MaxWong

\(z^5 = 1\\ z^5 - 1 = 0\\ (z - 1)(z^4 + z^3 + z^2 + z + 1) = 0\\ z^4 + z^3 + z^2 + z + 1 = 0 \text{ ($\because z \neq 1$)}\\ \dfrac{z^4 + z^3 + z^2 + z + 1}{z^2} = 0\\ z + \dfrac1z + z^2 + \dfrac1{z^2} = \boxed{-1}\)

\(\log_2 \left(\dfrac{2}1\right) + \log_2\left(\dfrac32\right)+\cdots + \log_2 \left(\dfrac{64}{63}\right) \\ = \log_2 \left(\dfrac21 \times \dfrac32 \times \cdots \times \dfrac{64}{63}\right)\\ = \log_2 64\\ = 6\)

The answer is obvious at this point.

Seriously? 25, 25, 70 cannot be the sides of a triangle!

Observation: \((1 + \sqrt 2)^8 + (\sqrt 2-1)^8\) is an integer, and \(0 < (\sqrt 2- 1)^8 < \dfrac12\). That integer is the required integer.

Using binomial theorem,

\(\quad(1 + \sqrt 2)^8 + (\sqrt 2-1)^8\\ =\displaystyle\sum_{k = 0}^8 \binom{8}k ((\sqrt 2)^k + (-\sqrt 2)^k)\\ =\displaystyle\binom{8}0 (2) + \binom82 (2(\sqrt 2)^2) + \binom84 (2(\sqrt 2)^4) + \binom86 (2(\sqrt 2)^6) + \binom88 (2(\sqrt 2)^8)\\ = 1154\)

Note that \(\sqrt 2 - 1 \approx 0.414\), so \(0 < (\sqrt 2 - 1)^8 < \sqrt 2 - 1 < \dfrac12\). That means \((1 + \sqrt 2)^8\) plus a real number way smaller than 1/2 gives 1154.

Therefore, the nearest integer to \((1 + \sqrt 2)^8\) is 1154.

Remarks: The error is \(\left|(1 + \sqrt2)^8 - 1154\right| = (\sqrt 2 - 1)^8 \approx 8.665517772 \times 10^{-4}\), which is really really small.

A similar problem is solved here:
https://web2.0calc.com/questions/inverse-mod#r4

Similar solved problem here, with only the dilation factor changed:

https://web2.0calc.com/questions/help-please_79855

Let x be the number. \(\dfrac{5x}{x + 7} = 8\)

Multiplying by (x + 7) on both sides gives 

\(5x = 8(x + 7)\\ 5x = 8x + 56\)

Now this is just a linear equation. Can you try to solve it?

Let r% be such rate of interest.

\(P\displaystyle\lim_{n\to\infty}\left(1 + \dfrac{r\%}n\right)^{3n} = 2P\\ e^{3r\%} = 2\\ 3r\% = \ln(2)\\ r\% = \dfrac{\ln(2)}3 \approx 0.231\)

The rate must be at least 23.1%.

\(\quad\displaystyle \sqrt{2}+\frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{2}} \\= \sqrt 2 + \sqrt 3 + \dfrac 2{\sqrt 2} \\= 2 \sqrt 2 + \sqrt3\)

Now, the values of a, b, and c are obvious.

Exact same question answered here: https://web2.0calc.com/questions/floor-function_20#r1

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