

### Step 1: Eliminate the Denominators

The denominators in the equation are 2 and 3. The least common multiple (LCM) of 2 and 3 is 6. To eliminate the fractions, we multiply both sides of the equation by 6:

$$
6 \left( \frac{x}{2} + \frac{x}{3} \right) = 6 \cdot 5
$$

Distribute the 6 across the terms inside the parentheses:

$$
6 \cdot \frac{x}{2} + 6 \cdot \frac{x}{3} = 30
$$

Simplify each term:

$$
3x + 2x = 30
$$

---

### Step 2: Combine Like Terms

$$
5x = 30
$$

---

### Step 3: Solve for $ x $

Divide both sides by 5:

$$
x = \frac{30}{5} = 6
$$

---

### Step 4: Verify the Solution

Substitute $ x = 6 $ back into the original equation:

$$
\frac{6}{2} + \frac{6}{3} = 3 + 2 = 5
$$

This matches the right-hand side of the equation, confirming that the solution is correct.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation (x/2) + (x/3) = 5. Hmm, let me think about how to approach this. I remember that when you have fractions in an equation, it's often helpful to eliminate them by finding a common denominator or multiplying both sides by the least common multiple of the denominators. Let me try that.

First, let me identify the denominators here. They are 2 and 3. The least common multiple of 2 and 3 is 6, right? So if I multiply both sides of the equation by 6, that should eliminate the fractions. Let me write that down:

6 * [(x/2) + (x/3)] = 6 * 5

Now, distributing the 6 to each term inside the brackets:

6*(x/2) + 6*(x/3) = 30

Calculating each term:

For the first term, 6 divided by 2 is 3, so 3x. For the second term, 6 divided by 3 is 2, so 2x. So the equation becomes:

3x + 2x = 30

Combining like terms:

3x + 2x is 5x, so:

5x = 30

Now, to solve for x, I just need to divide both sides by 5:

x = 30 / 5

Which simplifies to:

x = 6

Wait, let me check if that makes sense. Let me substitute x = 6 back into the original equation to verify.

Original equation: (6/2) + (6/3) = 5

Calculating each term:

6 divided by 2 is 3, and 6 divided by 3 is 2. Adding them together: 3 + 2 = 5. Which matches the right side of the equation. So that checks out.

But just to make sure I didn't make any mistakes in my steps, let me go through them again. Starting with multiplying both sides by 6. That's a valid step because 6 is not zero, so it's allowed. Then distributing the 6 correctly: 6*(x/2) is indeed 3x, and 6*(x/3) is 2x. Adding those gives 5x, which equals 30. Dividing by 5 gives x = 6. Yep, that all seems correct.

Alternatively, maybe I can solve it another way to confirm. Let me try combining the fractions first before multiplying. The original equation is (x/2) + (x/3) = 5. To combine the fractions, I need a common denominator, which is 6 again. So converting each term:

x/2 is equivalent to 3x/6, and x/3 is equivalent to 2x/6. Adding them together:

3x/6 + 2x/6 = (3x + 2x)/6 = 5x/6

So the equation becomes:

5x/6 = 5

Then, solving for x, multiply both sides by 6:

5x = 30

Divide by 5:

x = 6

Same result. So that confirms it again. So whether I eliminate the denominators first or combine the fractions first, I end up with x = 6.

Another way to think about it: maybe using decimal equivalents? Let me try that. Let's convert the fractions to decimals. 1/2 is 0.5 and 1/3 is approximately 0.333... So the equation becomes:

0.5x + 0.333...x = 5

Adding those decimals: 0.5 + 0.333... is approximately 0.833... So:

0.833...x = 5

Then, solving for x: x = 5 / 0.833...

Calculating that, 5 divided by 0.833... Well, 0.833... is 5/6, right? Because 5 divided by 6 is approximately 0.833... So 5 divided by (5/6) is 5 * (6/5) = 6. So again, x = 6.

So all methods lead to the same answer. Therefore, I can be confident that x = 6 is the correct solution.

I think that's thorough enough. I considered different methods to solve the equation and verified the answer by substitution. All steps check out, so I don't see any mistakes here.

**Final Answer**
The solution is \boxed{6}.

To solve the equation:

$$
\frac{x}{2} + \frac{x}{3} = 5
$$

we aim to isolate $ x $ by eliminating the fractions and simplifying the expression.

---

### Step 1: Eliminate the Denominators

The denominators in the equation are 2 and 3. The least common multiple (LCM) of 2 and 3 is 6. To eliminate the fractions, we multiply both sides of the equation by 6:

$$
6 \left( \frac{x}{2} + \frac{x}{3} \right) = 6 \cdot 5
$$

Distribute the 6 across the terms inside the parentheses:

$$
6 \cdot \frac{x}{2} + 6 \cdot \frac{x}{3} = 30
$$

Simplify each term:

$$
3x + 2x = 30
$$

---

### Step 2: Combine Like Terms

$$
5x = 30
$$

---

### Step 3: Solve for $ x $

Divide both sides by 5:

$$
x = \frac{30}{5} = 6
$$

---

### Step 4: Verify the Solution

Substitute $ x = 6 $ back into the original equation:

$$
\frac{6}{2} + \frac{6}{3} = 3 + 2 = 5
$$

This matches the right-hand side of the equation, confirming that the solution is correct.

---

### Final Answer

$$
\boxed{6}
$$

Solution: Writing the left-hand side with a common denominator, we have$$\dfrac{x}{2} + \dfrac{x}{3} = \dfrac{3x}{6} + \dfrac{2x}{6} = \dfrac{5x}{6},$$ so our equation is $$\dfrac{5x}{6} = 5.$$Multiplying both sides by $\dfrac{6}{5}$ gives $$x = 5\cdot \dfrac{6}{5} = \boxed{6}.$$


Explanation:
Okay, let's try to figure out Rosa's age. The problem says that when her age is divided by 2, 3, 4, or 6, the remainder is 1. Also, when divided by 7, there's no remainder. And she's less than 75 years old. Hmm, okay, let's break this down step by step.

First, let me note down the conditions:

1. Rosa's age divided by 2 leaves a remainder of 1.
2. Divided by 3, remainder 1.
3. Divided by 4, remainder 1.
4. Divided by 6, remainder 1.
5. Divided by 7, remainder 0.
6. Age is less than 75.

So, we need to find a number (Rosa's age) that satisfies all these conditions. Let me think about how to approach this.

Starting with the first condition: when divided by 2, remainder 1. That means Rosa's age is an odd number because if you divide an odd number by 2, the remainder is 1. Similarly, for division by 3, 4, 6, the remainder is 1. So, the age is 1 more than a multiple of 2, 3, 4, and 6.

Wait, so if I think about it, the age minus 1 should be divisible by 2, 3, 4, and 6. Let me write that down:

Let Rosa's age be N. Then:

N ≡ 1 mod 2
N ≡ 1 mod 3
N ≡ 1 mod 4
N ≡ 1 mod 6
N ≡ 0 mod 7
N < 75

So, N - 1 is divisible by 2, 3, 4, and 6. Therefore, N - 1 is a common multiple of these numbers. So, the least common multiple (LCM) of 2, 3, 4, 6. Let me compute that.

First, let's find LCM of 2, 3, 4, 6. Let's break them down into prime factors:

- 2 is 2
- 3 is 3
- 4 is 2²
- 6 is 2 × 3

The LCM is the product of the highest powers of all primes present. So, for prime 2, the highest power is 2² (from 4), and for prime 3, it's 3¹. Therefore, LCM is 2² × 3 = 4 × 3 = 12.

So, LCM of 2, 3, 4, 6 is 12. Therefore, N - 1 must be a multiple of 12. Therefore, N can be written as 12k + 1, where k is a positive integer.

Additionally, N must be divisible by 7. So, N = 12k + 1 must be divisible by 7. Therefore, 12k + 1 ≡ 0 mod 7. Let me write that equation:

12k + 1 ≡ 0 mod 7
Which simplifies to 12k ≡ -1 mod 7
But 12 mod 7 is 5, since 7*1=7, 12-7=5. So:

5k ≡ -1 mod 7
But -1 mod 7 is the same as 6 mod 7. So:

5k ≡ 6 mod 7
Now, we need to solve for k here. Let me find the value of k that satisfies this congruence.

To solve 5k ≡ 6 mod 7, we can find the multiplicative inverse of 5 modulo 7. The inverse of 5 mod 7 is a number x such that 5x ≡ 1 mod 7. Let me test numbers:

5*1=5≡5
5*2=10≡3
5*3=15≡1
Ah, so x=3. Therefore, multiplying both sides of 5k ≡6 mod7 by 3:

k ≡ 6*3 mod7
k ≡ 18 mod7
18 divided by 7 is 2*7=14, remainder 4. So k ≡4 mod7.

Therefore, k can be written as k=7m +4, where m is a non-negative integer.

Therefore, substituting back into N=12k +1:

N=12*(7m +4) +1= 84m +48 +1=84m +49.

So, N=84m +49. Now, we also know that N <75. Let's find possible values of m.

For m=0: N=0 +49=49. Which is less than 75.
For m=1: N=84 +49=133. Which is more than 75. So, m=0 is the only possible value here. Therefore, N=49.

Wait, but let me check if this satisfies all the original conditions. Let me verify.

First, check if 49 divided by 2 gives remainder 1. 49 divided by 2 is 24 with remainder 1. Correct.
Divided by 3: 3*16=48, so remainder 1. Correct.
Divided by 4: 4*12=48, remainder 1. Correct.
Divided by 6: 6*8=48, remainder 1. Correct.
Divided by 7: 49/7=7, remainder 0. Correct.
And 49 is less than 75. Perfect.

But wait, let me check if there are other possible values. Since m=0 gives 49, m=1 gives 133 which is over 75. So, only 49.

But let me think again. Is there another way to approach this? Maybe I made a mistake in calculating LCM?

Wait, the LCM of 2,3,4,6 is 12. So N-1 is multiple of 12. So N=12k +1. Then N must be divisible by 7. So 12k +1 divisible by 7. So 12k ≡ -1 mod7. Which is 5k ≡6 mod7. Then solving for k gives k≡4 mod7. Therefore, k=7m +4. Therefore N=12*(7m +4)+1=84m +49. So yes, same result.

So, with N <75, m=0 gives 49, m=1 gives 133. So 49 is the only possible.

But let me check if there's another number less than 75 that satisfies N-1 divisible by 12 and N divisible by 7. Let me think.

Numbers less than 75 where N-1 is divisible by 12: N-1=12,24,36,48,60,72. Therefore N=13,25,37,49,61,73. Now check which of these are divisible by 7.

Check 13: 13/7=1.85... no.
25/7≈3.57... no.
37/7≈5.28... no.
49/7=7. Yes!
61/7≈8.71... no.
73/7≈10.42... no.

So only 49 is in that list. Therefore, Rosa is 49.

Therefore, the answer is 49.

But let me check again if there's another way someone might approach this problem and maybe get confused. For example, someone might think that since N divided by 2,3,4,6 leaves remainder 1, then N is 1 more than LCM(2,3,4,6). But LCM(2,3,4,6)=12, so N=13? But 13 divided by 7 gives remainder 6, not 0. So that's not. Then next multiple would be 12*2 +1=25. 25 divided by 7 is 3*7=21, remainder 4. Not 0. Next is 37: 37/7=5*7=35, remainder 2. Not 0. Next is 49: 49/7=7, remainder 0. So that's the one. Then 61: 61/7=8*7=56, remainder 5. 73: 73/7=10*7=70, remainder 3. So yes, only 49.

Therefore, the answer is 49.

But let me check once more if there's a possibility that LCM(2,3,4,6) is different? Wait, LCM of 2,3,4,6. Let me recalculate.

Factors:

- 2: 2
- 3: 3
- 4: 2²
- 6: 2×3

So LCM is max exponent for each prime. So 2²×3=12. Correct. So N-1 is multiple of 12. So N=12k +1.

Alternatively, someone might think that since N divided by 4 leaves remainder 1, then N is 1 mod 4. Similarly, N is 1 mod 2, which is already covered by mod 4. Similarly, mod 3 and mod 6. But since 6 is a multiple of 2 and 3, so if N ≡1 mod 6, then automatically N ≡1 mod 2 and N≡1 mod 3. Therefore, the conditions N ≡1 mod 2,3,4,6 are equivalent to N≡1 mod LCM(2,3,4,6)=12. Therefore, N=12k +1.

Therefore, the rest of the reasoning holds.

Therefore, I think 49 is the correct answer.

But just to make sure, let me check all the conditions again with N=49:

Divided by 2: 49 /2 = 24*2 +1 → remainder 1 ✔
Divided by 3: 49 /3 = 16*3 +1 → remainder 1 ✔
Divided by 4: 49 /4 = 12*4 +1 → remainder 1 ✔
Divided by 6: 49 /6 = 8*6 +1 → remainder 1 ✔
Divided by 7: 49 /7 = 7*7 +0 → remainder 0 ✔
Age less than 75: 49 <75 ✔

All conditions are satisfied.

Therefore, Rosa is 49 years old.

**Final Answer**
Rosa is \boxed{49} years old.

To determine Rosa's age, we are given the following conditions:

1. When her age is divided by **2, 3, 4, or 6**, the **remainder is 1**.
2. When her age is divided by **7**, the **remainder is 0**.
3. Her age is **less than 75**.

---

### Step 1: Understanding the First Condition

If dividing Rosa's age $ N $ by 2, 3, 4, or 6 leaves a remainder of 1, then:

$$
N \equiv 1 \pmod{2}, \quad N \equiv 1 \pmod{3}, \quad N \equiv 1 \pmod{4}, \quad N \equiv 1 \pmod{6}
$$

This implies that:

$$
N - 1 \equiv 0 \pmod{2}, \quad N - 1 \equiv 0 \pmod{3}, \quad N - 1 \equiv 0 \pmod{4}, \quad N - 1 \equiv 0 \pmod{6}
$$

So, $ N - 1 $ is divisible by **2, 3, 4, and 6**. Therefore, $ N - 1 $ must be a **common multiple** of these numbers.

---

### Step 2: Finding the Least Common Multiple (LCM)

We compute the LCM of 2, 3, 4, and 6:

- Prime factorization:
  - 2 = $ 2 $
  - 3 = $ 3 $
  - 4 = $ 2^2 $
  - 6 = $ 2 \times 3 $

So, the LCM is:

$$
\text{LCM}(2, 3, 4, 6) = 2^2 \times 3 = 12
$$

Thus, $ N - 1 = 12k $ for some integer $ k $, and:

$$
N = 12k + 1
$$

---

### Step 3: Incorporating the Second Condition

We are also told that $ N $ is divisible by 7:

$$
N \equiv 0 \pmod{7}
$$

Substituting $ N = 12k + 1 $ into this:

$$
12k + 1 \equiv 0 \pmod{7} \Rightarrow 12k \equiv -1 \pmod{7}
$$

Since $ 12 \equiv 5 \pmod{7} $, this becomes:

$$
5k \equiv -1 \pmod{7} \Rightarrow 5k \equiv 6 \pmod{7}
$$

Now, we solve for $ k $:

- Find the **multiplicative inverse** of 5 modulo 7. That is, find $ x $ such that:

$$
5x \equiv 1 \pmod{7}
$$

Trying values, we find $ x = 3 $, since $ 5 \times 3 = 15 \equiv 1 \pmod{7} $

So, multiply both sides of $ 5k \equiv 6 \pmod{7} $ by 3:

$$
k \equiv 6 \times 3 \equiv 18 \equiv 4 \pmod{7}
$$

Thus, $ k = 7m + 4 $ for some integer $ m $

---

### Step 4: Expressing $ N $ in Terms of $ m $

Substitute $ k = 7m + 4 $ into $ N = 12k + 1 $:

$$
N = 12(7m + 4) + 1 = 84m + 48 + 1 = 84m + 49
$$

Now, we are told $ N < 75 $, so:

$$
84m + 49 < 75 \Rightarrow 84m < 26 \Rightarrow m < \frac{26}{84} \Rightarrow m = 0
$$

So, the only valid value is $ m = 0 $, which gives:

$$
N = 84(0) + 49 = 49
$$

---

### Step 5: Verifying the Solution

Let’s check if $ N = 49 $ satisfies all the original conditions:

- $ 49 \div 2 = 24 $ remainder 1 ✔
- $ 49 \div 3 = 16 $ remainder 1 ✔
- $ 49 \div 4 = 12 $ remainder 1 ✔
- $ 49 \div 6 = 8 $ remainder 1 ✔
- $ 49 \div 7 = 7 $ remainder 0 ✔
- $ 49 < 75 $ ✔

All conditions are satisfied.

---

### Final Answer

$$
\boxed{49}
$$

[Assistant]
Explanation:
Okay, let's try to figure out Rosa's age. The problem says that when her age is divided by 2, 3, 4, or 6, the remainder is 1. Also, when divided by 7, there's no remainder. And she's less than 75 years old. Hmm, okay, let's break this down step by step.

First, let me note down the conditions:

1. Rosa's age divided by 2 leaves a remainder of 1.
2. Divided by 3, remainder 1.
3. Divided by 4, remainder 1.
4. Divided by 6, remainder 1.
5. Divided by 7, remainder 0.
6. Age is less than 75.

So, we need to find a number (Rosa's age) that satisfies all these conditions. Let me think about how to approach this.

Starting with the first condition: when divided by 2, remainder 1. That means Rosa's age is an odd number because if you divide an odd number by 2, the remainder is 1. Similarly, for division by 3, 4, 6, the remainder is 1. So, the age is 1 more than a multiple of 2, 3, 4, and 6.

Wait, so if I think about it, the age minus 1 should be divisible by 2, 3, 4, and 6. Let me write that down:

Let Rosa's age be N. Then:

N ≡ 1 mod 2
N ≡ 1 mod 3
N ≡ 1 mod 4
N ≡ 1 mod 6
N ≡ 0 mod 7
N < 75

So, N - 1 is divisible by 2, 3, 4, and 6. Therefore, N - 1 is a common multiple of these numbers. So, the least common multiple (LCM) of 2, 3, 4, 6. Let me compute that.

First, let's find LCM of 2, 3, 4, 6. Let's break them down into prime factors:

- 2 is 2
- 3 is 3
- 4 is 2²
- 6 is 2 × 3

The LCM is the product of the highest powers of all primes present. So, for prime 2, the highest power is 2² (from 4), and for prime 3, it's 3¹. Therefore, LCM is 2² × 3 = 4 × 3 = 12.

So, LCM of 2, 3, 4, 6 is 12. Therefore, N - 1 must be a multiple of 12. Therefore, N can be written as 12k + 1, where k is a positive integer.

Additionally, N must be divisible by 7. So, N = 12k + 1 must be divisible by 7. Therefore, 12k + 1 ≡ 0 mod 7. Let me write that equation:

12k + 1 ≡ 0 mod 7
Which simplifies to 12k ≡ -1 mod 7
But 12 mod 7 is 5, since 7*1=7, 12-7=5. So:

5k ≡ -1 mod 7
But -1 mod 7 is the same as 6 mod 7. So:

5k ≡ 6 mod 7
Now, we need to solve for k here. Let me find the value of k that satisfies this congruence.

To solve 5k ≡ 6 mod 7, we can find the multiplicative inverse of 5 modulo 7. The inverse of 5 mod 7 is a number x such that 5x ≡ 1 mod 7. Let me test numbers:

5*1=5≡5
5*2=10≡3
5*3=15≡1
Ah, so x=3. Therefore, multiplying both sides of 5k ≡6 mod7 by 3:

k ≡ 6*3 mod7
k ≡ 18 mod7
18 divided by 7 is 2*7=14, remainder 4. So k ≡4 mod7.

Therefore, k can be written as k=7m +4, where m is a non-negative integer.

Therefore, substituting back into N=12k +1:

N=12*(7m +4) +1= 84m +48 +1=84m +49.

So, N=84m +49. Now, we also know that N <75. Let's find possible values of m.

For m=0: N=0 +49=49. Which is less than 75.
For m=1: N=84 +49=133. Which is more than 75. So, m=0 is the only possible value here. Therefore, N=49.

Wait, but let me check if this satisfies all the original conditions. Let me verify.

First, check if 49 divided by 2 gives remainder 1. 49 divided by 2 is 24 with remainder 1. Correct.
Divided by 3: 3*16=48, so remainder 1. Correct.
Divided by 4: 4*12=48, remainder 1. Correct.
Divided by 6: 6*8=48, remainder 1. Correct.
Divided by 7: 49/7=7, remainder 0. Correct.
And 49 is less than 75. Perfect.

But wait, let me check if there are other possible values. Since m=0 gives 49, m=1 gives 133 which is over 75. So, only 49.

But let me think again. Is there another way to approach this? Maybe I made a mistake in calculating LCM?

Wait, the LCM of 2,3,4,6 is 12. So N-1 is multiple of 12. So N=12k +1. Then N must be divisible by 7. So 12k +1 divisible by 7. So 12k ≡ -1 mod7. Which is 5k ≡6 mod7. Then solving for k gives k≡4 mod7. Therefore, k=7m +4. Therefore N=12*(7m +4)+1=84m +49. So yes, same result.

So, with N <75, m=0 gives 49, m=1 gives 133. So 49 is the only possible.

But let me check if there's another number less than 75 that satisfies N-1 divisible by 12 and N divisible by 7. Let me think.

Numbers less than 75 where N-1 is divisible by 12: N-1=12,24,36,48,60,72. Therefore N=13,25,37,49,61,73. Now check which of these are divisible by 7.

Check 13: 13/7=1.85... no.
25/7≈3.57... no.
37/7≈5.28... no.
49/7=7. Yes!
61/7≈8.71... no.
73/7≈10.42... no.

So only 49 is in that list. Therefore, Rosa is 49.

Therefore, the answer is 49.

But let me check again if there's another way someone might approach this problem and maybe get confused. For example, someone might think that since N divided by 2,3,4,6 leaves remainder 1, then N is 1 more than LCM(2,3,4,6). But LCM(2,3,4,6)=12, so N=13? But 13 divided by 7 gives remainder 6, not 0. So that's not. Then next multiple would be 12*2 +1=25. 25 divided by 7 is 3*7=21, remainder 4. Not 0. Next is 37: 37/7=5*7=35, remainder 2. Not 0. Next is 49: 49/7=7, remainder 0. So that's the one. Then 61: 61/7=8*7=56, remainder 5. 73: 73/7=10*7=70, remainder 3. So yes, only 49.

Therefore, the answer is 49.

But let me check once more if there's a possibility that LCM(2,3,4,6) is different? Wait, LCM of 2,3,4,6. Let me recalculate.

Factors:

- 2: 2
- 3: 3
- 4: 2²
- 6: 2×3

So LCM is max exponent for each prime. So 2²×3=12. Correct. So N-1 is multiple of 12. So N=12k +1.

Alternatively, someone might think that since N divided by 4 leaves remainder 1, then N is 1 mod 4. Similarly, N is 1 mod 2, which is already covered by mod 4. Similarly, mod 3 and mod 6. But since 6 is a multiple of 2 and 3, so if N ≡1 mod 6, then automatically N ≡1 mod 2 and N≡1 mod 3. Therefore, the conditions N ≡1 mod 2,3,4,6 are equivalent to N≡1 mod LCM(2,3,4,6)=12. Therefore, N=12k +1.

Therefore, the rest of the reasoning holds.

Therefore, I think 49 is the correct answer.

But just to make sure, let me check all the conditions again with N=49:

Divided by 2: 49 /2 = 24*2 +1 → remainder 1 ✔
Divided by 3: 49 /3 = 16*3 +1 → remainder 1 ✔
Divided by 4: 49 /4 = 12*4 +1 → remainder 1 ✔
Divided by 6: 49 /6 = 8*6 +1 → remainder 1 ✔
Divided by 7: 49 /7 = 7*7 +0 → remainder 0 ✔
Age less than 75: 49 <75 ✔

All conditions are satisfied.

Therefore, Rosa is 49 years old.

**Final Answer**
Rosa is \boxed{49} years old.

To determine Rosa's age, we are given the following conditions:

1. When her age is divided by **2, 3, 4, or 6**, the **remainder is 1**.
2. When her age is divided by **7**, the **remainder is 0**.
3. Her age is **less than 75**.

---

### Step 1: Understanding the First Condition

If dividing Rosa's age $ N $ by 2, 3, 4, or 6 leaves a remainder of 1, then:

$$
N \equiv 1 \pmod{2}, \quad N \equiv 1 \pmod{3}, \quad N \equiv 1 \pmod{4}, \quad N \equiv 1 \pmod{6}
$$

This implies that:

$$
N - 1 \equiv 0 \pmod{2}, \quad N - 1 \equiv 0 \pmod{3}, \quad N - 1 \equiv 0 \pmod{4}, \quad N - 1 \equiv 0 \pmod{6}
$$

So, $ N - 1 $ is divisible by **2, 3, 4, and 6**. Therefore, $ N - 1 $ must be a **common multiple** of these numbers.

---

### Step 2: Finding the Least Common Multiple (LCM)

We compute the LCM of 2, 3, 4, and 6:

- Prime factorization:
  - 2 = $ 2 $
  - 3 = $ 3 $
  - 4 = $ 2^2 $
  - 6 = $ 2 \times 3 $

So, the LCM is:

$$
\text{LCM}(2, 3, 4, 6) = 2^2 \times 3 = 12
$$

Thus, $ N - 1 = 12k $ for some integer $ k $, and:

$$
N = 12k + 1
$$

---

### Step 3: Incorporating the Second Condition

We are also told that $ N $ is divisible by 7:

$$
N \equiv 0 \pmod{7}
$$

Substituting $ N = 12k + 1 $ into this:

$$
12k + 1 \equiv 0 \pmod{7} \Rightarrow 12k \equiv -1 \pmod{7}
$$

Since $ 12 \equiv 5 \pmod{7} $, this becomes:

$$
5k \equiv -1 \pmod{7} \Rightarrow 5k \equiv 6 \pmod{7}
$$

Now, we solve for $ k $:

- Find the **multiplicative inverse** of 5 modulo 7. That is, find $ x $ such that:

$$
5x \equiv 1 \pmod{7}
$$

Trying values, we find $ x = 3 $, since $ 5 \times 3 = 15 \equiv 1 \pmod{7} $

So, multiply both sides of $ 5k \equiv 6 \pmod{7} $ by 3:

$$
k \equiv 6 \times 3 \equiv 18 \equiv 4 \pmod{7}
$$

Thus, $ k = 7m + 4 $ for some integer $ m $

---

### Step 4: Expressing $ N $ in Terms of $ m $

Substitute $ k = 7m + 4 $ into $ N = 12k + 1 $:

$$
N = 12(7m + 4) + 1 = 84m + 48 + 1 = 84m + 49
$$

Now, we are told $ N < 75 $, so:

$$
84m + 49 < 75 \Rightarrow 84m < 26 \Rightarrow m < \frac{26}{84} \Rightarrow m = 0
$$

So, the only valid value is $ m = 0 $, which gives:

$$
N = 84(0) + 49 = 49
$$

---

### Step 5: Verifying the Solution

Let’s check if $ N = 49 $ satisfies all the original conditions:

- $ 49 \div 2 = 24 $ remainder 1 ✔
- $ 49 \div 3 = 16 $ remainder 1 ✔
- $ 49 \div 4 = 12 $ remainder 1 ✔
- $ 49 \div 6 = 8 $ remainder 1 ✔
- $ 49 \div 7 = 7 $ remainder 0 ✔
- $ 49 < 75 $ ✔

All conditions are satisfied.

---

### Final Answer

$$
\boxed{49}
$$

Solution:
Since her age divided by 7 results in a remainder of 0, her age must be a multiple of 7. If her age is $n$, we notice that $n-1$ must be a multiple of 2, 3, 4, and 6. The least common multiple of those numbers is 12, so $n-1$ must be a multiple of 12. The multiples of 12 less than 75 are 12, 24, 36, 48, and 60. Adding 1 results in 13, 25, 37, 49, and 61, where 49 is the only multiple of 7. So Rosa is $\boxed{49}$ years old.

OR

We look for a multiple of 7 that is not divisible by 2, 3, 4, or 6. First we list all odd multiples of 7 less than 75, which are 7, 21, 35, 49, and 63. Since 21 and 63 are multiples of 3, we're left with 7, 35, and 49 as possibilities. Only $\boxed{49}$ has a remainder of 1 when divided by 2, 3, 4, or 6.