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7

Just rearrange them and notice that the bold triangle is right and isosceles: Another proof of $\arctan 1=\arctan\frac{1}{2}+\arctan\frac{1}{3}$ comes from: $$ (3+i)(2+i) = 5+5i $$ by switching to arguments.


4

A step of one meter perpendicular to the radius will increase the first friend's distance from the center, from $r$ to $\sqrt{r^2+1}$; and whether the step is clockwise or counterclockwise makes no difference. That is, it increases the square of the distance from the center by one square meter, and the second friend's instruction is irrelevant. It's easy ...


3

Isn't it much simpler to assign a value of 1 to the daughter, 2 to the wife and 4 to son? Adding those values totals to 7. The ratios are then easily derived by placing the assigned values over the total to yield: 1/7, 2/7 and 4/7


3

The corner tiles have 2 ways to move The side tiles have 3 ways to move The middle tiles have 4 ways to move 2 3 3 2 2 3 3 3 2 ... 3 4 4 3 3 4 4 4 3 3 4 4 3 3 4 4 4 3 2 3 3 2 3 4 4 4 3 2 3 3 3 2 Then the formula for different ways to make a $2048$ tile on $(n * n)$ square of $1024$ tiles is: $4 * 2 + (n - 2) *...


3

The following two questions are equivalent: "Can the cards be folded to put them in increasing order, when they are labeled in a specified arbitrary order?" "Can the cards be folded to produce a specified arbitrary order, when they are labeled in increasing order?" The relevant result was proved by Koehler in 1968 ("Folding a Strip of Stamps"): An ...


3

It is not entirely unambiguously clear to me what those expressions should mean. The dots suggest something of a limiting process, so I'd interpret them as limits. The expression for $s$ can be viewed as a limit as follows: Define a sequence $s_0,s_1,\ldots$ as $$s_0:=1\qquad\text{ and }\qquad s_{n+1}:=\sum_{i=1}^{s_n}1\ \text{ for }\ n\geq0.$$ Then $s$ can ...


3

There are five odd digits in total and four of them must occur in the four unit places. Also, some odd digit must occur among the tens places of the three summands as $2+4+6$ would be too big. Then the tens of the sum must be even, at most $8$. So the odd tens digit is at most $8-4-2$, hence $1$ is in some tens place and our four odd unit digits are $3,5,7,...


2

my hunch is this. Each weighing can turn out one of 3 ways -- be left side heavy, right side heavy or ballance. If we put the coins in a line, there are ${11\choose 7} = 28$ ways the counterfeits could be arranged. But each coin could be one of 8 different weights! I think there are $28\cdot8!$ different distributions to identify. $n > \log_3 28\...


2

Consider following triangle: As $BC = AC$ we have $\angle ABC = \angle CAB$ or $$ \pi - \gamma - \beta = \gamma - \frac{\pi}{2} + \beta \iff \gamma + \beta = \frac{3\pi}{4} $$ (here $\gamma$ is red angle from picture in question and $\beta$ is yellow one). It's obvious that green angle from question (detote it as $\alpha$) is equal to $\frac{\pi}{4}$. ...


2

Because $N=1$, the equation is: $300S + 3X = 2020 + 170I +2E$ Let's write it $R(S,X) = L(I,E)$. We notice that $L(5,0)=2870$ and $R(9,9)=2727 < L(5,0)$ so $I<5$. Let's check the remaining 5 possible values for $I$: $I=0$ Then $2020 \leq L(0,E) \leq 2038$ But $R(7,0)=2100$ and $R(6,9) = 1827$ so we cannot have $R(S,X) = L(0,E)$. Hence $I=0$ ...


2

$$3\times\overline{SIX}=2\times\overline{NINE}$$ As you wrote, we have $N=1$. Also, we know that $X$ is even. Let us separate it into cases. In the following, let $[n]$ be the right-most digit of $n$. Case 1 : $X=0$. Then, $E=0,5$ from $[2E]=[3X]=0$. If $E=0$, then since the right-most two digits of $3\times\overline{IX}$ has to be $20$, we have to have $[...


2

For this sort of problem, if possible, I will be lazy and ask a computer to find all solutions for me. If we ignore commutativity and associativity of addition and multiplication, there are $30965760$ ways to write down an arithmetic expression involving $6$ variables. Out of all these choices, $2400$ of them will return $828$. If we use the fact the ...


1

The problem is actually already solved by Jyrki's comment, since the only way to satisfy the given conditions is that the $3$ voters have the cyclical preferences given in his example, or the opposite preferences. Thus you just need to count the number of times each of the candidates wins under the proposed "solution", given such a preference set. As you ...


1

This question should be well-known and easy to find on the internet. related pages are for example; http://www.mathpages.com/home/kmath668/kmath668.htm http://physics.stackexchange.com/questions/2072/on-this-infinite-grid-of-resistors-whats-the-equivalent-resistance


1

@Jack D'Aurizio thanks for your solution and suggestions i found a little bit different solution from yours :)


1

He who starts with a multiple of three loses because his opponent can maintain this property until the end. He who starts with a non multiple of three wins by leaving the opponent with a multiple of three. $\color{blue}9\to8|7\to\color{blue}6\to5|4\to\color{blue}3\to2|1\to\color{blue}0$ $\color{blue}{10}\to9\to\color{blue}{8|7}\to6\to\color{blue}{5|4}\...


1

The key to this game is modular arithmetic. It is essentially the same game if it has $0,3,6,...,$ or $3k$ coins, $1,4,...,$ or $3k+1$ coins, and $2,5, ... ,$ or $3k+2$ coins. The optimal strategy is to always keep the number of coins $n$ on the table such that $n\equiv 0\pmod 3.$ This guarantees that when there are only $3$ coins left, it is the other ...


1

Unless there's some other unstated constraint, the problem doesn't have a unique solution. Even if you demand an integer at every step of the calculation, you get 301 distinct $(a,\ldots,i)$ tuples that qualify. My guess is that the constructor didn't realize (or care) that there is more than one solution. Writing a small script would be the quickest way to ...


1

Without thinking too much of how they would be defined, you are essentially looking for solutions to $$ s = \sum_{n=1}^s 1 $$ and $$ t = \int_0^t dx $$ which both can be any value at all (although they must be integral in the first case). So they are certainly not unique...


1

This is not an answer, but input for inspiration on the question. I've looked at bingo puzzles with the same rules, but with smaller cards. That is, I've looked at cards with the sizes $N*N$ where $N = 2$ (with the numbers $1-4$), $N=3$ (with the numbers $1 - 9$) and $N=4$ (with the numbers $1 - 16$). As a guideline for how many calls were made in each case,...


1

I guess my idea is similar to Christian Blatter above. I did a R simulation with the assumptions that buses arrive promptly every 10 and 15 minutes, respectively. However, the interval time among buses is fixed with uniformly distributed starting times. I simulate buses between [0,10000] minutes and the guy arriving at the bus stop at time t in [50,9950] and ...


1

More mathematically, it can be done as : The minute hand moves 360 degrees in 60 minutes. This means that the angle of the minute hand is given by 6t, where t is number of minutes past midnight. The hour hand moves 30 degrees in 60 minutes. This means that the angle of the hours hand is given by 0.5t. The hands start together at midnight. The first time ...



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