# Tag Info

0

Let $A(n)$ be the required number of strings of length $n$. For $n\geq 2,$ any such string begins with either $B$ or $AA$. The number of strings beginning with $B$ equals $A(n-1)$ since that's the number of strings to fill the rest of the string. The number of strings beginning with $AA$ equals $A(n-2)$ since that's the number of strings to fill the rest ...

0

Without loss of generality, we assume that $a\leq b$. Since $a=2\left(\frac{ab}{2b}\right)\leq2\left(\frac{ab}{a+b}\right)< 2\left(\frac{ab+1}{a+b}\right)<3$, we have $a=1$ or $a=2$. If $a=1$, then $b$ can be any natural number and $\frac{a^3b^3+1}{a^3+b^3}=1$. If $a=2$, then $\frac{2b+1}{b+2}=\frac{ab+1}{a+b}<\frac{3}{2}$ gives $4b+2<3b+6$, ...

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Try manipulating the first inequality to define $a$ in terms of $b$ (or vice versa) $$\frac{ab+1}{a+b} < \frac{3}{2} \Rightarrow 2ab+2 < 3a+3b$$ $$\Rightarrow 2ab - 3a < 3b-2$$ $$\Rightarrow a < \frac{3b-2}{2b-3}$$ Notice that if $a=1$, then the second fraction involving $a$ and $b$ would become $$\frac{(1)^3b^3+1}{(1)^3+b^3} = ... 0 There are (n-1)n ways for your chosen squares to be adjacent horizontally, and the same number of ways to be adjacent vertically. There are n^2(n^2-1)/2 ways to choose the two squares. Taking n=89 gives a probability of 1/2002.5 and n=90 gives 1/2047.5. So, you did get the correct answer. The argument you made works only for squares away from ... 11 There is no number x for which f(x)=x[x[x[x]]] equals 88. f(x) is an increasing function over \mathbb{R}^+ and:$$ f(3)=81,\qquad \lim_{x\to 3^+} f(x) = 120. $$f(x) is a decreasing function over (-\infty,-1] and:$$ f(-3) = 81, \qquad \lim_{x\to -3^-}f(x) = 90.$$2 The reason your argument doesn't work is that the corner and edge squares of the grid are NOT adjacent to 4 squares each. Edge squares are adjacent to 3, and corners are adjacent to 2. Instead, a good strategy would be: (1) count the number of PAIRS of squares that are adjacent (horizontally or vertically); (2) count the number of total pairs of ... 6 hint Note that if x=3, then x^4=81<88 and if x=4 then x^4 = 256 > 88. So you want to find numbers closer to 3. What happens, for example, if you look at x = 3.1 or 3.05? PLaying with these should give you an idea... 0 Note that$$\alpha = \angle DMN + \angle NMQ,$$where \angle DMN = 45^\circ and \tan \angle NMQ=0.5. Thus,$$\tan\alpha=\frac{1+0.5}{1-(1)(0.5)}=3$$0 a^2 + 3a + 4 \equiv a^2 - 4a + 4 \equiv (a-2)^2 \pmod 7 If a\equiv b \pmod 7, then a^2 + 3a + 4 \equiv (b-2)^2 \pmod 7 0 Let A=(0,0), C=(1,1). Then M=(1/2,1), Q=(1/4,1/4), R=(1/6,0), \tan \alpha=\frac {1}{1/3}=3 1 Hint: Start with A is simple, for then we need to have another A, and we append a good string of length n-2. Start with B is more complicated. (i) If we have an A next, then we need another, then a good string of length n-3. (ii) If we have a B next, we need another, and \dots. Added: We expand on the hint. Let F(n) be the number of good strings ... 0 This can be done with Maple by sol := RealDomain:-solve({-x*y*z+z^3 = 20, -x*y*z+y^3 = 6, x^3-x*y*z = 2}, explicit);$$ \left\{ x=-\frac 1 7\,{7}^{2/3},y=\frac 3 7\,{7}^{2/3},z=\frac 5 7\,{7}^{2/3} \right\},\,\left\{ x=-\sqrt [3]{2},y=\sqrt [3]{2},z=2\,\sqrt [3]{2} \right\}$$L := seq(rhs(sol[j][1])^3+rhs(sol[j][2])^3+rhs(sol[j][3])^3, j = 1 .. 2); ... 2 Hint. You have$$\left\{ \begin{aligned} x^3=2+t\\ y^3=6+t\\ z^3=20+t \end{aligned} \right.$$Hence$$x^3 y^3 z^3 = t^3 = (2+t)(6+t)(20+t)$$2 Your idea of putting t=xyz is quite correct. You then have x^3=t+6,y^3=t+6,z^3=t+20, whence t^3=(t+2)(t+6)(t+20). Expanding, you see that 7t^2+43t+60=0. The rest is easy and I leave it to you. (by the way, the answer is m+n=158). 2 Start with your idea a=6+\sqrt{x} and b=6-\sqrt{x}. You have$$ab=36-x \text{, } \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{3} \text{ and } a+b=12$$Raise the middle equation to power 3 you get$$a+b +3\sqrt[3]{ab}(\sqrt[3]{a} + \sqrt[3]{b})=3$$and using the initial assumption$$12 +3\sqrt[3]{ab}\sqrt[3]{3}=3 \text{ or } \sqrt[3]{ab}\sqrt[3]{3}=-3$$and ... 0 go to power of three both side$$(\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3})^3 \\\xrightarrow[(a+b)^3=a^3+b^3+3ab(a+b)]{} 6+\sqrt{x} +6-\sqrt{x} +3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})(\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} \sqrt[3] {3})=3\\ $$we can subsitute \sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3} so we have$$ ...

3

Let $a = \sqrt[3]{6+\sqrt x}$ and $b = \sqrt[3]{6-\sqrt x}$, then we have: $$\left\{\begin{matrix} a^3 + b^3 = 12 \\ a + b = \sqrt[3]{3} \end{matrix}\right.$$ Therefore, $$ab = \sqrt[3]{-9}$$ or, $$\sqrt[3]{36 - x} = \sqrt[3]{-9}$$ then, $$x = 45$$

1

Put $z:=e^{i\pi/(2n+1)}$. The points $$z_k:=z^k\qquad(1\leq k\leq 4n+2)$$ are the vertices of a regular $(4n+2)$-gon $P$ inscribed in the unit circle. Denote by $\phi_k:={\rm arg}(z_k)$ the polar angles of these vertices. We then are told to compute $$S:=\sum_{k=1}^n\cos^4\phi_k={1\over4}\left(\sum_{k=1}^{4n+2}\cos^4\phi_k \ -2\right)\ ,$$ whereby we have ...

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$$6-\sqrt{x} = t^3$$ then $$6+\sqrt{x} = t^3+2\sqrt{x}$$ The expression becomes: $$t + \sqrt[3]{t^3+2\sqrt{x}} = \sqrt[3]{3}$$ Substitute back $\sqrt{x}$ in terms of t, rewrite so the cube root is alone. Raise to 3, solve 3rd degree polynomial equation. Test all roots. Fun fact: The valid $t$ solution happens to become closely related to the golden ratio: ...

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\begin{align*} a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\ 6+\sqrt{x} + 6-\sqrt{x} + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\ 12 + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\ 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= -9 \end{align*} Divide by $3\sqrt[3]{ab}$: $$\sqrt[3]{b} + \sqrt[3]{a} = \frac{-9}{3\sqrt[3]{ab}}$$ Using the original ...

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Step 1 (conjecture): there is some $e>0$ such that $$6+\sqrt{x}=(e+\sqrt[3]{3}/2)^3,\quad 6-\sqrt{x}=(-e+\sqrt[3]{3}/2)^3.\tag{*}$$ You can see that if such an $e$ exists then the equality $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$ is satisfied. Solving for $e$ is simple: $$... 1 K will be like that so - DK\perp CD we know that DK\perp CD than DK||AB (because AC\perp AB), Also we know that CD=AD, because of that we can understand DK is median of triangle \Delta ABC so CK=KB=6. We know that CE=3, than CE=EK=3. \Delta CDK is a right triangle, than we can understand that - DE=CE=EK=3 (Because DE is ... 2 Let AD=y so that \cos C=\frac{2y}{12} Then using the cosine rule,$$x^2=y^2+3^2-6y\cos C$$So x=3 1 It's 3! To see it, draw the perpendicular From E to AC (let G be the foot of that perpendicular). Then triangle EGC is similar to ABC with scale factor \frac 14, whence GC is \frac 12 of DC. Hence the two right triangles EGC and EGD are congruent and the result follows. Note: this was corrected to reflect a typo-generated arithmetic error pointed out ... 2 Hint:$$\sum_{k=1}^{n}\cos^{4}\left(\frac{\pi k}{2n+1}\right)=\frac{1}{16}\sum_{k=1}^{n}\left(e^{-\pi ik/\left(2n+1\right)}+e^{\pi ik/\left(2n+1\right)}\right)^{4}= =\frac{1}{16}\sum_{k=1}^{n}\left(4e^{-2i\pi k/\left(2n+1\right)}+4e^{2i\pi k/\left(2n+1\right)}+e^{-4i\pi k/\left(2n+1\right)}+e^{4i\pi k/\left(2n+1\right)}+6\right). $$0 HINT: Consider the sum$$ \sum_{k=0}^n \cos k\alpha + i\sin k\alpha=\sum_{k=0}^n e^{ik\alpha}. $$The real part will be the sum of \cos's. 2 If the remainder when a divided by 7 is b, then a = 7n+b for some integer n. Hence, a^2+3a+4 = (7n+b)^2+3(7n+b)+4 = 49n^2 + 14nb + b^2 + 21n + 3b + 4 = 7(7n^2+2nb+3n) + (b^2+3b+4). So, the remainder when a^2+3a+4 is divided by 7 will be the same as the remainder when b^2+3b+4 is divided by 7. For the specific case when b = ... 1 The remainder of a^2+3a+4 divided by 7 is sum of the remainder of each terms, modulo 7. So a^2\equiv 1 \pmod{7} since a=7k+6 then a^2=7l+1; \quad 3a\equiv 4 \pmod{7} since 3a=21k+18=21k+14+4 and clearly 4\equiv 4 \pmod{7}. Finally 1+4+4 \equiv 2 \pmod{7} then the remainder is 2. 5 a = 6 \quad(\mathrm{mod} 7) a^2 = 36 = 1 \quad(\mathrm{mod} 7) 3a = 18 = 4\quad (\mathrm{mod} 7) a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7) 1 Set g(x):=f(x)-f(1), the we know that f(x)=f(2x^{2}) and so the function g (and f) can not be a polynomial), because g has infinitely roots. Now we show that the function g must be zero. Assume that there is x_{0}\in \mathbb{N} such that g(x_{0})\neq 0, since \mathbb{N} is a countable set, let x_{0} be an element s.t., g(x_{0}) has ... 1 Let \mathbb{N} = {0,1,2,\dots}. Determine all functions f : \mathbb{N} \rightarrow \mathbb{N} such that xf(y) + yf(x) = (x + y)f(x^2 + y^2) \tag{OP} for all x,y \in \mathbb{N}. Let k=x=y, then we get$$ k f(k) + k f(k) = (k+k) f( k^2 + k^2) f(x) = f(2x^2) $$which implies$$ f(k) = f( 2 k^2 ) \tag{1} $$Let ... 1 "Randomly chosen" isn't clearly defined, but let's take it to mean that we randomly choose the values of f(1),f(2),f(3), f(4) and similarly for g. We'll work case by case, indexed by the size of the range of f. I will not complete the calculation, it's a bit messy and I'll leave some of the arithmetic off. Case I: Range of f has exactly 1 element. ... 2 Hint: We have:$$A = \left(\frac{x}{x+y} \right)^{2007} + \left(\frac{y}{x+y} \right)^{2007} = \left( \frac{x}{y} \right)^{1003} \frac{x}{x+y} + \left( \frac{y}{x} \right)^{1003}\frac{y}{x+y}  = - \left[ \left( \frac{x}{y} \right)^{1002}+ \left( \frac{y}{x} \right)^{1002} \right]$$From the condition x^2 + xy + y^2 =0, we have$$\left( ...

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Usage of the multinomial coefficient $(k_1, k_2, \cdots, k_n)$!: $$\big( 1 + x^5 + x^7\big)^{20} = \sum_{k_1=1}^{20} \sum_{k_2=1}^{20-k_1} (k_1, k_2, 20 - k_1 - k_2)! x^{5k_1} x^{7k_2},$$ where $$(k_1, k_2, \cdots, k_n)! = \frac{ (k_1 + k_2 + \cdots + k_n )! } { k_1! k_2! \cdots k_n!}.$$ So we get $k_1=2$ and $k_2=1$, thus $$(2,1,17)! = ... 5 17 can only be obtained by using two 5s and one 7 . These two 5s can be obtained in \binom{20}2 ways which is 190 and the 7 can be got in from one of the remaining 18 brackets. So 190 x 18 = 3420 is the answer. 3 So if you think about$$ (1 + x^5 + x^7)^{20} $$That intuitively is just$$ ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) ... $$Which can be expanded out term by term. By the Binomial Theorem as$$ (1 + x^5)^{20} (x^7)^0 + \begin{pmatrix} 20 \\ 1\end{pmatrix}(1 + x^5)^{19}x^7 + \begin{pmatrix} 20 \\ 2\end{pmatrix}(1 + ...

3

$(1+x^5+x^7)^{20}=\{(1+x^5)+x^7\}^{20}$ $=(1+x^5)^{20}+\binom{20}1(1+x^5)^{20-1}(x^7)^1+\binom{20}2(1+x^5)^{20-2}(x^7)^2+\cdots+(x^7)^{20}$ So the required sum will be the coefficient of $x^{17}$ in $(1+x^5)^{20}$ $+\binom{20}1\cdot$ the coefficient of $x^{17-7}$ in $(1+x^5)^{20-1}$ $+\binom{20}2\cdot$ the coefficient of $x^{17-7\cdot2}$ in ...

1

Dividing $x^{2} + 2xy + y^{2}$ by $y^{2}$ gives an equation whose roots are the non-real cube roots of unity. That is, say $x/y$ = $\omega$ then $y/x$ = $\omega^{2}$. With $x + y = \sqrt{xy}$, the given equation can now be expressed conveniently in terms of these complex roots, and I think the individual terms will come out to 1 + 1 = 2 or -1 - 1 = -2 (I ...

2

Solving $$x^2+xy+y^2=0\Rightarrow \frac xy=e^{\pm\frac{2i\pi}{3}}\Rightarrow (\frac xy)^3=1$$ Since $x+y$ can be replaced with $-\frac{y^2}{x}$, the expression boils down to $$-1^{1338}-1^{669}=-2$$

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Set $x=ry$ $\implies y^2(r^2+r+1)=0\implies r^2+r+1=0\implies r^3-1=(r-1)(r^2+r+1)=0$ $\implies r^3=1\ \ \ \ (1)$ $\dfrac x{x+y}=\dfrac{ry}{y+ry}=\dfrac r{1+r}$ $\dfrac y{x+y}=\dfrac y{y+ry}=\dfrac 1{1+r}$ As $2007\equiv3\pmod6=6a+3$ where $a=334$( in fact $a$ can be any integer) The required sum ...

3

Add the first two $$\frac 1{1+\sqrt3+\sqrt2} + \frac 1{1+\sqrt3-\sqrt2}=2\frac{1+\sqrt3}{(1+\sqrt3)^2-2}=2\frac{1+\sqrt3}{2+2\sqrt3}=1.$$ Similarly, add the last two $$2\frac{1-\sqrt3}{2-2\sqrt3}=1.$$

0

Rearranging, we have $$\frac{1}{1+(a+b)}+\frac{1}{1-(a+b)}+\frac{1}{1-(a-b)}+\frac{1}{1+(a-b)},$$ giving $$\frac{2}{1-(a+b)^2}+\frac{2}{1-(a-b)^2}.$$ Then, $$\frac{4-2(a-b)^2-2(a+b)^2}{1-(a-b)^2-(a+b)^2+(a+b)^2(a-b)^2}.$$ Expanding, $$\frac{4(1-a^2-b^2)}{1-2a^2-2b^2+(a^2-b^2)^2}.$$ In your case $a=\sqrt{2}$ and $b=\sqrt{3}$, so that your sum is ...

15

We have that $\pm\sqrt{2}\pm\sqrt{3}$ are the roots of the polynomial $(x^2-5)^2-24$, hence: $$x^4-10\,x^2+1 = \prod_{\xi_i\in Z}(x-\xi)$$ and $1\pm\sqrt{2}\pm\sqrt{3}$ are the roots of the polynomial: $$(x-1)^4-10(x-1)^2+1 = x^4-4x^3-4x^2+16x-8.$$ By Viète's theorem, the sum of the roots of a polynomial $p(x)$ raised to the minus one power is given by ...

3

We have, $$\underbrace{\frac {1}{1+\sqrt2+\sqrt3} + \frac {1}{1-\sqrt2+\sqrt3}} + \underbrace{\frac {1}{1+\sqrt2-\sqrt3} + \frac {1}{1-\sqrt2-\sqrt3}}$$ $$=\left(\frac {1}{1+\sqrt2+\sqrt3} + \frac {1}{1-\sqrt2+\sqrt3}\right) +\left( \frac {1}{1+\sqrt2-\sqrt3} + \frac {1}{1-\sqrt2-\sqrt3}\right)$$ $$=\left(\frac ... 5 Another way :$$\begin{align}\\&\frac{1}{1+\sqrt 2+\sqrt 3}+\frac{1}{1-\sqrt 2+\sqrt 3}+\frac{1}{1+\sqrt 2-\sqrt 3}+\frac{1}{1-\sqrt 2-\sqrt 3}\\&=\left(\frac{1}{1+\sqrt 2+\sqrt 3}+\frac{1}{1+\sqrt 2-\sqrt 3}\right)+\left(\frac{1}{1-\sqrt 2+\sqrt 3}+\frac{1}{1-\sqrt 2-\sqrt 3}\right)\\&=\frac{1+\sqrt 2-\sqrt 3+1+\sqrt 2+\sqrt 3}{(1+\sqrt 2+\sqrt ...

1

from here, separating it so that they have same thing $$\frac 1{1+(\sqrt2+\sqrt3)} + \frac 1{1-(\sqrt2-\sqrt3)} + \frac 1{1+(\sqrt2-\sqrt3)} + \frac 1{1-(\sqrt2+\sqrt3)}$$ let $x = \sqrt 2 + \sqrt 3$, $y = \sqrt 2 - \sqrt 3$ $$\frac 1{1+x} + \frac 1{1-y} + \frac 1{1+y} + \frac 1{1-x}$$ combine fraction with x into one fraction, same thing as y $$=\frac ... 0 Let there be n women. To meet the criteria, men must either be in two blocks of 4 & 2 and positioned in two of the (n+1) gaps between women (including the ends), or in a block of 6 positioned in the (n+1) gaps, i.e. in [(n+1)\cdot n + (n+1)] patterns, having 4!2! and 6! permutations respectively, thus$$\frac{[(n+1)\cdot n!]\cdot(n\cdot 4!\cdot2! ...

1

Suppose there are $x$ women. I assume them all to be the same, like you seem to do in the question. Case 1. If there is a block of exactly 4 men, we have two possibilities. Case 1.1. If the block of 4 men is at the end or the beginning of the row, we only need one women enclosing it. So we have the block $B$ consisting of 4 men and 1 woman, which can ...

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Consider the men as $M, M,\ldots,M$ to get 4 aligned and no men alone you need to form 2 groups, one with $4$ men and one with $2$ men. let $G(4),G(2)$ be the group with $4$ and $2$ men respectively. A favorable occurrence is of the form $$W(n_1) G(4) W(n_2) G(2) W(n_3)$$ $$W(n_1) G(2) W(n_2) G(4) W(n_3)$$ or $$W(N_1) G(4) G(2) W(N_2)$$ where ...

1

Construction: Extend AM to cut the circle BXHC at P. Join BP and join CP. $\beta$ is the exterior angle of the cyclic quadrilateral $PCHX$. Therefore, $\alpha = \beta$. $\gamma = \beta$ because they are the angles on the same segment of the cyclic quadrilateral $AHXC’$. $\alpha = \gamma$ implies $AB // CP$. Similarly, $\theta = \delta$ implies \$BP // ...

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