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 $xy = 2$. That seems important.

So, if I can find a recurrence relation for $x^n + y^n$, maybe I can compute it step by step. Let me denote $S_n = x^n + y^n$. Then, since $x$ and $y$ are roots of the quadratic equation $t^2 - (x + y)t + xy = 0$, which would be $t^2 - 2\sqrt{7}t + 2 = 0$. Wait, but that equation has irrational coefficients. Maybe that complicates things?

Alternatively, maybe I can find a recurrence relation for $S_n$ using the relation between $x$ and $y$. Let me think. Since $x$ and $y$ satisfy the equation $t^2 - 2\sqrt{7}t + 2 = 0$, then for each $n \geq 2$, we have $x^n = 2\sqrt{7}x^{n-1} - 2x^{n-2}$ and similarly for $y^n$. Therefore, adding these two equations:

$S_n = x^n + y^n = 2\sqrt{7}(x^{n-1} + y^{n-1}) - 2(x^{n-2} + y^{n-2}) = 2\sqrt{7}S_{n-1} - 2S_{n-2}$

But this recurrence relation still involves $\sqrt{7}$, which is irrational. That might not be helpful for computing integer values. Wait, but maybe if I consider $S_n$ and another sequence?

Alternatively, maybe I can find a recurrence with integer coefficients. Let me think. Since $x$ and $y$ are conjugates, perhaps if I consider $S_n$ and another sequence involving $x^n - y^n$?

Alternatively, maybe I can square $x$ and $y$ to find a relation. Let me compute $x^2$ and $y^2$:

$x^2 = (\sqrt{7} + \sqrt{5})^2 = 7 + 2\sqrt{35} + 5 = 12 + 2\sqrt{35}$

Similarly, $y^2 = (\sqrt{7} - \sqrt{5})^2 = 7 - 2\sqrt{35} + 5 = 12 - 2\sqrt{35}$

So, $x^2 + y^2 = 24$, and $x^2 - y^2 = 4\sqrt{35}$. Hmm, not sure if that helps yet.

But maybe if I compute $S_1 = x + y = 2\sqrt{7}$, $S_2 = x^2 + y^2 = 24$, $S_3 = x^3 + y^3$, etc. But again, with the recurrence involving $\sqrt{7}$, which complicates things. Maybe I need another approach.

Wait, but if I consider that $x$ and $y$ are roots of the equation $t^2 - 2\sqrt{7}t + 2 = 0$, then perhaps I can find a recurrence relation for $S_n$ in terms of previous terms. However, since the coefficients are irrational, maybe I need to square the recurrence or something?

Alternatively, maybe I can find a relation between $S_n$ and $S_{n-2}$ or something else. Let me try computing $S_1$, $S_2$, $S_3$, etc., step by step.

We have:

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

$S_2 = x^2 + y^2 = 24$

Now, let's compute $S_3$. Using the identity $x^3 + y^3 = (x + y)^3 - 3xy(x + y)$. Let me verify that:

Yes, $x^3 + y^3 = (x + y)^3 - 3xy(x + y)$. So substituting:

$S_3 = (2\sqrt{7})^3 - 3 \times 2 \times 2\sqrt{7} = 8 \times 7^{3/2} - 12\sqrt{7}$. Wait, that seems messy. Maybe I need to compute it numerically? But I can't use a calculator. Hmm.

Alternatively, maybe I can use the recurrence relation I had earlier:

$S_n = 2\sqrt{7} S_{n-1} - 2 S_{n-2}$. Let me try that for n=3:

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

But then $S_3 = 44\sqrt{7}$. Hmm, that's still irrational. But if I compute $S_4$:

$S_4 = 2\sqrt{7} S_3 - 2 S_2 = 2\sqrt{7} \times 44\sqrt{7} - 2 \times 24 = 2 \times 44 \times 7 - 48 = 616 - 48 = 568$. Oh! That's an integer! So S_4 is 568. Interesting. Let me check that again:

$2\sqrt{7} \times 44\sqrt{7} = 2 \times 44 \times (\sqrt{7})^2 = 88 \times 7 = 616$. Then subtract 2*S_2 = 48. So 616 - 48 = 568. Yes, that's correct. So S_4 is 568.

Similarly, let's compute S_5:

$S_5 = 2\sqrt{7} S_4 - 2 S_3 = 2\sqrt{7} \times 568 - 2 \times 44\sqrt{7} = (1136\sqrt{7} - 88\sqrt{7}) = 1048\sqrt{7}$. Again, irrational.

Then S_6:

$S_6 = 2\sqrt{7} S_5 - 2 S_4 = 2\sqrt{7} \times 1048\sqrt{7} - 2 \times 568 = 2 \times 1048 \times 7 - 1136$. Let me compute that:

First, 2*1048 = 2096. Then 2096*7. Let me compute 2000*7 = 14000, 96*7=672, so total is 14000 + 672 = 14672. Then subtract 1136: 14672 - 1136. Let me compute that:

14672 - 1000 = 13672

13672 - 136 = 13536

So S_6 = 13536.

Wait, so S_6 is 13536? But S_6 is x^6 + y^6. Therefore, since y = sqrt(7) - sqrt(5), which is approximately... Well, sqrt(7) is about 2.6458, sqrt(5) is about 2.2361, so y is about 0.4097. Therefore, y^6 is a very small positive number. Therefore, x^6 = S_6 - y^6. Therefore, since y^6 is positive, x^6 = 13536 - y^6. Therefore, the value we need is x^6, which is less than 13536. Therefore, the greatest integer less than x^6 is 13536 - 1, if y^6 is less than 1. Wait, but is y^6 less than 1?

Let me check. Since y = sqrt(7) - sqrt(5) ≈ 2.6458 - 2.2361 ≈ 0.4097. Then y^2 ≈ 0.1678, y^3 ≈ 0.1678 * 0.4097 ≈ 0.0687, y^4 ≈ 0.0687 * 0.4097 ≈ 0.0281, y^5 ≈ 0.0281 * 0.4097 ≈ 0.0115, y^6 ≈ 0.0115 * 0.4097 ≈ 0.0047. So y^6 is approximately 0.0047, which is less than 1. Therefore, x^6 = S_6 - y^6 ≈ 13536 - 0.0047 ≈ 13535.9953. Therefore, the greatest integer less than x^6 is 13535.

But wait, is that correct? Let me check again. Because S_6 = x^6 + y^6 = 13536. Therefore, x^6 = 13536 - y^6. Since y^6 is positive, x^6 is less than 13536. Therefore, floor(x^6) is 13535. But is that the case? Wait, but maybe y^6 is so small that x^6 is just slightly less than 13536, so the floor is 13535. But is there a possibility that y^6 is more than 1? Wait, earlier approximation suggested it's about 0.0047, which is less than 1. So floor(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_1 = 2√7, S_2 = 24.

Then S_3 = 2√7 * S_2 - 2 * S_1 = 2√7 * 24 - 2*(2√7) = 48√7 - 4√7 = 44√7. Correct.

Then S_4 = 2√7 * S_3 - 2 * S_2 = 2√7 * 44√7 - 2*24 = 2*44*7 - 48 = 616 - 48 = 568. Correct.

Then S_5 = 2√7 * S_4 - 2 * S_3 = 2√7 * 568 - 2*(44√7) = 1136√7 - 88√7 = 1048√7. Correct.

Then S_6 = 2√7 * S_5 - 2 * S_4 = 2√7 * 1048√7 - 2*568 = 2*1048*7 - 1136.

Calculating 2*1048 = 2096. Then 2096*7: Let me compute 2000*7=14000, 96*7=672, so total 14000 + 672 = 14672. Then subtract 1136: 14672 - 1136. Let me do this subtraction again:

14672 - 1000 = 13672

13672 - 136 = 13536. Yes, that's correct. So S_6 = 13536.

Therefore, since y^6 is positive and less than 1, then x^6 = 13536 - y^6 is less than 13536 but greater than 13535. Therefore, the greatest integer less than x^6 is 13535. Therefore, the answer is 13535.

But wait, let me check if there's another way to compute x^6 directly, maybe using binomial theorem? Let me try that as a verification.

Compute $(\sqrt{7} + \sqrt{5})^6$. Let me expand this using 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 to compute without a calculator. However, maybe I can pair terms? Let me note that terms with k and 6 - k might combine in some way. Alternatively, maybe I can write this as:

Let me note that $(\sqrt{7} + \sqrt{5})^6 + (\sqrt{7} - \sqrt{5})^6 = S_6 = 13536$, as we found earlier. Therefore, if I can compute $(\sqrt{7} + \sqrt{5})^6$ as S_6 - $(\sqrt{7} - \sqrt{5})^6$, which is 13536 - y^6. Since y^6 is positive and less than 1, then floor(x^6) is 13535.

But maybe I can compute y^6 more accurately? Let me try.

First, compute y = sqrt(7) - sqrt(5). Let me compute y^2:

y^2 = (sqrt(7) - sqrt(5))^2 = 7 + 5 - 2*sqrt(35) = 12 - 2*sqrt(35).

Then y^3 = y^2 * y = (12 - 2√35)(sqrt(7) - sqrt(5)). Let me compute that:

Multiply 12*(sqrt(7) - sqrt(5)) - 2√35*(sqrt(7) - sqrt(5)).

First term: 12√7 - 12√5.

Second term: -2√35*sqrt(7) + 2√35*sqrt(5) = -2√(35*7) + 2√(35*5) = -2√245 + 2√175.

Simplify radicals:

√245 = √(49*5) = 7√5

√175 = √(25*7) = 5√7

Therefore, second term becomes: -2*7√5 + 2*5√7 = -14√5 + 10√7.

Therefore, combining first and second terms:

12√7 - 12√5 -14√5 + 10√7 = (12√7 + 10√7) + (-12√5 -14√5) = 22√7 - 26√5.

So y^3 = 22√7 - 26√5.

Hmm, that's still irrational. Let me compute y^4 = y^3 * y = (22√7 - 26√5)(sqrt(7) - sqrt(5)).

Multiply term by term:

22√7 * sqrt(7) = 22*7 = 154

22√7 * (-sqrt(5)) = -22√35

-26√5 * sqrt(7) = -26√35

-26√5 * (-sqrt(5)) = +26*5 = 130

So combining terms:

154 + 130 = 284

-22√35 -26√35 = -48√35

Therefore, y^4 = 284 - 48√35.

Similarly, compute y^5 = y^4 * y = (284 - 48√35)(sqrt(7) - sqrt(5)).

Let me compute that:

First, 284*sqrt(7) - 284*sqrt(5) - 48√35*sqrt(7) + 48√35*sqrt(5)

Simplify each term:

First term: 284√7

Second term: -284√5

Third term: -48√(35*7) = -48√245 = -48*7√5 = -336√5

Fourth term: +48√(35*5) = +48√175 = +48*5√7 = +240√7

Combine like terms:

For √7: 284√7 + 240√7 = 524√7

For √5: -284√5 - 336√5 = -620√5

Therefore, y^5 = 524√7 - 620√5.

Now compute y^6 = y^5 * y = (524√7 - 620√5)(sqrt(7) - sqrt(5)).

Again, multiply term by term:

524√7 * sqrt(7) = 524*7 = 3668

524√7 * (-sqrt(5)) = -524√35

-620√5 * sqrt(7) = -620√35

-620√5 * (-sqrt(5)) = +620*5 = 3100

Combine terms:

3668 + 3100 = 6768

-524√35 -620√35 = -1144√35

Therefore, y^6 = 6768 - 1144√35.

So, y^6 = 6768 - 1144√35. Now, we need to compute this value numerically? Wait, but we can't use a calculator, but maybe we can estimate √35?

Well, √35 is between 5.9 and 6.0 because 5.9² = 34.81 and 6.0² = 36. So √35 ≈ 5.916. Let me check:

5.916² = (5.9 + 0.016)^2 = 5.9² + 2*5.9*0.016 + 0.016² = 34.81 + 0.1888 + 0.000256 ≈ 34.81 + 0.1888 = 34.9988 + 0.000256 ≈ 34.999056. Close to 35. So √35 ≈ 5.916. Let me take √35 ≈ 5.916.

Therefore, 1144√35 ≈ 1144 * 5.916. Let me compute that:

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. So total 591.6 + 236.64 = 828.24 + 23.664 = 851.904. Therefore, total 1144*5.916 ≈ 5916 + 851.904 = 6767.904.

Therefore, y^6 ≈ 6768 - 6767.904 = 0.096. Wait, that's approximately 0.096. Earlier, my rough estimate was 0.0047, but this more precise calculation gives about 0.096? That's a big difference. Which one is correct?

Wait, let me check my calculation again. Wait, y^6 = 6768 - 1144√35. If √35 ≈ 5.916, then 1144 * 5.916 ≈ ?

Wait, let me check 1144 * 5.916:

First, note that 1144 * 5 = 5720

1144 * 0.916 = ?

Compute 1144 * 0.9 = 1029.6

1144 * 0.016 = 18.304

So total 1029.6 + 18.304 = 1047.904

Therefore, total 1144 * 5.916 ≈ 5720 + 1047.904 = 6767.904. Therefore, y^6 ≈ 6768 - 6767.904 = 0.096. So approximately 0.096.

But earlier, when I approximated y ≈ 0.4097, then y^6 ≈ (0.4097)^6. Let me compute that:

First, 0.4^6 = 0.004096. But 0.4097 is slightly larger than 0.4, so y^6 would be slightly larger than 0.004096. However, according to the exact expression, it's approximately 0.096. There's a discrepancy here. Which one is correct?

Wait, there must be a mistake in my calculation of y^6. Let me check again.

Earlier, I calculated y^6 = 6768 - 1144√35. Let me check the steps again.

Starting from y^5 = 524√7 - 620√5. Then y^6 = y^5 * y = (524√7 - 620√5)(sqrt(7) - sqrt(5)).

Expanding:

First term: 524√7 * sqrt(7) = 524 * 7 = 3668

Second term: 524√7 * (-sqrt(5)) = -524√35

Third term: -620√5 * sqrt(7) = -620√35

Fourth term: -620√5 * (-sqrt(5)) = +620 * 5 = 3100

So adding up the constants: 3668 + 3100 = 6768. Correct.

Adding up the radicals: -524√35 -620√35 = -(524 + 620)√35 = -1144√35. Correct. Therefore, y^6 = 6768 - 1144√35. So that's correct.

But then why is there a discrepancy between the two estimates? Let me check my approximate value of y. Earlier, I said y = sqrt(7) - sqrt(5) ≈ 2.6458 - 2.2361 ≈ 0.4097. Let me check that more accurately.

sqrt(7) is approximately 2.6457513110645906

sqrt(5) is approximately 2.2360679775

Therefore, y ≈ 2.6457513110645906 - 2.2360679775 ≈ 0.4096833335645906. So approximately 0.409683.

So y ≈ 0.409683. Let me compute y^6:

First, compute y^2 ≈ (0.409683)^2 ≈ 0.16783

Then y^3 ≈ y^2 * y ≈ 0.16783 * 0.409683 ≈ 0.0687

y^4 ≈ 0.0687 * 0.409683 ≈ 0.0281

y^5 ≈ 0.0281 * 0.409683 ≈ 0.0115

y^6 ≈ 0.0115 * 0.409683 ≈ 0.00471.

But according to the exact expression, y^6 ≈ 0.096. There's a big difference here. Which one is correct?

Wait, there must be a mistake in my exact calculation. Let me check again. Wait, if y^6 = 6768 - 1144√35, then let me compute 1144√35 numerically. Wait, if √35 ≈ 5.916079783, then 1144 * 5.916079783 ≈ ?

Let me compute 1144 * 5.916079783:

First, note that 1000 * 5.916079783 = 5916.079783

Then 144 * 5.916079783:

Compute 100 * 5.916079783 = 591.6079783

40 * 5.916079783 = 236.6431913

4 * 5.916079783 = 23.66431913

Adding those: 591.6079783 + 236.6431913 = 828.2511696 + 23.66431913 ≈ 851.9154887

Therefore, total 1144 * 5.916079783 ≈ 5916.079783 + 851.9154887 ≈ 6767.9952717

Therefore, y^6 = 6768 - 6767.9952717 ≈ 0.0047283.

Ah! So approximately 0.0047283, which matches the earlier approximation of ~0.0047. So my previous calculation had a miscalculation where I thought 1144*5.916 was 6767.904, but actually, it's approximately 6767.995, so 6768 - 6767.995 ≈ 0.005. Therefore, y^6 ≈ 0.0047. Therefore, my initial approximation was correct. So why did I get 0.096 before? Because I miscalculated 1144*5.916. I think I had a decimal error. So the exact value of y^6 is approximately 0.0047, which is less than 1. Therefore, x^6 = S_6 - y^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 once more. Since S_6 = x^6 + y^6 = 13536, and y^6 is positive and less than 1, then x^6 is between 13535 and 13536. Therefore, floor(x^6) = 13535.

But to be thorough, let me check if y^6 is less than 1. Since y ≈ 0.4097, then y^6 ≈ (0.4)^6 = 0.004096, and since 0.4097 > 0.4, y^6 is slightly larger than 0.004096, but still less than 1. Therefore, yes, x^6 is less than 13536 but greater than 13535. Therefore, the floor is 13535.

But let me check if there's another way to confirm this. For example, maybe compute x^6 directly using binomial expansion? Let me try that.

Compute $(\sqrt{7} + \sqrt{5})^6$. Let me write it as:

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

Let me compute each term:

For k=0: $\binom{6}{0} (\sqrt{7})^6 (\sqrt{5})^0 = 1 * 7^3 * 1 = 343$

For k=1: $\binom{6}{1} (\sqrt{7})^5 (\sqrt{5})^1 = 6 * 7^{2.5} * 5^{0.5}$. Wait, this seems messy. Maybe I can write them in terms of radicals:

Alternatively, note that $(\sqrt{7} + \sqrt{5})^6 + (\sqrt{7} - \sqrt{5})^6 = 13536$, so if I can compute $(\sqrt{7} + \sqrt{5})^6$ as 13536 - y^6, and since y^6 ≈ 0.0047, then x^6 ≈ 13535.9953, so floor(x^6) = 13535.

Alternatively, maybe compute x^6 using the recurrence relation? Wait, we already did that.

Alternatively, maybe compute x^2, x^3, x^4, x^5, x^6 step by step?

Let me try that.

First, compute x = sqrt(7) + sqrt(5).

Compute x^2 = (sqrt(7) + sqrt(5))^2 = 7 + 2*sqrt(35) + 5 = 12 + 2*sqrt(35).

Compute x^3 = x^2 * x = (12 + 2√35)(sqrt(7) + sqrt(5)). Let me compute that:

Multiply 12*sqrt(7) + 12*sqrt(5) + 2√35*sqrt(7) + 2√35*sqrt(5)

Simplify each term:

12√7 + 12√5 + 2√(35*7) + 2√(35*5)

Which is 12√7 + 12√5 + 2√245 + 2√175

Simplify radicals:

√245 = 7√5, √175 = 5√7

Therefore:

12√7 + 12√5 + 2*7√5 + 2*5√7 = 12√7 + 12√5 + 14√5 + 10√7 = (12√7 + 10√7) + (12√5 + 14√5) = 22√7 + 26√5.

So x^3 = 22√7 + 26√5.

Similarly, compute x^4 = x^3 * x = (22√7 + 26√5)(sqrt(7) + sqrt(5)).

Multiply term by term:

22√7*sqrt(7) + 22√7*sqrt(5) + 26√5*sqrt(7) + 26√5*sqrt(5)

Simplify:

22*7 + 22√35 + 26√35 + 26*5

Which is 154 + 22√35 + 26√35 + 130 = (154 + 130) + (22√35 + 26√35) = 284 + 48√35.

So x^4 = 284 + 48√35.

Then x^5 = x^4 * x = (284 + 48√35)(sqrt(7) + sqrt(5)).

Multiply term by term:

284*sqrt(7) + 284*sqrt(5) + 48√35*sqrt(7) + 48√35*sqrt(5)

Simplify:

284√7 + 284√5 + 48√(35*7) + 48√(35*5)

Which is 284√7 + 284√5 + 48√245 + 48√175

Again, √245 = 7√5, √175 = 5√7:

So:

284√7 + 284√5 + 48*7√5 + 48*5√7 = 284√7 + 284√5 + 336√5 + 240√7

Combine like terms:

(284√7 + 240√7) + (284√5 + 336√5) = 524√7 + 620√5.

Therefore, x^5 = 524√7 + 620√5.

Now compute x^6 = x^5 * x = (524√7 + 620√5)(sqrt(7) + sqrt(5)).

Multiply term by term:

524√7*sqrt(7) + 524√7*sqrt(5) + 620√5*sqrt(7) + 620√5*sqrt(5)

Simplify:

524*7 + 524√35 + 620√35 + 620*5

Which is 3668 + 524√35 + 620√35 + 3100 = (3668 + 3100) + (524√35 + 620√35) = 6768 + 1144√35.

Therefore, x^6 = 6768 + 1144√35.

Wait a second! Earlier, we had y^6 = 6768 - 1144√35, so x^6 + y^6 = 6768 + 1144√35 + 6768 - 1144√35 = 13536, which matches our previous result.

So x^6 = 6768 + 1144√35. Now, we need to find the greatest integer less than x^6.

So, if I can compute 1144√35 numerically, then I can find x^6.

Earlier, we found that √35 ≈ 5.916079783. Therefore, 1144 * √35 ≈ 1144 * 5.916079783 ≈ 6767.99527.

Therefore, x^6 ≈ 6768 + 6767.99527 ≈ 13535.99527.

Therefore, x^6 ≈ 13535.99527, so the greatest integer less than x^6 is 13535.

Therefore, the answer is 13535.

But let me check once more. Since x^6 = 6768 + 1144√35, and we know that √35 is irrational, but we can bound it.

We know that √35 is between 5.916 and 5.9161. Let me check:

5.916^2 = 34.996656

5.9161^2 = ?

Let me compute 5.916^2 = 34.996656

Then, 5.9161^2 = (5.916 + 0.0001)^2 = 5.916^2 + 2*5.916*0.0001 + (0.0001)^2 ≈ 34.996656 + 0.0011832 + 0.00000001 ≈ 34.99783921.

But 35 is larger than that. Therefore, √35 is between 5.9161 and 5.9162. Let me check 5.9162^2:

= (5.916 + 0.0002)^2 = 5.916^2 + 2*5.916*0.0002 + (0.0002)^2 ≈ 34.996656 + 0.0023664 + 0.00000004 ≈ 34.99902244. Still less than 35.

Next, 5.9163^2:

= 5.9162^2 + 2*5.9162*0.0001 + (0.0001)^2 ≈ 34.99902244 + 0.00118324 + 0.00000001 ≈ 35.00020569.

Ah, so √35 is between 5.9162 and 5.9163. Therefore, let me approximate √35 ≈ 5.9162 + (35 - 34.99902244)/(2*5.9162).

Wait, using linear approximation. Let me denote f(x) = x^2. We know that f(5.9162) ≈ 34.99902244, and we want to find x such that f(x) = 35.

Let me take x = 5.9162 + δ, where δ is small. Then:

f(x) ≈ f(5.9162) + 2*5.9162*δ = 35

So 34.99902244 + 11.8324*δ ≈ 35

Therefore, 11.8324*δ ≈ 35 - 34.99902244 = 0.00097756

Therefore, δ ≈ 0.00097756 / 11.8324 ≈ 0.0000826.

Therefore, √35 ≈ 5.9162 + 0.0000826 ≈ 5.9162826.

Therefore, √35 ≈ 5.9162826.

Therefore, 1144 * √35 ≈ 1144 * 5.9162826. Let me compute this:

First, compute 1000 * 5.9162826 = 5916.2826

Then 144 * 5.9162826. Let me compute 100 * 5.9162826 = 591.62826

40 * 5.9162826 = 236.651304

4 * 5.9162826 = 23.6651304

Adding those: 591.62826 + 236.651304 = 828.279564 + 23.6651304 ≈ 851.9446944

Therefore, total 1144 * √35 ≈ 5916.2826 + 851.9446944 ≈ 6768.2272944.

Therefore, x^6 = 6768 + 1144√35 ≈ 6768 + 6768.2272944 ≈ 13536.2272944? Wait, no! Wait, no, x^6 = 6768 + 1144√35. Wait, earlier we had:

Wait, no! Wait, earlier, we had x^6 = 6768 + 1144√35. But earlier, we also had y^6 = 6768 - 1144√35. Therefore, if √35 ≈ 5.9162826, then 1144√35 ≈ 6768.227. Therefore, x^6 ≈ 6768 + 6768.227 ≈ 13536.227? Wait, that can't be, because earlier we had S_6 = x^6 + y^6 = 13536. If x^6 ≈ 13536.227, then y^6 would be negative? Which is impossible. There must be a miscalculation here.

Wait, no! Wait, no, x^6 = 6768 + 1144√35. But earlier, we had S_6 = x^6 + y^6 = 13536. And y^6 = 6768 - 1144√35. Therefore, if x^6 = 6768 + 1144√35, then adding y^6 = 6768 - 1144√35 gives 13536. Therefore, if 1144√35 ≈ 6768.227, then x^6 ≈ 6768 + 6768.227 ≈ 13536.227? But that would make y^6 = 6768 - 6768.227 ≈ -0.227, which is impossible because y^6 is positive. Therefore, there must be a mistake in my calculation of 1144√35.

Wait, earlier, I had that y^6 = 6768 - 1144√35 ≈ 0.0047. Therefore, 1144√35 ≈ 6768 - 0.0047 ≈ 6767.9953. Therefore, 1144√35 ≈ 6767.9953. Therefore, x^6 = 6768 + 1144√35 ≈ 6768 + 6767.9953 ≈ 13535.9953. Which matches our previous result.

But why did I get 1144√35 ≈ 6768.227 earlier? Because I miscalculated. Wait, no. Wait, I think I confused the value of √35. Wait, earlier, I had that √35 ≈ 5.9162826, so 1144 * √35 ≈ 1144 * 5.9162826. But if √35 ≈ 5.916, then 1144 * 5.916 ≈ 6767.995, which is approximately 6768. Therefore, 1144√35 ≈ 6768 - 0.0047 ≈ 6767.9953. Therefore, x^6 = 6768 + 1144√35 ≈ 6768 + 6767.9953 ≈ 13535.9953.

Therefore, my previous calculation where I thought 1144√35 ≈ 6768.227 was wrong. Where did that come from? Probably a miscalculation when I added 5916.2826 + 851.9446944. Let me check that again:

Wait, 1000 * 5.9162826 = 5916.2826

Then 144 * 5.9162826. Let me compute 144 * 5.9162826:

First, note that 144 * 5 = 720

144 * 0.9162826 = ?

Compute 144 * 0.9 = 129.6

144 * 0.0162826 ≈ 144 * 0.016 = 2.304, and 144 * 0.0002826 ≈ 0.0406. So total ≈ 2.304 + 0.0406 ≈ 2.3446

Therefore, total 129.6 + 2.3446 ≈ 131.9446

Therefore, total 144 * 5.9162826 ≈ 720 + 131.9446 ≈ 851.9446

Therefore, total 1144 * √35 ≈ 5916.2826 + 851.9446 ≈ 6768.2272.

But this contradicts the previous result where y^6 = 6768 - 1144√35 ≈ 0.0047. Therefore, there must be a miscalculation here.

Wait, but if y^6 = 6768 - 1144√35 ≈ 0.0047, then 1144√35 ≈ 6768 - 0.0047 ≈ 6767.9953. Therefore, 1144√35 ≈ 6767.9953. But according to the calculation above, it's ≈ 6768.2272. There's a discrepancy of about 0.23. Where is this coming from?

Wait, I think I made a mistake in the value of √35. Earlier, I approximated √35 ≈ 5.9162826, but if 1144√35 ≈ 6767.9953, then √35 ≈ 6767.9953 / 1144 ≈ let me compute that:

Divide 6767.9953 by 1144:

First, 1144 * 5 = 5720

Subtract 5720 from 6767.9953: 6767.9953 - 5720 = 1047.9953

Now, 1144 * 0.9 = 1029.6

Subtract 1029.6 from 1047.9953: 1047.9953 - 1029.6 = 18.3953

Now, 1144 * 0.016 = 18.304

Subtract 18.304 from 18.3953: ≈ 0.0913

So total is 5 + 0.9 + 0.016 + ... ≈ 5.916 + ... So √35 ≈ 5.916 + 0.0913 / 1144 ≈ 5.916 + 0.0000798 ≈ 5.9160798. Which is consistent with the known value of √35 ≈ 5.916079783. Therefore, my previous calculation of 1144 * √35 ≈ 6768.2272 must be wrong.

Wait, let me check with the exact value of √35 ≈ 5.916079783. Then 1144 * 5.916079783 = ?

Let me compute 1144 * 5.916079783:

First, note that 1144 * 5 = 5720

1144 * 0.916079783 = ?

Let me compute 1144 * 0.9 = 1029.6

1144 * 0.016079783 ≈ 1144 * 0.016 = 18.304, and 1144 * 0.000079783 ≈ 1144 * 0.00008 ≈ 0.09152

So total ≈ 18.304 + 0.09152 ≈ 18.39552

Therefore, total 1144 * 0.916079783 ≈ 1029.6 + 18.39552 ≈ 1047.99552

Therefore, total 1144 * 5.916079783 ≈ 5720 + 1047.99552 ≈ 6767.99552.

Ah! So my previous calculation where I thought it was 6768.2272 was wrong. The correct value is approximately 6767.9955. Therefore, x^6 = 6768 + 1144√35 ≈ 6768 + 6767.9955 ≈ 13535.9955.

Therefore, x^6 ≈ 13535.9955, which is just slightly less than 13536. Therefore, the greatest integer less than x^6 is 13535.

Therefore, after all these calculations and checking, I can be confident that the answer 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 conjugates and recurrence relations.

---

### Step 1: Define the Conjugate Pair

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

Note that:
- $ x + y = 2\sqrt{7} $
- $ x - y = 2\sqrt{5} $
- $ xy = (\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5}) = 7 - 5 = 2 $

---

### Step 2: Define a Recurrence for $ S_n = x^n + y^n $

We define:
$$
S_n = x^n + y^n
$$

Using the identity:
$$
x^n + y^n = (x + y)(x^{n-1} + y^{n-1}) - xy(x^{n-2} + y^{n-2})
$$

We get the recurrence:
$$
S_n = 2\sqrt{7} \cdot S_{n-1} - 2 \cdot S_{n-2}
$$

We can compute the first few terms:

- $ S_1 = x + y = 2\sqrt{7} $
- $ S_2 = x^2 + y^2 = (\sqrt{7} + \sqrt{5})^2 + (\sqrt{7} - \sqrt{5})^2 = 12 + 2\sqrt{35} + 12 - 2\sqrt{35} = 24 $

Now compute the rest using the recurrence:

- $ 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 = 2 \cdot 44 \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 = 2 \cdot 1048 \cdot 7 - 1136 = 14672 - 1136 = 13536 $

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

---

### Step 3: Estimate $ y^6 $

We know that:
$$
x^6 = S_6 - y^6 = 13536 - y^6
$$

Now, estimate $ y = \sqrt{7} - \sqrt{5} \approx 2.6458 - 2.2361 = 0.4097 $

Then:
$$
y^6 \approx (0.4097)^6 \approx 0.0047
$$

So:
$$
x^6 = 13536 - y^6 \approx 13536 - 0.0047 = 13535.9953
$$

---

### Step 4: Find the Greatest Integer Less Than $ x^6 $

Since $ x^6 \approx 13535.9953 $, 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 $xy = 2$. That seems important.

So, if I can find a recurrence relation for $x^n + y^n$, maybe I can compute it step by step. Let me denote $S_n = x^n + y^n$. Then, since $x$ and $y$ are roots of the quadratic equation $t^2 - (x + y)t + xy = 0$, which would be $t^2 - 2\sqrt{7}t + 2 = 0$. Wait, but that equation has irrational coefficients. Maybe that complicates things?

Alternatively, maybe I can find a recurrence relation for $S_n$ using the relation between $x$ and $y$. Let me think. Since $x$ and $y$ satisfy the equation $t^2 - 2\sqrt{7}t + 2 = 0$, then for each $n \geq 2$, we have $x^n = 2\sqrt{7}x^{n-1} - 2x^{n-2}$ and similarly for $y^n$. Therefore, adding these two equations:

$S_n = x^n + y^n = 2\sqrt{7}(x^{n-1} + y^{n-1}) - 2(x^{n-2} + y^{n-2}) = 2\sqrt{7}S_{n-1} - 2S_{n-2}$

But this recurrence relation still involves $\sqrt{7}$, which is irrational. That might not be helpful for computing integer values. Wait, but maybe if I consider $S_n$ and another sequence?

Alternatively, maybe I can find a recurrence with integer coefficients. Let me think. Since $x$ and $y$ are conjugates, perhaps if I consider $S_n$ and another sequence involving $x^n - y^n$?

Alternatively, maybe I can square $x$ and $y$ to find a relation. Let me compute $x^2$ and $y^2$:

$x^2 = (\sqrt{7} + \sqrt{5})^2 = 7 + 2\sqrt{35} + 5 = 12 + 2\sqrt{35}$

Similarly, $y^2 = (\sqrt{7} - \sqrt{5})^2 = 7 - 2\sqrt{35} + 5 = 12 - 2\sqrt{35}$

So, $x^2 + y^2 = 24$, and $x^2 - y^2 = 4\sqrt{35}$. Hmm, not sure if that helps yet.

But maybe if I compute $S_1 = x + y = 2\sqrt{7}$, $S_2 = x^2 + y^2 = 24$, $S_3 = x^3 + y^3$, etc. But again, with the recurrence involving $\sqrt{7}$, which complicates things. Maybe I need another approach.

Wait, but if I consider that $x$ and $y$ are roots of the equation $t^2 - 2\sqrt{7}t + 2 = 0$, then perhaps I can find a recurrence relation for $S_n$ in terms of previous terms. However, since the coefficients are irrational, maybe I need to square the recurrence or something?

Alternatively, maybe I can find a relation between $S_n$ and $S_{n-2}$ or something else. Let me try computing $S_1$, $S_2$, $S_3$, etc., step by step.

We have:

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

$S_2 = x^2 + y^2 = 24$

Now, let's compute $S_3$. Using the identity $x^3 + y^3 = (x + y)^3 - 3xy(x + y)$. Let me verify that:

Yes, $x^3 + y^3 = (x + y)^3 - 3xy(x + y)$. So substituting:

$S_3 = (2\sqrt{7})^3 - 3 \times 2 \times 2\sqrt{7} = 8 \times 7^{3/2} - 12\sqrt{7}$. Wait, that seems messy. Maybe I need to compute it numerically? But I can't use a calculator. Hmm.

Alternatively, maybe I can use the recurrence relation I had earlier:

$S_n = 2\sqrt{7} S_{n-1} - 2 S_{n-2}$. Let me try that for n=3:

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

But then $S_3 = 44\sqrt{7}$. Hmm, that's still irrational. But if I compute $S_4$:

$S_4 = 2\sqrt{7} S_3 - 2 S_2 = 2\sqrt{7} \times 44\sqrt{7} - 2 \times 24 = 2 \times 44 \times 7 - 48 = 616 - 48 = 568$. Oh! That's an integer! So S_4 is 568. Interesting. Let me check that again:

$2\sqrt{7} \times 44\sqrt{7} = 2 \times 44 \times (\sqrt{7})^2 = 88 \times 7 = 616$. Then subtract 2*S_2 = 48. So 616 - 48 = 568. Yes, that's correct. So S_4 is 568.

Similarly, let's compute S_5:

$S_5 = 2\sqrt{7} S_4 - 2 S_3 = 2\sqrt{7} \times 568 - 2 \times 44\sqrt{7} = (1136\sqrt{7} - 88\sqrt{7}) = 1048\sqrt{7}$. Again, irrational.

Then S_6:

$S_6 = 2\sqrt{7} S_5 - 2 S_4 = 2\sqrt{7} \times 1048\sqrt{7} - 2 \times 568 = 2 \times 1048 \times 7 - 1136$. Let me compute that:

First, 2*1048 = 2096. Then 2096*7. Let me compute 2000*7 = 14000, 96*7=672, so total is 14000 + 672 = 14672. Then subtract 1136: 14672 - 1136. Let me compute that:

14672 - 1000 = 13672

13672 - 136 = 13536

So S_6 = 13536.

Wait, so S_6 is 13536? But S_6 is x^6 + y^6. Therefore, since y = sqrt(7) - sqrt(5), which is approximately... Well, sqrt(7) is about 2.6458, sqrt(5) is about 2.2361, so y is about 0.4097. Therefore, y^6 is a very small positive number. Therefore, x^6 = S_6 - y^6. Therefore, since y^6 is positive, x^6 = 13536 - y^6. Therefore, the value we need is x^6, which is less than 13536. Therefore, the greatest integer less than x^6 is 13536 - 1, if y^6 is less than 1. Wait, but is y^6 less than 1?

Let me check. Since y = sqrt(7) - sqrt(5) ≈ 2.6458 - 2.2361 ≈ 0.4097. Then y^2 ≈ 0.1678, y^3 ≈ 0.1678 * 0.4097 ≈ 0.0687, y^4 ≈ 0.0687 * 0.4097 ≈ 0.0281, y^5 ≈ 0.0281 * 0.4097 ≈ 0.0115, y^6 ≈ 0.0115 * 0.4097 ≈ 0.0047. So y^6 is approximately 0.0047, which is less than 1. Therefore, x^6 = S_6 - y^6 ≈ 13536 - 0.0047 ≈ 13535.9953. Therefore, the greatest integer less than x^6 is 13535.

But wait, is that correct? Let me check again. Because S_6 = x^6 + y^6 = 13536. Therefore, x^6 = 13536 - y^6. Since y^6 is positive, x^6 is less than 13536. Therefore, floor(x^6) is 13535. But is that the case? Wait, but maybe y^6 is so small that x^6 is just slightly less than 13536, so the floor is 13535. But is there a possibility that y^6 is more than 1? Wait, earlier approximation suggested it's about 0.0047, which is less than 1. So floor(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_1 = 2√7, S_2 = 24.

Then S_3 = 2√7 * S_2 - 2 * S_1 = 2√7 * 24 - 2*(2√7) = 48√7 - 4√7 = 44√7. Correct.

Then S_4 = 2√7 * S_3 - 2 * S_2 = 2√7 * 44√7 - 2*24 = 2*44*7 - 48 = 616 - 48 = 568. Correct.

Then S_5 = 2√7 * S_4 - 2 * S_3 = 2√7 * 568 - 2*(44√7) = 1136√7 - 88√7 = 1048√7. Correct.

Then S_6 = 2√7 * S_5 - 2 * S_4 = 2√7 * 1048√7 - 2*568 = 2*1048*7 - 1136.

Calculating 2*1048 = 2096. Then 2096*7: Let me compute 2000*7=14000, 96*7=672, so total 14000 + 672 = 14672. Then subtract 1136: 14672 - 1136. Let me do this subtraction again:

14672 - 1000 = 13672

13672 - 136 = 13536. Yes, that's correct. So S_6 = 13536.

Therefore, since y^6 is positive and less than 1, then x^6 = 13536 - y^6 is less than 13536 but greater than 13535. Therefore, the greatest integer less than x^6 is 13535. Therefore, the answer is 13535.

But wait, let me check if there's another way to compute x^6 directly, maybe using binomial theorem? Let me try that as a verification.

Compute $(\sqrt{7} + \sqrt{5})^6$. Let me expand this using 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 to compute without a calculator. However, maybe I can pair terms? Let me note that terms with k and 6 - k might combine in some way. Alternatively, maybe I can write this as:

Let me note that $(\sqrt{7} + \sqrt{5})^6 + (\sqrt{7} - \sqrt{5})^6 = S_6 = 13536$, as we found earlier. Therefore, if I can compute $(\sqrt{7} + \sqrt{5})^6$ as S_6 - $(\sqrt{7} - \sqrt{5})^6$, which is 13536 - y^6. Since y^6 is positive and less than 1, then floor(x^6) is 13535.

But maybe I can compute y^6 more accurately? Let me try.

First, compute y = sqrt(7) - sqrt(5). Let me compute y^2:

y^2 = (sqrt(7) - sqrt(5))^2 = 7 + 5 - 2*sqrt(35) = 12 - 2*sqrt(35).

Then y^3 = y^2 * y = (12 - 2√35)(sqrt(7) - sqrt(5)). Let me compute that:

Multiply 12*(sqrt(7) - sqrt(5)) - 2√35*(sqrt(7) - sqrt(5)).

First term: 12√7 - 12√5.

Second term: -2√35*sqrt(7) + 2√35*sqrt(5) = -2√(35*7) + 2√(35*5) = -2√245 + 2√175.

Simplify radicals:

√245 = √(49*5) = 7√5

√175 = √(25*7) = 5√7

Therefore, second term becomes: -2*7√5 + 2*5√7 = -14√5 + 10√7.

Therefore, combining first and second terms:

12√7 - 12√5 -14√5 + 10√7 = (12√7 + 10√7) + (-12√5 -14√5) = 22√7 - 26√5.

So y^3 = 22√7 - 26√5.

Hmm, that's still irrational. Let me compute y^4 = y^3 * y = (22√7 - 26√5)(sqrt(7) - sqrt(5)).

Multiply term by term:

22√7 * sqrt(7) = 22*7 = 154

22√7 * (-sqrt(5)) = -22√35

-26√5 * sqrt(7) = -26√35

-26√5 * (-sqrt(5)) = +26*5 = 130

So combining terms:

154 + 130 = 284

-22√35 -26√35 = -48√35

Therefore, y^4 = 284 - 48√35.

Similarly, compute y^5 = y^4 * y = (284 - 48√35)(sqrt(7) - sqrt(5)).

Let me compute that:

First, 284*sqrt(7) - 284*sqrt(5) - 48√35*sqrt(7) + 48√35*sqrt(5)

Simplify each term:

First term: 284√7

Second term: -284√5

Third term: -48√(35*7) = -48√245 = -48*7√5 = -336√5

Fourth term: +48√(35*5) = +48√175 = +48*5√7 = +240√7

Combine like terms:

For √7: 284√7 + 240√7 = 524√7

For √5: -284√5 - 336√5 = -620√5

Therefore, y^5 = 524√7 - 620√5.

Now compute y^6 = y^5 * y = (524√7 - 620√5)(sqrt(7) - sqrt(5)).

Again, multiply term by term:

524√7 * sqrt(7) = 524*7 = 3668

524√7 * (-sqrt(5)) = -524√35

-620√5 * sqrt(7) = -620√35

-620√5 * (-sqrt(5)) = +620*5 = 3100

Combine terms:

3668 + 3100 = 6768

-524√35 -620√35 = -1144√35

Therefore, y^6 = 6768 - 1144√35.

So, y^6 = 6768 - 1144√35. Now, we need to compute this value numerically? Wait, but we can't use a calculator, but maybe we can estimate √35?

Well, √35 is between 5.9 and 6.0 because 5.9² = 34.81 and 6.0² = 36. So √35 ≈ 5.916. Let me check:

5.916² = (5.9 + 0.016)^2 = 5.9² + 2*5.9*0.016 + 0.016² = 34.81 + 0.1888 + 0.000256 ≈ 34.81 + 0.1888 = 34.9988 + 0.000256 ≈ 34.999056. Close to 35. So √35 ≈ 5.916. Let me take √35 ≈ 5.916.

Therefore, 1144√35 ≈ 1144 * 5.916. Let me compute that:

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. So total 591.6 + 236.64 = 828.24 + 23.664 = 851.904. Therefore, total 1144*5.916 ≈ 5916 + 851.904 = 6767.904.

Therefore, y^6 ≈ 6768 - 6767.904 = 0.096. Wait, that's approximately 0.096. Earlier, my rough estimate was 0.0047, but this more precise calculation gives about 0.096? That's a big difference. Which one is correct?

Wait, let me check my calculation again. Wait, y^6 = 6768 - 1144√35. If √35 ≈ 5.916, then 1144 * 5.916 ≈ ?

Wait, let me check 1144 * 5.916:

First, note that 1144 * 5 = 5720

1144 * 0.916 = ?

Compute 1144 * 0.9 = 1029.6

1144 * 0.016 = 18.304

So total 1029.6 + 18.304 = 1047.904

Therefore, total 1144 * 5.916 ≈ 5720 + 1047.904 = 6767.904. Therefore, y^6 ≈ 6768 - 6767.904 = 0.096. So approximately 0.096.

But earlier, when I approximated y ≈ 0.4097, then y^6 ≈ (0.4097)^6. Let me compute that:

First, 0.4^6 = 0.004096. But 0.4097 is slightly larger than 0.4, so y^6 would be slightly larger than 0.004096. However, according to the exact expression, it's approximately 0.096. There's a discrepancy here. Which one is correct?

Wait, there must be a mistake in my calculation of y^6. Let me check again.

Earlier, I calculated y^6 = 6768 - 1144√35. Let me check the steps again.

Starting from y^5 = 524√7 - 620√5. Then y^6 = y^5 * y = (524√7 - 620√5)(sqrt(7) - sqrt(5)).

Expanding:

First term: 524√7 * sqrt(7) = 524 * 7 = 3668

Second term: 524√7 * (-sqrt(5)) = -524√35

Third term: -620√5 * sqrt(7) = -620√35

Fourth term: -620√5 * (-sqrt(5)) = +620 * 5 = 3100

So adding up the constants: 3668 + 3100 = 6768. Correct.

Adding up the radicals: -524√35 -620√35 = -(524 + 620)√35 = -1144√35. Correct. Therefore, y^6 = 6768 - 1144√35. So that's correct.

But then why is there a discrepancy between the two estimates? Let me check my approximate value of y. Earlier, I said y = sqrt(7) - sqrt(5) ≈ 2.6458 - 2.2361 ≈ 0.4097. Let me check that more accurately.

sqrt(7) is approximately 2.6457513110645906

sqrt(5) is approximately 2.2360679775

Therefore, y ≈ 2.6457513110645906 - 2.2360679775 ≈ 0.4096833335645906. So approximately 0.409683.

So y ≈ 0.409683. Let me compute y^6:

First, compute y^2 ≈ (0.409683)^2 ≈ 0.16783

Then y^3 ≈ y^2 * y ≈ 0.16783 * 0.409683 ≈ 0.0687

y^4 ≈ 0.0687 * 0.409683 ≈ 0.0281

y^5 ≈ 0.0281 * 0.409683 ≈ 0.0115

y^6 ≈ 0.0115 * 0.409683 ≈ 0.00471.

But according to the exact expression, y^6 ≈ 0.096. There's a big difference here. Which one is correct?

Wait, there must be a mistake in my exact calculation. Let me check again. Wait, if y^6 = 6768 - 1144√35, then let me compute 1144√35 numerically. Wait, if √35 ≈ 5.916079783, then 1144 * 5.916079783 ≈ ?

Let me compute 1144 * 5.916079783:

First, note that 1000 * 5.916079783 = 5916.079783

Then 144 * 5.916079783:

Compute 100 * 5.916079783 = 591.6079783

40 * 5.916079783 = 236.6431913

4 * 5.916079783 = 23.66431913

Adding those: 591.6079783 + 236.6431913 = 828.2511696 + 23.66431913 ≈ 851.9154887

Therefore, total 1144 * 5.916079783 ≈ 5916.079783 + 851.9154887 ≈ 6767.9952717

Therefore, y^6 = 6768 - 6767.9952717 ≈ 0.0047283.

Ah! So approximately 0.0047283, which matches the earlier approximation of ~0.0047. So my previous calculation had a miscalculation where I thought 1144*5.916 was 6767.904, but actually, it's approximately 6767.995, so 6768 - 6767.995 ≈ 0.005. Therefore, y^6 ≈ 0.0047. Therefore, my initial approximation was correct. So why did I get 0.096 before? Because I miscalculated 1144*5.916. I think I had a decimal error. So the exact value of y^6 is approximately 0.0047, which is less than 1. Therefore, x^6 = S_6 - y^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 once more. Since S_6 = x^6 + y^6 = 13536, and y^6 is positive and less than 1, then x^6 is between 13535 and 13536. Therefore, floor(x^6) = 13535.

But to be thorough, let me check if y^6 is less than 1. Since y ≈ 0.4097, then y^6 ≈ (0.4)^6 = 0.004096, and since 0.4097 > 0.4, y^6 is slightly larger than 0.004096, but still less than 1. Therefore, yes, x^6 is less than 13536 but greater than 13535. Therefore, the floor is 13535.

But let me check if there's another way to confirm this. For example, maybe compute x^6 directly using binomial expansion? Let me try that.

Compute $(\sqrt{7} + \sqrt{5})^6$. Let me write it as:

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

Let me compute each term:

For k=0: $\binom{6}{0} (\sqrt{7})^6 (\sqrt{5})^0 = 1 * 7^3 * 1 = 343$

For k=1: $\binom{6}{1} (\sqrt{7})^5 (\sqrt{5})^1 = 6 * 7^{2.5} * 5^{0.5}$. Wait, this seems messy. Maybe I can write them in terms of radicals:

Alternatively, note that $(\sqrt{7} + \sqrt{5})^6 + (\sqrt{7} - \sqrt{5})^6 = 13536$, so if I can compute $(\sqrt{7} + \sqrt{5})^6$ as 13536 - y^6, and since y^6 ≈ 0.0047, then x^6 ≈ 13535.9953, so floor(x^6) = 13535.

Alternatively, maybe compute x^6 using the recurrence relation? Wait, we already did that.

Alternatively, maybe compute x^2, x^3, x^4, x^5, x^6 step by step?

Let me try that.

First, compute x = sqrt(7) + sqrt(5).

Compute x^2 = (sqrt(7) + sqrt(5))^2 = 7 + 2*sqrt(35) + 5 = 12 + 2*sqrt(35).

Compute x^3 = x^2 * x = (12 + 2√35)(sqrt(7) + sqrt(5)). Let me compute that:

Multiply 12*sqrt(7) + 12*sqrt(5) + 2√35*sqrt(7) + 2√35*sqrt(5)

Simplify each term:

12√7 + 12√5 + 2√(35*7) + 2√(35*5)

Which is 12√7 + 12√5 + 2√245 + 2√175

Simplify radicals:

√245 = 7√5, √175 = 5√7

Therefore:

12√7 + 12√5 + 2*7√5 + 2*5√7 = 12√7 + 12√5 + 14√5 + 10√7 = (12√7 + 10√7) + (12√5 + 14√5) = 22√7 + 26√5.

So x^3 = 22√7 + 26√5.

Similarly, compute x^4 = x^3 * x = (22√7 + 26√5)(sqrt(7) + sqrt(5)).

Multiply term by term:

22√7*sqrt(7) + 22√7*sqrt(5) + 26√5*sqrt(7) + 26√5*sqrt(5)

Simplify:

22*7 + 22√35 + 26√35 + 26*5

Which is 154 + 22√35 + 26√35 + 130 = (154 + 130) + (22√35 + 26√35) = 284 + 48√35.

So x^4 = 284 + 48√35.

Then x^5 = x^4 * x = (284 + 48√35)(sqrt(7) + sqrt(5)).

Multiply term by term:

284*sqrt(7) + 284*sqrt(5) + 48√35*sqrt(7) + 48√35*sqrt(5)

Simplify:

284√7 + 284√5 + 48√(35*7) + 48√(35*5)

Which is 284√7 + 284√5 + 48√245 + 48√175

Again, √245 = 7√5, √175 = 5√7:

So:

284√7 + 284√5 + 48*7√5 + 48*5√7 = 284√7 + 284√5 + 336√5 + 240√7

Combine like terms:

(284√7 + 240√7) + (284√5 + 336√5) = 524√7 + 620√5.

Therefore, x^5 = 524√7 + 620√5.

Now compute x^6 = x^5 * x = (524√7 + 620√5)(sqrt(7) + sqrt(5)).

Multiply term by term:

524√7*sqrt(7) + 524√7*sqrt(5) + 620√5*sqrt(7) + 620√5*sqrt(5)

Simplify:

524*7 + 524√35 + 620√35 + 620*5

Which is 3668 + 524√35 + 620√35 + 3100 = (3668 + 3100) + (524√35 + 620√35) = 6768 + 1144√35.

Therefore, x^6 = 6768 + 1144√35.

Wait a second! Earlier, we had y^6 = 6768 - 1144√35, so x^6 + y^6 = 6768 + 1144√35 + 6768 - 1144√35 = 13536, which matches our previous result.

So x^6 = 6768 + 1144√35. Now, we need to find the greatest integer less than x^6.

So, if I can compute 1144√35 numerically, then I can find x^6.

Earlier, we found that √35 ≈ 5.916079783. Therefore, 1144 * √35 ≈ 1144 * 5.916079783 ≈ 6767.99527.

Therefore, x^6 ≈ 6768 + 6767.99527 ≈ 13535.99527.

Therefore, x^6 ≈ 13535.99527, so the greatest integer less than x^6 is 13535.

Therefore, the answer is 13535.

But let me check once more. Since x^6 = 6768 + 1144√35, and we know that √35 is irrational, but we can bound it.

We know that √35 is between 5.916 and 5.9161. Let me check:

5.916^2 = 34.996656

5.9161^2 = ?

Let me compute 5.916^2 = 34.996656

Then, 5.9161^2 = (5.916 + 0.0001)^2 = 5.916^2 + 2*5.916*0.0001 + (0.0001)^2 ≈ 34.996656 + 0.0011832 + 0.00000001 ≈ 34.99783921.

But 35 is larger than that. Therefore, √35 is between 5.9161 and 5.9162. Let me check 5.9162^2:

= (5.916 + 0.0002)^2 = 5.916^2 + 2*5.916*0.0002 + (0.0002)^2 ≈ 34.996656 + 0.0023664 + 0.00000004 ≈ 34.99902244. Still less than 35.

Next, 5.9163^2:

= 5.9162^2 + 2*5.9162*0.0001 + (0.0001)^2 ≈ 34.99902244 + 0.00118324 + 0.00000001 ≈ 35.00020569.

Ah, so √35 is between 5.9162 and 5.9163. Therefore, let me approximate √35 ≈ 5.9162 + (35 - 34.99902244)/(2*5.9162).

Wait, using linear approximation. Let me denote f(x) = x^2. We know that f(5.9162) ≈ 34.99902244, and we want to find x such that f(x) = 35.

Let me take x = 5.9162 + δ, where δ is small. Then:

f(x) ≈ f(5.9162) + 2*5.9162*δ = 35

So 34.99902244 + 11.8324*δ ≈ 35

Therefore, 11.8324*δ ≈ 35 - 34.99902244 = 0.00097756

Therefore, δ ≈ 0.00097756 / 11.8324 ≈ 0.0000826.

Therefore, √35 ≈ 5.9162 + 0.0000826 ≈ 5.9162826.

Therefore, √35 ≈ 5.9162826.

Therefore, 1144 * √35 ≈ 1144 * 5.9162826. Let me compute this:

First, compute 1000 * 5.9162826 = 5916.2826

Then 144 * 5.9162826. Let me compute 100 * 5.9162826 = 591.62826

40 * 5.9162826 = 236.651304

4 * 5.9162826 = 23.6651304

Adding those: 591.62826 + 236.651304 = 828.279564 + 23.6651304 ≈ 851.9446944

Therefore, total 1144 * √35 ≈ 5916.2826 + 851.9446944 ≈ 6768.2272944.

Therefore, x^6 = 6768 + 1144√35 ≈ 6768 + 6768.2272944 ≈ 13536.2272944? Wait, no! Wait, no, x^6 = 6768 + 1144√35. Wait, earlier we had:

Wait, no! Wait, earlier, we had x^6 = 6768 + 1144√35. But earlier, we also had y^6 = 6768 - 1144√35. Therefore, if √35 ≈ 5.9162826, then 1144√35 ≈ 6768.227. Therefore, x^6 ≈ 6768 + 6768.227 ≈ 13536.227? Wait, that can't be, because earlier we had S_6 = x^6 + y^6 = 13536. If x^6 ≈ 13536.227, then y^6 would be negative? Which is impossible. There must be a miscalculation here.

Wait, no! Wait, no, x^6 = 6768 + 1144√35. But earlier, we had S_6 = x^6 + y^6 = 13536. And y^6 = 6768 - 1144√35. Therefore, if x^6 = 6768 + 1144√35, then adding y^6 = 6768 - 1144√35 gives 13536. Therefore, if 1144√35 ≈ 6768.227, then x^6 ≈ 6768 + 6768.227 ≈ 13536.227? But that would make y^6 = 6768 - 6768.227 ≈ -0.227, which is impossible because y^6 is positive. Therefore, there must be a mistake in my calculation of 1144√35.

Wait, earlier, I had that y^6 = 6768 - 1144√35 ≈ 0.0047. Therefore, 1144√35 ≈ 6768 - 0.0047 ≈ 6767.9953. Therefore, 1144√35 ≈ 6767.9953. Therefore, x^6 = 6768 + 1144√35 ≈ 6768 + 6767.9953 ≈ 13535.9953. Which matches our previous result.

But why did I get 1144√35 ≈ 6768.227 earlier? Because I miscalculated. Wait, no. Wait, I think I confused the value of √35. Wait, earlier, I had that √35 ≈ 5.9162826, so 1144 * √35 ≈ 1144 * 5.9162826. But if √35 ≈ 5.916, then 1144 * 5.916 ≈ 6767.995, which is approximately 6768. Therefore, 1144√35 ≈ 6768 - 0.0047 ≈ 6767.9953. Therefore, x^6 = 6768 + 1144√35 ≈ 6768 + 6767.9953 ≈ 13535.9953.

Therefore, my previous calculation where I thought 1144√35 ≈ 6768.227 was wrong. Where did that come from? Probably a miscalculation when I added 5916.2826 + 851.9446944. Let me check that again:

Wait, 1000 * 5.9162826 = 5916.2826

Then 144 * 5.9162826. Let me compute 144 * 5.9162826:

First, note that 144 * 5 = 720

144 * 0.9162826 = ?

Compute 144 * 0.9 = 129.6

144 * 0.0162826 ≈ 144 * 0.016 = 2.304, and 144 * 0.0002826 ≈ 0.0406. So total ≈ 2.304 + 0.0406 ≈ 2.3446

Therefore, total 129.6 + 2.3446 ≈ 131.9446

Therefore, total 144 * 5.9162826 ≈ 720 + 131.9446 ≈ 851.9446

Therefore, total 1144 * √35 ≈ 5916.2826 + 851.9446 ≈ 6768.2272.

But this contradicts the previous result where y^6 = 6768 - 1144√35 ≈ 0.0047. Therefore, there must be a miscalculation here.

Wait, but if y^6 = 6768 - 1144√35 ≈ 0.0047, then 1144√35 ≈ 6768 - 0.0047 ≈ 6767.9953. Therefore, 1144√35 ≈ 6767.9953. But according to the calculation above, it's ≈ 6768.2272. There's a discrepancy of about 0.23. Where is this coming from?

Wait, I think I made a mistake in the value of √35. Earlier, I approximated √35 ≈ 5.9162826, but if 1144√35 ≈ 6767.9953, then √35 ≈ 6767.9953 / 1144 ≈ let me compute that:

Divide 6767.9953 by 1144:

First, 1144 * 5 = 5720

Subtract 5720 from 6767.9953: 6767.9953 - 5720 = 1047.9953

Now, 1144 * 0.9 = 1029.6

Subtract 1029.6 from 1047.9953: 1047.9953 - 1029.6 = 18.3953

Now, 1144 * 0.016 = 18.304

Subtract 18.304 from 18.3953: ≈ 0.0913

So total is 5 + 0.9 + 0.016 + ... ≈ 5.916 + ... So √35 ≈ 5.916 + 0.0913 / 1144 ≈ 5.916 + 0.0000798 ≈ 5.9160798. Which is consistent with the known value of √35 ≈ 5.916079783. Therefore, my previous calculation of 1144 * √35 ≈ 6768.2272 must be wrong.

Wait, let me check with the exact value of √35 ≈ 5.916079783. Then 1144 * 5.916079783 = ?

Let me compute 1144 * 5.916079783:

First, note that 1144 * 5 = 5720

1144 * 0.916079783 = ?

Let me compute 1144 * 0.9 = 1029.6

1144 * 0.016079783 ≈ 1144 * 0.016 = 18.304, and 1144 * 0.000079783 ≈ 1144 * 0.00008 ≈ 0.09152

So total ≈ 18.304 + 0.09152 ≈ 18.39552

Therefore, total 1144 * 0.916079783 ≈ 1029.6 + 18.39552 ≈ 1047.99552

Therefore, total 1144 * 5.916079783 ≈ 5720 + 1047.99552 ≈ 6767.99552.

Ah! So my previous calculation where I thought it was 6768.2272 was wrong. The correct value is approximately 6767.9955. Therefore, x^6 = 6768 + 1144√35 ≈ 6768 + 6767.9955 ≈ 13535.9955.

Therefore, x^6 ≈ 13535.9955, which is just slightly less than 13536. Therefore, the greatest integer less than x^6 is 13535.

Therefore, after all these calculations and checking, I can be confident that the answer 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 conjugates and recurrence relations.

---

### Step 1: Define the Conjugate Pair

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

Note that:
- $ x + y = 2\sqrt{7} $
- $ x - y = 2\sqrt{5} $
- $ xy = (\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5}) = 7 - 5 = 2 $

---

### Step 2: Define a Recurrence for $ S_n = x^n + y^n $

We define:
$$
S_n = x^n + y^n
$$

Using the identity:
$$
x^n + y^n = (x + y)(x^{n-1} + y^{n-1}) - xy(x^{n-2} + y^{n-2})
$$

We get the recurrence:
$$
S_n = 2\sqrt{7} \cdot S_{n-1} - 2 \cdot S_{n-2}
$$

We can compute the first few terms:

- $ S_1 = x + y = 2\sqrt{7} $
- $ S_2 = x^2 + y^2 = (\sqrt{7} + \sqrt{5})^2 + (\sqrt{7} - \sqrt{5})^2 = 12 + 2\sqrt{35} + 12 - 2\sqrt{35} = 24 $

Now compute the rest using the recurrence:

- $ 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 = 2 \cdot 44 \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 = 2 \cdot 1048 \cdot 7 - 1136 = 14672 - 1136 = 13536 $

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

---

### Step 3: Estimate $ y^6 $

We know that:
$$
x^6 = S_6 - y^6 = 13536 - y^6
$$

Now, estimate $ y = \sqrt{7} - \sqrt{5} \approx 2.6458 - 2.2361 = 0.4097 $

Then:
$$
y^6 \approx (0.4097)^6 \approx 0.0047
$$

So:
$$
x^6 = 13536 - y^6 \approx 13536 - 0.0047 = 13535.9953
$$

---

### Step 4: Find the Greatest Integer Less Than $ x^6 $

Since $ x^6 \approx 13535.9953 $, 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 the same as 6 in base 10. Wait, is that right? Let me confirm. Yes, because the value of a digit in base 8 is its face value if it's in the units place. So 6_8 is 6*8^0 = 6*1 = 6. So that's 6 in base 10.

Now the second number is 7_8. Similarly, this is just 7 in base 10. Because again, it's a single digit in the units place. So 7_8 is 7*8^0 = 7.

So multiplying them together in base 10 would be 6 * 7. Let me calculate that. 6*7 is 42. So the product in base 10 is 42. Now I need to convert 42 from base 10 to base 8.

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

First, divide 42 by 8. Let me do that division. 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, are 5 and 2. Therefore, 42 in base 10 is 52 in base 8.

Wait, let me check that. Let me verify by converting 52_8 back to base 10.

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

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?

So, if I try to multiply 6_8 by 7_8 without converting to base 10. Let me recall how multiplication in base 8 works.

In base 8, the digits go from 0 to 7. So when multiplying, if the product of digits is 8 or more, we need to carry over to the next higher place value.

So, let me write it out like a standard multiplication problem.

       6
     x 7
     -----
       ?

So, multiplying 6 by 7. In base 10, that's 42, but in base 8, how does that work?

Wait, but if I do this in base 8, I need to perform the multiplication in base 8. Let me think.

First, 6_8 * 7_8. Let me think of 6 and 7 as base 8 digits.

But 6*7 in base 10 is 42, but how do I express 42 in base 8? Which is what I did earlier. So that gives me 52_8.

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

Let me try that.

So, 6_8 multiplied by 7_8. Let me think of it as 6*7.

But in base 8, the multiplication table might be different. Wait, but actually, the multiplication of digits is the same as in base 10, but then you have to convert the result to base 8.

Wait, maybe not. Let me think.

For example, if I have 6_8 * 7_8, the product is 6*7 = 42 in base 10. Then, converting 42 to base 8 gives 52_8.

Alternatively, if I try to do the multiplication in base 8, I need to remember that each digit is in base 8.

But since both numbers are single digits, the multiplication is straightforward.

But maybe I need to think about how to handle carries.

Wait, if I multiply 6_8 by 7_8, let me think of it as 6*7.

But in base 8, 6 is 6, 7 is 7. So 6*7 is 42. Now, since we are working in base 8, we need to express 42 in base 8.

Which is the same as before.

So, 42 divided by 8 is 5 with remainder 2, so 5*8 + 2 = 42. Therefore, 52_8.

So that seems consistent.

Alternatively, if I wanted to do the multiplication in base 8 without converting to base 10, maybe I can think of it as follows:

First, 6_8 * 7_8. Let me recall that in base 8, the digits go up to 7, so when multiplying, if the product is more than 7, we need to carry over.

But since 6*7 is 42 in base 10, which is 5*8 + 2, so in base 8, that would be 5*8^1 + 2*8^0, so 52_8.

Therefore, the answer is 52_8.

But let me check with another approach. Let me convert both numbers to base 10, multiply, convert back.

As before, 6_8 is 6, 7_8 is 7. 6*7=42. 42 in base 8 is 52.

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

Let me try that.

So, 6_8 multiplied by 7_8.

First, let me recall that in base 8, the multiplication is similar to base 10, but with carries happening every 8.

So, 6 * 7. Let me compute 6*7 in base 8.

But 6*7 is 42 in base 10. Now, how do I convert 42 into base 8?

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

So, same answer.

Alternatively, if I think of multiplying 6_8 by 7_8 as follows:

Let me think of 7_8 as 7 in base 8. So, 6_8 * 7_8.

But maybe I can think of 7_8 as 8 - 1. So, 6_8 * (8 - 1)_8.

But in base 8, 8 is 10_8. So, 6_8 * (10_8 - 1_8) = 6_8 * 10_8 - 6_8 * 1_8.

In base 8, multiplying by 10_8 is like shifting left by one digit, which is equivalent to multiplying by 8 in base 10.

So, 6_8 * 10_8 is 60_8. Then subtract 6_8.

So, 60_8 - 6_8.

Let me compute that.

First, 60_8 minus 6_8.

In base 8, subtracting 6 from 60.

Let me write it down:

   60
 -  6
-----
?

Starting from the rightmost digit: 0 - 6. Since 0 is less than 6, I need to borrow from the next digit.

The next digit is 6, so I borrow 1 from the 6, which is worth 8 in the units place.

So, the 6 becomes 5, and the 0 becomes 8.

Now, subtract 6 from 8: 8 - 6 = 2.

So the result is 52_8.

Which matches the previous results.

Therefore, this method also gives 52_8.

So, all methods lead to the same answer.

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

But just to make sure, let me try another way. Let me convert both numbers to base 10, multiply, convert back.

As before, 6_8 is 6, 7_8 is 7. 6*7=42.

Convert 42 to base 8.

Divide 42 by 8: 42 ÷ 8 = 5 with remainder 2.

So, 5*8 + 2 = 42. Therefore, 52_8.

Same answer.

Alternatively, if I wanted to do the multiplication in base 8 using the standard algorithm, like how we do in base 10.

Let me try that.

Multiplying 6_8 by 7_8.

Since both numbers are single digits, it's straightforward.

But let me think of it as:

First, multiply 6 by 7.

But in base 8, 6*7 is 42 in base 10.

But how do I represent 42 in base 8?

As before, 42 divided by 8 is 5 with remainder 2. So, 5*8 + 2 = 42.

Therefore, the result is 52_8.

So, again, same answer.

Alternatively, if I think of the multiplication table for base 8.

But I don't remember the exact multiplication table for base 8, but I can compute it.

For example, 6*7 in base 8.

But since 6 and 7 are both less than 8, their product is 42 in base 10.

But in base 8, 42 is 5*8 + 2, so 52.

Alternatively, if I think of 6*7 in base 8, maybe I can do it step by step.

But I think that's the same as before.

Alternatively, maybe I can think of 6_8 as 6 in base 8, and 7_8 as 7 in base 8.

But since they are single digits, the multiplication is straightforward.

But perhaps I can think of 6_8 * 7_8 as follows:

In base 8, 6 is 6, 7 is 7.

So, 6*7 = 42.

But since we are in base 8, we need to write 42 in base 8.

Which is 5*8 + 2, so 52.

Therefore, the answer is 52_8.

So, all methods lead to the same conclusion.

Therefore, I think the answer is 52 in base 8.

But just to be thorough, let me check with another example.

Suppose I have 6_8 * 6_8. What would that be?

In base 10, that's 6*6=36. Converting 36 to base 8: 36 divided by 8 is 4 with remainder 4. So 44_8.

Alternatively, in base 8, 6*6.

But 6*6=36 in base 10. 36 divided by 8 is 4 with remainder 4. So 44_8.

Which is correct.

Similarly, if I do 7_8 * 7_8.

In base 10, that's 7*7=49. Converting 49 to base 8: 49 divided by 8 is 6 with remainder 1. So 61_8.

Which is correct.

So, if I do 6_8 * 7_8, which is 42 in base 10, which is 52_8.

So, that seems consistent.

Alternatively, if I think of 6_8 * 7_8 as 6*7 in base 8.

But since 6 and 7 are both less than 8, their product is 42 in base 10, which is 52 in base 8.

Therefore, I think that's correct.

Another way to check is to convert 52_8 back to base 10 and see if it's 42.

So, 5*8 + 2 = 40 + 2 = 42.

Yes, that's correct.

Therefore, I can be confident that the answer is 52_8.

So, after going through multiple methods and checking each step, I believe the product of 6_8 and 7_8 in base 8 is 52_8.

**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 the same as 6 in base 10. Wait, is that right? Let me confirm. Yes, because the value of a digit in base 8 is its face value if it's in the units place. So 6_8 is 6*8^0 = 6*1 = 6. So that's 6 in base 10.

Now the second number is 7_8. Similarly, this is just 7 in base 10. Because again, it's a single digit in the units place. So 7_8 is 7*8^0 = 7.

So multiplying them together in base 10 would be 6 * 7. Let me calculate that. 6*7 is 42. So the product in base 10 is 42. Now I need to convert 42 from base 10 to base 8.

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

First, divide 42 by 8. Let me do that division. 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, are 5 and 2. Therefore, 42 in base 10 is 52 in base 8.

Wait, let me check that. Let me verify by converting 52_8 back to base 10.

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

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?

So, if I try to multiply 6_8 by 7_8 without converting to base 10. Let me recall how multiplication in base 8 works.

In base 8, the digits go from 0 to 7. So when multiplying, if the product of digits is 8 or more, we need to carry over to the next higher place value.

So, let me write it out like a standard multiplication problem.

       6
     x 7
     -----
       ?

So, multiplying 6 by 7. In base 10, that's 42, but in base 8, how does that work?

Wait, but if I do this in base 8, I need to perform the multiplication in base 8. Let me think.

First, 6_8 * 7_8. Let me think of 6 and 7 as base 8 digits.

But 6*7 in base 10 is 42, but how do I express 42 in base 8? Which is what I did earlier. So that gives me 52_8.

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

Let me try that.

So, 6_8 multiplied by 7_8. Let me think of it as 6*7.

But in base 8, the multiplication table might be different. Wait, but actually, the multiplication of digits is the same as in base 10, but then you have to convert the result to base 8.

Wait, maybe not. Let me think.

For example, if I have 6_8 * 7_8, the product is 6*7 = 42 in base 10. Then, converting 42 to base 8 gives 52_8.

Alternatively, if I try to do the multiplication in base 8, I need to remember that each digit is in base 8.

But since both numbers are single digits, the multiplication is straightforward.

But maybe I need to think about how to handle carries.

Wait, if I multiply 6_8 by 7_8, let me think of it as 6*7.

But in base 8, 6 is 6, 7 is 7. So 6*7 is 42. Now, since we are working in base 8, we need to express 42 in base 8.

Which is the same as before.

So, 42 divided by 8 is 5 with remainder 2, so 5*8 + 2 = 42. Therefore, 52_8.

So that seems consistent.

Alternatively, if I wanted to do the multiplication in base 8 without converting to base 10, maybe I can think of it as follows:

First, 6_8 * 7_8. Let me recall that in base 8, the digits go up to 7, so when multiplying, if the product is more than 7, we need to carry over.

But since 6*7 is 42 in base 10, which is 5*8 + 2, so in base 8, that would be 5*8^1 + 2*8^0, so 52_8.

Therefore, the answer is 52_8.

But let me check with another approach. Let me convert both numbers to base 10, multiply, convert back.

As before, 6_8 is 6, 7_8 is 7. 6*7=42. 42 in base 8 is 52.

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

Let me try that.

So, 6_8 multiplied by 7_8.

First, let me recall that in base 8, the multiplication is similar to base 10, but with carries happening every 8.

So, 6 * 7. Let me compute 6*7 in base 8.

But 6*7 is 42 in base 10. Now, how do I convert 42 into base 8?

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

So, same answer.

Alternatively, if I think of multiplying 6_8 by 7_8 as follows:

Let me think of 7_8 as 7 in base 8. So, 6_8 * 7_8.

But maybe I can think of 7_8 as 8 - 1. So, 6_8 * (8 - 1)_8.

But in base 8, 8 is 10_8. So, 6_8 * (10_8 - 1_8) = 6_8 * 10_8 - 6_8 * 1_8.

In base 8, multiplying by 10_8 is like shifting left by one digit, which is equivalent to multiplying by 8 in base 10.

So, 6_8 * 10_8 is 60_8. Then subtract 6_8.

So, 60_8 - 6_8.

Let me compute that.

First, 60_8 minus 6_8.

In base 8, subtracting 6 from 60.

Let me write it down:

   60
 -  6
-----
?

Starting from the rightmost digit: 0 - 6. Since 0 is less than 6, I need to borrow from the next digit.

The next digit is 6, so I borrow 1 from the 6, which is worth 8 in the units place.

So, the 6 becomes 5, and the 0 becomes 8.

Now, subtract 6 from 8: 8 - 6 = 2.

So the result is 52_8.

Which matches the previous results.

Therefore, this method also gives 52_8.

So, all methods lead to the same answer.

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

But just to make sure, let me try another way. Let me convert both numbers to base 10, multiply, convert back.

As before, 6_8 is 6, 7_8 is 7. 6*7=42.

Convert 42 to base 8.

Divide 42 by 8: 42 ÷ 8 = 5 with remainder 2.

So, 5*8 + 2 = 42. Therefore, 52_8.

Same answer.

Alternatively, if I wanted to do the multiplication in base 8 using the standard algorithm, like how we do in base 10.

Let me try that.

Multiplying 6_8 by 7_8.

Since both numbers are single digits, it's straightforward.

But let me think of it as:

First, multiply 6 by 7.

But in base 8, 6*7 is 42 in base 10.

But how do I represent 42 in base 8?

As before, 42 divided by 8 is 5 with remainder 2. So, 5*8 + 2 = 42.

Therefore, the result is 52_8.

So, again, same answer.

Alternatively, if I think of the multiplication table for base 8.

But I don't remember the exact multiplication table for base 8, but I can compute it.

For example, 6*7 in base 8.

But since 6 and 7 are both less than 8, their product is 42 in base 10.

But in base 8, 42 is 5*8 + 2, so 52.

Alternatively, if I think of 6*7 in base 8, maybe I can do it step by step.

But I think that's the same as before.

Alternatively, maybe I can think of 6_8 as 6 in base 8, and 7_8 as 7 in base 8.

But since they are single digits, the multiplication is straightforward.

But perhaps I can think of 6_8 * 7_8 as follows:

In base 8, 6 is 6, 7 is 7.

So, 6*7 = 42.

But since we are in base 8, we need to write 42 in base 8.

Which is 5*8 + 2, so 52.

Therefore, the answer is 52_8.

So, all methods lead to the same conclusion.

Therefore, I think the answer is 52 in base 8.

But just to be thorough, let me check with another example.

Suppose I have 6_8 * 6_8. What would that be?

In base 10, that's 6*6=36. Converting 36 to base 8: 36 divided by 8 is 4 with remainder 4. So 44_8.

Alternatively, in base 8, 6*6.

But 6*6=36 in base 10. 36 divided by 8 is 4 with remainder 4. So 44_8.

Which is correct.

Similarly, if I do 7_8 * 7_8.

In base 10, that's 7*7=49. Converting 49 to base 8: 49 divided by 8 is 6 with remainder 1. So 61_8.

Which is correct.

So, if I do 6_8 * 7_8, which is 42 in base 10, which is 52_8.

So, that seems consistent.

Alternatively, if I think of 6_8 * 7_8 as 6*7 in base 8.

But since 6 and 7 are both less than 8, their product is 42 in base 10, which is 52 in base 8.

Therefore, I think that's correct.

Another way to check is to convert 52_8 back to base 10 and see if it's 42.

So, 5*8 + 2 = 40 + 2 = 42.

Yes, that's correct.

Therefore, I can be confident that the answer is 52_8.

So, after going through multiple methods and checking each step, I believe the product of 6_8 and 7_8 in base 8 is 52_8.

**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 the formula or concept related to rotational symmetry. For a regular polygon with n sides, the order of rotational symmetry is n, right? So, the angle of rotation would be 360 degrees divided by n. Is that correct? Let me verify.

For example, a square has 4 sides, so 360 divided by 4 is 90 degrees. And indeed, rotating a square 90 degrees brings it back to its original position. Similarly, an equilateral triangle has 3 sides, so 360 divided by 3 is 120 degrees. That seems right too. So, applying the same logic, a regular pentagon has 5 sides, so the angle should be 360 divided by 5. Let me calculate that.

360 divided by 5 is 72. So, 72 degrees? Wait, is that the answer? But let me make sure I'm not missing something here.

Let me visualize a regular pentagon. Each vertex is equally spaced around the center. So, if I rotate it by 72 degrees, each vertex should move to the position of the next vertex. Since there are 5 vertices, rotating by 72 degrees would cycle them through each position. Therefore, after 5 such rotations (each 72 degrees), it would have rotated a full 360 degrees and be back to the original position.

But the question is asking for the minimum number of degrees. So, 72 degrees is the smallest angle where it coincides with its original position? Let me check if there's a smaller angle.

Suppose I try 36 degrees. If I rotate the pentagon by 36 degrees, would it look the same? Let me think. Since 36 degrees is 360 divided by 10, but the pentagon has 5 sides. If I divide the circle into 5 equal parts, each part is 72 degrees. So, 36 degrees is half of that. If I rotate by 36 degrees, the vertices wouldn't align with the original positions because the spacing between vertices is 72 degrees. Therefore, rotating by 36 degrees would not map the pentagon onto itself.

Similarly, trying 72 degrees, which is 360/5, would map each vertex to the next one. So, that should work. What about 144 degrees? That's 2 times 72. If you rotate by 144 degrees, it would map the pentagon to itself, but since 72 is smaller, 72 is the minimum.

Wait, but let me think again. Maybe there's a different way to think about it. For rotational symmetry, the angle has to be such that when you rotate the shape by that angle, every point maps to another point of the shape. For a regular pentagon, the rotational symmetry is of order 5, meaning there are 5 positions where it looks the same. Therefore, the angle between each of these positions is 360/5 = 72 degrees.

Therefore, the minimum angle is 72 degrees.

But let me check with coordinates. Suppose I have a regular pentagon with vertices at angles 0°, 72°, 144°, 216°, 288°, and back to 360° (which is 0°). If I rotate the pentagon by 72°, each vertex moves to the next one. So, the vertex at 0° moves to 72°, which is the next vertex. Similarly, the one at 72° moves to 144°, etc. Therefore, after a rotation of 72°, the pentagon coincides with its original position.

If I tried a smaller angle, say 36°, then the vertex at 0° would move to 36°, which is not one of the original vertices. Therefore, the pentagon wouldn't coincide with itself.

Alternatively, maybe there's a different symmetry? For example, maybe some other angle? But since the pentagon is regular, all the rotational symmetries are multiples of 72°, right?

Another way to think about it: the number of times the pentagon maps onto itself when rotated around its center is equal to the number of sides, which is 5. Therefore, the angle between each of these symmetries is 360° divided by 5, which is 72°.

Therefore, the minimum angle is 72 degrees.

But just to be thorough, let me think if there's a possibility of a smaller angle. For example, if the pentagon had some other symmetry. But since it's regular, all the symmetries are determined by the number of sides. For example, a regular pentagon doesn't have any other rotational symmetries except multiples of 72°.

Wait, but maybe if you rotate by some fraction? For example, if there was a divisor of 360° that is smaller than 72°, but which divides 360° and also divides into the angles of the pentagon.

But the angles between the vertices are 72°, so the rotation angle must be a multiple of 72° to align the vertices. Therefore, the minimal angle is 72°.

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 aren't considered here.

Therefore, the answer should be 72 degrees.

But let me check with an example. Let me imagine a regular pentagon. If I rotate it by 72 degrees, each vertex moves to the next one. So, the whole shape is the same. If I rotate it by 144 degrees, it's like two steps, so it's still the same. But 72 is smaller.

Alternatively, if I think about the central angles. Each central angle (the angle at the center between two adjacent vertices) is 72 degrees. Therefore, rotating by that angle would bring one vertex to the next. Therefore, the minimal rotation is 72 degrees.

Therefore, I think the answer is 72 degrees.

But just to make sure, let me check if 360 divided by 5 is indeed 72. 5 times 72 is 360. Yes, that's correct.

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

Therefore, the minimum number of degrees is 72.

I think that's solid. I can't think of any other angle smaller than 72 that would work. So, I think 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** when it can be rotated about its center by a certain angle and still appear identical to its original position. The number of such distinct positions (excluding the identity rotation of 0°) is equal to the number of sides of the polygon.

For a **regular pentagon**, which has **5 sides**, the **order of rotational symmetry** is 5. This means that the pentagon can be rotated into 5 different positions that look the same, including the original position.

---

### Calculating the Rotation Angle

To find the **smallest angle** of rotation that maps the pentagon onto itself, we divide the full circle (360°) by the number of sides:

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

Where $ n $ is the number of sides. For a pentagon:

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

This means that rotating the pentagon by **72 degrees** will align it with its original position.

---

### Verifying the Result

- A regular pentagon has **five vertices** equally spaced around the center.
- The angle between adjacent vertices (from the center) is $ \frac{360^\circ}{5} = 72^\circ $.
- Rotating the pentagon by 72° moves each vertex to the position of the next one, and the entire shape remains unchanged.
- Any smaller angle (e.g., 36°, 18°, etc.) would not align the vertices with their original positions, so the pentagon would not coincide with its original form.

---

### Conclusion

The **minimum number of degrees** the regular pentagon must be rotated counterclockwise about its center to coincide with its original position is:

$$
\boxed{72}
$$

[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 the formula or concept related to rotational symmetry. For a regular polygon with n sides, the order of rotational symmetry is n, right? So, the angle of rotation would be 360 degrees divided by n. Is that correct? Let me verify.

For example, a square has 4 sides, so 360 divided by 4 is 90 degrees. And indeed, rotating a square 90 degrees brings it back to its original position. Similarly, an equilateral triangle has 3 sides, so 360 divided by 3 is 120 degrees. That seems right too. So, applying the same logic, a regular pentagon has 5 sides, so the angle should be 360 divided by 5. Let me calculate that.

360 divided by 5 is 72. So, 72 degrees? Wait, is that the answer? But let me make sure I'm not missing something here.

Let me visualize a regular pentagon. Each vertex is equally spaced around the center. So, if I rotate it by 72 degrees, each vertex should move to the position of the next vertex. Since there are 5 vertices, rotating by 72 degrees would cycle them through each position. Therefore, after 5 such rotations (each 72 degrees), it would have rotated a full 360 degrees and be back to the original position.

But the question is asking for the minimum number of degrees. So, 72 degrees is the smallest angle where it coincides with its original position? Let me check if there's a smaller angle.

Suppose I try 36 degrees. If I rotate the pentagon by 36 degrees, would it look the same? Let me think. Since 36 degrees is 360 divided by 10, but the pentagon has 5 sides. If I divide the circle into 5 equal parts, each part is 72 degrees. So, 36 degrees is half of that. If I rotate by 36 degrees, the vertices wouldn't align with the original positions because the spacing between vertices is 72 degrees. Therefore, rotating by 36 degrees would not map the pentagon onto itself.

Similarly, trying 72 degrees, which is 360/5, would map each vertex to the next one. So, that should work. What about 144 degrees? That's 2 times 72. If you rotate by 144 degrees, it would map the pentagon to itself, but since 72 is smaller, 72 is the minimum.

Wait, but let me think again. Maybe there's a different way to think about it. For rotational symmetry, the angle has to be such that when you rotate the shape by that angle, every point maps to another point of the shape. For a regular pentagon, the rotational symmetry is of order 5, meaning there are 5 positions where it looks the same. Therefore, the angle between each of these positions is 360/5 = 72 degrees.

Therefore, the minimum angle is 72 degrees.

But let me check with coordinates. Suppose I have a regular pentagon with vertices at angles 0°, 72°, 144°, 216°, 288°, and back to 360° (which is 0°). If I rotate the pentagon by 72°, each vertex moves to the next one. So, the vertex at 0° moves to 72°, which is the next vertex. Similarly, the one at 72° moves to 144°, etc. Therefore, after a rotation of 72°, the pentagon coincides with its original position.

If I tried a smaller angle, say 36°, then the vertex at 0° would move to 36°, which is not one of the original vertices. Therefore, the pentagon wouldn't coincide with itself.

Alternatively, maybe there's a different symmetry? For example, maybe some other angle? But since the pentagon is regular, all the rotational symmetries are multiples of 72°, right?

Another way to think about it: the number of times the pentagon maps onto itself when rotated around its center is equal to the number of sides, which is 5. Therefore, the angle between each of these symmetries is 360° divided by 5, which is 72°.

Therefore, the minimum angle is 72 degrees.

But just to be thorough, let me think if there's a possibility of a smaller angle. For example, if the pentagon had some other symmetry. But since it's regular, all the symmetries are determined by the number of sides. For example, a regular pentagon doesn't have any other rotational symmetries except multiples of 72°.

Wait, but maybe if you rotate by some fraction? For example, if there was a divisor of 360° that is smaller than 72°, but which divides 360° and also divides into the angles of the pentagon.

But the angles between the vertices are 72°, so the rotation angle must be a multiple of 72° to align the vertices. Therefore, the minimal angle is 72°.

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 aren't considered here.

Therefore, the answer should be 72 degrees.

But let me check with an example. Let me imagine a regular pentagon. If I rotate it by 72 degrees, each vertex moves to the next one. So, the whole shape is the same. If I rotate it by 144 degrees, it's like two steps, so it's still the same. But 72 is smaller.

Alternatively, if I think about the central angles. Each central angle (the angle at the center between two adjacent vertices) is 72 degrees. Therefore, rotating by that angle would bring one vertex to the next. Therefore, the minimal rotation is 72 degrees.

Therefore, I think the answer is 72 degrees.

But just to make sure, let me check if 360 divided by 5 is indeed 72. 5 times 72 is 360. Yes, that's correct.

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

Therefore, the minimum number of degrees is 72.

I think that's solid. I can't think of any other angle smaller than 72 that would work. So, I think 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** when it can be rotated about its center by a certain angle and still appear identical to its original position. The number of such distinct positions (excluding the identity rotation of 0°) is equal to the number of sides of the polygon.

For a **regular pentagon**, which has **5 sides**, the **order of rotational symmetry** is 5. This means that the pentagon can be rotated into 5 different positions that look the same, including the original position.

---

### Calculating the Rotation Angle

To find the **smallest angle** of rotation that maps the pentagon onto itself, we divide the full circle (360°) by the number of sides:

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

Where $ n $ is the number of sides. For a pentagon:

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

This means that rotating the pentagon by **72 degrees** will align it with its original position.

---

### Verifying the Result

- A regular pentagon has **five vertices** equally spaced around the center.
- The angle between adjacent vertices (from the center) is $ \frac{360^\circ}{5} = 72^\circ $.
- Rotating the pentagon by 72° moves each vertex to the position of the next one, and the entire shape remains unchanged.
- Any smaller angle (e.g., 36°, 18°, etc.) would not align the vertices with their original positions, so the pentagon would not coincide with its original form.

---

### Conclusion

The **minimum number of degrees** the regular pentagon must be rotated counterclockwise about its center to coincide with its original position is:

$$
\boxed{72}
$$

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]