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We have the identity$$(n+1)\, \text{lcm}\left(\binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n-1}\right) = \text{lcm}(1,2,\ldots, n+1).$$If $p$ is a prime number that does not divide $n + 1$ and $p^r \le n$, then the right-hand side is divisible by $p^r$ which forces some $\binom{n}{k}$ to be divisible by $p^r$. Such a prime $p$ exists for all large $n$ ...

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While i haven't found a definite answer yet, let me at least provide some quick n' dirty bounds where the solution must lie. Let me rename the number of moves for the most complicated configuration possible $f(C_n)$. It's easy enough to prove that: $$n \le f(C_n) \le 3n$$ Because: $n \le f(C_n)$ since you need $n$ operations to perform a "flip every ...

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Clearly $k > 1$; otherwise $p(x) = x^5$ and $m_1, \ldots, m_5$ are not distinct. Assume $k = 2$, so $p(x) = x^5 + ax^j$ with $0 \le j \le 4$. We can not have $j \ge 2$ since then at least two of the $m_l$'s are equal to $0$. Hence $p(x) = x^5 + a$ or $p(x) = x(x^4 + a)$ with $a \neq 0$. But $x^5 + a$ and $x^4 + a$ have at most two real zeros. Therefore $k ... 0 Here's my personal favorite 1.) Consider a polygonal line$P_0P_1...P_n$such that$\angle P_0P_1P_2 = \angle P_1P_2P_3 = ... = \angle P_{n-2}P_{n-1}P_n$, all measured clockwise. If$P_0P_1 > P_1P_2 > ... > P_{n-1}P_n$, show that$P_0$and$P_n$cannot coincide. 2.) sorry I don't know how to insert a diagram 3.) We will consider the complex ... 0 This question of mine is, I think, related: How many different ways can the signs be chosen so that$\pm 1\pm 2\pm 3 ... \pm (n-1) \pm n = n+1$? Since$1+2+...+(2n) =\frac12(2n)(2n+1) =(2n+1)n $, then if a subset sums to$\frac12 n(2n+1) $, the rest also sum to$\frac12 n(2n+1)(n+2) $so the sum$\text{sum(original subset)-sum(others)} =0 $. Therefore ... 0 Let$x=a/by=b/c$and$z=c/a$then$1=xyz=\frac{x+y+x}{x+y+z}=\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}$but each of the fractions on the far right are less than$1$. So it follows$x=y=z$which is to say$a/b=b/c=c/a$so$a^2=b^2=c^2$. 2 Let$m=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$and$n=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$. Then$\frac{a}{b},\frac{b}{c},\frac{c}{a}$are the roots of$x^3-mx^2+nx-1=0$. Hence we have$a^3-ma^2b+nab^2-b^3=0$, where$a,b,m,n$are all integers, with$a,b$positive. Let$p^r$be the highest power of$p$dividing$a$, and$p^s$be the highest power of$p$... -1 Suppose$g$is continuous in$t$and$h$is discontinuous in$t. By definition of differentiability $$\lim_{k\to 0^+} f'(t+k) = \lim_{k\to 0^-} f'(t+k)$$ So that: $$\lim_{k\to 0^+} f(t+k)+g(t+k)+h(t+k)=\lim_{k\to 0^-} f(t+k)+g(t+k)+h(t+k)$$ However, we know: $$\lim_{k\to 0^+} f(t+k) = \lim_{k\to 0^-} f(t+k) \\ \lim_{k\to 0^+} g(t+k) = \lim_{k\to 0^-} ... 1 As I said above, given any polynomial f that's bound correctly, we can construct infinitely more by writing x^n f(x). Moreover, given f,g that are both bound correctly, their product fg will also be bound correctly. Now, to see that there isn't some finite set \{f_0,f_1,...,f_n\} such that all other polynomials bound this way are a product of ... 0 Let AQ be a chord perpendicular to BC cutting BC at P. If PD = x, then BP = 1 – x, QP = PA = \sqrt 3 x and AD = 2x [\triangle APD is special angled.] At P, (1 – x)(x + 3) = (\sqrt 3 x)^2. This gives x = \dfrac {\sqrt {13} - 1}{4}, after rejecting the negative length From \triangle APO, (\sqrt 3 x)^2 = 2^2 – (1 – x)^2 … (*) From ... 0 constuct a circle of radius 2, centerd at 0. D = (1,0), B = (2,0), c = (-2,0) constuct a vector v at a 60 degree angle with tail at D. A = (1 + |v| \cos 60, |v| \sin 60)\\ |A|^2 = 1 + |v| + \frac 14 |v|^2 + \frac34 |v|^2 = 2^2\\ |v|^2 + |v| - 3 = 0 use the binomial theorm. |v| = \frac{-1 + \sqrt{13}}{2}\\ |AC| = \sqrt{(3 + |v| \cos 60)^2 + |v|^2 ... 0 The circumcircle provides a good approach. |DM|=1, |AM|=2, and |MC|=2. Because \angle ADB = 60^{\circ}, \angle ADM = 120^{\circ}. We may then find \angle DAM using the law of sines:$$\frac{\sin{\angle DAM}}{|DM|} = \frac{\sin{\angle ADM}}{|AM|} \implies \sin{\angle DAM} = \frac12 \sin{120^{\circ}} = \frac{\sqrt{3}}{4} $$We may therefore ... 2 Here is an algebraic solution to the problem. I suppose that the coefficients are complex numbers here. Suppose, therefore, that the complex polynomial P\in\mathbf C[x] satisfies the equation$$ P(x)^2+P(\tfrac1x)^2=P(x^2)P(\tfrac1{x^2}) $$in the fraction field \mathbf C(x). Let us first prove that P(1)=0 and P(-1)=0. Indeed, evaluating the ... 4 Step 1: Let p(t) = at^n + \cdots + b. What happens for large x? Step 2: Your turn. 2 We want solutions to k^2=n^2+(n+1)^2. So (k,n,n+1) is a Pythagorean triple. Moreover it is a primitive triple since n,n+1 are coprime. So we must be able to write it as k=a^2+b^2,n=2ab,n+1=a^2-b^2 for n even or k=a^2+b^2,n=a^2-b^2,n+1=2ab for n odd. If n is odd, then a^2-b^2+1=2ab and a>b, so we have (a+b)^2-2a^2=1. If n is even, ... 3 Well,$$\begin{align}\dfrac{\sqrt{2}+\sqrt{a}}{\sqrt{3}+\sqrt{b}} ~=~& \dfrac{(\sqrt{2}+\sqrt{a})(\sqrt{3}-\sqrt{b})}{(\sqrt{3}+\sqrt{b})(\sqrt{3}-\sqrt{b})} & \textsf{except when }b=3 \\[2ex] =~&\dfrac{\sqrt{6}+\sqrt{3a}-\sqrt{2b}-\sqrt{ab}}{3-b} \end{align}$$Then, when might this be rational? ... 0$$ p(X^k Y^m=t)=\int_0^1dx\int_0^1 dy\ \delta(t-x^ky^m)\ , $$and using the Dirac delta to kill the x-integral$$ p(X^k Y^m=t)=\int_0^1 dy\ \Theta\left(0\leq (t/y^m)^{1/k}\leq 1\right) \frac{1}{y^m k ( (t/y^m)^{1/k})^{k-1}}\ ,  =\Theta(0\leq t\leq 1)\int_{(1/t)^{-1/m}}^1 \frac{dy}{y^m k ( (t/y^m)^{1/k})^{k-1}}=\boxed{\Theta(0\leq t\leq ... 1 The geometry is the same as in 3 dimensions, when taking the plane sectionx+y+z = c$of a sphere$x^2+y^2+z^2=C$. Extrema of$z$are when$x=y$. The only remaining thing to understand is why the answer for the parameters given in the problem does not involve square roots. Let$n$, which equals$4$in the posted problem, be the number of variables other ... 5 Physics (Classical Mechanics) Solution: Consider a$1$-dimensional elastic collision of a particle$X$of mass$4$moving at velocity$2$into a particle$E$of mass$1$, initially at rest. Due to this collision,$X$breaks into$4$smaller particles$A$,$B$,$C$, and$D$(of course, we are ignoring their binding energy, which would lead to an inelastic ... 6 By the Cauchy-Schwarz inequality, $$(8-e)^2 = (a+b+c+d)^2 \leq 4(a^2+b^2+c^2+d^2) = 4(16-e^2)$$ from which it follows that$e\leq \color{red}{\frac{16}{5}}$. Now it is enough to show that the inequality holds as an equality for some$(a,b,c,d,e)\in\mathbb{R}^5$, pretty easy. In the same way you may also show that$e\geq 0$. 1 The repunits are$\frac{10^1-1}{9}=1,\frac{10^2-1}{9}=11,\frac{10^3-1}{9}=111,\dots$. So if$n$is a power of$10$, then$\frac{n-1}{9}$is a repunit, so$f(\frac{n-1}{9})$is a repunit, and hence$9f(\frac{n-1}{9})+1$is a power of$10$. But if$f(x)$is a polynomial, then$g(x)=9f(\frac{x-1}{9})+1$is also a polynomial. Suppose its leading term is ... 0 Recall that arc length =$\theta r$and that the area of a sector =$\frac{\theta}{2} r^2$where the angle theta is in radians EDIT: I left out a bit of information that might help you answer this question. C =$2\pi r$30 =$2\pi r$You are also given the volume of the cut to be$50cm^3$. You can use this and the uniform height given to calculate the ... 2 Too long for a comment. The Engel form of Cauchy-Schwarz is not the right way: $$\frac{(x^2)^2}{8x^3+5y^3}+\frac{(y^2)^2}{8y^3+5z^3}+\frac{(z^2)^2}{8z^3+5x^3} \geq \frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}$$ So we should prove that $$\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}\geq\frac{x+y+z}{13}$$ which is equivalent to ... 2 Hint: Take the Maclaurin series for$\sin(x)$and then replace$x$with$1/z$. Then take the Maclaurin series for$\cos(x)$and replace$x$with$z$. Then do some arithmetic with these to find all terms of order$-1$in the Laurent series expansion of$(1-\cos z)\sin \frac{1}{z}$at$z=0$. -3$Hint$: Given expression is also equal to : $$(xi+yj+zk).(\frac{1}{8+5\frac{y^3}{x^3}}i+\frac{1}{8+5\frac{z^3}{y^3}}j+\frac{1}{8+5\frac{x^3}{z^3}}k)$$ where i,j and k are unit vectors in 3d plane. Now minimise second VECTOR's MODULUS( not its components), Thus, we get$$(xi+yj+zk).(\frac{i+j+k}{13})$$and get the required result. 2 This is a question of the symmetric type, such as listed in: Why does Group Theory not come in here? With a constraint$\;x+y+z=1\;$and$\;x,y,z > 0$. Sort of a general method to transform such a constraint into the inside of a triangle in 2-D has been explained at length in: How prove this inequality$(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$Our ... 3 The problem is false as stated. Here is a counterexample that works for any$n=4k$: $$\vec a = (\color{red}{4k-1}, \color{blue}{2k-1}, \color{red}{4k-3}, \color{blue}{2k-3}, \ldots, \color{red}{2k+3}, \color{blue}{3}, \color{red}{2k+1}, \color{blue}{1}; \color{green}{4k}, \color{brown}{2k}, \color{green}{4k-2}, \color{brown}{2k-2}, \ldots, ... 0 I write a start for a full answer (this is an idea that @Starfall first proposed in comment). If someone wants to use it to end the proof, she/he is welcome! Let$$f(x,y,z):=\frac{x^4}{ax^3+by^3}+\frac{y^4}{ay^3+bz^3}+\frac{z^4}{az^3+bx^3}.$$Since f is homogeneous of degree 1, it is sufficient to consider x,y,z on the plane P:=\{x+y+z=1\}. Let ... 2 1) (8-\sqrt{63})^{2012}=N.0000000...00ABC... 2) (8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}=A \in Z Then 3) (8+\sqrt{63})^{2012}=K.9999999...99DCE... 1 As André Nicolas has kindly pointed out,$$(8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}$$is equal to an integral value. Also, (8-\sqrt{63})^{2012} is equal to a fractional value. Thus, we can write:$$I + f = (8+\sqrt{63})^{2012}f = (8-\sqrt{63})^{2012}$$Where I means integer while f means fractional part. Addding the two, we get:$$I + f + f = ... 9 Hints: (i)$(8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}$is an integer. One way to prove this is to use the binomial theorem. (ii)$(8-\sqrt{63})^{2012}$is positive and quite a bit less than$10^{-1000}$. (The fact that$\sqrt{63}$is close to$8$is useful for this part). 1 Your "reverse" reasoning looks correct, but I think a more direct approach would be easier. As$CA$and$CB$are tangent to$\omega_2$then$CO_2$is the bisector of$\angle ACB$. It follows that arcs$AO_2$and$BO_2$are congruent and so are their chords:$AO_2\cong BO_2$. But then triangles$AO_1O_2$and$BO_1O_2$are congruent and it follows immediately ... 0 An ellipse centred at$(4,3)$with major and minor axes parallel to the$x-$and$y-$axes respectively has the following equation: $$\frac {(x-4)^2}{a^2}+\frac{(y-3)^2}{b^2}=1\qquad (a>b>0)\qquad\cdots(1)$$ Applying the rotation matrix for$\theta=-\pi/4$and simplifying: $$\frac{(x+y-7)^2}{2a^2}+\frac{(x-y-1)^2}{2b^2}=1\qquad\cdots(2)$$ ... 1 You can also use the Spherical Law of Cosines. We imagine a sphere centered at$O$, and we take$A$,$B$,$C$the points where the rays from$O$meet that sphere as vertices of a curvilinear triangle on the surface of that sphere. The "sides" of that triangle are arcs of great circles, and the lengths of those arcs are directly proportional to the measures ... 0 If you try generating palindromes using the given definition, you'll get a set that looks something like:$\{\lambda, 00, 11, 0000, 1001, 0110, 1111, 000000, 100001, 010010, 110011, 001100, 101101, 011110, 111111, \dots \}$. In fact, all of the strings in this list have an even length. The reason why should be pretty clear - you're starting with a 0-length ... 1 Your starting was right, so we have $$O = (0,0,0)\quad A = (1,0,0)\quad B = {1 \over 2}\left( {1,\sqrt 3 ,0} \right)$$ for point C instead, we shall have $$\left\{ \matrix{ \mathop {OB}\limits^ \to \, \cdot \;\mathop {OC}\limits^ \to = \cos \left( {90^\circ } \right) = 0 \hfill \cr \mathop {OA}\limits^ \to \, \cdot \;\mathop {OC}\limits^ \to = ... 2 Not anywhere near as elegant as the five-equidistant-point proof, but I've a nine-point brute-force proof. Premise: any point on a circle will form an icoseles triangle with points placed equidistantly either side. Pick any number of points separated by distance X. If any three sequential points are red (111) or blue (000) then there's an isosceles ... 11 I think that the function that sends any irrational number on 1 and sends a rational \frac{p}q (where the fraction is irreducible) on 1-\frac1q does the job. Given any rational number \frac{p}q, you can always find a neighborhood so that any rational number has a denominator bigger than q 11 Here's a pretty direct hint. If there were only rationals, the function could be defined as the denominator in lowest terms, because everything nearby enough would have a larger denominator. If we want to also define it on the irrationals this wouldn't work. However, we can compress the positive integers into [0,1) in an order preserving way, then we can ... 3 In the wonderfully clever solution given, starting with that pentagon, the two possible monochromatic triangles are not congruent. It may be interesting to note that this is of necessity. Say an \alpha-triangle is an isoceles triangle in which the two sides of equal length meet at an angle \alpha. Fix an angle \alpha. There exists a coloring of the ... 33 Hint: Take regular pentagon. The three points are the same color. 0$$x_{n}=\sum_{i=0}^{n-1} \frac{(-1)^{i}}{i+1}=\sum_{i=0}^{n-1} (-1)^{i}\int_0^1x^i\,dx=\int_0^1\sum_{i=0}^{n-1}(-1)^ix^i\,dx=\int_0^1\frac{1-(-x)^n}{1-(-x)}\,dx=\int_0^1\frac{1}{1+x}\,dx+\int_0^1\frac{-(-x)^n}{1+x}\,dx$$The first term is \log2, and the second term can be bounded in absolute value by noting that: ... 1 No, unfortunately your solution is not correct. Your proof breaks down when you make the claim: Now, if \gcd(A_n, x10^n+x-y) \neq 1, that must mean x=0, an impossibility. If say n = 2, x = 1, y = 2, and z = 3, then we have \gcd(A_n, x10^n+x-y) = \gcd(11,99) = 11 \neq 1, but x = 1 \neq 0. So your claim is false in this case. Hint ... 1 Well, P(0) must be 0 or 1, and as you note, if P(0)=0 then P(x)=xQ(x) and it follows that Q(x^2)=Q(x)^2 as well. Great so far. We can repeat this until we get a Q that does not have Q(0)=0. So there exists j\ge0 and Q so that P(x)=x^jQ(x) and$$Q(x^2)=Q(x)^2,\quad Q(0)=1.$$But there are problems with your argument about roots of ... 5 Hint If p \neq 0, then P has the form$$P(x) = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 ,$$where a_n \neq 0. Comparing the leading terms of P(x^2) and P(x)^2 gives that a_n = 1. If P(x) \neq x^n, then there is some largest index m < n such that a_m \neq 0, and so P has the form$$P(x) = x^n + a_m x^m + O(x^{m - 1}) .$$... 1 Hint: Assume that P(x) is nonconstant. If r is a root of P(x), then it also follows that any 2^n-th root of r is a root of P(x). The only complex number r that will allow P(x) to have finitely many roots is r=0. By the way, this hint also works if \mathbb{C} is replaced by a field of characteristic not equal to 2 as well (since ... 0 In fact \sum_{k=3}^{10}P(k,10-k)=968, not 1976; this is simply the number of subsets of A with at least 3 elements, so it’s$$2^{10}-\left(\binom{10}0+\binom{10}1+\binom{10}2\right)=1024-(1+10+45)=968\;.$$That’s merely a computational error; unfortunately, what follows it embodies an error in logic. The$100$chosen subsets of$A$produce ... 1 We want to find the image of$f(\theta)=\sin\theta - p\cos\theta = \sqrt{1+p^2}\sin(\theta-\tan^{-1}p)$. The inequality can be satisfied only if$\max f-\min f \le \sqrt2-1$in$\theta\in[0,\frac12\pi]$. Then we could choose$q$around$\frac12(\max+\min)$to fit in. Note that$f(0)=-p$and$f(\frac12\pi)=1$, and$\sqrt2-1=0.414\dots<1$. We separate ... 4 The first part is obvious. Since$\alpha,\beta, P(\alpha)$and$P(\beta)$are all integers,$\mid \beta - \alpha \mid$divides$\mid P(\beta) - P(\alpha) \mid = 2$. For the second part, notice that$(P(x))^2-1=(P(x)+1)(P(x)-1)$. So, all the roots of those two polynomials would be a root of this polynomial. There are$2d$such (not necessarily distinct) ... 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 ...

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