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16

Hint If (this was the original post) $$2f(x) - f(1-x) = x^2$$ then $$2f(1-x) - f(x) = (1-x)^2$$ So, multiplying the first by $2$ and adding to the second, we have $$3f(x)=2x^2+(1-x)^2$$ Edit By the way, the answer for $x=-5$ is not in the list of choices and your answer is perfectly correct. You would get $\frac {34}3$ for $f(-3)$. All other cases ...


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 Vi├Ęte's theorem, the sum of the roots of a polynomial $p(x)$ raised to the minus one power is given by ...


14

Note that $$b^2+ac=c^2+ab\iff b^2-c^2+ac-ab=0\iff (b-c)(b+c)-a(b-c)=0$$ $$\iff (b-c)(b+c-a)=0\iff a=b+c.$$ Now $$a^2+b^2+c^2=(b+c)^2+b^2+c^2=2(b^2+c^2+bc)$$ $$=2((b+c)^2-bc)=2(a^2-bc)=2\times 2014$$


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: $$ f(-3) = 81, \qquad \lim_{x\to -3^-}f(x) = 90.$$


11

Hint: $$a^2 + b^2 = (a+b)^2-2ab$$ Going forward, $$ \left(x + \frac{x}{x-1}\right)^2 - \frac{2x^2}{x-1} = 2010$$ $$ \left(\frac{x^2}{x-1}\right)^2 - \frac{2x^2}{x-1} = 2010$$ Let $\frac{x^2}{x-1} = n$ $$ n^2 -2n -2010 =0 $$ You should be able to solve this using the quadratic formula or otherwise to get a vale for $ \frac{x^2}{x-1}$ Indeed, ...


10

Hint: Try expanding $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right)^2$.


8

Intuitively speaking, you are putting "mass" along the line $[1,3]$ such that the total mass is zero, and the density at any point must only be between $1$ and $-1$. Since you are trying to maximize the value of $\int_1^3 f(x) / x \; dx$, mass closer to $1$ is weighted higher and mass closer to $3$ is weighted lower. So you should put as much mass as ...


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 ...


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...


6

We have, $$2f(x)-f(1-x)=x^2$$ $$\implies 2f(6)-f(1-6)=6^2$$ $$\implies 2f(6)-f(-5)=36\tag 1$$ $$ 2f(-5)-f(1-(-5))=(-5)^2$$ $$\implies 2f(-5)-f(6)=25$$ $$\implies 4f(-5)-2f(6)=50\tag 2$$ Now, adding (1) & (2), we get $$3f(-5)=36+50=86$$ $$\implies f(-5)=\frac{86}{3}$$ Your answer is correct. There may be some printing mistake in the options provided in ...


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: $$ ...


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.


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 ...


5

$a = 6 \quad(\mathrm{mod} 7)$ $a^2 = 36 = 1 \quad(\mathrm{mod} 7)$ $3a = 18 = 4\quad (\mathrm{mod} 7)$ $a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$


5

Since their product is a power of $3$, both $x-2$ and $x-10$ must be powers of $3$, perhaps with minus signs (as Andre Nicolas pointed out). Note, however, that $x-2$ and $x-10$ cannot simultaneously be divisible by $3$. The only power of $3$ which is not divisible by $3$ is $1$, hence we must have that $x-2 = \pm 1$ or $x-10 = \pm 1$. With this in mind, ...


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 ...


4

We have to use the Polignac Legendre formula twice. The answer is: $\lfloor\frac{2008}{11}\rfloor+\lfloor\frac{2008}{121}\rfloor+\lfloor\frac{2008}{1331}\rfloor-(\lfloor\frac{1000}{11}\rfloor+\lfloor\frac{1000}{121}\rfloor+\lfloor\frac{1000}{1331}\rfloor)$ So $182+16+1-(90+8+0)=101$


4

Recognize the squares of binomials in both the numerator and denominator to rewrite the equation as $$\begin{align}\frac{x^2-(y-1)^2}{y^2-(x-1)^2} &= 2, \end{align}$$ and thus, factoring, $$\require\cancel \begin{align}\frac{(x-y+1)\cancel{(x+y-1)}}{(y-x+1)\cancel{(y+x-1)}}&=2. \end{align}$$ Finally, let $t=x-y$ and multiply both sides by $1-t$ to ...


4

It happens that $987$ is the product of three distinct primes: $\,3\cdot7\cdot47$. If $\,a\le b\le c$, $\,abc=987\,$ and $a+b+c=149$, we know that $$1\le a,b, c\le 147,$$ hence $c$ must be in $\,\{1,3, 7,21,47, 141\}$. If $c=141$, then $b=7, \ a=1$. Indeed $a+b+c=149$ in that case. If $c=47$, then $ab=21$, so that $a,b\le 21$, and $a+b+c\le89$. If $c\le ...


4

rac{1}{d})^2=\\(\frac{1}{a})^2+(\frac{1}{b})^2+(\frac{1}{c})^2+(\frac{1}{d})^2+2(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})\\65^2=209+2(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})\\(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})=\frac{1}{2}(65^2-209)=2008$$ ...


4

If $2^i\equiv 2^j\pmod{1000}$, then $2^i\equiv 2^j\pmod{125}$ and $2^i\equiv 2^j\pmod{8}$. But the latter holds for all $i,j$ that are at least $3$. So instead focus on the former. It turns out that the order of $2$, mod $125$, is $100$. That is, $2^{100}\equiv 1\pmod{125}$, but not for any smaller positive (integer) power.


4

$${ x }^{ 2 }+\frac { { x }^{ 2 } }{ { \left( x-1 \right) }^{ 2 } } =2010\\ { \left( x+\frac { x }{ x-1 } \right) }^{ 2 }-2\frac { { x }^{ 2 } }{ x-1 } =2010\\ { \left( \frac { { x }^{ 2 } }{ x-1 } \right) }^{ 2 }-2\left( \frac { { x }^{ 2 } }{ x-1 } \right) -2010=0\\ \left( \frac { { x }^{ 2 } }{ x-1 } \right) =t\\ { t }^{ 2 }-2t-2010=0\\ t=1\pm ...


3

Continuing vadim123's answer, we have that $n\geq 3$ implies $2^n\equiv 0\pmod{8}$. The subgroup generated by $2$ in $\mathbb{Z}_{/125\mathbb{Z}}^*$ is the whole $\mathbb{Z}_{/125\mathbb{Z}}^*$, since: $$ \left|\mathbb{Z}_{/125\mathbb{Z}}^*\right|=100,\quad 2^{50}\equiv -1\pmod{125},\quad 2^{20}\equiv 76\pmod{125} $$ so a power of two $\pmod{125}$ is allowed ...


3

$(1/a+1/b+1/c+1/d)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}+2\times \frac{1}{a}\times\frac{1}{b}+2\times\frac{1}{a}\times\frac{1}{c}+2\times\frac{1}{a}\times\frac{1}{d}+2\times\frac{1}{b}\times\frac{1}{c}+2\times\frac{1}{b}\times\frac{1}{d}+2\times\frac{1}{c}\times\frac{1}{d}$ ...


3

$f(n)=\sum_{k=1}^{m}(r_k n - \lfloor {r_k n}\rfloor)$ $\lfloor {r_k n}\rfloor \leq r_k n$ So, $f(n) \geq 0$ $r_k n - \lfloor {r_k n}\rfloor <1 $ $f(n) < \sum_{k=1}^{m}1$ So, $f(n) < m$ $r_k = p_k / q_k$ where $p_k < q_k$ If $n = q_1.q_2.....q_m - 1$ Then $n= (q_1.q_2.....q_m-q_k)+(q_k-1)$ So, $r_k n - \lfloor {r_k n}\rfloor = ...


3

Observe that from $AB^2+AC^2=BC^2$, the sum of the area of the two smaller circles is the area of the biggest circle. Then $A_1+$white part$+A_2+A_2+A_3 = A_2+A_4+$white part, $\Rightarrow A_1+A_2+A_3 = A_4$.


3

We can say, at least, that the option (C) is correct. Let $p,q,r,s,t,u$ be the areas of white parts : $p$ is the upper left area of $AB$, inside the circle whose diameter is $BC$. $q$ is lower of $AB$, upper of $BC$, outside the circle whose diameter is $AC$. $r$ is lower of $BC$, inside the circle whose diameter is $AB$. $s$ is lower of $BC$, inside ...


3

By Holder's Inequality, $$S^2\sum_{cyc} a^2b^2\ge (a^2+b^2+c^2+d^2)^3$$ So it remains to note that by AM-GM $$\sum_{cyc}a^2b^2 = (a^2+c^2)(b^2+d^2) \le \frac{\left((a^2+c^2)+(b^2+d^2)\right)^2}4$$


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 $$2\frac{1-\sqrt3}{2-2\sqrt3}=1.$$



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