# Tag Info

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+bxa(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^nof 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 Ifx,y,z\$ are not the sides of a triangle, then the inequality is trivial because the left hand side would negative.

Top 50 recent answers are included