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1

I propose that the question has a typo and that the inequalities in the index of the summation cannot be strictly less than and need to be less than or equal to. Here is why. Consider the case where $n=2$ and $r=3$. Clearly, the left hand side is an application of binomial theorem and we have $a_1^3+3a_1^2a_2+3a_1a_2^2+a_2^3$ The right hand side is (from ...


1

The first two problems ask for Steiner Triple Systems. It is known that a Steiner Triple System exists if and only if $n \equiv 1 \text{ or } 3 \pmod 6$. For $K_7$ we have $7 \equiv 1 \pmod 6$, so one exists. An example is drawn on the Wikipedia page (with triangles are drawn as lines): (see also Fano plane). For $K_{12}$ we have $12 \equiv 0 \pmod ...


1

first of all you should find $A$, $B$ and $C$ with initial $y(-2)=41$, $y(5)=20$ and quadratic function is minimum at $x_{min}=-\frac{-B}{2A}=\frac{B}{2A}$. so you have three equations and three : $41=4A+2B+C$ $20=25A-5B+C$ $2=\frac{B}{2A}$ Solve this for $A$, $B$, $C$ and then you can find $D=y_{min}=y(x_{min})$ or ...


2

Hint: A quadratic function that has its minumum (or extremum) at $x=x_0$ is of the form $y=a(x-x_0)^2 +b$.


0

Firs off: notice the race lasts 40 minutes. Note Dexter and Prexter are only simultaneously at the edge of the pool at minute 26. Since Dexter is at the edge after an even amount of minutes and Prexter is at the edge after a multiple of $3.25$ minutes, the only even multiple of $3.25$ under $40$ is $26$. So they are only at the edge at the same time at ...


3

This is not a separate answer, but rather a (too long) comment in response to the proper answer by boywholived. Sure, there exists exactly one such value of $x_1$, but the comment by barak manos still stands: any chance you can tell us what that value of $x_1$ is? Here is a little computer program that does the job: program IMO_1985; procedure find; const ...


2

One solution is the number 7454. The solution can potentially start 29, 38, 47, 56, 65, 74, 83, 92. Since the sum of the first and second digit is the same as the sum of the first and last digit, the possible solutions look like this: 29?9, 38?8, 47?7, 56?6, 65?5, 74?4, 83?3, 92?2. But the constraint that the first digit be 3 more than the last rules out ...


3

HINT: For $\displaystyle(abcd)_{10}, a+b+c+d=20, a+b=11\iff c+d=9$ $\displaystyle a+b=a+d\iff b=d$ $\displaystyle a=d+3\implies a=b+3, (b+3)+b=11$ Hope you can take it home from here


4

Let $a,b,c$ be $5/2,5/4,1/4.$ Then there are two solutions of the system giving different values for $x(1-x)$, so that one cannot in general determine it from the equations. The two solutions are $(x,y,z)=(3/2,1/2,0)$ and $(x,y,z)=(1/2,-1/2,0).$ Then note that $x(1-x)$ is $-3/4$ for the first solution, but is $+1/4$ for the second solution. Perhaps the OP ...


6

Let $ABCD$ be some tetrahedron with incenter $O$, and let $E$ be the incenter of the tetrahedron $OBCD$. Draw a small sphere around $O$ and select $B'C'D'$ such that this sphere is inscribed in $AB'C'D'$. This is always possible, by selecting three tangent planes through $A$ that enclose the sphere, and intersecting them with a tangent plane behind the ...


9

Assume that there is some $x_1$ such that $0<x_n<x_{n+1}<1$ then $\{x_n\}$ is monotonically increasing and bounded. Hence converges by monotone convergence theorem. Let $\lim\limits_{n\rightarrow \infty}{x_n}=l$, then obtain $l^2=l \Rightarrow l=0,1$, we can readily omit $0$ and hence $x_n\rightarrow 1$. Proof for uniqueness of $x_1$. Assume ...


1

Put $g(t)=\exp(-t)f(t)$. We have: $\displaystyle\int_0^1 tf(t)dt=\int_0^1 t\exp(t) g(t)dt$ and integrating by parts: $$\int_0^1 tf(t)dt=\big [(t-1)\exp(t)g(t)\big ]_0^1 -\int_0^1 (t-1) \exp(t)g^{\prime}(t) dt=f(0)-\int_0^1 (t-1) \exp(t)g^{\prime}(t) dt$$ In the same manner: $$\int_1^2 tf(t)dt=\big [(t-1)\exp(t)g(t)\big ]_1^2 -\int_1^2 (t-1) ...


1

Hint: For $y\in \mathbb{R}$, define $$ G(t) := \int_y^t f(x)dx $$ Then, $G$ is continuous, $G(y) = 0$ and prove that $$ \lim_{t\to \infty} G(t) = +\infty $$ Now apply Intermediate Value theorem.


1

Hint: For a fixed $y$, you can show that the function $$F(t) = \int_y^t f(x)dx$$ is a continuous function which has a value $F(y)=0$ and a limit of $\displaystyle\lim_{t\to\infty} F(t) = \infty$. What you are trying to show then follows from the IVT.


0

Fix $\varepsilon>0$. Assume that there exist $x$ such that $f(x)$ is positive. Consider $h$ so small so that $f(x+2h)$ is positive Then since $f$ is continuous there exist $h>0$ such that $f(x+h) < f(x)+\varepsilon$ Then $$f(x+2h)-f(x+h)>f(x+2h)-f(x)-\varepsilon$$ So if we take the limit as h tends to zero we get $$\lim_{h\to0} ...


4

Hint: for a given $h$ one has $${f(x + h) - f(x) \over h} = {f(x + h) - f(x + h/2) \over h} + {f(x + h/2) - f(x + h/4) \over h} + ....$$ $$= {1 \over 2}{f(x + h) - f(x + h/2) \over h/2} + {1 \over 4}{f(x + h/2) - f(x + h/4) \over h/4 } + ....$$ You actually need continuity of $f(x)$ already for the above. Now take limits as $h$ goes to zero in the above ...


1

As shown by Gottfried Helms in a linked question, a solution over $[-1,1]$ is given by a function defined over $(-2,+\infty)$: $$ 2\cdot T_{\sqrt{2}}(x/2) $$ where $T_n$ is a solution to the Chebyshev differential equation $$ (1-x^2)\frac{d^2 y}{dx^2}-x\frac{dy}{dx}+ n^2 y = 0.$$ The first terms of the Taylor series in zero are: $$2 \cos\left(\frac{\pi ...


1

The condition $x^2+y^2+z^2=1$ is redundant. Let $a=\sqrt{y+z},b=\sqrt{z+x},c=\sqrt{x+y}$ we have $c\ge b \ge a \ge 0$ and $$x=\frac{b^2 + c^2 - a^2}{2},y=\frac{c^2 + a^2 - b^2}{2},z=\frac{a^2 + b^2 - c^2}{2}.$$ Replacing into the original inequality, it becomes $$\sum_{a,b,c} \frac{b^2 + c^2 - a^2}{2c} \ge \sum_{a,b,c} \frac{c^2 + a^2 - b^2}{2c},$$ which is ...


2

If we let the four-digit number be XXYY, then this number can be expressed as: $$1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) = k^2$$ (since it's a perfect square) In order for this to be true, $100X + Y$ must be the product of $11$ and a perfect square, and looks like $X0Y$. So now our question is "which product of $11$ and a perfect square looks ...


1

Let $a_j$ and $a_k$, with $j<k$, be two lowest terms of the sequence, and suppose that $k>j+2$. Without loss of generality, we can also assume that all terms $a_i$ with $j<i<k$ are higher than both $a_j$ and $a_k$. If we consider $a_{j+2}$, there are two possibilities: 1) $a_{j+2}$ is equal to the difference between the two preceding terms, i.e. ...


1

Hint: Write the sum of days in binary. Any integer $n$ from $0$ to $127$ can be uniquely written as $$n=a_0\cdot 1 + a_1\cdot 2 + a_2\cdot 4 + a_3\cdot 8 + a_5\cdot 16 + a_6\cdot 32 + a_7\cdot 64$$ where $a_i\in\{0,1\}$. In binary, $n$ is then written as $$n=a_7a_6a_5a_4a_3a_2a_1a_0.$$ Simply looking at the digits of the binary expression of $n$ then gives ...


0

As stated in the previous answers, once you put 6 marbles, 1 in each bag, the question is reduced to how many ways can you distribute 6 marbles into 6 bags, with no restrictions. The way to solve this is imagining 5 dividers and 6 circles, where the circles represent the marbles, and the amount of marbles in between the dividers represents the amount of ...


0

Your solution is correct, but the answer in the link you have provided is wrong, and here's why. If runners $P$ and $Q$ both ran for the same amount of time at the same speed, then $P$ and $Q$ would've traveled each 75 meters upon their first meeting, and because totally they traveled half the tracks distance, the total distance of the circular track is 300 ...


1

The cool procedure is a linear transformation on the vector space $\mathbb F_2^n$, given by a matrix like this $$ \begin{pmatrix} 1&0&0&\cdots&1&1\\ 1&1&0&\cdots&1&1\\ 0&1&1&\cdots&1&1\\ 0&1&1&\cdots&1&1\\ 0&0&1&\cdots&1&1\\ ...


1

Dividing these into two separate runs on a straight track: When the distance between them is $d$, then $P$ has to run $75$ meters before they meet. When the distance between them is $2d$ then naturally, $P$ has to run twice as far before they meet, so he runs $150$ meters. $Q$ runs $100$ meters, and their sum is the total circumference $250$. The solution ...


0

We can solve it by generating function. Since each bag can have a minimum of one marble and maximum of twelve, and there are six bags, the gf is: \begin{align*} G(x) &= \left(x^{1}+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}+x^{9}+x^{10}+x^{11}+x^{12}\right)^6 \end{align*} and we need $[x^{12}]$, which can be extracted easily: \begin{align*} ...


2

Hint: imagine the twelve marbles in a line. Place 5 vertical lines in the 11 interior gaps between pairs of adjacent marbles, no more than one line in any gap.


0

Hint: First put a marble in each bag. Then the question is reduced to how many ways are there of putting $6$ marbles into $6$ bags with no restrictions.


2

You already know that 103|n. This is also sufficient. First, suppose the minimum value of S1 (obtained by taking $1, 2, ... floor(n^2/2)$ in black squares is x. Also, call sum of all numbers from 1 to $n^2$ as y. In other words, maximum value of S1 is y - x. I will show that S1 can take any value from x to y - x. Here is how: Suppose the sequence of ...


1

For number theory I would suggest "An introduction to the theory of numbers" by Niven, Zuckerman and Montgomery. It has a very good theory and problems. Everything is nicely explained with elegant proofs. An easier book would be "Elementary Number Theory" by Burton. However, it doesn't have any difficult problems. However, for olympiad preparation you ...


2

A sketch showing Alice can win. Alice makes a directed chain, extending it by one step (or two if there is a stray edge she can use) at each turn. When the chain has $n$ steps, Bob needs to have directed a total of $\frac 12(n)(n-1)$ edges or Alice can win the next time. We therefore need $\frac 12(n)(n-1) \gt 1000n$, which happens at $n=2001$, so Alice ...


6

Fix $k$ arbitrarily and denote the sum $k! + (2k)! + \cdots + (nk)!$ by $I_n$. It is enough to show that the set of primes $p$ that divide at least one of the $I_n$ is infinite—here is an argument that this is so. Suppose otherwise that there are only finitely many primes that divide any of the $I_n$. For a given prime $p$ and integer $a$, let $e_p(a)$ be ...


1

Suppose such a positive sequence $\{a_n\}_{n \in \mathbb{N}}$ exists. Then by repeated application of the property $a_n < a_{n+1}+a_{n^2}$ on each term appearing starting from $a_2$, gives: $a_2 < a_3 + a_4 < a_4 + a_5 + a_9 +a_{16} < \ldots $ and so on. Let, $\{S_k\}_{k\ge 1}$ be the resulting sequence of subscripts formed by such repeated ...


2

(the question is uncleared, so I assume that the outer edges do not count, otherwise it would be much harder) There are a total of $7\times 8\times 2=112$ edges. Each way of tiling the dominoes, since there are exactly $32$ dominoes used, each failed to involve $1$ edge, there are a total of $32$ edge not involved in each way of tiling. So each way of ...


1

This is not a full solution, but this may help: The name of the structure we get at the end is tournament. A transitive tournament is one where $a\to b$ and $b\to c$ implies $a\to c$, thus they give us a total ordering of the vertices. Tournaments are only acyclic if they are transitive, in particular a tournament $T$ has no $3$-cycle if and only if it is ...


2

Proof. Three circumcircles of $\triangle AFE$,$\triangle BDF$,$\triangle CED$ concur at $M$ (Miquel point) $\angle BB'M+\angle AA'M=\angle BFM+\angle AFM=180^{\circ}$ and then $\square IA'MB'$ is concyclic. Likewise, $\square IB'MC'$ is concyclic and then $I$,$A'$,$B'$,$C'$,$M$ are concyclic. Remark. (1) $I$ need not be incenter. (2) $BF+BD=AC$, ...


1

Any function $f_{base}$ satisfying: (1) Is defined on $[0,r)$ for some $0<r<2$. (2) Is continuous and strictly increasing. (3) $f_{base}(0)=r$ and $\lim\limits_{x\rightarrow r^{-}}f_{base}(x)=2$. can be extended to a function $f$ on nonnegative real satisfying the above functional equation and is strictly increasing. This can be proved by ...


0

You would there would be a direct binary way to calculate this, but currently there is none discovered. The fastest way is somewhere around $O(\log^2 n)$ where a binary way may look more like $O(\log n)$. This comparison is like the difference between how fast computers can add numbers, and how fast computers can multiply (multiplying use normal ...


0

In Computability Theory, a very important class of fucntions is the class of Primitive recursive function. This class is defined starting form some initial (very "simple") functions and using few methods to build up new functions from already existing ones. One of the methods allowed is that of composition [not "combination"] : Given $f$, a $k$-ary ...


2

I tried a proof by contradiction without succes. Here is my approach nonetheless. I will use the same notation Aaron has used. Write $J_{k,n} = \frac{k!}{k!} + \dots + \frac{nk!}{k!}$ and $I_{k,n} = k! + \dots + (nk)!$. Suppose the statement is not true. Then a $k \in \mathbb{N}_{>0}$ exists with the following property: for every $n \in ...


1

Lemma. Given $\overline{XP} \cong \overline{XQ}$ and $\overline{XY}\cong\overline{ZQ}$ (and $\overline{XZ} \cong\overline{YP}$) as in the diagram, let $K$ be isosceles $\triangle XPQ$'s circumcenter (which necessarily lies on the bisector of $\angle X$). Then $X$, $Y$, $Z$, $K$ are concyclic. Proof of Lemma. Evidently, $\triangle XYK \cong \triangle ...


2

Here are a few observations and thoughts. Up to $k=30$ one needs $n=3$ for $k=7,11,12,15,16,18,19,20,21,23$ but $n=2$ is sufficient the other $20$ times. Let $I_{k,n}=I=k!+(2k)!+(3k)!+\cdots+(nk)!=k!J_{k,n}$ where $J_{k,n}=1+\frac{2k!}{k!}+\frac{3k!}{k!}+\cdots+\frac{nk!}{k!}$. A prime less than $2k$ divides all the summands of $J_{k,n}$ except $1$ and ...


2

Let's do this problem in two different ways. First, this question screams pigeon-hole principle to me, so we'll start with that. There are 16000 committees, each with 80 people. So there are 1280000 different committee members across all the committees. Since there are only 1600 people, and 1280000 memberships, at least one committee member (call him Bob) ...


1

For completeness, I'll post my solution here. By looking at the power of $J$ with respect to $\Gamma$, we see that: $|JD|^2 = |JP|\cdot|JA|$. On the other hand, let $\angle BAD = \alpha$ and $\angle PAD = \beta$. From this we can see that $\angle BAJ = \alpha - \beta$. It is also true that $\angle DAM = \alpha$ (since $AD$ is the angle bisector of $\angle ...


3

P is not a nice point, so let's try and avoid it. Hint: show that $JB^2=JP \times JA = JD^2$. Hint: Show that 2 triangles are similar to prove the first. Hint: define N in a similar manner to M


2

Hint $\ $ Specialize $\ c,a,b = 4,3,2\ $ below Lemma $\ \ 5\nmid n\pm 1,\,\ ab\mid n,\,\ \color{#c00}{b^2\!\mid 5c\!-\!4}\ \Rightarrow\ {\rm lcm}(5,a^2,b^4)\mid n^2(n^2\!+5c\!-\!4)$ $\begin{eqnarray}{\bf Proof}\quad\! &&a\mid n\,\Rightarrow\ a^2\!\mid n^2,\ \ \ {\rm and}\ \ \ \ b\mid n\,\Rightarrow\, b^2\mid n^2,\ n^2\!+\color{#c00}{5c\!-\!4}\\\ ...


0

$n=6k$ implies $n\equiv k \bmod 5$. If $k\equiv 0 \bmod 5$, then you're done. Othwerwise, $n-1$ and $n+1$ prime implies $k\equiv n\equiv \pm 2 \bmod 5$ and so $9k^2+4\equiv -k^2+4 \equiv 0 \bmod 5$.


3

HINT (since this is worth solving yourself): You need to use that $n-1, n+1$ are prime. If either of these are equal to $5$, you have the excluded values $n=6$ or $n=4$. Otherwise, note that the product of any five successive integers is divisible by $5$. You are given $n+1$ and $n-1$ - can you see which five successive integers you might use? Further to ...


1

Okay, after thinking about this for a while, I don't have a final answer, but I think I've made some progress. I'm not positive that we can extend this to $k!$, but I'll walk through it anyway. It starts by recognizing that we can rewrite $I$ as $$I=k![1+(k+1)(k+2)\cdots (2k)+\cdots+(k+1)(k+2)\cdots (nk)]$$ Now, $k!$ has some prime factorization: $$k! = ...


0

Hint $ $ If $\,n\,$ is odd, and $\ a^{n-1}\! + n b = (n-1)^{2j+1} $ then every prime $\,p\,$ dividing $\,n\,$ is $\,\equiv 1\pmod 4 $ since $\ n\!-\!1 = 2k,\,$ so $\, {\rm mod}\ p\!:\ (a^k)^2 \equiv\, -1,\ $ so by Euler or reciprocity, $ $ we infer $\ p\equiv 1\pmod 4$



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