New answers tagged

0

I would simplify the problem as follows: Let $e$ = Expected number of flips until $5$ consecutive $H$, i.e., $E[5H]$ Let $f$ = Expected number of flips until $5$ consecutive $H$ when we have seen one $H$, i.e., $E[5H|H]$ Let $g$ = Expected number of flips until $5$ consecutive $H$ when we have seen two $H$, i.e., $E[5H|2H]$ Let $h$ = Expected number of ...


1

As dez pointed such a number exists if and only if $gcd(n,10)=1$. Case 1 $3 \nmid x$. Then $$x |111...1 \Leftrightarrow x|999...9 \Leftrightarrow 10^n \equiv 1 \pmod(x) \Leftrightarrow ord_x(10)|n$$ Therefore, the smallest $n$ is $$n=ord_x(10)$$ that is the order of $10$ modulo $n$. Case 2 $3 \mid x$. Then $$x |111...1 \Leftrightarrow 9x|999...9 ...


0

Your question is equivalent to asking what is the smallest $n$ such that $10^n-1$ is divisible by given number $x$. Such a number exists if and only if $\gcd(10,x) = 1$. So what you need to do is solve the problem for $2^n-1$ and $5^n-1$, and take their least common multiple. You only need to consider the prime power divisors of $n$, not all the divisors of ...


1

To know the nature of the graph you will have to have some ideas about graphs of $y=x^2$ and $y=x^3$ individually. Look closely...It's given $x^2=y^3$. So,$x$ must be greater that $y$ for all integer values. SO, for any increase $\delta x$ in x-axis the increase $\delta y$ in y-axis must be smaller than it as you have a power of $3$ in y but $2$ in x.So,a ...


2

It helps to investigate the range of the function. Calculating on $\mathbb{R}$, you know that $x^2$ will be nonnegative: $x^2 \geq 0$. Since $y^3 = x^2$, it must hold that $y^3$ is nonnegative. By the nature of the cube-root, it follows that $y\geq 0$. From this you can conclude there are no points below the $x$-axis. You know there is symmetry in the ...


2

Here's my idea: taking into account zero (or $\;0\pmod p\;$ , if you prefer), there are $\;\frac{p+1}2\;$ quadratic residues modulo $\;p\;$ , and now define for $\;k\in\Bbb Z/p\Bbb Z\;$ fixed: $$T:=\left\{\,x^2\;:\;x\in\Bbb Z/p\Bbb Z\,\right\}\;,\;\;S_k:=\left\{\,k-x^2\;:\;x\in\Bbb Z/p\Bbb Z\,\right\}$$ and observe that again $\;|S_k|=\frac{p+1}2\;$ , of ...


0

Setting $c=-a-b$ gives just $$ (l-1)(l+1)=0. $$


0

Hint the least value of AM and max value of-GM is obtained when all numbers are equal so $x=y=z$ so least value is $6$ by putting all as $1$


1

If $x,y,z>0$, then by AM-GM: $$\frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{1/2}}=\frac{x+y+z+xy+yz+zx}{(xyz)^{1/2}}$$ $$\ge \frac{6\sqrt[6]{x\cdot y\cdot z\cdot xy\cdot yz\cdot zx}}{(xyz)^{1/2}}=6$$ Equality holds if and only if (iff) $x=y=z=xy=yz=zx$, i.e. iff $x=y=z=1$.


5

Circular inversion is not really needed, Ceva's theorem is enough. Let $O_X$ be the centre of the incircle of $XBC$. Clearly $O_X$ lies on the perpendicular to $BC$ through $D$, and $CD=CY=CE$ as well as $BD=BZ=BF$. $EFZY$ is a cyclic quadrilateral iff the perpendicular bisectors of $FZ,YE$ and $EF$ concur (the last line is just the angle bisector of ...


1

Let $a=b=2$ and $c=1$. Hence, $k\geq100$. We'll prove that $100$ it's an answer. Indeed, let there are positives $a$, $b$ and $c$ for which $ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$ and $k<100$. But it's impossible because we'll prove now that $ \frac{100abc}{a+b+c}\leq (a+b)^2+(a+b+4c)^2$. Let $c=(a+b)x$. Hence, by AM-GM ...


5

Note that this is $$ (x + y)^3 + y^3 = 37^3; $$ by Wiles' theorem the only integral solutions are $(37,0)$ and $(-37,37)$.


0

First, all convex function over $\mathbb{R}$ is continuous. (See Is every convex function on an open interval continuous?, https://en.wikipedia.org/wiki/Convex_function, the other answers here, and also possibly convex function in open interval is continuous) Thus $\displaystyle \lim_{x\to 0^-} f(x) = \lim_{x\to 0^+} f(x) \iff c = 1$. $f$ is convex over ...


2

Effectively, you want to show $$\frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c) \geqslant 100$$ and you already have a case of equality. Using homogeneity, we may set $a+b+c=5$, to equivalently show $$(5-c)^2+(5+3c)^2 \geqslant 20 abc$$ Now $a+b = 5-c$, so for any $c$, we have $ab$ maximized when $a=b$. Thus it is enough to show $$(5-c)^2+(5+3c)^2 \geqslant 20 ...


0

Yes. Take a look at Flash Anzan competitions in Japan. Basically, you are given 15 numbers, each between 100 and 999. The numbers are shown for a total of 2 seconds. You must then give the sum of the numbers in the next second.


0

Perhaps this can make things easier. Let $AP = a, BP = b, CP = c$ and $AA’ = BB’ = CC’ = L$. Then, $PA’ = L – a, PB’ = L – b, PC’ = L – c$. Note that we can jump to the conclusion directly if $a = b = c$. The easiest case is when $a, b, c$ are distinct. We choose to prove the case when $a = c$ but $c \ne b$ and therefore $a \ne b$. For the chords ...


2

$\cos 3 y=4\cos^3 y-3\cos y.$ When $\cos 3 y=-1/2$ and $x=\cos y$, we have $8 x^3-6 x +1=0,$ with $3$ solutions $x=\cos [2\pi(1+3 n)/9]$ for $n\in \{0,1,2\}.$


2

Hint: for $y=8x^3-6x+1$ and $y'=24x^2-6$ . So ve have two stationary points for $x=\pm1/2$. Now you can see that there is a positive valued max at $x=-1/2$ and a negative valued min at $x=1/2$ and the function is negative for $x=-1$ and positive for $x=1$. Use continuity (intermediate value theorem) to show that there are three roots in $[-1,1]$


0

For $f(x)=8x^3-6x+1$ the gradient function is $f'(x)=24x^2-6=6(4x^2-1).$ This leads to an observation about the ordinates of the stationary points. Then find $f(-1)$ and $f(1)$. The conclusion follows.


2

DISCLAIMER: I posted this answer when the first equation read $2x^{\log 2}=3y^{\log 3}$ without parenthesis, but I see that that has been changed to $(2x)^{\log 2}=(3y)^{\log 3}$ in which case my answer below does not fit the problem. For this new equation $(1)$ it becomes $$ as-bt=b^2-a^2 $$ leading through the same steps as below to $$ s=-a\quad t=-b $$ so ...


1

That's from Kedlaya I believe? First of all, let's mention a useful theorem: "For any two triples of non-collinear points, we can find a unique affine transformation sending the points of the first triple to then corresponding points of the second triple". That's a known theorem. So, in our case we apply the affine transformation that sends $A,C,D$ (which ...


1

We know that $ab = cd$ by the power of the point $P$, and also that $a + b = c + d$ because the lengths of the chords $AB$ and $CD$ are equal. So $$ b = \frac{cd}{a} \\ \implies a + \frac{cd}{a} = c + d \\ \implies a^2 + cd = ac + ad \\ \implies a^2 - ac + cd - ad = 0 \\ \implies (a-c)(a-d) = 0 \\ \implies a = c\quad \text{OR}\quad a = d $$ We obtain that ...


1

Assume $2\leq A<B$. Then $B=qA+r$ with $q\geq1$, $\>1\leq r\leq A-1$, and $r$ prime to $A$. Imagine an "abstract" regular $A$-gon with vertex set $V\sim{\mathbb Z}/A$. If we draw the $A$ chords $\{k,k+r\}$ we obtain a cyclic graph $\Gamma$. Omitting an edge from $\Gamma$ leaves a connected graph $\Gamma'$, but if we omit $\geq2$ edges from $\Gamma$ ...


0

Just some thoughts to address further answers. Let we say that a colouring is maximal if $n$ is maximal. As stated in the question, in a maximal colouring $c$, $A$ and $B$ cannot have the same colour: otherwise we may define $c'$ on $\{1,\ldots,n+1\}$ such that $c'(1)=c(A)=c(B)$ and $c'(m)=c(m-1)$ for any $m\geq 1$, leading to a valid colouring and ...


0

You should think of what convexity of a function means. For convex functions it must hold that for every two points located above the convec function a line between those points does not intersect the function. In mathematical terms this means that the function should be continuous ánd that the tangent is strictly increasing. If $x \geq 0$ then the value ...


0

You have to study just the convexity for $x<0$ really. By continuity it is immediately $c=1$ so you need the convexity of $ax^2+bx+1$ for $x<0$ with the constraint left-side derivative at $x=0$ (as a limit position in the domain $x<0$ and equal to b) less than or equal to zero because if the tangent to the curve has a positive slope then, by ...


2

A convex function is automatically continuous and has one-sided derivatives at each point $x$ in its domain, whereby $f'(x-)\leq f'(x+)$, and of course $f''(x)\geq0$ at all points where the second derivative is defined. These facts enforce $c=1$, $b\leq0$, and $a\geq0$ in your problem. That an $f$ fulfilling these conditions is in fact convex on ${\mathbb ...


0

When $x=0$, $f(xf(0))=0$. When we put $f(0)=k,f(kx)=0.$So , $f(x)=0$ or $f(0)=k=0.$ put $y=\frac1x$,$f(\frac{f(x)}x)=x^2f(1)$ if put $f(x)=ax^2+bx$ $a(ax+b)^2+b(ax+b)=x^2f(1)=x^2(a+b)$ $⇔a^3x^2+(2a^2b+ab)x+ab^2+b^2=x^2(a+b)$ $⇒b^2(a+1)=0$ , $ab(2a+1)=0$ , $a^3=a+b$ $⇒b=0 , a=±1,0$ therefore by $f:\mathbb N^* \to \mathbb N^*$ $f(x)=x^2$


0

B.J.Venkatachala for inequality is a very good book for what you are searching.You may see this book.


1

Old and New Inequalities Volume 1 - Titu Andreescu Old and New Inequalities Volume 2 - Vo Quoc Ba Can et.al. Algebraic Inequalities - Vasile Cirtoaje Secrets in Inequalities - Pham Kim Hung [Volume 1 and 2] Inequalities with Beautiful Solutions - Vo Quoc Ba Can et.al. To my best knowledge, all the problems presented in the above mentioned books are ...


3

$$x^2 + y^2 + z^2 + 4(xy + yz + zx) - 4(x + y + z) =(x+y+z-2)^2-4+2(xy + yz + zx)$$ Now, by AM-GM: $$x+y+z\ge3\sqrt[3]{xyz}\ge3\tag{1}$$ $$xy+yz+zx\ge3\sqrt[3]{x^2y^2z^2}\ge3\tag{2}$$ which pretty much settles the matter: $$\underbrace{(x+y+z-2)^2}_{\ge1}-4+2(\;\underbrace{xy + yz + zx}_{\ge3}\;)\ge1-4+6=3$$


1

Your direction is correct, now you only need to think about the "connection" between the two conditions on X , which is at X = 0. For the function to be convex it also must be continuous , so C = 1 and then the lim f(x) when x is approaching 0 from either sides is 1 , which equals to f(0) so the function is continous


1

I haven't any idea what means are available to the students for making the calculation, but I'm wondering whether they are supposed to notice that in the sequence of numbers for $ \ _{2n}C_{n} \ = \ \binom{2n}{n} \ , $ $$ 1 \ \ 2 \ \ 6 \ \ 20 \ \ 70 \ \ 252 \ \ 924 \ \ 3432 \ \ \ldots \ \ , $$ each entry is $ \ 3 \ + \ \frac{n-2}{n} \ \ = \ \ ...


1

Since the $1^{st}$ row are the binomial coefficients of $(a+b)^0$, the $29^{th}$ row are the coefficients of $(a+b)^{28}.$ Here is a fairly easy rule to generate that row. $1, 28, \frac{28*27}{2}, \frac{28*27}{2}\frac{26}{3}$.... each entry is the entry before it, times one number less, divided by one number greater. The middle number will be the last ...


4

It's a slightly odd question. Although it's for primary school, it is also Olympiad training, so I would say it's reasonable to expect that students will know that the number is $$\binom{28}{14}=\frac{28\times27\times\cdots\times15}{14\times13\times\cdots\times1}\ .$$ If you now cancel lots of common factors it reduces to ...


0

Reasoning by contradiction. If $G$ not centroid, then at least one of the points on the sides is not in the middle. Let $AE > EC$. By property chevian,product $(AE:EC)(CD:DB)(BF:FA)$ is unity. Hence, at least one of the last two relations less than 1. Let us consider two cases. 1) $CD < DB$. According to Van Aubel's Theorem, $CG:GF=CE:EA+CD:DB < ...


0

Let $\Delta$ = area BGD = area CGE = area AGF. Let $p=\frac{DC}{BD},q=\frac{EA}{CE},r=\frac{FB}{AF}$. By Ceva we have $pqr=1$. So taking area BGD=1 the areas of the other small triangles are as shown. Now AG/GD = area AGB/area DGB = $r+1$. But AG/GD = area AGC/area DGC = $\frac{1+q}{p}$, so $p(1+r)=1+q$. Similarly $q(1+p)=1+r,r(1+q)=1+p$. Adding we get ...


2

Note that $2005\equiv3$ in modulo $7$. Since $2005\equiv1$ in modulo $6$, you can work out that $2005^{2005}\equiv3$ in modulo $7$. A perfect cube can only take values $1$, $0$ and $-1$ in modulo $7$. Therefore sum of two perfect cubes, namely sum of two of these numbers, can never be equivalent to $3$ in modulo $7$.


1

If you multiply together all the terms, you'll get a sum. Think of each term in this sum as a $400$ letter long word, where the $j$th letter is either an $x$ or the number $j$ (for $1\leq j \leq 400)$. You can of course simplify this expression into the form of $l \cdot x^k$ where k and l is some numbers. Now, for example the only way to get $x^{400}$ in ...


3

Case 1: Let $k\gt0$ Then, the graph will be a concave-up parabola, cutting the $x$-axis once on the positive side and once on the negative. Clearly, at $x=0$, the function should take a negative value. $$k-2015\lt 0$$ Thus, $$0\lt k\lt 2015$$ Case 2: Let $k\lt 0$. Now, the graph will be a concave-down parabola. The value of the function at $x=0$ must ...


0

Here's something similar to Ewan's except it may be more transparent for some. Firstly if you want $\det=0$ take the matrix of all $1$'s, except of course in the $1\times1$ case. Then for $\det\neq 0$ take the diagonal matrix $D=(d_{ij})_{1\leq i,\,j\leq n}$ where $$d_{ij}=\begin{cases} 0 &\text{if} & i\neq j \\ 1 &\text{if} & i=j \text{ ...


0

If there are $n$ chairs in total then there are $\frac7{5\cdot 31}$ rows of the first kind, which implies $n$ is divisible by $5\cdot 31$. Similarly, $n$ is divisible by $13\cdot 31$, hence by $5\cdot13\cdot 31=2015$. The only positive multiple of $2015$ below $4000$ is $2015$ itself.


2

There are $2015$ chairs in the hall. Let the number of chairs be $x$. Then there are $\frac{7}{31}x$ chairs arranged in rows of $5$. Since this is a whole number of chairs, $x$ must be divisible by $31$. Also since $\frac{7}{31}x$ must be divisible by 5, $x$ has to be divisible by $5$ (because $7$ is not divisible by $5$). Similarly, we know that ...


0

I would go about this problem in the following way: Let $A$ be the person who is in exactly 2 clubs (which we can call $Club1$ and $Club2$), and let $B$ be the couple partner of $A$. Since $A$ must be in exactly one club with every person other than B, the remaining 8 people must each belong to exactly one of $A$'s 2 clubs. Furthermore, since those 8 ...


2

By AM/GM applied to $1^a,2^a,\dots,n^a$, we have $d_n>b_n$. The inequality is strict because the terms are obviously unequal. [AM/GM = Arithmetic Mean/Geometric Mean inequality]


0

From various sources, https://math.dartmouth.edu/archive/m8f02/public_html/pauls_mws/boxeg.pdf http://www.leadinglesson.com/problem-on-finding-the-rectangular-prism-of-maximal-volume I confirmed that to obtain the max volume, the prism would be a cube. So in this case, we would try to find the volume assuming that the prism is a cube. 1 Face of the cube ...


3

The answer is $\mathbf{k = 14}$. Consider the more general question where the number of couples is a parameter. In the case of just one couple the minimal system has $k=3$, with one person belonging to two clubs, and the other person belonging to a different club. Therefore suppose there are $n\ge 2$ couples. Condition 3 implies that there is some person ...


3

Let $A=(a_{ij})_{1\leq i\leq j}$ where $$a_{ij}=\left\lbrace\begin{array}{lcl}1 &\text{if} & i\neq j & \text{or} & i=j=1\\2 &\text{if} & 1<i= j<n, \\ x+1 &\text{if} & i= j=n, \\\end{array}\right.$$ Then you can check that the determinant of $A$ is exactly $x$. For example, when $n=5$, $A$ is $$ ...


1

Partial fraction decomposition is a proper method to answer the question. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We obtain \begin{align*} [x^n]\frac{2-3x}{1-3x+2x^2}&=[x^n]\left(\frac{1}{1-2x}+\frac{1}{1-x}\right)\tag{1}\\ ...


2

If $x,y,z$ are not the sides of a triangle, then the inequality is trivial because the left hand side would negative.



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