# Tag Info

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Vladimir starts with daughter 1 on the motorbike. After 48 km (that's one hour and 12 min at 40 kph), he stops and daughter 1 continues on foot. She reaches the dacha after exactly three hours. Vladimir drives back to meet daughter 2, who has already covered 12 km. They meet 36 min later (one hour and 48 min after the start), 18 km from the starting point (...

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The answer is I imagine the line of reasoning the author wants is as follows:

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Shkeil got all of the nontrivial answers, but he forgot the simplest, so I'll add it here: For completeness, I'll add the rest of my thought process.

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The answer is yes, this is possible. Fix $t = 100$. A big key is found in this excellent puzzle:

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The problem in modern English might be stated as follows: A farmer wanted to buy three hundred oxen. He bought them at a price of $3$ oxen for $\$63$. Afterwards, he sold the oxen at$3$for$\$63$ as well, but managed to make a profit of $\$787.50$. How did he manage this? At first blush, as you said. it appears that the farmer should have ... 15 To illustrate the puzzle I made the following image: All angles I entered can be simply calculated using the fact any n-gon has a total of$180 + (n-3) \cdot 180 °$. Now let's call the intersection of line$b$and$aX_1$and the intersection of$b$and$cX_2$. To prove that$X_1 = X_2$I'm going to show that$P_1X_1 = P_1X_2$First, let's look at$...

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Assume that a bumble bee costs $x$ cents and that a honey bee costs $y$ cents, where $x$ and $y$ are integers. Then the problem statement gives $125x<175y$, which yields $5x<7y$ and hence $5x+1\le7y$ $175y<126x$, which yields $25y<18x$ and hence $25y+1\le18x$ We multiply the first inequality $5x+1\le7y$ by $18$ and the second inequality $25y+1\... 14 There is Reasonning. 13 (I'm assuming throughout that the concentration of wine at the end needs to be 50%.) Unless I'm missing something: There are$n$pints of wine to begin with. After the first 3 pints are taken, you have an$(n-3)/n$fraction of wine after the water is put back. After the second 3 pints are taken and replaced, you just multiply by the same fraction again, so ... 12 And here is the number you are probably thinking of: It works only for$ab$where$a \le b$. I suppose that it is a mistake in the problem statement. Others have proven that as it is, the problem is unsolvable. And here is how I came to that number. PS: I have been playing with this problem. You can extend it to$ab$with$a > b$with the ... 11 The minimum greater than 1000 is: Because: 11 As for why it is the only solution: 11 By brute-force search, yes. I started by searching over all tuples$(r,s,t)$less than$1000$, stopping at the first example I found: $$\left(138 + \sqrt{320}\right)^{570} \approx 10^{1249.9041}$$ In order to find the smallest example, I used the following strategy to search all examples smaller than the previous best (which I'll call$x$). Since the ... 11 @Tony Ruth's answer provides this alternate form of the equation: Adding 2 to both sides and factoring, Then Substituting this back in, So Add 2 to both sides again: So And For every$y>1$, this is true for some integer$n$, so we can choose the$y$that gives the biggest value of$m$less than 1 million. This is: So the answer is 10 This picture shows the best strategy to get to the Dacha. 10 Inscribe the original$n$-gon in a circle of radius 1. The apothem of the large$n$-gon is$\cos(\frac{\pi}{n})$and the apothem of the small$n$-gon is$\sin(\frac{\pi}{2n})$. Therefore the ratio of their areas is$\left(\frac{\cos(\frac{\pi}{n})}{\sin(\frac{\pi}{2n})}\right)^2$. For$n=99$this is about$3968.53$. Demonstration on a heptagon:$OA=1$,$\...

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The doesn't belong, because

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Following up on the suggestion by Dennis Meng, I'm posting my suggestion as an answer instead of as part of the question. So here are my thoughts on the problem. (1) As Joe Z comments, the cattle owner needs to increase prices by one eighth in order to make the stated profit. (2) For some reason, the price is given as 63 daler for 3 oxen, rather than the ...

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Note: $51.5625 = 825/16$ To calculate the sum for a given $n$ we can do $\frac{n \cdot (n+1)}{2}$. Then we need to have that $\frac{n \cdot (n+1)}{2} - \frac{825 \cdot (n - 4)}{16}$ is a number that can be expressed as four even consecutive numbers. Let $x$ be the lowest of the four consecutive numbers. The four even consecutive numbers are of the form $... 9 The smallest possible value of$n$is Claim: We can get every non-negative integer$n\leq 2016$on the board. Proof: By induction. We start with$n=0$on the board. We can get$1$using Lord of the Dark's method:$2016!x+2016!=0$has$-1$as a root,$-x^2+2016!=0$has$\pm\sqrt{2016!}$as roots, and$\sqrt{2016!}x-\sqrt{2016!}$has$1$as a root. Now ... 9 Edit: 8 First things first: we will declare some constants:$\text{COW_GRASS_PER_WEEK}=$quantity of grass eaten by a cow in one week$\text{ACRE_GRASS_PER_WEEK}=$quantity of grass that grows in one acre in one week$\text{INITIAL_GRASS_PER_ACRE}=$initial quantity of grass per acre We will call them respectively$CGW$,$AGW$and$IGA$. The variables are going ... 8 Let$x$be the starting amount of pints of wine. Now just focus on the amount of water in the barrel. After the first step there is 3 pints of water in the barrel. Because the content of the barrel is still$x$pints the amount of water per pint is therefore$3/x$So in the second step we take$3$times$3/x$water so we take away$9/x$water and we add$...

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Same answer as Shkeil shown in a different way.

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The answer is $14.542$ pints if he replaces wine with water after taking 3 pints of wine. Let say you have 300 units of wine and you take off 3 pints of wine and add 3 pints of wine 3 times. I. Theft After you take off 3 pints of wine (3p) and add 3 pints of wine and add water the new concentration of wine would be; $\frac{300-3p}{300}$ or \$\frac{100-...

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sum of: we need to know avg of n-4 so sum of 4 numbers must be: numbers to be removed are :

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Explanation:

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The odd one out is the This is because Safe Cracked!

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I am going to prove that Indeed Let me dump here previous thoughts that turned out not to be useful but might be in the future. First, Second

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