Explanation:
Okay, so I need to find the greatest integer less than $(\sqrt{7} + \sqrt{5})^6$. Hmm, that seems a bit intimidating at first because of the exponents and the square roots. But maybe there's a way to simplify this without actually calculating the entire value. Let me think...

I remember that expressions like $(\sqrt{a} + \sqrt{b})^n$ can sometimes be simplified using binomial expansion or by finding a recurrence relation. But since the exponent here is 6, maybe expanding it directly would be too tedious? Let me check.

Alternatively, I recall that sometimes when you have something like $(\sqrt{a} + \sqrt{b})^n + (\sqrt{a} - \sqrt{b})^n$, the irrational parts cancel out, leaving an integer. Maybe that's useful here? Let me try that idea.

Let me denote $x = \sqrt{7} + \sqrt{5}$ and $y = \sqrt{7} - \sqrt{5}$. Then, if I compute $x^6 + y^6$, maybe that will be an integer? Let me check if that's true.

First, let me compute $x + y$ and $x - y$:

$x + y = (\sqrt{7} + \sqrt{5}) + (\sqrt{7} - \sqrt{5}) = 2\sqrt{7}$

$x - y = (\sqrt{7} + \sqrt{5}) - (\sqrt{7} - \sqrt{5}) = 2\sqrt{5}$

But maybe more useful is $x \times y$:

$x \times y = (\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5}) = 7 - 5 = 2$. Oh, that's nice! So $xy = 2$.

So, if I can find $x^6 + y^6$, since $y = \sqrt{7} - \sqrt{5}$, which is a positive number less than 1 because $\sqrt{7} \approx 2.645$ and $\sqrt{5} \approx 2.236$, so their difference is about 0.409. Therefore, $y^6$ is a small positive number. Therefore, $x^6 = (x^6 + y^6) - y^6$, so the integer part of $x^6$ would be $x^6 + y^6 - 1$, since $y^6$ is less than 1. Therefore, the greatest integer less than $x^6$ is $x^6 + y^6 - 1$. Therefore, if I can compute $x^6 + y^6$, then subtract 1, I will get the answer.

So, how do I compute $x^6 + y^6$? Maybe using recurrence relations or some recursive formula?

Let me recall that for expressions like $x^n + y^n$, we can use the recurrence relation based on $x + y$ and $xy$. Let me denote $S_n = x^n + y^n$. Then, we can find a recurrence relation for $S_n$.

We know that $x$ and $y$ are roots of the quadratic equation $t^2 - (x + y)t + xy = 0$. Wait, but we already have $x + y = 2\sqrt{7}$ and $xy = 2$. However, since $x + y$ is irrational, maybe this complicates things? But perhaps we can still find a recurrence relation?

Alternatively, maybe we can find a recurrence relation for $S_n$ in terms of previous terms. Let me think. For any $n$, we have:

$x^n + y^n = (x + y)(x^{n-1} + y^{n-1}) - xy(x^{n-2} + y^{n-2})$

Yes, that's a standard recurrence relation. Let me verify that:

Let me expand $(x + y)(x^{n-1} + y^{n-1})$:

That would be $x^n + x y^{n-1} + y x^{n-1} + y^n$. Then subtract $xy(x^{n-2} + y^{n-2})$:

Which is $x^n + y^n + xy(x^{n-1} + y^{n-1}) - xy(x^{n-2} + y^{n-2})$. Wait, maybe I need to check again.

Wait, actually, the standard recurrence is:

If $S_n = x^n + y^n$, then $S_n = (x + y) S_{n-1} - (xy) S_{n-2}$. Let me check for n=2:

$S_2 = x^2 + y^2 = (x + y)^2 - 2xy$. Which is equal to $(x + y) S_1 - 2xy$. Since $S_1 = x + y$, so $(x + y) S_1 - 2xy$ is indeed $S_2$. So yes, the recurrence is correct.

Therefore, in our case, since $x + y = 2\sqrt{7}$ and $xy = 2$, we can write:

$S_n = 2\sqrt{7} \cdot S_{n-1} - 2 \cdot S_{n-2}$

But this recurrence involves irrational numbers, which might complicate things. However, since we need to compute $S_6$, maybe we can compute each term step by step?

But since $S_1$ and $S_2$ involve radicals, maybe we can compute them and see if they simplify?

Let me try:

First, compute $S_0 = x^0 + y^0 = 1 + 1 = 2$

$S_1 = x + y = 2\sqrt{7}$

$S_2 = (x + y)^2 - 2xy = (2\sqrt{7})^2 - 2*2 = 4*7 - 4 = 28 - 4 = 24$

Wait, that's nice! So $S_2 = 24$. Let me check with the recurrence:

Using the recurrence $S_n = 2\sqrt{7} S_{n-1} - 2 S_{n-2}$

For n=2: $S_2 = 2\sqrt{7} * S_1 - 2 * S_0 = 2\sqrt{7} * 2\sqrt{7} - 2*2 = 4*7 - 4 = 28 - 4 = 24$. Correct.

So, moving on:

Compute $S_3 = 2\sqrt{7} * S_2 - 2 * S_1 = 2\sqrt{7} * 24 - 2 * 2\sqrt{7} = 48\sqrt{7} - 4\sqrt{7} = 44\sqrt{7}$

Hmm, so $S_3 = 44\sqrt{7}$. Interesting. Let me check if that's correct by expanding $(\sqrt{7} + \sqrt{5})^3 + (\sqrt{7} - \sqrt{5})^3$:

Using the formula $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$. Let me compute that:

Let $a = \sqrt{7} + \sqrt{5}$, $b = \sqrt{7} - \sqrt{5}$. Then $a + b = 2\sqrt{7}$, $ab = 2$. Therefore, $a^3 + b^3 = (2\sqrt{7})^3 - 3*2*(2\sqrt{7}) = 8*7\sqrt{7} - 12\sqrt{7} = 56\sqrt{7} - 12\sqrt{7} = 44\sqrt{7}$. Yes, that matches. So $S_3 = 44\sqrt{7}$.

Okay, moving on to $S_4$:

Using the recurrence: $S_4 = 2\sqrt{7} * S_3 - 2 * S_2 = 2\sqrt{7} * 44\sqrt{7} - 2 * 24$

Compute each term:

First term: $2\sqrt{7} * 44\sqrt{7} = 2*44*(\sqrt{7})^2 = 88*7 = 616$

Second term: 2*24 = 48

Therefore, $S_4 = 616 - 48 = 568$. Nice! So $S_4 = 568$. Let me check with another method to be safe.

Alternatively, compute $S_4 = (x^2)^2 + (y^2)^2$. Since $x^2 = (\sqrt{7} + \sqrt{5})^2 = 7 + 5 + 2\sqrt{35} = 12 + 2\sqrt{35}$, similarly $y^2 = 12 - 2\sqrt{35}$. Therefore, $x^4 + y^4 = (12 + 2\sqrt{35})^2 + (12 - 2\sqrt{35})^2$. Let me compute that:

First, expand each square:

$(12 + 2\sqrt{35})^2 = 144 + 48\sqrt{35} + 4*35 = 144 + 48\sqrt{35} + 140 = 284 + 48\sqrt{35}$

Similarly, $(12 - 2\sqrt{35})^2 = 144 - 48\sqrt{35} + 140 = 284 - 48\sqrt{35}$

Adding them together: 284 + 48√35 + 284 - 48√35 = 568. Perfect, matches the previous result. So $S_4 = 568$.

Great, moving on to $S_5$:

Using the recurrence: $S_5 = 2\sqrt{7} * S_4 - 2 * S_3 = 2\sqrt{7} * 568 - 2 * 44\sqrt{7}$

Compute each term:

First term: $2\sqrt{7} * 568 = 1136\sqrt{7}$

Second term: 2 * 44√7 = 88√7

Therefore, $S_5 = 1136\sqrt{7} - 88\sqrt{7} = (1136 - 88)\sqrt{7} = 1048\sqrt{7}$

Let me check with another method. Let me compute $S_5 = x^5 + y^5$. Alternatively, using the recurrence relation again, but maybe that's too tedious. Alternatively, using the previous terms. But maybe I can trust the recurrence here since it worked for previous terms. Let me just note that $S_5 = 1048\sqrt{7}$.

Now, moving on to $S_6$:

Using the recurrence: $S_6 = 2\sqrt{7} * S_5 - 2 * S_4 = 2\sqrt{7} * 1048\sqrt{7} - 2 * 568$

Compute each term:

First term: $2\sqrt{7} * 1048\sqrt{7} = 2*1048*(\sqrt{7})^2 = 2096*7 = let's compute that. 2000*7=14000, 96*7=672, so total is 14000 + 672 = 14672

Second term: 2*568 = 1136

Therefore, $S_6 = 14672 - 1136 = 13536$.

Wait, so $S_6 = 13536$. Therefore, since $x^6 + y^6 = 13536$, and since $y = \sqrt{7} - \sqrt{5} \approx 0.409$, so $y^6$ is a small positive number less than 1. Therefore, $x^6 = 13536 - y^6$, so the greatest integer less than $x^6$ is $13536 - 1 - 1$? Wait, no. Wait, since $y^6$ is between 0 and 1, then $x^6 = 13536 - y^6$ is between 13536 - 1 and 13536. Therefore, the greatest integer less than $x^6$ is 13536 - 1 = 13535? Wait, but hold on. Wait, if $x^6 = 13536 - y^6$, and since $0 < y^6 < 1$, then $x^6$ is between 13535 and 13536. Therefore, the floor of $x^6$ is 13535. Therefore, the answer is 13535?

Wait, but let me check if that's correct. Let me verify with approximate values. Maybe I can estimate $y^6$ to check if it's less than 1. Let me compute $y = \sqrt{7} - \sqrt{5}$. Let me approximate sqrt(7) ≈ 2.6458, sqrt(5) ≈ 2.2361, so y ≈ 2.6458 - 2.2361 ≈ 0.4097. Then y^6 ≈ (0.4097)^6. Let me compute that step by step:

First, 0.4097^2 ≈ 0.1678

Then, 0.1678 * 0.4097 ≈ 0.0687 (which is 0.4097^3)

Then, 0.0687 * 0.4097 ≈ 0.0281 (which is 0.4097^4)

Then, 0.0281 * 0.4097 ≈ 0.0115 (which is 0.4097^5)

Then, 0.0115 * 0.4097 ≈ 0.00471 (which is 0.4097^6)

So approximately 0.00471. Therefore, y^6 ≈ 0.0047, which is indeed less than 1. Therefore, x^6 = 13536 - 0.0047 ≈ 13535.9953. Therefore, the greatest integer less than x^6 is 13535. Therefore, the answer is 13535.

But let me check if my calculation of S_6 is correct. Let me retrace the steps:

Starting with S_0 = 2, S_1 = 2√7, S_2 = 24, S_3 = 44√7, S_4 = 568, S_5 = 1048√7, S_6 = 14672 - 1136 = 13536.

Yes, that seems correct. Let me check S_5 again. S_5 = 2√7 * S_4 - 2 * S_3. S_4 is 568, so 2√7 * 568 is 1136√7. Then subtract 2 * S_3, which is 2 * 44√7 = 88√7. Therefore, 1136√7 - 88√7 = 1048√7. Correct. Then S_6 is 2√7 * S_5 - 2 * S_4. So 2√7 * 1048√7 = 2*1048*7 = 2096*7. Let me compute 2000*7=14000, 96*7=672, so total 14672. Then subtract 2*S_4 = 2*568=1136. 14672 - 1136: 14672 - 1000 = 13672, then -136 = 13536. Correct.

Therefore, S_6 = 13536. Therefore, since x^6 + y^6 = 13536, and y^6 ≈ 0.0047, then x^6 ≈ 13536 - 0.0047 ≈ 13535.9953, so floor(x^6) = 13535. Therefore, the answer is 13535.

But let me check if there's another way to compute this, just to be sure. Maybe expanding $(\sqrt{7} + \sqrt{5})^6$ directly? Although that might be tedious, but let me try.

First, note that $(\sqrt{7} + \sqrt{5})^6$ can be expanded using the binomial theorem:

$$
(\sqrt{7} + \sqrt{5})^6 = \sum_{k=0}^6 \binom{6}{k} (\sqrt{7})^{6 - k} (\sqrt{5})^k
$$

Which simplifies to:

$$
\sum_{k=0}^6 \binom{6}{k} 7^{(6 - k)/2} 5^{k/2}
$$

But this seems complicated. However, note that each term can be written as $\binom{6}{k} \cdot 7^{3 - k/2} \cdot 5^{k/2}$. Alternatively, maybe grouping terms with radicals?

Alternatively, since we already have that $x^6 + y^6 = 13536$, and we know that $y^6$ is positive and less than 1, so $x^6 = 13536 - y^6$, so floor(x^6) = 13535. Therefore, unless there's a miscalculation in S_6, which I think is correct, then the answer is 13535.

But let me check with another approach. Let me compute $(\sqrt{7} + \sqrt{5})^2$ first, which is 7 + 5 + 2\sqrt{35} = 12 + 2\sqrt{35}. Then, $(\sqrt{7} + \sqrt{5})^4 = (12 + 2\sqrt{35})^2 = 144 + 48\sqrt{35} + 4*35 = 144 + 48\sqrt{35} + 140 = 284 + 48\sqrt{35}$. Then, $(\sqrt{7} + \sqrt{5})^6 = (\sqrt{7} + \sqrt{5})^4 * (\sqrt{7} + \sqrt{5})^2 = (284 + 48\sqrt{35})(12 + 2\sqrt{35})$. Let me compute that:

Multiply 284 * 12 = 3408

284 * 2√35 = 568√35

48√35 * 12 = 576√35

48√35 * 2√35 = 96*(√35)^2 = 96*35 = 3360

Therefore, adding all terms:

3408 + 3360 + (568√35 + 576√35) = 3408 + 3360 + 1144√35

Compute 3408 + 3360: 3408 + 3360 = 6768

So total is 6768 + 1144√35. Therefore, $(\sqrt{7} + \sqrt{5})^6 = 6768 + 1144\sqrt{35}$.

Now, we need to find the floor of this value. Let me compute 1144√35. Let me approximate √35. Since √36 = 6, so √35 ≈ 5.916. Let me check:

5.916^2 = (5.9)^2 + 2*5.9*0.016 + (0.016)^2 ≈ 34.81 + 0.1888 + 0.000256 ≈ 34.999, which is very close to 35. Therefore, √35 ≈ 5.916. Therefore, 1144 * 5.916 ≈ ?

First, compute 1000 * 5.916 = 5916

Then 144 * 5.916. Let me compute 100*5.916 = 591.6, 40*5.916 = 236.64, 4*5.916 = 23.664. Adding those: 591.6 + 236.64 = 828.24 + 23.664 = 851.904. Therefore, total 1144 * 5.916 ≈ 5916 + 851.904 ≈ 6767.904.

Therefore, $(\sqrt{7} + \sqrt{5})^6 ≈ 6768 + 6767.904 ≈ 13535.904$. Therefore, the value is approximately 13535.904, so the greatest integer less than that is 13535. Which matches our previous result. Therefore, this confirms that the answer is 13535.

Therefore, I can be confident that the answer is 13535.

But just to make sure, let me check if my approximation of √35 is accurate enough. Let me compute √35 more precisely. Let me use the Newton-Raphson method. Let me take an initial guess of 5.916. Let me compute 5.916^2:

As above, 5.916^2 ≈ 34.999, which is very close to 35. Let me compute 5.916^2:

First, 5.916 * 5.916:

Break it down as (5.9 + 0.016)^2 = 5.9^2 + 2*5.9*0.016 + 0.016^2

5.9^2 = 34.81

2*5.9*0.016 = 11.8 * 0.016 = 0.1888

0.016^2 = 0.000256

Total: 34.81 + 0.1888 + 0.000256 ≈ 34.999056. So, 5.916^2 ≈ 34.999056, which is 35 - 0.000944. Therefore, √35 ≈ 5.916 + ε, where ε is small. Let me compute ε:

Let me use linear approximation. Let f(x) = √x, x = 35, f(x) = √35 ≈ 5.916. Let me take x = 35 - 0.000944, so f(x) ≈ f(35) - (0.000944)/(2√35). Therefore, √(35 - 0.000944) ≈ 5.916 - 0.000944/(2*5.916). Compute denominator: 2*5.916 ≈ 11.832. Therefore, 0.000944 / 11.832 ≈ 0.0000798. Therefore, √35 ≈ 5.916 + 0.0000798 ≈ 5.9160798. Therefore, √35 ≈ 5.91608. Therefore, more accurately, √35 ≈ 5.91608.

Therefore, 1144 * √35 ≈ 1144 * 5.91608. Let me compute this more accurately.

First, note that earlier, we had 1144 * 5.916 ≈ 6767.904. Now, since √35 ≈ 5.91608, which is 5.916 + 0.00008. Therefore, the additional term is 1144 * 0.00008 = 0.09152. Therefore, total is approximately 6767.904 + 0.09152 ≈ 6767.9955. Therefore, $(\sqrt{7} + \sqrt{5})^6 ≈ 6768 + 6767.9955 ≈ 13535.9955$. Therefore, approximately 13535.9955, which is just slightly less than 13536. Therefore, the floor is indeed 13535.

Therefore, this confirms again that the answer is 13535.

Alternatively, since we have two different methods leading to the same conclusion, I think this is solid.

Therefore, the greatest integer less than $(\sqrt{7} + \sqrt{5})^6$ is 13535.

**Final Answer**
\boxed{13535}

To find the greatest integer less than $(\sqrt{7} + \sqrt{5})^6$, we can use a clever algebraic trick involving the conjugate of the expression. Let:

$$
x = \sqrt{7} + \sqrt{5}, \quad y = \sqrt{7} - \sqrt{5}
$$

Note that:

$$
x + y = 2\sqrt{7}, \quad xy = (\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5}) = 7 - 5 = 2
$$

We define the sequence:

$$
S_n = x^n + y^n
$$

This sequence satisfies the recurrence relation:

$$
S_n = (x + y) S_{n-1} - xy S_{n-2}
$$

Substituting the known values:

$$
S_n = 2\sqrt{7} \cdot S_{n-1} - 2 \cdot S_{n-2}
$$

We compute the first few terms:

- $S_0 = x^0 + y^0 = 1 + 1 = 2$
- $S_1 = x + y = 2\sqrt{7}$
- $S_2 = (x + y)^2 - 2xy = (2\sqrt{7})^2 - 2 \cdot 2 = 28 - 4 = 24$
- $S_3 = 2\sqrt{7} \cdot S_2 - 2 \cdot S_1 = 2\sqrt{7} \cdot 24 - 2 \cdot 2\sqrt{7} = 48\sqrt{7} - 4\sqrt{7} = 44\sqrt{7}$
- $S_4 = 2\sqrt{7} \cdot S_3 - 2 \cdot S_2 = 2\sqrt{7} \cdot 44\sqrt{7} - 2 \cdot 24 = 88 \cdot 7 - 48 = 616 - 48 = 568$
- $S_5 = 2\sqrt{7} \cdot S_4 - 2 \cdot S_3 = 2\sqrt{7} \cdot 568 - 2 \cdot 44\sqrt{7} = 1136\sqrt{7} - 88\sqrt{7} = 1048\sqrt{7}$
- $S_6 = 2\sqrt{7} \cdot S_5 - 2 \cdot S_4 = 2\sqrt{7} \cdot 1048\sqrt{7} - 2 \cdot 568 = 2096 \cdot 7 - 1136 = 14672 - 1136 = 13536$

So, we have:

$$
x^6 + y^6 = S_6 = 13536
$$

Now, observe that $y = \sqrt{7} - \sqrt{5} \approx 0.4097$, so $y^6$ is a small positive number less than 1. Therefore:

$$
x^6 = 13536 - y^6
$$

Since $0 < y^6 < 1$, it follows that:

$$
13535 < x^6 < 13536
$$

Thus, the greatest integer less than $x^6$ is:

$$
\boxed{13535}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest integer less than $(\sqrt{7} + \sqrt{5})^6$. Hmm, that seems a bit intimidating at first because of the exponents and the square roots. But maybe there's a way to simplify this without actually calculating the entire value. Let me think...

I remember that expressions like $(\sqrt{a} + \sqrt{b})^n$ can sometimes be simplified using binomial expansion or by finding a recurrence relation. But since the exponent here is 6, maybe expanding it directly would be too tedious? Let me check.

Alternatively, I recall that sometimes when you have something like $(\sqrt{a} + \sqrt{b})^n + (\sqrt{a} - \sqrt{b})^n$, the irrational parts cancel out, leaving an integer. Maybe that's useful here? Let me try that idea.

Let me denote $x = \sqrt{7} + \sqrt{5}$ and $y = \sqrt{7} - \sqrt{5}$. Then, if I compute $x^6 + y^6$, maybe that will be an integer? Let me check if that's true.

First, let me compute $x + y$ and $x - y$:

$x + y = (\sqrt{7} + \sqrt{5}) + (\sqrt{7} - \sqrt{5}) = 2\sqrt{7}$

$x - y = (\sqrt{7} + \sqrt{5}) - (\sqrt{7} - \sqrt{5}) = 2\sqrt{5}$

But maybe more useful is $x \times y$:

$x \times y = (\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5}) = 7 - 5 = 2$. Oh, that's nice! So $xy = 2$.

So, if I can find $x^6 + y^6$, since $y = \sqrt{7} - \sqrt{5}$, which is a positive number less than 1 because $\sqrt{7} \approx 2.645$ and $\sqrt{5} \approx 2.236$, so their difference is about 0.409. Therefore, $y^6$ is a small positive number. Therefore, $x^6 = (x^6 + y^6) - y^6$, so the integer part of $x^6$ would be $x^6 + y^6 - 1$, since $y^6$ is less than 1. Therefore, the greatest integer less than $x^6$ is $x^6 + y^6 - 1$. Therefore, if I can compute $x^6 + y^6$, then subtract 1, I will get the answer.

So, how do I compute $x^6 + y^6$? Maybe using recurrence relations or some recursive formula?

Let me recall that for expressions like $x^n + y^n$, we can use the recurrence relation based on $x + y$ and $xy$. Let me denote $S_n = x^n + y^n$. Then, we can find a recurrence relation for $S_n$.

We know that $x$ and $y$ are roots of the quadratic equation $t^2 - (x + y)t + xy = 0$. Wait, but we already have $x + y = 2\sqrt{7}$ and $xy = 2$. However, since $x + y$ is irrational, maybe this complicates things? But perhaps we can still find a recurrence relation?

Alternatively, maybe we can find a recurrence relation for $S_n$ in terms of previous terms. Let me think. For any $n$, we have:

$x^n + y^n = (x + y)(x^{n-1} + y^{n-1}) - xy(x^{n-2} + y^{n-2})$

Yes, that's a standard recurrence relation. Let me verify that:

Let me expand $(x + y)(x^{n-1} + y^{n-1})$:

That would be $x^n + x y^{n-1} + y x^{n-1} + y^n$. Then subtract $xy(x^{n-2} + y^{n-2})$:

Which is $x^n + y^n + xy(x^{n-1} + y^{n-1}) - xy(x^{n-2} + y^{n-2})$. Wait, maybe I need to check again.

Wait, actually, the standard recurrence is:

If $S_n = x^n + y^n$, then $S_n = (x + y) S_{n-1} - (xy) S_{n-2}$. Let me check for n=2:

$S_2 = x^2 + y^2 = (x + y)^2 - 2xy$. Which is equal to $(x + y) S_1 - 2xy$. Since $S_1 = x + y$, so $(x + y) S_1 - 2xy$ is indeed $S_2$. So yes, the recurrence is correct.

Therefore, in our case, since $x + y = 2\sqrt{7}$ and $xy = 2$, we can write:

$S_n = 2\sqrt{7} \cdot S_{n-1} - 2 \cdot S_{n-2}$

But this recurrence involves irrational numbers, which might complicate things. However, since we need to compute $S_6$, maybe we can compute each term step by step?

But since $S_1$ and $S_2$ involve radicals, maybe we can compute them and see if they simplify?

Let me try:

First, compute $S_0 = x^0 + y^0 = 1 + 1 = 2$

$S_1 = x + y = 2\sqrt{7}$

$S_2 = (x + y)^2 - 2xy = (2\sqrt{7})^2 - 2*2 = 4*7 - 4 = 28 - 4 = 24$

Wait, that's nice! So $S_2 = 24$. Let me check with the recurrence:

Using the recurrence $S_n = 2\sqrt{7} S_{n-1} - 2 S_{n-2}$

For n=2: $S_2 = 2\sqrt{7} * S_1 - 2 * S_0 = 2\sqrt{7} * 2\sqrt{7} - 2*2 = 4*7 - 4 = 28 - 4 = 24$. Correct.

So, moving on:

Compute $S_3 = 2\sqrt{7} * S_2 - 2 * S_1 = 2\sqrt{7} * 24 - 2 * 2\sqrt{7} = 48\sqrt{7} - 4\sqrt{7} = 44\sqrt{7}$

Hmm, so $S_3 = 44\sqrt{7}$. Interesting. Let me check if that's correct by expanding $(\sqrt{7} + \sqrt{5})^3 + (\sqrt{7} - \sqrt{5})^3$:

Using the formula $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$. Let me compute that:

Let $a = \sqrt{7} + \sqrt{5}$, $b = \sqrt{7} - \sqrt{5}$. Then $a + b = 2\sqrt{7}$, $ab = 2$. Therefore, $a^3 + b^3 = (2\sqrt{7})^3 - 3*2*(2\sqrt{7}) = 8*7\sqrt{7} - 12\sqrt{7} = 56\sqrt{7} - 12\sqrt{7} = 44\sqrt{7}$. Yes, that matches. So $S_3 = 44\sqrt{7}$.

Okay, moving on to $S_4$:

Using the recurrence: $S_4 = 2\sqrt{7} * S_3 - 2 * S_2 = 2\sqrt{7} * 44\sqrt{7} - 2 * 24$

Compute each term:

First term: $2\sqrt{7} * 44\sqrt{7} = 2*44*(\sqrt{7})^2 = 88*7 = 616$

Second term: 2*24 = 48

Therefore, $S_4 = 616 - 48 = 568$. Nice! So $S_4 = 568$. Let me check with another method to be safe.

Alternatively, compute $S_4 = (x^2)^2 + (y^2)^2$. Since $x^2 = (\sqrt{7} + \sqrt{5})^2 = 7 + 5 + 2\sqrt{35} = 12 + 2\sqrt{35}$, similarly $y^2 = 12 - 2\sqrt{35}$. Therefore, $x^4 + y^4 = (12 + 2\sqrt{35})^2 + (12 - 2\sqrt{35})^2$. Let me compute that:

First, expand each square:

$(12 + 2\sqrt{35})^2 = 144 + 48\sqrt{35} + 4*35 = 144 + 48\sqrt{35} + 140 = 284 + 48\sqrt{35}$

Similarly, $(12 - 2\sqrt{35})^2 = 144 - 48\sqrt{35} + 140 = 284 - 48\sqrt{35}$

Adding them together: 284 + 48√35 + 284 - 48√35 = 568. Perfect, matches the previous result. So $S_4 = 568$.

Great, moving on to $S_5$:

Using the recurrence: $S_5 = 2\sqrt{7} * S_4 - 2 * S_3 = 2\sqrt{7} * 568 - 2 * 44\sqrt{7}$

Compute each term:

First term: $2\sqrt{7} * 568 = 1136\sqrt{7}$

Second term: 2 * 44√7 = 88√7

Therefore, $S_5 = 1136\sqrt{7} - 88\sqrt{7} = (1136 - 88)\sqrt{7} = 1048\sqrt{7}$

Let me check with another method. Let me compute $S_5 = x^5 + y^5$. Alternatively, using the recurrence relation again, but maybe that's too tedious. Alternatively, using the previous terms. But maybe I can trust the recurrence here since it worked for previous terms. Let me just note that $S_5 = 1048\sqrt{7}$.

Now, moving on to $S_6$:

Using the recurrence: $S_6 = 2\sqrt{7} * S_5 - 2 * S_4 = 2\sqrt{7} * 1048\sqrt{7} - 2 * 568$

Compute each term:

First term: $2\sqrt{7} * 1048\sqrt{7} = 2*1048*(\sqrt{7})^2 = 2096*7 = let's compute that. 2000*7=14000, 96*7=672, so total is 14000 + 672 = 14672

Second term: 2*568 = 1136

Therefore, $S_6 = 14672 - 1136 = 13536$.

Wait, so $S_6 = 13536$. Therefore, since $x^6 + y^6 = 13536$, and since $y = \sqrt{7} - \sqrt{5} \approx 0.409$, so $y^6$ is a small positive number less than 1. Therefore, $x^6 = 13536 - y^6$, so the greatest integer less than $x^6$ is $13536 - 1 - 1$? Wait, no. Wait, since $y^6$ is between 0 and 1, then $x^6 = 13536 - y^6$ is between 13536 - 1 and 13536. Therefore, the greatest integer less than $x^6$ is 13536 - 1 = 13535? Wait, but hold on. Wait, if $x^6 = 13536 - y^6$, and since $0 < y^6 < 1$, then $x^6$ is between 13535 and 13536. Therefore, the floor of $x^6$ is 13535. Therefore, the answer is 13535?

Wait, but let me check if that's correct. Let me verify with approximate values. Maybe I can estimate $y^6$ to check if it's less than 1. Let me compute $y = \sqrt{7} - \sqrt{5}$. Let me approximate sqrt(7) ≈ 2.6458, sqrt(5) ≈ 2.2361, so y ≈ 2.6458 - 2.2361 ≈ 0.4097. Then y^6 ≈ (0.4097)^6. Let me compute that step by step:

First, 0.4097^2 ≈ 0.1678

Then, 0.1678 * 0.4097 ≈ 0.0687 (which is 0.4097^3)

Then, 0.0687 * 0.4097 ≈ 0.0281 (which is 0.4097^4)

Then, 0.0281 * 0.4097 ≈ 0.0115 (which is 0.4097^5)

Then, 0.0115 * 0.4097 ≈ 0.00471 (which is 0.4097^6)

So approximately 0.00471. Therefore, y^6 ≈ 0.0047, which is indeed less than 1. Therefore, x^6 = 13536 - 0.0047 ≈ 13535.9953. Therefore, the greatest integer less than x^6 is 13535. Therefore, the answer is 13535.

But let me check if my calculation of S_6 is correct. Let me retrace the steps:

Starting with S_0 = 2, S_1 = 2√7, S_2 = 24, S_3 = 44√7, S_4 = 568, S_5 = 1048√7, S_6 = 14672 - 1136 = 13536.

Yes, that seems correct. Let me check S_5 again. S_5 = 2√7 * S_4 - 2 * S_3. S_4 is 568, so 2√7 * 568 is 1136√7. Then subtract 2 * S_3, which is 2 * 44√7 = 88√7. Therefore, 1136√7 - 88√7 = 1048√7. Correct. Then S_6 is 2√7 * S_5 - 2 * S_4. So 2√7 * 1048√7 = 2*1048*7 = 2096*7. Let me compute 2000*7=14000, 96*7=672, so total 14672. Then subtract 2*S_4 = 2*568=1136. 14672 - 1136: 14672 - 1000 = 13672, then -136 = 13536. Correct.

Therefore, S_6 = 13536. Therefore, since x^6 + y^6 = 13536, and y^6 ≈ 0.0047, then x^6 ≈ 13536 - 0.0047 ≈ 13535.9953, so floor(x^6) = 13535. Therefore, the answer is 13535.

But let me check if there's another way to compute this, just to be sure. Maybe expanding $(\sqrt{7} + \sqrt{5})^6$ directly? Although that might be tedious, but let me try.

First, note that $(\sqrt{7} + \sqrt{5})^6$ can be expanded using the binomial theorem:

$$
(\sqrt{7} + \sqrt{5})^6 = \sum_{k=0}^6 \binom{6}{k} (\sqrt{7})^{6 - k} (\sqrt{5})^k
$$

Which simplifies to:

$$
\sum_{k=0}^6 \binom{6}{k} 7^{(6 - k)/2} 5^{k/2}
$$

But this seems complicated. However, note that each term can be written as $\binom{6}{k} \cdot 7^{3 - k/2} \cdot 5^{k/2}$. Alternatively, maybe grouping terms with radicals?

Alternatively, since we already have that $x^6 + y^6 = 13536$, and we know that $y^6$ is positive and less than 1, so $x^6 = 13536 - y^6$, so floor(x^6) = 13535. Therefore, unless there's a miscalculation in S_6, which I think is correct, then the answer is 13535.

But let me check with another approach. Let me compute $(\sqrt{7} + \sqrt{5})^2$ first, which is 7 + 5 + 2\sqrt{35} = 12 + 2\sqrt{35}. Then, $(\sqrt{7} + \sqrt{5})^4 = (12 + 2\sqrt{35})^2 = 144 + 48\sqrt{35} + 4*35 = 144 + 48\sqrt{35} + 140 = 284 + 48\sqrt{35}$. Then, $(\sqrt{7} + \sqrt{5})^6 = (\sqrt{7} + \sqrt{5})^4 * (\sqrt{7} + \sqrt{5})^2 = (284 + 48\sqrt{35})(12 + 2\sqrt{35})$. Let me compute that:

Multiply 284 * 12 = 3408

284 * 2√35 = 568√35

48√35 * 12 = 576√35

48√35 * 2√35 = 96*(√35)^2 = 96*35 = 3360

Therefore, adding all terms:

3408 + 3360 + (568√35 + 576√35) = 3408 + 3360 + 1144√35

Compute 3408 + 3360: 3408 + 3360 = 6768

So total is 6768 + 1144√35. Therefore, $(\sqrt{7} + \sqrt{5})^6 = 6768 + 1144\sqrt{35}$.

Now, we need to find the floor of this value. Let me compute 1144√35. Let me approximate √35. Since √36 = 6, so √35 ≈ 5.916. Let me check:

5.916^2 = (5.9)^2 + 2*5.9*0.016 + (0.016)^2 ≈ 34.81 + 0.1888 + 0.000256 ≈ 34.999, which is very close to 35. Therefore, √35 ≈ 5.916. Therefore, 1144 * 5.916 ≈ ?

First, compute 1000 * 5.916 = 5916

Then 144 * 5.916. Let me compute 100*5.916 = 591.6, 40*5.916 = 236.64, 4*5.916 = 23.664. Adding those: 591.6 + 236.64 = 828.24 + 23.664 = 851.904. Therefore, total 1144 * 5.916 ≈ 5916 + 851.904 ≈ 6767.904.

Therefore, $(\sqrt{7} + \sqrt{5})^6 ≈ 6768 + 6767.904 ≈ 13535.904$. Therefore, the value is approximately 13535.904, so the greatest integer less than that is 13535. Which matches our previous result. Therefore, this confirms that the answer is 13535.

Therefore, I can be confident that the answer is 13535.

But just to make sure, let me check if my approximation of √35 is accurate enough. Let me compute √35 more precisely. Let me use the Newton-Raphson method. Let me take an initial guess of 5.916. Let me compute 5.916^2:

As above, 5.916^2 ≈ 34.999, which is very close to 35. Let me compute 5.916^2:

First, 5.916 * 5.916:

Break it down as (5.9 + 0.016)^2 = 5.9^2 + 2*5.9*0.016 + 0.016^2

5.9^2 = 34.81

2*5.9*0.016 = 11.8 * 0.016 = 0.1888

0.016^2 = 0.000256

Total: 34.81 + 0.1888 + 0.000256 ≈ 34.999056. So, 5.916^2 ≈ 34.999056, which is 35 - 0.000944. Therefore, √35 ≈ 5.916 + ε, where ε is small. Let me compute ε:

Let me use linear approximation. Let f(x) = √x, x = 35, f(x) = √35 ≈ 5.916. Let me take x = 35 - 0.000944, so f(x) ≈ f(35) - (0.000944)/(2√35). Therefore, √(35 - 0.000944) ≈ 5.916 - 0.000944/(2*5.916). Compute denominator: 2*5.916 ≈ 11.832. Therefore, 0.000944 / 11.832 ≈ 0.0000798. Therefore, √35 ≈ 5.916 + 0.0000798 ≈ 5.9160798. Therefore, √35 ≈ 5.91608. Therefore, more accurately, √35 ≈ 5.91608.

Therefore, 1144 * √35 ≈ 1144 * 5.91608. Let me compute this more accurately.

First, note that earlier, we had 1144 * 5.916 ≈ 6767.904. Now, since √35 ≈ 5.91608, which is 5.916 + 0.00008. Therefore, the additional term is 1144 * 0.00008 = 0.09152. Therefore, total is approximately 6767.904 + 0.09152 ≈ 6767.9955. Therefore, $(\sqrt{7} + \sqrt{5})^6 ≈ 6768 + 6767.9955 ≈ 13535.9955$. Therefore, approximately 13535.9955, which is just slightly less than 13536. Therefore, the floor is indeed 13535.

Therefore, this confirms again that the answer is 13535.

Alternatively, since we have two different methods leading to the same conclusion, I think this is solid.

Therefore, the greatest integer less than $(\sqrt{7} + \sqrt{5})^6$ is 13535.

**Final Answer**
\boxed{13535}

To find the greatest integer less than $(\sqrt{7} + \sqrt{5})^6$, we can use a clever algebraic trick involving the conjugate of the expression. Let:

$$
x = \sqrt{7} + \sqrt{5}, \quad y = \sqrt{7} - \sqrt{5}
$$

Note that:

$$
x + y = 2\sqrt{7}, \quad xy = (\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5}) = 7 - 5 = 2
$$

We define the sequence:

$$
S_n = x^n + y^n
$$

This sequence satisfies the recurrence relation:

$$
S_n = (x + y) S_{n-1} - xy S_{n-2}
$$

Substituting the known values:

$$
S_n = 2\sqrt{7} \cdot S_{n-1} - 2 \cdot S_{n-2}
$$

We compute the first few terms:

- $S_0 = x^0 + y^0 = 1 + 1 = 2$
- $S_1 = x + y = 2\sqrt{7}$
- $S_2 = (x + y)^2 - 2xy = (2\sqrt{7})^2 - 2 \cdot 2 = 28 - 4 = 24$
- $S_3 = 2\sqrt{7} \cdot S_2 - 2 \cdot S_1 = 2\sqrt{7} \cdot 24 - 2 \cdot 2\sqrt{7} = 48\sqrt{7} - 4\sqrt{7} = 44\sqrt{7}$
- $S_4 = 2\sqrt{7} \cdot S_3 - 2 \cdot S_2 = 2\sqrt{7} \cdot 44\sqrt{7} - 2 \cdot 24 = 88 \cdot 7 - 48 = 616 - 48 = 568$
- $S_5 = 2\sqrt{7} \cdot S_4 - 2 \cdot S_3 = 2\sqrt{7} \cdot 568 - 2 \cdot 44\sqrt{7} = 1136\sqrt{7} - 88\sqrt{7} = 1048\sqrt{7}$
- $S_6 = 2\sqrt{7} \cdot S_5 - 2 \cdot S_4 = 2\sqrt{7} \cdot 1048\sqrt{7} - 2 \cdot 568 = 2096 \cdot 7 - 1136 = 14672 - 1136 = 13536$

So, we have:

$$
x^6 + y^6 = S_6 = 13536
$$

Now, observe that $y = \sqrt{7} - \sqrt{5} \approx 0.4097$, so $y^6$ is a small positive number less than 1. Therefore:

$$
x^6 = 13536 - y^6
$$

Since $0 < y^6 < 1$, it follows that:

$$
13535 < x^6 < 13536
$$

Thus, the greatest integer less than $x^6$ is:

$$
\boxed{13535}
$$

Solution:
Let $x = \sqrt{7} + \sqrt{5}$ and $y = \sqrt{7} - \sqrt{5}.$

First, we can square $x = \sqrt{7} + \sqrt{5}$ and $y = \sqrt{7} - \sqrt{5},$ to get
\begin{align*}
x^2 &= (\sqrt{7} + \sqrt{5})^2 = 7 + 2 \sqrt{35} + 5 = 12 + 2 \sqrt{35}, \\
y^2 &= (\sqrt{7} - \sqrt{5})^2 = 7 - 2 \sqrt{35} + 5 = 12 - 2 \sqrt{35}.
\end{align*}Note that $x^2$ and $y^2$ are radical conjugates.  Also, $x^2 y^2 = (12 + 2 \sqrt{35})(12 - 2 \sqrt{35}) = 12^2 - 2^2 \cdot 35 = 4,$ so
\[y^2 = \frac{4}{x^2} = \frac{4}{12 + 2 \sqrt{35}} < 1.\]Then
\[x^4 = (12 + 2 \sqrt{35})^2 = 12^2 + 2 \cdot 12 \cdot 2 \sqrt{35} + 2^2 \cdot 35 = 284 + 48 \sqrt{35},\]and
\begin{align*}
x^6 &= x^2 \cdot x^4 \\
&= (12 + 2 \sqrt{35})(284 + 48 \sqrt{35}) \\
&= 12 \cdot 284 + 12 \cdot 48 \sqrt{35} + 2 \sqrt{35} \cdot 284 + 2 \cdot \sqrt{35} \cdot 48 \cdot \sqrt{35} \\
&= 6768 + 1144 \sqrt{35}.
\end{align*}Then $y^6$ is the radical conjugate of $x^6,$ so $y^6 = 6768 - 1144 \sqrt{35}.$  Hence,
\[x^6 + y^6 = (6768 + 1144 \sqrt{35}) + (6768 - 1144 \sqrt{35}) = 13536.\]Since $0 < y^6 < 1,$ the greatest integer less than $x^6$ is $\boxed{13535}.$


Explanation:
Okay, so I need to find the product of 6 base 8 and 7 base 8, and then express the answer in base 8. Hmm, let me think about how to approach this.

First, I remember that when multiplying numbers in different bases, it might be easier to convert them to base 10 first, do the multiplication there, and then convert the result back to base 8. Let me try that method.

So, starting with the first number, which is 6 in base 8. Since it's a single digit, converting it to base 10 is straightforward. In base 8, each digit represents a power of 8. But since it's just 6, it's 6 * 8^0, which is 6 * 1 = 6. So, 6 base 8 is 6 in base 10.

Now the second number is 7 base 8. Similarly, this is also a single digit, so converting to base 10 is just 7 * 8^0 = 7 * 1 = 7. So, 7 base 8 is 7 in base 10.

Now, multiplying them in base 10: 6 * 7. Let me calculate that. 6 times 7 is 42. So, the product in base 10 is 42. Now, I need to convert 42 from base 10 back to base 8.

To convert a base 10 number to base 8, I need to divide the number by 8 and find the remainders. Let me recall the process.

First, divide 42 by 8. Let me do that division. 8 goes into 42 how many times? 8*5 is 40, which is less than 42. 8*5=40, so the quotient is 5 and the remainder is 42 - 40 = 2.

So, the first remainder is 2. Then, take the quotient, which is 5, and divide that by 8. 5 divided by 8 is 0 with a remainder of 5.

So, the remainders, from last to first, give the base 8 number. So, the first remainder we got was 2, and then the next remainder was 5. Therefore, 42 in base 10 is 52 in base 8.

Wait, let me check that. Let me verify by converting 52 base 8 back to base 10.

5*8^1 + 2*8^0 = 5*8 + 2*1 = 40 + 2 = 42. Yes, that's correct. So, 52 base 8 is indeed 42 base 10.

Therefore, the product of 6_8 and 7_8 is 52_8.

But let me try another method just to make sure I didn't make a mistake. Maybe multiplying directly in base 8?

Multiplying numbers in base 8. Let me recall how multiplication works in different bases.

So, similar to base 10, but with base 8. Let me write it out:

      6_8
    x 7_8
    -------

So, multiplying 6 by 7 in base 8. Wait, but in base 8, the digits go from 0 to 7, so 6 and 7 are valid digits.

But when multiplying, I need to remember that if the product is more than 7, I need to carry over.

So, 6 * 7. Let me compute that in base 10 first, which is 42. But since we are working in base 8, I need to express 42 in base 8. Wait, that's exactly what I did earlier. So, 42 in base 10 is 52 in base 8.

Alternatively, if I try to do the multiplication directly in base 8, let me think.

But maybe I need to do the multiplication step by step. Let me try that.

So, 6_8 multiplied by 7_8.

First, 6 * 7. In base 8, what is 6 * 7? Well, 6*7 is 42 in base 10. Now, converting 42 to base 8, as before, gives 52. So, the result is 52_8.

Alternatively, if I try to do the multiplication without converting to base 10, maybe I can do it. Let me think.

In base 8, each digit is multiplied, and then carries are handled. Let me recall that.

But since both numbers are single digits, it's straightforward. So, 6 * 7.

But in base 8, the multiplication is similar to base 10, except that when you get a product that is equal to or larger than 8, you have to carry over.

So, 6 * 7 = 42. Now, 42 divided by 8 is 5 with a remainder of 2. So, the digit is 2, and carry over 5.

But since there are no more digits to multiply, the carry over becomes the next digit. So, the result is 52_8.

So, same result. Therefore, regardless of the method, I get 52_8.

Wait, but let me check once more. Maybe I can convert both numbers to base 10, multiply, convert back.

As before, 6_8 is 6, 7_8 is 7. 6*7=42. 42 divided by 8 is 5, remainder 2. So, 5*8 + 2 = 42. So, 52_8.

Alternatively, maybe I can use another approach. Let me think.

Alternatively, maybe I can use base 8 multiplication tables. Let me try to recall or construct the multiplication table for base 8.

But since I don't have it memorized, maybe I can compute 6*7 in base 8.

But 6*7 in base 8. Let me think of 6 as 6 and 7 as 7.

But in base 8, 6*7 is equal to... Well, 6*7 is 42 in decimal. But how do I express that in base 8?

Well, 8*5=40, so 42-40=2. So, 5*8 + 2 = 42. Therefore, 52 in base 8.

So, same result.

Alternatively, maybe I can think of 6_8 as 6 and 7_8 as 7, and multiply them as if they were base 10 numbers, but then adjust for base 8.

But that seems similar to the first method.

Alternatively, maybe I can do the multiplication in base 8 step by step.

Let me try that.

So, multiplying 6_8 by 7_8.

First, write down the multiplication:

      6
    x 7
    ----

In base 8.

So, 6 * 7. Let me think of 7 as 7 in base 8.

But in base 8, 7 is just 7. So, 6*7.

But in base 8, the multiplication is similar to base 10. Let me think of 6*7.

But since 6*7 is 42 in base 10, which is 5*8 + 2, so 52 in base 8.

Therefore, the result is 52_8.

Alternatively, if I didn't convert to base 10, how would I do it?

Well, maybe I can think of 6*7 in base 8. Let me think of 6 as 6 and 7 as 7.

But since 6*7 is 42, which is more than 8, so I need to carry over.

So, 42 divided by 8 is 5 with remainder 2. So, the result is 52.

Therefore, same answer.

Alternatively, maybe I can use the distributive property.

For example, 6_8 * 7_8.

But 7_8 is equal to 8_10 - 1_10. Wait, but in base 8, 7 is 7, but if I think in base 10, 7_8 is 7.

Alternatively, maybe I can think of 7_8 as 7 in base 8.

Alternatively, maybe I can think of 6_8 * 7_8 as 6_8 * (6_8 + 1_8).

So, 6_8 * 6_8 + 6_8 * 1_8.

But then, I need to compute 6_8 * 6_8 and add 6_8.

But maybe this is more complicated. Let me try.

First, compute 6_8 * 6_8.

Again, converting to base 10, 6*6=36. Then convert 36 to base 8.

Divide 36 by 8: 8*4=32, remainder 4. Then divide 4 by 8: 0 with remainder 4. So, 44_8.

So, 6_8 * 6_8 is 44_8. Then add 6_8.

So, 44_8 + 6_8.

Adding in base 8. Let me do that.

44_8 + 6_8.

Starting from the rightmost digit: 4 + 6.

In base 8, 4 + 6 = 10 in base 10. But since we are in base 8, 10 in base 10 is 12 in base 8 (since 8 + 2 = 10).

Wait, 4 + 6 = 10 decimal. To convert to base 8, divide 10 by 8: 1 with remainder 2. So, the digit is 2, carry over 1.

Then, add the carry over to the next digit. The next digit is 4 + 0 (since there is no digit in the second number) plus the carry over 1.

So, 4 + 1 = 5.

Therefore, the result is 52_8.

Which matches the previous result.

So, that's another way to do it.

Therefore, regardless of the method, I get 52_8.

But just to make sure, let me check once again.

Alternatively, maybe I can convert both numbers to base 2 and do the multiplication? Although that might be more complicated, but let me try.

First, converting 6_8 to binary.

Each octal digit is 3 binary digits.

6 in octal is 110 in binary.

Similarly, 7 in octal is 111 in binary.

So, multiplying 110_2 * 111_2.

Let me compute that.

First, 110 * 111.

In binary, let me recall that 110 is 6, 111 is 7. 6*7=42.

But converting 42 to binary.

Divide 42 by 2: 21, remainder 0

21 /2 = 10, rem 1

10 /2 = 5, rem 0

5 /2 = 2, rem 1

2 /2 = 1, rem 0

1 /2 = 0, rem 1

So, writing the remainders from last to first: 101010.

So, 101010 in binary.

Now, converting 101010 binary to octal.

Since each octal digit is 3 bits, let me group the binary number into groups of 3 starting from the right.

101010 is 101 010.

101 is 5 in octal, 010 is 2 in octal. So, 52_8.

Which is the same result.

Therefore, that confirms it again.

So, all methods lead to 52_8.

Therefore, I can be confident that the product of 6_8 and 7_8 is 52_8.

But just to recap, the steps were:

1. Convert both numbers from base 8 to base 10: 6 and 7.

2. Multiply them in base 10: 6*7=42.

3. Convert 42 from base 10 to base 8: 52_8.

Alternatively, multiplying directly in base 8 by using base 8 arithmetic, which also gives 52_8.

Another method was converting to binary, multiplying, then converting back to octal, which also gives 52_8.

Therefore, all methods concur.

I think that's thorough enough. I don't see any mistakes in my reasoning.

**Final Answer**
The product of $6_8 \cdot 7_8$ in base 8 is \boxed{52_8}.

To find the product of $6_8 \cdot 7_8$ and express the result in base 8, we can follow a systematic approach by first converting the numbers from base 8 to base 10, performing the multiplication in base 10, and then converting the result back to base 8.

---

### Step 1: Convert from Base 8 to Base 10

- $6_8 = 6 \times 8^0 = 6$
- $7_8 = 7 \times 8^0 = 7$

So, the numbers in base 10 are:
- $6_8 = 6$
- $7_8 = 7$

---

### Step 2: Multiply in Base 10

$$
6 \times 7 = 42
$$

---

### Step 3: Convert the Result Back to Base 8

To convert $42_{10}$ to base 8, we perform successive division by 8 and record the remainders:

- $42 \div 8 = 5$ remainder $2$
- $5 \div 8 = 0$ remainder $5$

Reading the remainders from bottom to top gives us the base 8 representation:

$$
42_{10} = 52_8
$$

---

### Final Answer

$$
\boxed{52}
$$

[Assistant]
Explanation:
Okay, so I need to find the product of 6 base 8 and 7 base 8, and then express the answer in base 8. Hmm, let me think about how to approach this.

First, I remember that when multiplying numbers in different bases, it might be easier to convert them to base 10 first, do the multiplication there, and then convert the result back to base 8. Let me try that method.

So, starting with the first number, which is 6 in base 8. Since it's a single digit, converting it to base 10 is straightforward. In base 8, each digit represents a power of 8. But since it's just 6, it's 6 * 8^0, which is 6 * 1 = 6. So, 6 base 8 is 6 in base 10.

Now the second number is 7 base 8. Similarly, this is also a single digit, so converting to base 10 is just 7 * 8^0 = 7 * 1 = 7. So, 7 base 8 is 7 in base 10.

Now, multiplying them in base 10: 6 * 7. Let me calculate that. 6 times 7 is 42. So, the product in base 10 is 42. Now, I need to convert 42 from base 10 back to base 8.

To convert a base 10 number to base 8, I need to divide the number by 8 and find the remainders. Let me recall the process.

First, divide 42 by 8. Let me do that division. 8 goes into 42 how many times? 8*5 is 40, which is less than 42. 8*5=40, so the quotient is 5 and the remainder is 42 - 40 = 2.

So, the first remainder is 2. Then, take the quotient, which is 5, and divide that by 8. 5 divided by 8 is 0 with a remainder of 5.

So, the remainders, from last to first, give the base 8 number. So, the first remainder we got was 2, and then the next remainder was 5. Therefore, 42 in base 10 is 52 in base 8.

Wait, let me check that. Let me verify by converting 52 base 8 back to base 10.

5*8^1 + 2*8^0 = 5*8 + 2*1 = 40 + 2 = 42. Yes, that's correct. So, 52 base 8 is indeed 42 base 10.

Therefore, the product of 6_8 and 7_8 is 52_8.

But let me try another method just to make sure I didn't make a mistake. Maybe multiplying directly in base 8?

Multiplying numbers in base 8. Let me recall how multiplication works in different bases.

So, similar to base 10, but with base 8. Let me write it out:

      6_8
    x 7_8
    -------

So, multiplying 6 by 7 in base 8. Wait, but in base 8, the digits go from 0 to 7, so 6 and 7 are valid digits.

But when multiplying, I need to remember that if the product is more than 7, I need to carry over.

So, 6 * 7. Let me compute that in base 10 first, which is 42. But since we are working in base 8, I need to express 42 in base 8. Wait, that's exactly what I did earlier. So, 42 in base 10 is 52 in base 8.

Alternatively, if I try to do the multiplication directly in base 8, let me think.

But maybe I need to do the multiplication step by step. Let me try that.

So, 6_8 multiplied by 7_8.

First, 6 * 7. In base 8, what is 6 * 7? Well, 6*7 is 42 in base 10. Now, converting 42 to base 8, as before, gives 52. So, the result is 52_8.

Alternatively, if I try to do the multiplication without converting to base 10, maybe I can do it. Let me think.

In base 8, each digit is multiplied, and then carries are handled. Let me recall that.

But since both numbers are single digits, it's straightforward. So, 6 * 7.

But in base 8, the multiplication is similar to base 10, except that when you get a product that is equal to or larger than 8, you have to carry over.

So, 6 * 7 = 42. Now, 42 divided by 8 is 5 with a remainder of 2. So, the digit is 2, and carry over 5.

But since there are no more digits to multiply, the carry over becomes the next digit. So, the result is 52_8.

So, same result. Therefore, regardless of the method, I get 52_8.

Wait, but let me check once more. Maybe I can convert both numbers to base 10, multiply, convert back.

As before, 6_8 is 6, 7_8 is 7. 6*7=42. 42 divided by 8 is 5, remainder 2. So, 5*8 + 2 = 42. So, 52_8.

Alternatively, maybe I can use another approach. Let me think.

Alternatively, maybe I can use base 8 multiplication tables. Let me try to recall or construct the multiplication table for base 8.

But since I don't have it memorized, maybe I can compute 6*7 in base 8.

But 6*7 in base 8. Let me think of 6 as 6 and 7 as 7.

But in base 8, 6*7 is equal to... Well, 6*7 is 42 in decimal. But how do I express that in base 8?

Well, 8*5=40, so 42-40=2. So, 5*8 + 2 = 42. Therefore, 52 in base 8.

So, same result.

Alternatively, maybe I can think of 6_8 as 6 and 7_8 as 7, and multiply them as if they were base 10 numbers, but then adjust for base 8.

But that seems similar to the first method.

Alternatively, maybe I can do the multiplication in base 8 step by step.

Let me try that.

So, multiplying 6_8 by 7_8.

First, write down the multiplication:

      6
    x 7
    ----

In base 8.

So, 6 * 7. Let me think of 7 as 7 in base 8.

But in base 8, 7 is just 7. So, 6*7.

But in base 8, the multiplication is similar to base 10. Let me think of 6*7.

But since 6*7 is 42 in base 10, which is 5*8 + 2, so 52 in base 8.

Therefore, the result is 52_8.

Alternatively, if I didn't convert to base 10, how would I do it?

Well, maybe I can think of 6*7 in base 8. Let me think of 6 as 6 and 7 as 7.

But since 6*7 is 42, which is more than 8, so I need to carry over.

So, 42 divided by 8 is 5 with remainder 2. So, the result is 52.

Therefore, same answer.

Alternatively, maybe I can use the distributive property.

For example, 6_8 * 7_8.

But 7_8 is equal to 8_10 - 1_10. Wait, but in base 8, 7 is 7, but if I think in base 10, 7_8 is 7.

Alternatively, maybe I can think of 7_8 as 7 in base 8.

Alternatively, maybe I can think of 6_8 * 7_8 as 6_8 * (6_8 + 1_8).

So, 6_8 * 6_8 + 6_8 * 1_8.

But then, I need to compute 6_8 * 6_8 and add 6_8.

But maybe this is more complicated. Let me try.

First, compute 6_8 * 6_8.

Again, converting to base 10, 6*6=36. Then convert 36 to base 8.

Divide 36 by 8: 8*4=32, remainder 4. Then divide 4 by 8: 0 with remainder 4. So, 44_8.

So, 6_8 * 6_8 is 44_8. Then add 6_8.

So, 44_8 + 6_8.

Adding in base 8. Let me do that.

44_8 + 6_8.

Starting from the rightmost digit: 4 + 6.

In base 8, 4 + 6 = 10 in base 10. But since we are in base 8, 10 in base 10 is 12 in base 8 (since 8 + 2 = 10).

Wait, 4 + 6 = 10 decimal. To convert to base 8, divide 10 by 8: 1 with remainder 2. So, the digit is 2, carry over 1.

Then, add the carry over to the next digit. The next digit is 4 + 0 (since there is no digit in the second number) plus the carry over 1.

So, 4 + 1 = 5.

Therefore, the result is 52_8.

Which matches the previous result.

So, that's another way to do it.

Therefore, regardless of the method, I get 52_8.

But just to make sure, let me check once again.

Alternatively, maybe I can convert both numbers to base 2 and do the multiplication? Although that might be more complicated, but let me try.

First, converting 6_8 to binary.

Each octal digit is 3 binary digits.

6 in octal is 110 in binary.

Similarly, 7 in octal is 111 in binary.

So, multiplying 110_2 * 111_2.

Let me compute that.

First, 110 * 111.

In binary, let me recall that 110 is 6, 111 is 7. 6*7=42.

But converting 42 to binary.

Divide 42 by 2: 21, remainder 0

21 /2 = 10, rem 1

10 /2 = 5, rem 0

5 /2 = 2, rem 1

2 /2 = 1, rem 0

1 /2 = 0, rem 1

So, writing the remainders from last to first: 101010.

So, 101010 in binary.

Now, converting 101010 binary to octal.

Since each octal digit is 3 bits, let me group the binary number into groups of 3 starting from the right.

101010 is 101 010.

101 is 5 in octal, 010 is 2 in octal. So, 52_8.

Which is the same result.

Therefore, that confirms it again.

So, all methods lead to 52_8.

Therefore, I can be confident that the product of 6_8 and 7_8 is 52_8.

But just to recap, the steps were:

1. Convert both numbers from base 8 to base 10: 6 and 7.

2. Multiply them in base 10: 6*7=42.

3. Convert 42 from base 10 to base 8: 52_8.

Alternatively, multiplying directly in base 8 by using base 8 arithmetic, which also gives 52_8.

Another method was converting to binary, multiplying, then converting back to octal, which also gives 52_8.

Therefore, all methods concur.

I think that's thorough enough. I don't see any mistakes in my reasoning.

**Final Answer**
The product of $6_8 \cdot 7_8$ in base 8 is \boxed{52_8}.

To find the product of $6_8 \cdot 7_8$ and express the result in base 8, we can follow a systematic approach by first converting the numbers from base 8 to base 10, performing the multiplication in base 10, and then converting the result back to base 8.

---

### Step 1: Convert from Base 8 to Base 10

- $6_8 = 6 \times 8^0 = 6$
- $7_8 = 7 \times 8^0 = 7$

So, the numbers in base 10 are:
- $6_8 = 6$
- $7_8 = 7$

---

### Step 2: Multiply in Base 10

$$
6 \times 7 = 42
$$

---

### Step 3: Convert the Result Back to Base 8

To convert $42_{10}$ to base 8, we perform successive division by 8 and record the remainders:

- $42 \div 8 = 5$ remainder $2$
- $5 \div 8 = 0$ remainder $5$

Reading the remainders from bottom to top gives us the base 8 representation:

$$
42_{10} = 52_8
$$

---

### Final Answer

$$
\boxed{52}
$$

Solution: Multiplying, we see that $6_8 \cdot 7_8 = 42_{10} = 52_8.$ Written out, $$\begin{array}{@{}c@{\;}c@{}c@{}c} && & 6_8 \\ & \times & & 7_8 \\ \cline{2-4} & & 5 & 2_8 \\ \end{array} $$ Thus, the answer is $\boxed{52_8}.$


Explanation:
Okay, so I need to figure out the minimum number of degrees a regular pentagon has to be rotated counterclockwise about its center so that it coincides with its original position. Hmm, let me think.

First, I remember that regular polygons have rotational symmetry. That means if you rotate them by certain angles, they look the same as before. The question is asking for the smallest such angle for a pentagon.

Let me recall, for a regular polygon with n sides, the order of rotational symmetry is n. So, for a pentagon, n is 5. The angle of rotation would be 360 degrees divided by n, right? So, 360 divided by 5. Let me check that.

Wait, if I have a pentagon, each vertex is equally spaced around the center. So, the angle between each vertex from the center is 360/5 degrees. So, if you rotate the pentagon by that angle, it should align with itself again.

Let me verify with an example. Let's say I have a regular pentagon. If I rotate it by 72 degrees, which is 360/5, then each vertex should move to the position of the next vertex. Since all sides and angles are equal, this should make the pentagon look the same.

But wait, is 72 degrees the smallest such angle? Let me think. If I rotate it by a smaller angle, say 36 degrees, would it coincide? Probably not, because 36 is a factor of 72. But since the pentagon has five sides, the symmetry is every 72 degrees.

Wait, but maybe there's a smaller angle? Let me think. For example, if you have a square, which has four sides, the rotation angle is 90 degrees. But if you rotate it by 180 degrees, it also coincides. But 90 is smaller than 180. So, the minimal angle is 360 divided by the number of sides.

Similarly, for a pentagon, the minimal angle should be 360/5, which is 72 degrees. So, rotating by 72 degrees would bring it back to the original position.

But let me think again. Suppose I have a regular pentagon. Let me try to visualize. Each vertex is separated by 72 degrees. So, if I rotate it by 72 degrees, each vertex moves to the next one. Since all the sides and angles are equal, the shape is the same.

But if I rotate it by a smaller angle, like 36 degrees, which is half of 72, would that work? Let me think. If I rotate by 36 degrees, then each vertex would have moved 36 degrees. But since the angle between each vertex is 72 degrees, rotating by 36 degrees would not align any vertex with another. Therefore, the pentagon would not coincide with its original position.

Therefore, 72 degrees is the minimal angle.

But let me check another way. The order of rotational symmetry is the number of times the shape maps onto itself when rotated around the center by 360 degrees. For a regular pentagon, the order is 5, meaning it maps onto itself 5 times as you rotate 360 degrees. Therefore, each rotation angle is 360/5 = 72 degrees.

Alternatively, if I think about the central angles. Each central angle (the angle at the center between two adjacent vertices) is 360/5 = 72 degrees. Therefore, rotating by that angle would bring each vertex to the position of the next one, so the whole pentagon coincides with itself.

Therefore, the minimal angle is 72 degrees.

But wait, is there a smaller angle? For example, if the pentagon has some other symmetry? For example, maybe if you rotate by 144 degrees, which is 2*72, but that's larger.

Alternatively, maybe if there's a smaller angle that divides 360 degrees? Let me think. The divisors of 360 degrees are numbers that can divide 360 evenly. For example, 1, 2, 3, 4, 5, 6, etc. But for the pentagon, the rotational symmetry is only multiples of 72 degrees.

Wait, but 72 is 360 divided by 5. So, if you want the minimal angle, it's 360 divided by the number of sides.

Alternatively, if you think about the dihedral group of the pentagon, which includes rotations and reflections. But the question is only about rotation, so reflections are not considered here.

Therefore, the minimal rotation angle is 72 degrees.

But let me check with an example. Let me think of a regular pentagon. Let me label the vertices A, B, C, D, E. If I rotate it by 72 degrees, then vertex A moves to where B is, B moves to C, etc., and E moves to A. Therefore, the entire pentagon is in the same position.

If I rotate by 72 degrees, it's the same as the original. If I rotate by less than 72 degrees, say 36 degrees, then vertex A would be somewhere between A and B. Since the angle between A and B is 72 degrees, rotating by 36 degrees would not align any vertex with another. Therefore, the pentagon would not coincide with its original position.

Therefore, 72 degrees is indeed the minimal angle.

Alternatively, if I think about the formula for rotational symmetry. For a regular polygon with n sides, the angle of rotation is 360/n degrees. Therefore, for n=5, it's 360/5=72 degrees.

Therefore, the answer should be 72 degrees.

But let me check if there's any other way this could be approached. For example, using the concept of the order of rotation. The order is the number of times the shape maps onto itself in a full rotation. For a regular pentagon, the order is 5, so the angle is 360 divided by order, which is 72.

Alternatively, if I think about the internal angles of the pentagon. Wait, the internal angle of a regular pentagon is 108 degrees. But that's different from the rotational symmetry angle. The internal angle is about the angles at the vertices, whereas the rotational symmetry is about the angle you rotate the whole shape.

So, that's different. So, the internal angle is not directly related to the rotational symmetry angle.

Another way: if I imagine the pentagon on a coordinate system, with its center at the origin. Each vertex is at an angle of 0°, 72°, 144°, 216°, 288°, and then back to 360°=0°. So, if you rotate the pentagon by 72°, each vertex moves to the next one. Therefore, the pentagon coincides with itself.

Therefore, the minimal angle is 72 degrees.

I think that's solid. So, the answer is 72 degrees.

**Final Answer**
The minimum number of degrees the pentagon must be rotated is \boxed{72}.

To determine the **minimum number of degrees** a regular pentagon must be rotated counterclockwise about its center to coincide with its original position, we need to understand the concept of **rotational symmetry** in regular polygons.

---

### Understanding Rotational Symmetry

A **regular polygon** has **rotational symmetry** if it can be rotated about its center by a certain angle and still appear identical to its original position. The **order of rotational symmetry** for a regular polygon with $ n $ sides is $ n $, meaning the shape maps onto itself exactly $ n $ times as it completes a full 360° rotation.

For a **regular pentagon**, which has 5 sides, the **order of rotational symmetry** is 5. This means that the shape will align with itself every time it is rotated by an angle that is a **multiple of** $ \frac{360^\circ}{5} $.

---

### Calculating the Minimum Rotation Angle

To find the **smallest angle** that results in the pentagon coinciding with its original position, we compute:

$$
\text{Minimum rotation angle} = \frac{360^\circ}{n} = \frac{360^\circ}{5} = 72^\circ
$$

This means that rotating the pentagon by **72 degrees** will bring each vertex to the position of the next one, and the entire shape will look exactly the same as before.

---

### Verifying the Result

- If we rotate the pentagon by **72°**, each vertex moves to the position of the next vertex.
- Since all sides and angles are equal, the shape remains unchanged.
- Any smaller angle (e.g., 36°, 18°, etc.) would not align the vertices with each other, and the pentagon would not coincide with its original position.

Thus, **72° is the smallest angle** for which the pentagon maps onto itself.

---

### Final Answer

$$
\boxed{72}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the minimum number of degrees a regular pentagon has to be rotated counterclockwise about its center so that it coincides with its original position. Hmm, let me think.

First, I remember that regular polygons have rotational symmetry. That means if you rotate them by certain angles, they look the same as before. The question is asking for the smallest such angle for a pentagon.

Let me recall, for a regular polygon with n sides, the order of rotational symmetry is n. So, for a pentagon, n is 5. The angle of rotation would be 360 degrees divided by n, right? So, 360 divided by 5. Let me check that.

Wait, if I have a pentagon, each vertex is equally spaced around the center. So, the angle between each vertex from the center is 360/5 degrees. So, if you rotate the pentagon by that angle, it should align with itself again.

Let me verify with an example. Let's say I have a regular pentagon. If I rotate it by 72 degrees, which is 360/5, then each vertex should move to the position of the next vertex. Since all sides and angles are equal, this should make the pentagon look the same.

But wait, is 72 degrees the smallest such angle? Let me think. If I rotate it by a smaller angle, say 36 degrees, would it coincide? Probably not, because 36 is a factor of 72. But since the pentagon has five sides, the symmetry is every 72 degrees.

Wait, but maybe there's a smaller angle? Let me think. For example, if you have a square, which has four sides, the rotation angle is 90 degrees. But if you rotate it by 180 degrees, it also coincides. But 90 is smaller than 180. So, the minimal angle is 360 divided by the number of sides.

Similarly, for a pentagon, the minimal angle should be 360/5, which is 72 degrees. So, rotating by 72 degrees would bring it back to the original position.

But let me think again. Suppose I have a regular pentagon. Let me try to visualize. Each vertex is separated by 72 degrees. So, if I rotate it by 72 degrees, each vertex moves to the next one. Since all the sides and angles are equal, the shape is the same.

But if I rotate it by a smaller angle, like 36 degrees, which is half of 72, would that work? Let me think. If I rotate by 36 degrees, then each vertex would have moved 36 degrees. But since the angle between each vertex is 72 degrees, rotating by 36 degrees would not align any vertex with another. Therefore, the pentagon would not coincide with its original position.

Therefore, 72 degrees is the minimal angle.

But let me check another way. The order of rotational symmetry is the number of times the shape maps onto itself when rotated around the center by 360 degrees. For a regular pentagon, the order is 5, meaning it maps onto itself 5 times as you rotate 360 degrees. Therefore, each rotation angle is 360/5 = 72 degrees.

Alternatively, if I think about the central angles. Each central angle (the angle at the center between two adjacent vertices) is 360/5 = 72 degrees. Therefore, rotating by that angle would bring each vertex to the position of the next one, so the whole pentagon coincides with itself.

Therefore, the minimal angle is 72 degrees.

But wait, is there a smaller angle? For example, if the pentagon has some other symmetry? For example, maybe if you rotate by 144 degrees, which is 2*72, but that's larger.

Alternatively, maybe if there's a smaller angle that divides 360 degrees? Let me think. The divisors of 360 degrees are numbers that can divide 360 evenly. For example, 1, 2, 3, 4, 5, 6, etc. But for the pentagon, the rotational symmetry is only multiples of 72 degrees.

Wait, but 72 is 360 divided by 5. So, if you want the minimal angle, it's 360 divided by the number of sides.

Alternatively, if you think about the dihedral group of the pentagon, which includes rotations and reflections. But the question is only about rotation, so reflections are not considered here.

Therefore, the minimal rotation angle is 72 degrees.

But let me check with an example. Let me think of a regular pentagon. Let me label the vertices A, B, C, D, E. If I rotate it by 72 degrees, then vertex A moves to where B is, B moves to C, etc., and E moves to A. Therefore, the entire pentagon is in the same position.

If I rotate by 72 degrees, it's the same as the original. If I rotate by less than 72 degrees, say 36 degrees, then vertex A would be somewhere between A and B. Since the angle between A and B is 72 degrees, rotating by 36 degrees would not align any vertex with another. Therefore, the pentagon would not coincide with its original position.

Therefore, 72 degrees is indeed the minimal angle.

Alternatively, if I think about the formula for rotational symmetry. For a regular polygon with n sides, the angle of rotation is 360/n degrees. Therefore, for n=5, it's 360/5=72 degrees.

Therefore, the answer should be 72 degrees.

But let me check if there's any other way this could be approached. For example, using the concept of the order of rotation. The order is the number of times the shape maps onto itself in a full rotation. For a regular pentagon, the order is 5, so the angle is 360 divided by order, which is 72.

Alternatively, if I think about the internal angles of the pentagon. Wait, the internal angle of a regular pentagon is 108 degrees. But that's different from the rotational symmetry angle. The internal angle is about the angles at the vertices, whereas the rotational symmetry is about the angle you rotate the whole shape.

So, that's different. So, the internal angle is not directly related to the rotational symmetry angle.

Another way: if I imagine the pentagon on a coordinate system, with its center at the origin. Each vertex is at an angle of 0°, 72°, 144°, 216°, 288°, and then back to 360°=0°. So, if you rotate the pentagon by 72°, each vertex moves to the next one. Therefore, the pentagon coincides with itself.

Therefore, the minimal angle is 72 degrees.

I think that's solid. So, the answer is 72 degrees.

**Final Answer**
The minimum number of degrees the pentagon must be rotated is \boxed{72}.

To determine the **minimum number of degrees** a regular pentagon must be rotated counterclockwise about its center to coincide with its original position, we need to understand the concept of **rotational symmetry** in regular polygons.

---

### Understanding Rotational Symmetry

A **regular polygon** has **rotational symmetry** if it can be rotated about its center by a certain angle and still appear identical to its original position. The **order of rotational symmetry** for a regular polygon with $ n $ sides is $ n $, meaning the shape maps onto itself exactly $ n $ times as it completes a full 360° rotation.

For a **regular pentagon**, which has 5 sides, the **order of rotational symmetry** is 5. This means that the shape will align with itself every time it is rotated by an angle that is a **multiple of** $ \frac{360^\circ}{5} $.

---

### Calculating the Minimum Rotation Angle

To find the **smallest angle** that results in the pentagon coinciding with its original position, we compute:

$$
\text{Minimum rotation angle} = \frac{360^\circ}{n} = \frac{360^\circ}{5} = 72^\circ
$$

This means that rotating the pentagon by **72 degrees** will bring each vertex to the position of the next one, and the entire shape will look exactly the same as before.

---

### Verifying the Result

- If we rotate the pentagon by **72°**, each vertex moves to the position of the next vertex.
- Since all sides and angles are equal, the shape remains unchanged.
- Any smaller angle (e.g., 36°, 18°, etc.) would not align the vertices with each other, and the pentagon would not coincide with its original position.

Thus, **72° is the smallest angle** for which the pentagon maps onto itself.

---

### Final Answer

$$
\boxed{72}
$$

Solution:
Each of the five marked angles measures $360/5=72$ degrees, so $\boxed{72}$ degrees is the minimum angle through which the pentagon may be rotated so that it coincides with its original position.

[asy]
size(150);
defaultpen(linewidth(0.7));
int i;
for(i=0;i<=4;++i)

{
draw(origin--dir(18+72*i)--dir(18+72*(i+1)));
draw(anglemark(dir(18+72*i),origin,dir(18+72*(i+1)),3+fmod(i,3)));
}
[/asy]


Explanation:
Okay, so I need to find the greatest common factor (GCF) of 6432 and 132, and then add 11 to that result. Let me think about how to approach this.

First, I remember that the GCF of two numbers is the largest number that divides both of them without leaving a remainder. To find the GCF, one method is to use the prime factorization of both numbers and then multiply the common prime factors with the lowest exponents. Alternatively, I can use the Euclidean algorithm, which might be more efficient for larger numbers. Let me try both methods and see which one works better here.

Starting with prime factorization. Let me factorize 132 first because it seems smaller.

132 can be divided by 2 first. 132 ÷ 2 = 66. Then 66 ÷ 2 = 33. Now 33 can be divided by 3: 33 ÷ 3 = 11. And 11 is a prime number. So the prime factors of 132 are 2² × 3 × 11.

Now let me factorize 6432. Hmm, this might take a bit longer. Let me start dividing by small primes.

First, check if it's even. 6432 ÷ 2 = 3216. Still even, so divide by 2 again: 3216 ÷ 2 = 1608. Again, divide by 2: 1608 ÷ 2 = 804. Again by 2: 804 ÷ 2 = 402. Again by 2: 402 ÷ 2 = 201. Now, 201 is not even. Let me check divisibility by 3. 2 + 0 + 1 = 3, which is divisible by 3. So 201 ÷ 3 = 67. Now, 67 is a prime number.

So putting it all together, the prime factors of 6432 are 2⁵ × 3 × 67. Wait, let me check that again. Let me recount the divisions:

Starting with 6432:

Divided by 2 five times: 6432 ÷ 2 = 3216, ÷2=1608, ÷2=804, ÷2=402, ÷2=201. So that's 2⁵. Then 201 ÷ 3 = 67. So yes, 2⁵ × 3 × 67.

So prime factors of 6432: 2⁵ × 3 × 67.

Prime factors of 132: 2² × 3 × 11.

Now, to find the GCF, we take the minimum exponents of the common prime factors.

Common prime factors are 2 and 3.

For 2: minimum exponent between 5 and 2 is 2.

For 3: minimum exponent between 1 and 1 is 1.

So GCF is 2² × 3 = 4 × 3 = 12.

Wait, but let me check if that's correct. Let me verify with another method to be sure. Maybe using the Euclidean algorithm.

The Euclidean algorithm says that GCF(a, b) = GCF(b, a mod b). So let's apply that.

First, take 6432 and 132.

Compute 6432 ÷ 132. Let me find how many times 132 goes into 6432.

Well, 132 × 50 = 6600, which is more than 6432. Let me try 132 × 49.

132 × 40 = 5280.

132 × 9 = 1188.

So 5280 + 1188 = 6468. That's still more than 6432.

So 132 × 48 = 132 × 40 + 132 × 8 = 5280 + 1056 = 6336.

Subtract that from 6432: 6432 - 6336 = 96.

So GCF(6432, 132) = GCF(132, 96).

Now compute GCF(132, 96).

Again, divide 132 by 96. 96 × 1 = 96. Subtract from 132: 132 - 96 = 36.

So GCF(132, 96) = GCF(96, 36).

Now GCF(96, 36). Divide 96 by 36. 36 × 2 = 72. Subtract from 96: 96 - 72 = 24.

So GCF(96, 36) = GCF(36, 24).

GCF(36, 24). Divide 36 by 24. 24 × 1 = 24. Subtract: 36 - 24 = 12.

GCF(36, 24) = GCF(24, 12).

GCF(24, 12). Divide 24 by 12. 12 × 2 = 24. Remainder 0.

So GCF is 12.

Okay, so both methods give me 12. So the GCF of 6432 and 132 is 12.

Therefore, increasing that by 11 would be 12 + 11 = 23.

Wait, but let me just make sure I didn't make any calculation errors. Let me check the prime factorization again.

For 6432:

Divided by 2 five times: 6432 / 2 = 3216, /2=1608, /2=804, /2=402, /2=201. Then 201 divided by 3 is 67. So yes, 2^5 * 3 * 67.

For 132: 2^2 * 3 * 11.

Common factors: 2^2 and 3^1. So 4*3=12. Correct.

Euclidean algorithm steps:

6432 divided by 132 gives remainder 96. Then 132 divided by 96 gives remainder 36. Then 96 divided by 36 gives remainder 24. Then 36 divided by 24 gives remainder 12. Then 24 divided by 12 gives remainder 0. So GCF is 12. Correct.

So adding 11 to 12 gives 23.

Wait, but let me check if there's a possibility that I miscalculated the remainder in the Euclidean algorithm. Let me verify the first step.

Compute 6432 divided by 132.

Let me compute 132 × 48 = 132 × 50 - 132 × 2 = 6600 - 264 = 6336.

Then 6432 - 6336 = 96. Correct.

Then 132 divided by 96: 96 × 1 = 96, remainder 36. Correct.

Then 96 divided by 36: 36 × 2 = 72, remainder 24. Correct.

Then 36 divided by 24: 24 × 1 = 24, remainder 12. Correct.

Then 24 divided by 12: remainder 0. Correct.

So yes, GCF is 12.

Therefore, 12 + 11 = 23.

But just to be thorough, let me check if 12 is indeed the GCF.

Check if 12 divides both 6432 and 132.

132 ÷ 12 = 11, which is an integer.

6432 ÷ 12: Let me compute that. 6432 ÷ 12.

Divide 6432 by 12: 12 × 500 = 6000. 6432 - 6000 = 432.

432 ÷ 12 = 36. So total is 500 + 36 = 536. So 12 × 536 = 6432. Correct.

So yes, 12 divides both numbers.

Is there a larger number that divides both? Let's check.

Suppose there is a larger common factor. Let's check 24.

Does 24 divide 132? 132 ÷ 24 = 5.5. Not an integer. So no.

How about 6? Well, 6 is smaller than 12, so 12 is larger.

So 12 is indeed the greatest common factor.

Therefore, adding 11 gives 23.

Wait, but just to make sure, let me check if there's a mistake in the question. The question says "the greatest common factor of 6432 and 132 is increased by 11". So GCF is 12, 12 + 11 = 23.

Alternatively, maybe the question is a trick question? But I don't think so.

Alternatively, maybe I made a mistake in the prime factors? Let me check 6432 again.

Let me compute 2^5 is 32. 32 × 3 = 96. 96 × 67. Let me compute 96 × 67.

Well, 100 × 67 = 6700. Subtract 4 × 67 = 268. So 6700 - 268 = 6432. Yes! So 2^5 × 3 × 67 = 32 × 3 × 67 = 96 × 67 = 6432. Correct.

So that's correct.

Therefore, I think my answer is correct.

**Final Answer**
The result is \boxed{23}.

To solve the problem, we are asked to find the **greatest common factor (GCF)** of the numbers **6432** and **132**, and then **add 11** to that result.

---

### Step 1: Find the GCF of 6432 and 132

We can use the **Euclidean Algorithm**, which is efficient for larger numbers.

#### Apply the Euclidean Algorithm:

1. $ 6432 \div 132 = 48 $ remainder $ 96 $
   - $ \text{GCF}(6432, 132) = \text{GCF}(132, 96) $

2. $ 132 \div 96 = 1 $ remainder $ 36 $
   - $ \text{GCF}(132, 96) = \text{GCF}(96, 36) $

3. $ 96 \div 36 = 2 $ remainder $ 24 $
   - $ \text{GCF}(96, 36) = \text{GCF}(36, 24) $

4. $ 36 \div 24 = 1 $ remainder $ 12 $
   - $ \text{GCF}(36, 24) = \text{GCF}(24, 12) $

5. $ 24 \div 12 = 2 $ remainder $ 0 $
   - $ \text{GCF}(24, 12) = 12 $

So, the **GCF of 6432 and 132 is 12**.

---

### Step 2: Add 11 to the GCF

Now, we simply add 11 to the GCF:

$$
12 + 11 = 23
$$

---

### Final Answer

$$
\boxed{23}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest common factor (GCF) of 6432 and 132, and then add 11 to that result. Let me think about how to approach this.

First, I remember that the GCF of two numbers is the largest number that divides both of them without leaving a remainder. To find the GCF, one method is to use the prime factorization of both numbers and then multiply the common prime factors with the lowest exponents. Alternatively, I can use the Euclidean algorithm, which might be more efficient for larger numbers. Let me try both methods and see which one works better here.

Starting with prime factorization. Let me factorize 132 first because it seems smaller.

132 can be divided by 2 first. 132 ÷ 2 = 66. Then 66 ÷ 2 = 33. Now 33 can be divided by 3: 33 ÷ 3 = 11. And 11 is a prime number. So the prime factors of 132 are 2² × 3 × 11.

Now let me factorize 6432. Hmm, this might take a bit longer. Let me start dividing by small primes.

First, check if it's even. 6432 ÷ 2 = 3216. Still even, so divide by 2 again: 3216 ÷ 2 = 1608. Again, divide by 2: 1608 ÷ 2 = 804. Again by 2: 804 ÷ 2 = 402. Again by 2: 402 ÷ 2 = 201. Now, 201 is not even. Let me check divisibility by 3. 2 + 0 + 1 = 3, which is divisible by 3. So 201 ÷ 3 = 67. Now, 67 is a prime number.

So putting it all together, the prime factors of 6432 are 2⁵ × 3 × 67. Wait, let me check that again. Let me recount the divisions:

Starting with 6432:

Divided by 2 five times: 6432 ÷ 2 = 3216, ÷2=1608, ÷2=804, ÷2=402, ÷2=201. So that's 2⁵. Then 201 ÷ 3 = 67. So yes, 2⁵ × 3 × 67.

So prime factors of 6432: 2⁵ × 3 × 67.

Prime factors of 132: 2² × 3 × 11.

Now, to find the GCF, we take the minimum exponents of the common prime factors.

Common prime factors are 2 and 3.

For 2: minimum exponent between 5 and 2 is 2.

For 3: minimum exponent between 1 and 1 is 1.

So GCF is 2² × 3 = 4 × 3 = 12.

Wait, but let me check if that's correct. Let me verify with another method to be sure. Maybe using the Euclidean algorithm.

The Euclidean algorithm says that GCF(a, b) = GCF(b, a mod b). So let's apply that.

First, take 6432 and 132.

Compute 6432 ÷ 132. Let me find how many times 132 goes into 6432.

Well, 132 × 50 = 6600, which is more than 6432. Let me try 132 × 49.

132 × 40 = 5280.

132 × 9 = 1188.

So 5280 + 1188 = 6468. That's still more than 6432.

So 132 × 48 = 132 × 40 + 132 × 8 = 5280 + 1056 = 6336.

Subtract that from 6432: 6432 - 6336 = 96.

So GCF(6432, 132) = GCF(132, 96).

Now compute GCF(132, 96).

Again, divide 132 by 96. 96 × 1 = 96. Subtract from 132: 132 - 96 = 36.

So GCF(132, 96) = GCF(96, 36).

Now GCF(96, 36). Divide 96 by 36. 36 × 2 = 72. Subtract from 96: 96 - 72 = 24.

So GCF(96, 36) = GCF(36, 24).

GCF(36, 24). Divide 36 by 24. 24 × 1 = 24. Subtract: 36 - 24 = 12.

GCF(36, 24) = GCF(24, 12).

GCF(24, 12). Divide 24 by 12. 12 × 2 = 24. Remainder 0.

So GCF is 12.

Okay, so both methods give me 12. So the GCF of 6432 and 132 is 12.

Therefore, increasing that by 11 would be 12 + 11 = 23.

Wait, but let me just make sure I didn't make any calculation errors. Let me check the prime factorization again.

For 6432:

Divided by 2 five times: 6432 / 2 = 3216, /2=1608, /2=804, /2=402, /2=201. Then 201 divided by 3 is 67. So yes, 2^5 * 3 * 67.

For 132: 2^2 * 3 * 11.

Common factors: 2^2 and 3^1. So 4*3=12. Correct.

Euclidean algorithm steps:

6432 divided by 132 gives remainder 96. Then 132 divided by 96 gives remainder 36. Then 96 divided by 36 gives remainder 24. Then 36 divided by 24 gives remainder 12. Then 24 divided by 12 gives remainder 0. So GCF is 12. Correct.

So adding 11 to 12 gives 23.

Wait, but let me check if there's a possibility that I miscalculated the remainder in the Euclidean algorithm. Let me verify the first step.

Compute 6432 divided by 132.

Let me compute 132 × 48 = 132 × 50 - 132 × 2 = 6600 - 264 = 6336.

Then 6432 - 6336 = 96. Correct.

Then 132 divided by 96: 96 × 1 = 96, remainder 36. Correct.

Then 96 divided by 36: 36 × 2 = 72, remainder 24. Correct.

Then 36 divided by 24: 24 × 1 = 24, remainder 12. Correct.

Then 24 divided by 12: remainder 0. Correct.

So yes, GCF is 12.

Therefore, 12 + 11 = 23.

But just to be thorough, let me check if 12 is indeed the GCF.

Check if 12 divides both 6432 and 132.

132 ÷ 12 = 11, which is an integer.

6432 ÷ 12: Let me compute that. 6432 ÷ 12.

Divide 6432 by 12: 12 × 500 = 6000. 6432 - 6000 = 432.

432 ÷ 12 = 36. So total is 500 + 36 = 536. So 12 × 536 = 6432. Correct.

So yes, 12 divides both numbers.

Is there a larger number that divides both? Let's check.

Suppose there is a larger common factor. Let's check 24.

Does 24 divide 132? 132 ÷ 24 = 5.5. Not an integer. So no.

How about 6? Well, 6 is smaller than 12, so 12 is larger.

So 12 is indeed the greatest common factor.

Therefore, adding 11 gives 23.

Wait, but just to make sure, let me check if there's a mistake in the question. The question says "the greatest common factor of 6432 and 132 is increased by 11". So GCF is 12, 12 + 11 = 23.

Alternatively, maybe the question is a trick question? But I don't think so.

Alternatively, maybe I made a mistake in the prime factors? Let me check 6432 again.

Let me compute 2^5 is 32. 32 × 3 = 96. 96 × 67. Let me compute 96 × 67.

Well, 100 × 67 = 6700. Subtract 4 × 67 = 268. So 6700 - 268 = 6432. Yes! So 2^5 × 3 × 67 = 32 × 3 × 67 = 96 × 67 = 6432. Correct.

So that's correct.

Therefore, I think my answer is correct.

**Final Answer**
The result is \boxed{23}.

To solve the problem, we are asked to find the **greatest common factor (GCF)** of the numbers **6432** and **132**, and then **add 11** to that result.

---

### Step 1: Find the GCF of 6432 and 132

We can use the **Euclidean Algorithm**, which is efficient for larger numbers.

#### Apply the Euclidean Algorithm:

1. $ 6432 \div 132 = 48 $ remainder $ 96 $
   - $ \text{GCF}(6432, 132) = \text{GCF}(132, 96) $

2. $ 132 \div 96 = 1 $ remainder $ 36 $
   - $ \text{GCF}(132, 96) = \text{GCF}(96, 36) $

3. $ 96 \div 36 = 2 $ remainder $ 24 $
   - $ \text{GCF}(96, 36) = \text{GCF}(36, 24) $

4. $ 36 \div 24 = 1 $ remainder $ 12 $
   - $ \text{GCF}(36, 24) = \text{GCF}(24, 12) $

5. $ 24 \div 12 = 2 $ remainder $ 0 $
   - $ \text{GCF}(24, 12) = 12 $

So, the **GCF of 6432 and 132 is 12**.

---

### Step 2: Add 11 to the GCF

Now, we simply add 11 to the GCF:

$$
12 + 11 = 23
$$

---

### Final Answer

$$
\boxed{23}
$$

Solution: We first recognize that $132=11\times 12$, so its prime factorization is $132 = 2^2 \cdot 3 \cdot 11$. We only need to see if these three prime factors will divide into $6432$. Indeed, $6432$ will satisfy the divisibility properties for both $3$ and $4$, and we can long divide to see that $11$ does not divide into $6432$. Thus, the greatest common factor is $3 \times 4 = 12$. The greatest common factor increased by 11 is $12+11 = \boxed{23}$.


Explanation:
Okay, so I need to find tan A in this right triangle ABC where angle B is 90 degrees, and they told me that sin A is equal to 2 times cos A. Hmm, let me think.

First, let me recall some basic trigonometric identities and properties of right triangles. In a right triangle, the sine of an angle is the opposite side over the hypotenuse, and the cosine is the adjacent side over the hypotenuse. Also, tangent is sine over cosine, so tan A would be opposite over adjacent.

But here, they gave me a relationship between sin A and cos A. Specifically, sin A = 2 cos A. Maybe I can use this to find the ratio of the sides, and then figure out tan A?

Let me denote the sides. Let me call the sides of the triangle. Since angle B is the right angle, then sides AB and BC are the legs, and AC is the hypotenuse. Let me label them properly. Let me think: angle A is at vertex A, so the sides opposite and adjacent to angle A would be BC and AB respectively. So, sin A is opposite over hypotenuse, which is BC / AC, and cos A is adjacent over hypotenuse, which is AB / AC.

So according to the problem, sin A = 2 cos A. Therefore, BC / AC = 2 * (AB / AC). If I multiply both sides by AC, that cancels out the denominators, so BC = 2 AB. So that tells me that the side opposite angle A is twice the length of the side adjacent to angle A.

So if I let AB be some length, say, let me assign variables. Let me let AB = x. Then BC = 2x. Then, since it's a right triangle, I can find AC using the Pythagorean theorem.

So AC² = AB² + BC² = x² + (2x)² = x² + 4x² = 5x². Therefore, AC = sqrt(5x²) = x*sqrt(5).

So now, if I want tan A, which is opposite over adjacent, that would be BC / AB = (2x)/x = 2. Wait, so tan A is 2? But let me check if that's consistent with the given condition sin A = 2 cos A.

Let me verify. If tan A is 2, then that means sin A / cos A = 2, so sin A = 2 cos A, which is exactly the given condition. So that seems to check out.

But let me go through it again step by step to make sure I didn't skip anything.

Starting with sin A = 2 cos A. Since in a right triangle, sin A = opposite / hypotenuse, cos A = adjacent / hypotenuse. So if I let the sides be AB, BC, AC with angle A, then:

sin A = BC / AC

cos A = AB / AC

Therefore, BC / AC = 2 * (AB / AC) => BC = 2 AB.

So BC is twice AB. Then, using Pythagoras, AC² = AB² + BC² = AB² + (2 AB)² = AB² + 4 AB² = 5 AB². Therefore, AC = AB * sqrt(5).

Therefore, tan A = BC / AB = 2 AB / AB = 2.

So tan A is 2.

But let me think if there's another way to approach this, maybe using trigonometric identities.

Since sin A = 2 cos A, and we know that sin² A + cos² A = 1. Maybe I can substitute sin A with 2 cos A in that identity.

So substituting, (2 cos A)^2 + cos² A = 1 => 4 cos² A + cos² A = 1 => 5 cos² A = 1 => cos² A = 1/5 => cos A = 1/√5 or -1/√5. But since angle A is in a right triangle, it must be between 0 and 90 degrees, so cosine is positive. Therefore, cos A = 1/√5.

Then sin A = 2 cos A = 2/√5.

Therefore, tan A = sin A / cos A = (2/√5) / (1/√5) = 2.

Same result. So that confirms it.

Alternatively, maybe using the definition of tan A directly. Since tan A = sin A / cos A, and they told us sin A = 2 cos A, so tan A = 2 cos A / cos A = 2.

Wait, that's even simpler. If sin A = 2 cos A, then dividing both sides by cos A (assuming cos A ≠ 0, which it isn't in a right triangle because angle A is acute), we get tan A = 2.

So that's straightforward.

But maybe I need to check if angle A is acute. Since it's a right triangle with angle B = 90°, angles A and C must be acute, so yes, angle A is between 0 and 90°, so tan A is positive, which matches our result.

So all methods lead to tan A = 2.

But let me think again if there's any possible mistake here. For example, could there be a case where sin A = 2 cos A is not possible? Let me check with actual values.

Suppose angle A is such that tan A = 2. Then angle A is arctangent of 2, which is approximately 63.43 degrees. Then sin(arctan 2) is 2 / sqrt(1 + 2²) = 2 / sqrt(5), and cos(arctan 2) is 1 / sqrt(5). Therefore, sin A is indeed twice cos A.

So that works.

Alternatively, if I think of a 1-2-sqrt(5) triangle. If AB = 1, BC = 2, then AC = sqrt(1 + 4) = sqrt(5). Then sin A = 2 / sqrt(5), cos A = 1 / sqrt(5), so sin A = 2 cos A.

Therefore, tan A = 2.

So all these different approaches confirm that tan A is 2.

Therefore, I think the answer is 2.

But let me check once more if I misread the problem. The problem says "right triangle ABC with angle B = 90°", so angle B is the right angle. So angles A and C are acute. The question is about tan A.

Yes, and we found tan A = 2.

Alternatively, if someone thought angle A was the right angle, but no, angle B is the right angle. So angle A is one of the acute angles.

Therefore, I think my answer is correct.

**Final Answer**
\boxed{2}

To solve the problem, we are given a right triangle $ ABC $ with $ \angle B = 90^\circ $, and the condition:

$$
\sin A = 2 \cos A
$$

We are asked to find $ \tan A $.

---

### Step 1: Use the definition of tangent

Recall the fundamental trigonometric identity:

$$
\tan A = \frac{\sin A}{\cos A}
$$

Given that $ \sin A = 2 \cos A $, we can substitute this into the expression for $ \tan A $:

$$
\tan A = \frac{2 \cos A}{\cos A} = 2
$$

This is a direct and elegant solution. However, to ensure correctness, let's verify this using another method.

---

### Step 2: Use the Pythagorean identity

We know from the Pythagorean identity:

$$
\sin^2 A + \cos^2 A = 1
$$

Substitute $ \sin A = 2 \cos A $ into the identity:

$$
(2 \cos A)^2 + \cos^2 A = 1 \\
4 \cos^2 A + \cos^2 A = 1 \\
5 \cos^2 A = 1 \\
\cos^2 A = \frac{1}{5}
$$

Since $ \cos A > 0 $ (as angle $ A $ is acute in a right triangle), we have:

$$
\cos A = \frac{1}{\sqrt{5}}
$$

Then:

$$
\sin A = 2 \cos A = \frac{2}{\sqrt{5}}
$$

Now compute $ \tan A $:

$$
\tan A = \frac{\sin A}{\cos A} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2
$$

---

### Step 3: Geometric interpretation

Let’s assign side lengths to the triangle to confirm the result. Let:

- $ AB = x $ (adjacent to angle $ A $)
- $ BC = 2x $ (opposite to angle $ A $)
- $ AC = \sqrt{x^2 + (2x)^2} = \sqrt{5x^2} = x\sqrt{5} $ (hypotenuse)

Then:

$$
\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{2x}{x} = 2
$$

This confirms the result again.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to find tan A in this right triangle ABC where angle B is 90 degrees, and they told me that sin A is equal to 2 times cos A. Hmm, let me think.

First, let me recall some basic trigonometric identities and properties of right triangles. In a right triangle, the sine of an angle is the opposite side over the hypotenuse, and the cosine is the adjacent side over the hypotenuse. Also, tangent is sine over cosine, so tan A would be opposite over adjacent.

But here, they gave me a relationship between sin A and cos A. Specifically, sin A = 2 cos A. Maybe I can use this to find the ratio of the sides, and then figure out tan A?

Let me denote the sides. Let me call the sides of the triangle. Since angle B is the right angle, then sides AB and BC are the legs, and AC is the hypotenuse. Let me label them properly. Let me think: angle A is at vertex A, so the sides opposite and adjacent to angle A would be BC and AB respectively. So, sin A is opposite over hypotenuse, which is BC / AC, and cos A is adjacent over hypotenuse, which is AB / AC.

So according to the problem, sin A = 2 cos A. Therefore, BC / AC = 2 * (AB / AC). If I multiply both sides by AC, that cancels out the denominators, so BC = 2 AB. So that tells me that the side opposite angle A is twice the length of the side adjacent to angle A.

So if I let AB be some length, say, let me assign variables. Let me let AB = x. Then BC = 2x. Then, since it's a right triangle, I can find AC using the Pythagorean theorem.

So AC² = AB² + BC² = x² + (2x)² = x² + 4x² = 5x². Therefore, AC = sqrt(5x²) = x*sqrt(5).

So now, if I want tan A, which is opposite over adjacent, that would be BC / AB = (2x)/x = 2. Wait, so tan A is 2? But let me check if that's consistent with the given condition sin A = 2 cos A.

Let me verify. If tan A is 2, then that means sin A / cos A = 2, so sin A = 2 cos A, which is exactly the given condition. So that seems to check out.

But let me go through it again step by step to make sure I didn't skip anything.

Starting with sin A = 2 cos A. Since in a right triangle, sin A = opposite / hypotenuse, cos A = adjacent / hypotenuse. So if I let the sides be AB, BC, AC with angle A, then:

sin A = BC / AC

cos A = AB / AC

Therefore, BC / AC = 2 * (AB / AC) => BC = 2 AB.

So BC is twice AB. Then, using Pythagoras, AC² = AB² + BC² = AB² + (2 AB)² = AB² + 4 AB² = 5 AB². Therefore, AC = AB * sqrt(5).

Therefore, tan A = BC / AB = 2 AB / AB = 2.

So tan A is 2.

But let me think if there's another way to approach this, maybe using trigonometric identities.

Since sin A = 2 cos A, and we know that sin² A + cos² A = 1. Maybe I can substitute sin A with 2 cos A in that identity.

So substituting, (2 cos A)^2 + cos² A = 1 => 4 cos² A + cos² A = 1 => 5 cos² A = 1 => cos² A = 1/5 => cos A = 1/√5 or -1/√5. But since angle A is in a right triangle, it must be between 0 and 90 degrees, so cosine is positive. Therefore, cos A = 1/√5.

Then sin A = 2 cos A = 2/√5.

Therefore, tan A = sin A / cos A = (2/√5) / (1/√5) = 2.

Same result. So that confirms it.

Alternatively, maybe using the definition of tan A directly. Since tan A = sin A / cos A, and they told us sin A = 2 cos A, so tan A = 2 cos A / cos A = 2.

Wait, that's even simpler. If sin A = 2 cos A, then dividing both sides by cos A (assuming cos A ≠ 0, which it isn't in a right triangle because angle A is acute), we get tan A = 2.

So that's straightforward.

But maybe I need to check if angle A is acute. Since it's a right triangle with angle B = 90°, angles A and C must be acute, so yes, angle A is between 0 and 90°, so tan A is positive, which matches our result.

So all methods lead to tan A = 2.

But let me think again if there's any possible mistake here. For example, could there be a case where sin A = 2 cos A is not possible? Let me check with actual values.

Suppose angle A is such that tan A = 2. Then angle A is arctangent of 2, which is approximately 63.43 degrees. Then sin(arctan 2) is 2 / sqrt(1 + 2²) = 2 / sqrt(5), and cos(arctan 2) is 1 / sqrt(5). Therefore, sin A is indeed twice cos A.

So that works.

Alternatively, if I think of a 1-2-sqrt(5) triangle. If AB = 1, BC = 2, then AC = sqrt(1 + 4) = sqrt(5). Then sin A = 2 / sqrt(5), cos A = 1 / sqrt(5), so sin A = 2 cos A.

Therefore, tan A = 2.

So all these different approaches confirm that tan A is 2.

Therefore, I think the answer is 2.

But let me check once more if I misread the problem. The problem says "right triangle ABC with angle B = 90°", so angle B is the right angle. So angles A and C are acute. The question is about tan A.

Yes, and we found tan A = 2.

Alternatively, if someone thought angle A was the right angle, but no, angle B is the right angle. So angle A is one of the acute angles.

Therefore, I think my answer is correct.

**Final Answer**
\boxed{2}

To solve the problem, we are given a right triangle $ ABC $ with $ \angle B = 90^\circ $, and the condition:

$$
\sin A = 2 \cos A
$$

We are asked to find $ \tan A $.

---

### Step 1: Use the definition of tangent

Recall the fundamental trigonometric identity:

$$
\tan A = \frac{\sin A}{\cos A}
$$

Given that $ \sin A = 2 \cos A $, we can substitute this into the expression for $ \tan A $:

$$
\tan A = \frac{2 \cos A}{\cos A} = 2
$$

This is a direct and elegant solution. However, to ensure correctness, let's verify this using another method.

---

### Step 2: Use the Pythagorean identity

We know from the Pythagorean identity:

$$
\sin^2 A + \cos^2 A = 1
$$

Substitute $ \sin A = 2 \cos A $ into the identity:

$$
(2 \cos A)^2 + \cos^2 A = 1 \\
4 \cos^2 A + \cos^2 A = 1 \\
5 \cos^2 A = 1 \\
\cos^2 A = \frac{1}{5}
$$

Since $ \cos A > 0 $ (as angle $ A $ is acute in a right triangle), we have:

$$
\cos A = \frac{1}{\sqrt{5}}
$$

Then:

$$
\sin A = 2 \cos A = \frac{2}{\sqrt{5}}
$$

Now compute $ \tan A $:

$$
\tan A = \frac{\sin A}{\cos A} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2
$$

---

### Step 3: Geometric interpretation

Let’s assign side lengths to the triangle to confirm the result. Let:

- $ AB = x $ (adjacent to angle $ A $)
- $ BC = 2x $ (opposite to angle $ A $)
- $ AC = \sqrt{x^2 + (2x)^2} = \sqrt{5x^2} = x\sqrt{5} $ (hypotenuse)

Then:

$$
\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{2x}{x} = 2
$$

This confirms the result again.

---

### Final Answer

$$
\boxed{2}
$$

Solution:
The triangle is shown below:

[asy]
pair A,B,C;
A = (0,0);
B = (5,0);
C = (5,10);
draw(A--B--C--A);
draw(rightanglemark(C,B,A,16));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
[/asy]

We have $\sin A = \frac{BC}{AC}$ and $\cos A = \frac{AB}{AC}$, so $\sin A = 2\cos A$ gives us $\frac{BC}{AC} = 2\cdot\frac{AB}{AC}$.  Multiplying both sides by $AC$ gives $BC = 2AB$, so $\frac{BC}{AB} = 2$.  Finally, we have $\tan A = \frac{BC}{AB} = \boxed{2}$.

We also could have noted that $\tan A = \frac{\sin A}{\cos A} = \frac{2\cos A}{\cos A } =\boxed{2}$.