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16

I suspect any adequate answer to this question is going to be extremely computation-heavy. Here's one. First we claim that $$ \sqrt{3} > \frac{3691}{2131} \tag{1} $$ This fraction is not pulled out of a hat; rather, it is found by taking the continued fraction $\sqrt{3} = [1; 1, 2, 1, 2, 1, 2, \ldots]$ and carrying it out a while. Anyway, the above is ...


10

Certainly $p\ne q$ (as we cannot have $pq^2-19=0$). Consider the equation modulo $p$ and $q$, using Fermat's little theorem: $$ -q\equiv -19\pmod p$$ $$ p\equiv -19\pmod q$$ Hence $$-19\equiv p-q\pmod{pq}.$$ Let $p-q=kpq-19$ with $k\in\mathbb Z$. With $p=2$ we arrive at $(2k+1)q=21$, so $q=3$ or $q=7$. This gives us two candidate solutions ...


9

Here is another mythical answer: We want to prove $(\log_2 \log_2 10)^2 < 3$. One way is to use the procedure outlined here to compute $\log_2$. The only mild complication is knowing when to stop. Since $\log_2$ is non-decreasing, it suffices to find $x\ge\log_2 10 $ and $y \ge \log_2 x$ such that $y^2 <3$. The procedure is straightforward, I am ...


7

The product is equal to $$\frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} $$ which behaves as $\sqrt{\pi n}$ as $n \to \infty$. As for the integral, rewrite as $$\int_{-1}^0 dx \, 2^{-x} \cos^{2 n}{x} + \sum_{k=0}^{\infty} 2^{-\pi k} \int_0^{\pi} dx \, 2^{-x} \cos^{2 n}{x}$$ Rewrite in a form useful for Watson's Lemma: $$\int_{-1}^0 dx \, 2^{-x} e^{2 n ...


6

Let $$\sum_{i=1}^{5} \frac{a_i}{x+i} = \frac{P(x)}{Q(x)}$$ where $P(x)$ is a polynomial of degree $4$, and $Q$ of degree $5$. The coefficent of $x^4$ in $P(x)$ is $\sum a_i$ and $Q(x) = \prod_{k=1}^{5} (x+k)$ We get that $$x P(x) - Q(x) = 0$$ as $0 = \dfrac{P(x)}{Q(x)} - \dfrac{1}{x} = \dfrac{x P(x) - Q(x)}{xQ(x)}$ for $x = 2^2, 3^2, \dots, 6^2$. for $x ...


6

Solution 1: Only use Cauchy-Schwarz inequality : $$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$ let $$a_{1}=x^{\frac{3}{2}},a_{2}=y^{\frac{1}{2}},a_{3}=1$$ $$b_{1}=x^{\frac{1}{2}},b_{2}=y^{\frac{3}{2}},b_{3}=1$$ so we have $$(x^3+y+1)(x+y^3+1)\ge (x^2+y^2+1)^2$$ so $$\sqrt{(x^3+y+1)(y^3+x+1)}\ge x^2+y^2+1$$ ...


6

Since,$$\frac{x^5-x^2}{x^5+y^2+z^2} - \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{x^2(x^3-1)^2(y^2+z^2)}{x^3(x^2+y^2+z^2)(x^5+y^2+z^2)} \ge 0$$ Hence, it suffices to prove $\displaystyle \sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} \ge 0$ $$\sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{\sum\limits_{cyc} \left(x^2 - ...


6

The absolute least value you can get is a rectangle topped by a half circle (the circle has the best area to arc length ratio of any shape) with a total arc length of $2 \big(1 - \frac{\pi}{8}\big) + \frac{\pi}{2} \approx 2.78539$. If you use Fourier approximation, you can come arbitrarily close to this limit. (I assume the fun of this challenge is to find ...


5

First of all, welcome to Math Stackexchange! There is no such thing as a single vector representation over all of these four points. Mathematically, a vector in the Euclidean $\mathbb{R}^n$ is a tuple of $n$ numbers — for example $$\begin{pmatrix}a\\b\end{pmatrix},\quad a,b\in\mathbb{R} $$ If you consider the 2-dimensional case, just as in your question. ...


5

In total, there are $8 \times 8 = 64$ possible combinations of two moves. In $8$ of these combinations we end up at the starting square. There are $8$ fields that can be reached using a unique combination (twice the exact same move). We can divide the other $64-16=48$ combinations in pairs with the same destination, by changing the order of the moves. ...


4

You should ignore the "divisible by $4$", since that is covered by "divisible by $2$". So reword the question as How many positive integers less than $2013$ are divisible by none of $2,3,5$? This is close to a standard combinatorics problem. Find how many numbers are divisible by $2$ (namely $\left\lfloor\frac{2012}2\right\rfloor$), how many by $3$, ...


4

First, note the $4$ is redundant. Then, a number $n$ is divisble by none of $2,3,5$ if and only if $\gcd (n, 30)= 1$. So, $n$ is in a residue class prime to $30$. There are $\phi(30)=8$ of those. So among any conscutive $30$ numbers you have $8$. This gives you the count from $1$ to $2010$. Then check the rest by hand.


4

Note that any two distinct elements $5^{p_1} 13^{q_1} 31^{r_1}$ and $5^{p_2} 13^{q_2} 31^{r_2}$ are incomparable (neither divides the other) if $p_1 + q_1 + r_1 = p_2 + q_2 + r_2$. So a mutually incomparable set is given by $$ S_n = \left\{5^p 13^q 31^r \;\big\vert\; p+q+r=n;\; 0\le p,q,r \le 200\right\} $$ for any $n$. The size of this set is maximal when ...


4

Well you've got 9 one digit numbers, then 90 two digit numbers and so on. You don't need to use modular arithmetic. Can you take it from here?


4

I think that you are over-thinking it. Since the balls are identical, there are only $3$ arrangements: $$ \dots WWRRRR \dots\quad \dots WRWRRR \dots\quad \dots WRRWRR \dots $$ where $\dots$ means that the circle wraps around.


3

After doing what he did he has three ropes: $1,2,3$. Suppose he puts them on the table, He begins tying rope $1$, the probability he ties it with another rope is $\frac{4}{5}$ since there are $5$ other rope ends and only $1$ is the same rope. If it ties with another rope there are now two ropes, proceed to tie the rope that is not made up of two ropes, the ...


3

As $(2^{2013},10)=2$ let us find $2^{2013-1}\pmod{\dfrac{10}2}$ Now $2^2\equiv-1\pmod5\implies2^{2012}=(2^2)^{1006}\equiv(-1)^{1006}\equiv1\pmod5$ Now as $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c}$ $\implies2^{2012}\cdot2\equiv1\cdot2\pmod{5\cdot2}$


3

Yes because if $n$ is the number of digits, then $$9^{2013}+9^{2013}+\ldots+9^{2013}=n\cdot9^{2013}<10^n$$ for sufficiently large $n$, above that point the number of digits won't increase anymore. Can you fill in the details?


3

One of my favorite "gateway problems" is from the 2001 Bay Area Mathematics Olympiad: Let $a_n$ be the number of permutations $\tau$ on $n$ letters such that $\tau \circ \tau \circ \tau$ is the identity. Prove that $3^{334}$ divides $a_{2001}$. This is the question that originally got me interested in $p$-adic analysis. The "high level" approach is ...


3

You're exactly right. If you add $f(p)$ for all permutations then on the one hand you should get something that is not a multiple of $n!$ of $n!$ since $1+2+3\dots +n!=\frac{n!(n!+1)}{2}$ and $n!+1$ is odd. So you are short by a multiple of $2$. Hence what you have is congruent to $n!/2$ On the other hand each $a_i$ appears $(n-1)!$ times, hence you want ...


3

Let's add the excircle of $\triangle ADE$ to the picture. $X, Y$ and $G$ are the tangent points. $AX$ is equal to semiperimeter of $\triangle ADE$ so $AX = AY = 2$. Moreover, $AB = AC = 1$ so $B$ and $C$ are the middle points of $AX$ and $AY$ respectively. Hence $BC$ is radical axis of that excircle and point $A$. So $FA = FG$. Finally, your sum becomes ...


3

Let $f(x)=\frac{x^x}{\left ( \lfloor x \rfloor \right )^{\lfloor x \rfloor}}$. If $\lim_{x \to \infty} f(x)$ exists, it must be $1$, because we can go to $\infty$ along the sequence $x_n = n$ where $f(x_n)=1$. On each interval $[n,n+1)$, the furthest $f$ gets from $1$ will occur near $n+1$. More precisely, we can make $f$ be arbitrarily close to but less ...


3

Your argument that $23$ is optimal is not convincing. In particular, at the end a certain number of wolves survives, not unicorns. Denote the momentaneous state by $(w,u,s)$. There are three kinds of moves, namely $$\eqalign{ T_1:\quad &(w,u,s)\to (w+1,u-1,s-1),\cr T_2:\quad &(w,u,s)\to (w-1,u+1,s-1),\cr T_3:\quad &(w,u,s)\to ...


3

There is nothing magical about a calculator: any computation performed on a calculator can be done by hand (and perhaps with extra rigor). (*) Most special functions have series approximations that are known to converge particularly quickly; but Taylor's theorem with remainder can be applied to almost all important special functions even without knowing ...


3

since $\sqrt{x} \ge 0$ so is $\sqrt{(B+3)^2}$, minimum at $B=-3$, similarly for the first term we get $A=B=-3$ For the subsequent part in $C$ use completing the square to arive at $(C-2)^2$ we know, $x^2 \ge 0$ so $C=2$ Thus, $A+B+C=-3+-3+2=-4$


3

if $a=1$, for the rest $9$ places, $1$ can be taken anywhere, and also the rest $8$ places must be $0$ or the sum would exceed $2$. Also if $a=2$, then the rest $9$ places must be $0$, or sum would exceed $2$. Hence there are $9+1=10$ such numbers.


3

The idea is that this polynomial has some expanded form $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$$ Now when we plug in $1$, we get $$f(1)=a_n+a_{n-1}+\dots+a_1+a_0$$ which is exactly what we want. Plugging in $0$ on the other hand gets us $$f(0)=a_0$$ so this just gives us the constant term.


3

For any polynomial with integer coefficients and any integer $x$, $f(x+2) \equiv f(x) \mod 2$.


2

$x = 0$ gives you the constant coefficient of the expansion. For example, the expansion of $(1+x)\cdot(2+x^2)$ is $2+2x+x^2+x^3$. The coefficients of this expansion are $(2, 2, 1, 1)$ : they are the coefficients of the polynomial. If you take $x=1$, you get $2+2+1+1$ which is the sum of the coefficients. $x=0$ would give you $2$, which is only the ...


2

We have a diophantine equation involving prime numbers. Therefore it makes sense to try to solve the problem with the help of techniques and theorems from this particular field of mathematics. I refer to the answer posted by Hagen von Eitzen. However, we can also see the problem as a fairly simple math puzzle, and solve it with common sense and elementary ...



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