MaxWong

Where is AC? Where is AB? Are they legs or hypotenuse? 

Here is a way without calculus :)

Let x = sin θ, y = cos θ.

\(\displaystyle \max_{x,y\in\{x,y:x^2 +y^2 = 1\}} (x+y) = \max_{\theta \in [0,2\pi]} (\sin \theta + \cos \theta) = \max_{\theta \in [0,2\pi]}\left(\sqrt2 \sin\left(\theta+\dfrac{\pi}{4}\right)\right) = \sqrt2\).

So \(\displaystyle \max_{x,y\in\{x,y:x^2 +y^2 = 1\}} (x+y)^2 = \left(\sqrt2\right)^2 = 2\).

Let r be the common ratio.

The series starts like this: 1, r, r², r³, ...

\(r^n = r^{n+1} + r^{n+2}\\ r^2 + r - 1 = 0\\ r = \dfrac{-1\pm\sqrt{5}}{2}\)

Now we found the possible common ratios. 

Denote \(r_1 = \dfrac{-1-\sqrt5}{2},\;r_2 = \dfrac{\sqrt 5 -1}{2}\)

Denote the sum of the series with common ratio r₁ by S₁.

Denote the sum of the series with common ratio r₂ by S₂.

\(S_1 = \dfrac{1}{1-\dfrac{-1-\sqrt 5}{2}} = \dfrac{3-\sqrt 5}{2}\)

\(S_2 = \dfrac{1}{1-\dfrac{\sqrt5-1}{2}} = \dfrac{3+\sqrt5}{2}\)

\(S_1 - r_1 = \dfrac{3-\sqrt5}{2} - \dfrac{-1-\sqrt5}{2} = 2\)

\(S_2 - r_2 = \dfrac{3+\sqrt5}{2} - \dfrac{\sqrt5 - 1}{2} = 2\)

For both cases, we get S-R = 2.

1.

Let a be the first term, r be the common ratio(which is 7/4).

\(b_1+b_2+\cdots+b_{10} = \dfrac{a(r^{10}-1)}{r-1}=180\\ a\cdot\dfrac{\left(\dfrac{7}{4}\right)^{10} -1}{\dfrac{7}{4}-1} = 180\\ a = \dfrac{47185920}{93808891}\)

So b₁ = 47185920/93808891.

b₂ = b₁ * 7 / 4 = 82575360/93808891.

\(b_2 + b_4 + b_6 + b_8+b_{10} = \dfrac{82575360}{93808891} \cdot \left(\dfrac{\left(\dfrac{7}{4}\right)^{10}-1}{\left(\dfrac{7}{4}\right)^2-1}\right)=\dfrac{1260}{11}\)

Therefore the required answer is 1260/11.

\(\displaystyle \binom {10^9}{k} < \binom{10^9+1}{k-1}\\ \dfrac{(10^9)!}{k!(10^9-k)!}<\dfrac{(10^9+1)!}{(k-1)!((10^9+1)-(k-1))!}\\ \dfrac{1}{k(10^9-k)!} < \dfrac{10^9+1}{(10^9-k+2)!}\\ (10^9-k+1)(10^9-k+2) < (10^9+1)k\\ 1500000002-\sqrt{1250000003000000002} < k < 1500000002 + \sqrt{1250000003000000002}\)

Smallest positive integer k that satisfies the inequality is 381966012. (Used some computational intelligence :P)

10a + b = 38. :)

2.

\(AA_5 = 5A + A = 6A \quad \forall \; A \in [0,4] \land A\in \mathbb Z\\ BB_7 = 7B + B = 8B \quad \forall \; B \in [0,6] \land B\in \mathbb Z\)

From this, we know that the required base-10 integer must be divisible by both 6 and 8.

Least common multiple of 6 and 8 = 24.

So the answer is 24.

Check: \(24_{10} = 44_5 = 33_7\)

1.

Obviously, \(n\geq 7\).

\(\quad 235236_n\\ =2n^5 + 3n^4 + 5n^3 + 2n^2 + 3n + 6\) 

This holds true for any \(n\geq 7\).

Then, clearly, \(2n^5+3n^4+5n^3+2n^2+3n\equiv 1\pmod7\).

Define \(P(n) = 2n^5+3n^4+5n^3+2n^2+3n-1\).

Consider all possible values of P(n) mod 7 when \(0 \leq n \leq 6, n\in \mathbb Z\).

\(P(0) \equiv 6\pmod 7\\ P(1) \equiv 0 \pmod 7\\ P(2) \equiv 4 \pmod 7\\ P(3) \equiv 1 \pmod 7\\ P(4) \equiv 1 \pmod 7\\ P(5) \equiv 1\pmod 7\\ P(6) \equiv 1 \pmod 7\\ \)

Generally, \(P(7k + 1) \equiv 0 \pmod 7 \quad\forall \;k\in \mathbb Z^+\).

All possible values of n are: 8,15,22,29,36,43,50,57,64,71,78,85,92,99.

So there are 14 possible values of n such that \(235236_n\) is divisible by 7.

\(\quad\dfrac{1}{4}+6-3\cdot 14\\ = \dfrac{-143}{4}\)

If you only consider the integer values, then the range is [3,8].

Consider a number infinitesimally close to 8. (7.9999999999999999999999...)

p(7.9999999999999999999...) = x + 1 = 8.999999999999999999999... = 9

So the range is actually [3,9].