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Here is almost a similar question. But in that case, $3-\sqrt{5}$ is a proper fraction (i.e. $0<3-\sqrt{5}<1$). So, if we expand both $(3-\sqrt{5})^{34}$ and $(3+\sqrt{5})^{34}$ using binomial theorm and do some calculations (including cancellations), we will reach to a conclusion.

My problem which is different. It should be solved without using calculators. Average solving time for each problem in the exam (that includes this problem) is $3$ minutes:

What is the remainder when dividing $\left \lfloor (6+\sqrt{7})^8 \right \rfloor$ by $9$?

$\text{(A) 0}$

$\text{(B) 1}$

$\text{(C) 2}$

$\text{(D) 4}$

$\text{(E) 8}$

Any help/hint would be appreciated.

Hussain-Alqatari
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  • Use binomial theorem – DSD Aug 16 '21 at 15:47
  • Less advanced than the binomial theorem, even less advanced than expansions using Pascal's triangle, is to simply square a binomial -- $(a+b)^2 = a^2 + 2ab + b^2$ -- three times in succession. However, the result will be of the form $P + Q\sqrt{7}$ where $P$ and $Q$ are integers, so the question seems incorrectly formulated to me. – Dave L. Renfro Aug 16 '21 at 16:18
  • @DaveL.Renfro , let P = 1, Q = 11. (1 + 11 sqrt(7)) ≈ 30.11, so (1 + 11 sqrt(7)) / 9 = 3, remainder ≈ 3.11. (I suppose using a spreadsheet when told "no calculator" is cheating :-( ) – gnasher729 Aug 16 '21 at 17:25
  • (an hour later) Returning to this ... Regarding "the question seems incorrectly formulated to me", I notice now that the greatest integer (least integer? I never remember which half-bracket version is for which) bracket notion is being used. I thought this was a simple parentheses usage, missing the tiny corner portions . . . (Reminder to self: If I ever use this notation, make sure I alert the reader to its use, first time it shows up, in some way.) @gnasher --- FYI, your comment appeared while I writing my comment. – Dave L. Renfro Aug 16 '21 at 17:27
  • Where is this problem from? Knowing this might provide insight as to how it is to be solved. – Dave L. Renfro Aug 16 '21 at 18:02
  • Cheating, with a calculator, gives the remainder 1. – John Bentin Aug 16 '21 at 19:30

2 Answers2

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It's not actually an answer, but it's to long for a comment. I am sorry. I tried to use the binomial theorem, but it seems to be very hard. Here is, what i tried.

We have \begin{align*} (6+\sqrt 7)^8&=\sum_{2\mid k}\binom{8}{k}6^{8-k}7^{k/2} +\sqrt 7\cdot \sum_{2\nmid k}\binom{8}{k}6^{8-k}7^{(k-1)/2} \\ &= \sum_{k=0}^4\binom{8}{2k}6^{8-2k}\cdot 7^{k}+\sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\\ &=7^4+\underbrace{6^2\cdot\sum_{k=0}^3\binom{8}{2k}6^{6-2k}\cdot 7^{k}}_{\text{divisible by $9$}}+\sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k \end{align*} One can calulate with much patience, that $$ \lfloor(6+\sqrt 7)^8\rfloor\equiv (-2)^4+\left\lfloor \sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\right\rfloor\equiv -2+\left\lfloor \sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\right\rfloor\equiv -2 + \lfloor 5896848 \sqrt 7\rfloor.$$ Maybe, you can get until here without a calculator. But now I stuck. Sorry sorry

Jochen
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  • Alternatively (and essentially getting no further than you did), from $(6 + \sqrt{7})^2 = 43 + 12\sqrt{7}$ we have $(6 + \sqrt{7})^8 = (43 + 12\sqrt{7})^4,$ which when expanded (use 1-4-6-4-1 coefficient pattern) gives the following 5-term expansion: $43^4 + 4(43^3\cdot 12\sqrt{7}) + 6(43^2 \cdot 12^2 \cdot 7) + 4(43\cdot12^3 \cdot 7\sqrt{7}) + 12^4 \cdot 49.$ The 3rd and 5th terms are each divisible by $9$ and thus can be ignored. Using basic modular arithmetic, it follows that when divided by $9,$ $43^4$ has the same remainder as $7^4 = 49^2,$ (continued) – Dave L. Renfro Aug 16 '21 at 18:30
  • hence the same remainder as $4^2 = 16,$ which is $7.$ Thus, it comes comes down to dealing with the sum of the 2nd and 4th terms, or $4\cdot 43\cdot 12(43^2 + 12^2\cdot 7)\sqrt{7}.$ I notice that $64/9 = 7.111\ldots,$ so $8/3 \approx \sqrt 7,$ but I don't see this helping much. – Dave L. Renfro Aug 16 '21 at 18:30
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$\left[\left(\mathrm{6}+\sqrt{\mathrm{7}}\right)^{\mathrm{8}} \right]{mod}\mathrm{9} \\ $ $\equiv\left[\left(-\mathrm{6}−\sqrt{\mathrm{7}}\right)^{\mathrm{8}} \right]{mod}\mathrm{9} \\ $ $\equiv\left[\left(\mathrm{3}−\sqrt{\mathrm{7}}\right)^{\mathrm{8}} \right]{mod}\mathrm{9}\equiv\mathrm{0} \\ $ ${BECAUSE}: \\ $ $\left(\mathrm{3}−\sqrt{\mathrm{7}}\right)\in\left[\mathrm{0},\mathrm{1}\right) \\ $ $\Rightarrow\left(\mathrm{3}−\sqrt{\mathrm{7}}\right)^{\mathrm{8}} \in\left[\mathrm{0},\mathrm{1}\right) \\ $

yoga mulyadi
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