# Tag Info

1

Let's consider a large number $N$ and the number of values in $B$ less than $N$. Suppose $k>10$, otherwise $b_n>n^{10}\geq n^k$. We can fit at most $\log_{a_i}N$ powers of each prime. We can bound the total number of values in $B$ to less than $(\log_2 N)^{|A|}$. Let $n = \left\lceil(\log_2 N)^{|A|}\right\rceil$. We have shown $b_n>N$. It suffices ...

0

scubasteve623 is correct in that there will most likely be adjustments in the future because the calendar is not perfect, but the answer you're looking for is Tuesday, Thursday and Sunday. Assuming those adjustments don't happen anytime soon... 1-1-2100 will be Friday 1-1-2200 will be Wednesday 1-1-2300 will be Monday 1-1-2400 will be Saturday ...and the ...

0

Consider the underlying directed graph, where $X \to Y$ if and only if $X$ beats $Y$. The only way that there is no directed triangle is if there is a total ordering, in particular, if there is a team which wins against all other teams. Now I show that every team must lose at least once. For each element $X$, let $v(x)$ be the number of victories that it ...

1

Like what Jesko said: There are ${9 \choose 3}$ ways to pick the 3 spots for the 3 white marbles and ${6 \choose 6}$ ways to pick the 6 remaining spots for the black marbles = ${9 \choose 3}$ * ${6 \choose 6}$ = 84 * 1 = 84 ways

0

There are $\binom 93$ ways to choose $3$ spots on the grid for the white marbles. The black ones automatically go to the other spots.

0

Let $x$ denote positive number of guesses. For positive score: $Mx-m(N-x) >0 \implies x>\frac{mN}{M+m}$ Probability that his guess is correct is $\frac{1}{c}$ Then, the required probability for scoring positive marks is $(\frac{1}{c})^{\lfloor\frac{mN}{M+m}\rfloor}$

0

If $A= a^2 +b^2$, where $a,b \in \mathbb{N}$. Then this is easy, take $x = a$. Of course $0 \in \mathbb{N}$ as we are doing algebra / number theory stuff. Suppose $A \neq a^2 + b^2$ for any $a,b \in \mathbb{N}$. Suppose for this $A$ $\exists y,q \in \mathbb{N}$ such that $A -y^2 = q^2$ So $A = q^2 + y^2$. This is a contradiction.

4

You just have to find the area of the curvilinear quadrilateral having its vertices in the four red points.

1

Solution: Let $S$ be the set of numbers not divisible by 23. I claim that $|S| \leq 22$; indeed in each residue class mod 23, there can be at most one number in $S$, else we have:$lcm(23a+r,23b+r) \geq 529ab/gcd(b-a) \geq 529$. The number of numbers which are divisible by 23 and at most 400 is at most 18. Thus the total number is at most 40.

-1

It's far easier than step 1. Step 2 : same definition than step 1, but with: $\forall x∈[0,1), \space f(x)=x$ instead of $\forall x∈(0,1), \space f(x)=x$. Step 3 : $\forall x, \space f(x)=1-x.$

2

All prime divisors of $\frac{(pr)^{p}-1}{pr-1}$ for any positive integer $r$ are of the form $pk+1$. (and so all positive divisors are of the form $pk+1$, but it's not needed). Proof: Define $\text{ord}_n(a)$ to be the least positive integer $m$ such that $a^m\equiv 1\pmod{n}$. First a lemma: if $x^k\equiv 1\pmod{n}$, then $\text{ord}_n(x)\mid k$. To ...

1

A constant solution $f(x)={1\over2}$ will do as well. Also, any other solution coincides with one of these at integer $x$. Let's see if we can extend that to rationals... Upd. OK, I got it. It's clumsy and boring, but anyway. Let $f(1)=a$; now we will repeatedly apply the formula for $f(x+1)$ to obtain: $$\begin{array}{l|ccccccc} n& 1& 2& ... 0 Let \mathcal{D}_{1999} consist of the single 2^{1999} \times 2^{1999}-square. For 1 \le n \le 1999, define the family \mathcal{D}_{n-1} recursively to contain the squares obtained from the squares of \mathcal{D}_n by splitting them into four equal size squares. For each 0 \le k \le 1999 there is a unique dyadic cube Q_k in \mathcal{D}_k that ... 4 Here is a slightly different point of view: First, we note that f(x)=f(0)+\int_0^x f'(t)\,dt and hence (here we also use that \int_0^1f(x)\,dx=0)$$ \int_0^1 (f(x))^2\,dx=\int_0^1 f(x)\biggl[f(0)+\int_0^x f'(t)\,dt\biggr]\,dx=\int_0^1\int_0^x f(x)f'(t)\,dt\,dx.\tag{1} $$Changing order of integration in the right-hand side of (1), using the fact that ... 5 I am assuming these are real-valued functions. Define$$g(x) = \int_0^x f(t)\,dt,$$and note that$$g(0) = g(1) = 0.$$Integrating by parts,$$\int_0^1 f(x)^2\,dx = g(x)f(x)\big|_0^1 - \int_0^1 g(x)f'(x)\,dx = -\int_0^1 g(x)f'(x)\,dx.$$A straightforward estimate then shows that$$\int_0^1 f(x)^2\,dx \le \sup_{0 \le x \le 1} \left|g(x)\right| \int_0^1 ...

0

A sketch of a possible proof: for any $a$ we may find some $d$ such that $f(a+d)>f(a)$. Moreover, every $d\geq d_0$ does that. If some $d\geq d_0$ is such that $f(a+2d)>f(a+d)$, we are done. Otherwise we may assume that for any $d\geq d_0$ we have $f(a+2d)<f(a+d)$. However, that implies: $$\ldots < f(a+8d)<f(a+4d)<f(a+d)$$ so ...

2

Let $D$ be the intersection of line $XL$ with $AB$, $F$ the intersection of $XM$ and $BC$, and $E$ the midpoint of $BC$ (so $A$, $G$, and $E$ are colinear). Since $XL\parallel AC$, $\triangle XLE$ is similar to $\triangle ACE$, and we have $$AE:EX=EC:EL=AC:XL=3:1.\tag{1}$$ Similarly, $\triangle XEF$ is similar to $\triangle ABE$ and ...

1

$ABC$ and its medial triangle are similar triangles: in particular, they have the same centroid $G$, and the medial triangle is the image of $ABC$ with respect to a dilation with centre $G$ and ratio $-\frac{1}{2}$. Since homothetic transformations preserve parallelism, by assuming that $AD,BE,CF$ concur in $P$ we have that the parallel lines to $AD,BE,CF$ ...

3

A simple way to see if a double radical $\sqrt{a\pm \sqrt{b}}$ can be denested is to check if $a^2-b$ is a perfect square. In this case we have: $$\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}$$ (you can easely verify this identity). In this case $a^2-b=35$ is not a perfect square. Note that if ...

1

Draw a picture. Let $M$, $N$, and $P$ be the midpoints of $BC$, $CA$, and $AB$ respectively. Then $\triangle MNP$ is similar to $\triangle ABC$ with parallel pair of edges. You can compare the angles $\angle BAD$, $\angle DAC$ with the corresponding angles at $M$. Similarly for the other angles at $B$ and $C$.

6

Hint: Multiplying the first three expressions we get $$abc+\left(a+\dfrac{1}{b}\right)+\left(b+\dfrac{1}{c}\right)+\left(c+\dfrac{1}{a}\right)+\dfrac{1}{abc}=\dfrac{28}{3}.$$

3

Suppose otherwise; then we may write$$P(x) = (x-r)^\ell H(x)$$for some $r\in\mathbb{C}$, integer $\ell\ge2$, and nonconstant $H$ with $H(r)\ne0$. The condition now translates to$$(x-r)^\ell H(x) = Q(x)((x-r)^\ell H''(x) + 2\ell(x-r)^{\ell-1}H'(x) + \ell(\ell-1)(x-r)^{\ell-2}H(x)),$$ or $$(x-r)^2 H(x) = Q(x)((x-r)^2 H''(x) + ... 1 If your domain S is countably infinite, you could use infinite summation to find the "average." No integral is needed. However, there is no fully additive uniform distribution over a countably infinite set. So any "average" would need to be done with a weighting of the points. If no weight is given, implied, or obvious, we could say that there is no such ... 1 using the information in the equation$$ \sum_{i=1}^3 r_i^3 = -\sum_{i=1}^3 (r_i^2 - 2r_i + 1) \\ = -5 -\sum_{i=1}^3 r_i^2 $$also, from the well-known expressions giving the elementary symmetric functions of the roots in terms of the coefficients,$$ \sum_{i=1}^3 r_i^2 = (\sum_{i=1}^3 r_i)^2 - 2(r_1r_2+r_2r_3+r_3r_1) \\ = 5 $$so the sum of cubes of the ... 1 Hint: use Newton's formulas and Vieta's formulas https://brilliant.org/wiki/newtons-identities/ 3 You know, a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca) a^3+b^3+c^3=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc Now I hope you know the relation between roots. 2 Suppose a,b,c\in\mathbb{C} are the roots to your equation. There is this algebraic identity you should know:$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$Do some manipulations to get this form:$$a^3+b^3+c^3=3abc+(a+b+c)\left((a+b+c)^2-3(ab+bc+ca)\right)$$Vieta's formulas are formulas that relate the coefficients of a polynomial to sums and ... 1 Hint: the perpendicular to AB through C\not\in AB is the locus of points P for which:$$ PA^2-PB^2 = CA^2-CB^2.$$0 The first point is that the question is not quite as you quoted it. You have a couple of obvious typos in the last two ranges. More important, the original question does not have the absolute value of h(x), but just h(x). I assume you want the original question solving. The first thing to try is linear f,g,h. The two points where the derivative is ... 3 Consider the equation in module 4. Every odd number(n) is always:$$n\equiv \pm 1\pmod 4$$therefore this expression is:$$p^{q+1} + q^{p+1}\equiv \pm1^{q+1} \pm 1^{p+1}\equiv 1+1\equiv 2 \pmod 4$$But the quadratic residue mod 4 are 0 and 1. Therefore contradiction. From here p or q or both must be equal 2. You can control easily that there ... 0 Hint: Suppose p is odd and it is a perfect square. Then$$p^{q+1}+q^{p+1}=a^2,p^{q+1}=a^2-q^{p+1}p^{q+1}= \left(a-q^{\frac{p+1}{2}}\right)\left(a+q^{\frac{p+1}{2}}\right)$$So then both \left(a-q^{\frac{p+1}{2}}\right) and \left(a+q^{\frac{p+1}{2}}\right) must be powers of p. 3 Bob can win using the strategy you hint at. After Alice makes the first move, he will be able to move in at least three corners as you describe in the next to last paragraph. He should move in one on his first move. After that he refuses to play in that corner until he has no other choice. Anywhere Alice can move, Bob can move too. If he comes to a point ... 1 The function is basically a combination of two shrinked f(x), so nothing at the two ends or interior of each half domain is worrisome. You need and only need to prove:$$\lim_{x\rightarrow0.5^-}g(x)=\lim_{x\rightarrow0.5^+}g(x)\lim_{x\rightarrow0.5^-}g'(x)=\lim_{x\rightarrow0.5^+}g'(x)$$Which means f(0)=f(1) and f'(0)=f'(1) 2 You might shorten the proof noting that, as x^2=\lvert x\rvert^2, one may as well suppose x\ge 0, and extend the results to the case  x\le 0  by symmetry. For the case x=0, you can incorporate it to the first case, i.e. consider the case x\ge 0 instead of x>0. You get the solution: ... 1 It looks like for the second case x < 0, the 3rd line in your equivalences has a typo, but otherwise it looks fine. Since you solved for A explicitly, you can determine whether it's an interval and whether it's bounded. An interval set must contain all intermediate points between x and y if x and y are in your set, it's easy to see 0 is not ... 0 That A is open does not mean it is an interval -- you set is a union of intervals, but not an interval. A is certainly bounded, so (2) is true and others are false. 0 Your arguments are good. Here's a different approach The derivative is, for x\ne0,$$ f'(x)=\frac{x\dfrac{1}{1+x^2}-\arctan x}{x^2}= \frac{1}{x(1+x^2)}-\frac{\arctan x}{x^2} $$In particular this shows that f has derivatives of all order in every point x\ne0. For deciding what happens at 0, we can consider the Taylor series of \arctan x, that ... 6 By the fundamental theorem of calculus, since f(0)=0,$$ f(t)=\int_0^tf'(s)\,ds. $$Hence, by Cauchy-Scwarz inequality, and domain monotonicity of integrals of non-negative functions,$$ \begin{aligned} \int_0^1 (f(t))^2\,dt&=\int_0^1\biggl[\int_0^t 1\cdot f'(s)\,ds\biggr]^2\,dt\\ &\leq \int_0^1\biggl[\int_0^t1^2\,ds\int_0^t (f'(s))^2\,ds\biggr]\\ ...

0

$f(x)=x$ allows to exclude 1] and 4]. $f(x)=\ln(x+1)$ allows to exclude 2]. So the right answer must be 3].

2

We will use the following results $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\text{ (via LHR)}$$ and $$\lim_{x \to 0}\frac{\arcsin x}{x} = \lim_{t \to 0}\frac{t}{\sin t} = 1\text{ (by putting }t = \arcsin x)$$ We can then proceed as follows \begin{align} L &= \lim_{x \to 0}\frac{\log(1 + \sin(x^{2})) - ...

4

You are correct; it is neither. Observe that $A_1 = 8, A_2 = 263, A_3 = 6568$. $A_2/A_1 \not= A_3/A_2$, so it is not geometric. $A_2-A_1 \not= A_3-A_2$, so it is not arithmetic.

1

You must use power series development at order $4$: $\sin x= x+o(x^2)$, hence $\sin x^2=x^2+o(x4)$ $\ln(1+u)=u-\dfrac{u^2}2+o(u^2)$, hence $$\ln(1+\sin x^2)=\ln\bigl(1+ x^2+o(x^4)\bigr)=x^2-\frac{x^4}2+o(x^4)$$ $\arcsin x=x+\dfrac12\dfrac{x^3}3+o(x^4)$, hence $(\arcsin x)^2=x^2+\dfrac{x^4}3+o(x^4)$ Grouping all the results we get: \frac{\ln ... 0 \begin{align} \lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2} &= \lim _{x\to \:0}\frac{\sin \left(x^2\right)-\frac{\sin^2(x^2)}{2} + O(x^6)-x^2}{\left(\arcsin \:x\right)^2-x^2} \end{align} (The above is true as \ln(1+x) \approx x-x^2/2 +O(x^3) when x is small) \begin{align} \lim _{x\to ... 2 The substitution x=e^{-u} gives\int_0^1{x\ln x\over(x^2+1)^2}dx=-{1\over4}\int_0^\infty u\,\text{sech}^2u\,du$$Integration by parts tells us the indefinite integral is$$\int u\,\text{sech}^2u\,du=u\tanh u-\int\tanh u\,du=u\tanh u-\ln\cosh u+C$$so, on noting that 0\tanh0-\ln\cosh0=0, it remains to evaluate the limit$$\begin{align} ...

0

From question we can form the equation $$z = 1/2(x+y-2)(x+y-1)$$ let $x+y = a$ $$2z = (a - 2)(a - 1)$$ $$2z = a^2 -3a + 2$$ $$a^2 -3a -2(z - 1)=0$$ apply $b^2 - 4ac$; $9+8(z-1)$ Now the discriminant have to be greater than or equal to $0$. Let $z = 1$ the smallest positive integer. Since $9>0$ then any positive integer value for $z$ will satisfy and ...

5

Using $$\sum_{k\geq0}\left(-1\right)^{k}x^{2k}=\frac{1}{1+x^{2}},\,\left|x\right|<1$$ we have, taking the derivative, $$\sum_{k\geq1}\left(-1\right)^{k}kx^{2k-1}=-\frac{x}{\left(1+x^{2}\right)^{2}}$$ hence ...

1

It looks fine to me. We may notice that it is possible to save a step by taking $\frac{1}{2}\left(1-\frac{1}{(x^2+1)}\right)$ as a primitive for $\frac{x}{(x^2+1)^2}$. In such a way, the original integral equals: ...

1

\begin{align} \lim_{n\to\infty}n\left(\tan\left(\frac{\pi}{3}+\frac 1n\right)-\tan\frac{\pi}{3} \right)&=\lim_{n\to\infty}n\left(\tan(\frac{\pi}{3}+\frac 1n-\frac{\pi}{3})\Big(1+\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big) \right)\\ &=\lim_{n\to\infty}\frac{n\sin\frac 1n}{\cos \frac 1n}\times \lim_{n\to\infty}\Big(1+\tan(\frac{\pi}{3}+\frac ...

2

$$\lim_{n\to \infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n}\right)-\sqrt 3\right)$$ $$=\lim_{n\to \infty}\frac{\tan\left(\frac{\pi}{3}+\frac{1}{n}\right)-\sqrt 3}{\frac{1}{n}}$$ $$=\lim_{n\to \infty}\frac{\frac{\tan\frac{\pi}{3}+\tan \frac{1}{n}}{1-\tan \frac{\pi}{3}\tan \frac{1}{n}}-\sqrt 3}{\frac{1}{n}}$$ $$=\lim_{n\to \infty}\frac{\frac{\sqrt 3+\tan ... 3$$\begin{align}\lim_{n\to\infty}n\left(\tan\left(\frac{\pi}{3}+\frac 1n\right)-\sqrt 3\right)&=\lim_{n\to\infty}n\left(\frac{\sqrt 3+\tan\frac 1n}{1-\sqrt 3\tan\frac 1n}-\sqrt 3\right)\\&=\lim_{n\to\infty}\frac{4n\tan\frac 1n}{1-\sqrt 3\tan\frac 1n}\\&=\lim_{n\to\infty}\frac{4\tan\frac 1n/(1/n)}{1-\sqrt 3\tan\frac 1n}\\&=\frac{4\cdot ...

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