Hot answers tagged

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 pairs $$4 + 3 + 2 + 1 = 10, \; \; 4 \cdot 3 \cdot 2 \cdot 1 = 24$$ 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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