# Tag Info

7

A simple draw : $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$

5

Let the first number in the sequence of positive integers be $m+1$ and the last number be $n$. Then $$\dfrac{n(n+1)}{2}-\dfrac{m(m+1)}{2}=2014$$ $$n^2+n-m^2-m=4028$$ $$(n-m)(n+m)+(n-m)=4028$$ $$(n-m)(n+m+1)=4028$$ As $n$ and $m$ are positive integers, we have $n+m+1$ greater than $n-m$. Now $$4028=2*2014=4*1007=19*212=38*106=53*76$$ For the sequence to ...

4

In the diagram below (borrowed from this answer), we have: $\sin(\theta) < PQ$, because $PQ$ is the hypothenuse of the right triangle $PQR$, $PQ < \theta$, because $PQ$ is the shortest distance from $P$ to $Q$, and so $\sin(\theta) < PQ < \theta$.

4

We know that $$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$ So now $$\gcd(2^{21}-1,2^{27}-1)=2^{\gcd(21,27)}-1=7$$ We'll now prove the theorem I used here. We can do the following. Assume $n>m$, then: \begin{align} \gcd(a^n-1,a^m-1)&=\gcd((a^n-1)-(a^m-1),a^m-1)\\ &=\gcd(a^m(a^{n-m}-1),a^m-1)\\ \end{align} since $\gcd(a^m,a^m-1)=1$, we now know ...

3

You can find a bijection between the set of partitions of $n$ into $r$ non-negative integers and of $n+r$ into $r$ positive integers. Let $(\alpha_1, \alpha_2, \dots, \alpha_r)$ be a partition of $n$ such that $\alpha_i \geq 0, \forall i \in \{1,2,\dots,r\}.$ Then $(\alpha_1 + 1, \alpha_2 + 1, \dots, \alpha_r + 1)$ is a partition of (\alpha_1 + 1) + ( ... 3 Link AC, BD and denote O as their intersection point. Since PQ \bot QR ,PQ//AC,AC = 2 PQ  and QR//BD,BD = 2QR\Rightarrow AC \bot BD,AC = 6,BD =8  then the quadrilateral is divided into to triangle ABC, ACD which share the same edge AC.Then the area of the quadrilateral equals to the sum area of these two triangle. Solve the problem using ... 3 The flaw in your reasoning is that as the disk rotates along the edge of the clock face, there are two components to its orientation: the disk's own rotation, which you accounted for, but also a second rotational movement corresponding to its changing position relative to the clock. To understand this, take two coins of equal size, and roll one around the ... 2 Hint: There is a four-coloring of the board so that Any straight tetranimo covers one of each color, and Any zig-zag covers either two of two colors or one each of all four coors. The square must color two colors exactly once, and another twice. 2 The generating function for the central binomial coefficients is \begin{align*} \sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad |z|<1 \end{align*} This is an application of the binomial series \begin{align*} (1+z)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}z^n\qquad |z|<1, \alpha\in\mathbb{C} \end{align*} and the ... 2 No, you can't. For example, the pairs4 + 3 + 2 + 1 = 10, \; \; 4 \cdot 3 \cdot 2 \cdot 1 = 24$$and$$12 + 1 + (-1) + (-2) = 10, \; \; 12 \cdot 1 \cdot (-1) \cdot (-2) = 24$$give different results for \frac{1}{\pi_1} + \frac{1}{\pi_2} + \frac{1}{\pi_3} + \frac{1}{\pi_4}. 2 Clearly N must have a 9 in its ones place or else the sum of digits from N and from N+1 will differ by exactly 1, and therefore could not both be divisible by 7. Now when d_k\cdots d_29 has 1 added to it, you maybe get d_k\cdots (d_2+1)0 if there is no carrying into the hundreds place. But this N and N+1 have digits summing to values ... 1 Take the Young diagram of a partition of n into at most r parts, and add an extra box to the end of each row. If there were less than r rows, then add additional rows of length one so that the diagram has r rows. This way, we get the Young diagram of a partition of n + r with exactly r rows. To see that this is a bijection, it is enough to show ... 1 Let l(x) be the length of the decimal representation of x. Use induction on l(x). Suppose the identity holds for l(x)\le n. Let x=10y+r where l(y)=n and r\in\{0,1,\cdots,9\}. S(10z)=z\;\forall\;z. If r<5\implies N(x)=N(y)$$ S(2x)=S(20y+2r)=S(20y)+2r=S(2y)+2r\\ =2S(y)-9N(y)+2r=2S(10y)-9N(x)+2r\\ =2S(10y+r)-9N(x)=2S(x)-9N(x) $$... 1 Here's a quite straightforward method with two coins, bounded number of throws, and both coins have rational values. First coin is unbiased; second coin has the following probability for a Head:$$p = \frac{2^m}{n!}; m = \lfloor{\log_2(n!)}\rfloor With the unbiased coin you can emulate a biased coin which has $\frac{k}{2^m}$ probability for a Head ($k$ is ...

1

An alternate way to think about your function $f(x)$ is as: Double the $x$ value. If it is less than 1 (i.e. $2x\leq1$ from $x\leq\frac12$) then leave it. If it is more than one then reflect it in the line $y=1$ (from second half being $2(1-x)$ ). So double your input and fold it down if its over 1. Using this definition and having knowledge of the first ...

1

We claim that in a class of $6$ students, one can always find $3$ people who are either pairwise friends of pairwise enemies. Let $v_1, \ldots, v_6$ be all the people in the class. Out of the five people $v_2, \ldots, v_6$, at least $3$ of them have the same feelings towards $v_1$, that is, either all three are enemies or all three are friends of $v_1$. ...

1

Model the group as a complete graph, where every line between students is either blue (friend) or red (enemy). The Ramsey number $R(3,3)$ equals 6. This already gives an upper bound, if you think about it.

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