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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 $4b+2<3b+6$, ...


0

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 ...


0

$$6-\sqrt{x} = t^3$$ 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: ...


6

\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 ...


5

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( ...


5

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 $5$s and one $7$ . These two $5$s 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 $$-1^{1338}-1^{669}=-2$$


4

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 ...


1

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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