# Tag Info

7

It is trivially possible. Just spill water from the $16$ litre bottle until it is empty. By the intermediate value theorem, there is an instant when the there is exactly $11$ litre left. Nothing in the question indicates that this is about a construction, so an abstract existence proof will do.

6

Hint : look at the picture and consider what is the front and back of each volume.

6

With a bit of lateral thinking, this is possible like this: Spill away water from the 16 liter bottle until it is half full. You can check for half-fullness by marking the water level on the side of the bottle and turning it upside down. If you spill away too much, you can refill from the 6 liter bottle. Now 8 liters remain in the 16 liter bottle, so just ...

5

If you have an extra reservoir that can contain at least 16 liters, it is possible: Start with 16-6-3-0 Empty bottles of 6 and 3 into reservoir: 16-0-0-9 Fill bottle of 6 from bottle of 16: 10-6-0-9 Empty bottle of 6 into reservoir: 10-0-0-15 Fill bottle of 6 with bottle of 16: 4-6-0-15 Empty bottle of 6 into reservoir: 4-0-0-21 Move content of bottle of ...

5

I assume that the Archer Tower cannot split its $35$ damage over several archers. Then a group of $n$ archers can survive for $n$ seconds and cause $6(n^2+n)$ damage (this is because $1+2+\ldots+n=\frac{n^2+n}{2}$). So $9$ archers can cause $540$ damage and $10$ can cause $660$ damage before they are wiped out one by one.

5

This is the sequence of numerators of taylor series expansion of $\log(x/\sin(x))$ as can be seen in the online encyclopedia of integer sequences: OEIS.org The next term is $43867$.

4

It is impossible. $16x+3y+6z=11$. Where $x$ and $y$ and $z$ are integers. By little inspection you'll observe that there are no such multiples of $6$ and $3$ and $16$ which add up to $11$. (Because adding more water is not allowed) If it is, then $(-10,3,27)$ is a possible solution. You have to have unlimited access to water to solve this.

3

If you ever empty the $16 L$ bottle, then you don't have enough water left in the other bottles to get $11 L$. If you don't empty that bottle, then you can only change any amount of any bottle by a multiple of $3 L$, and so you can't get $11 L$ in this way.

3

I assume that (as usual in this type of problem) the operations are executed left toright (not by arithemtic priority). Here are the solutions: [9, 4, 8, 6, 7, 3, 1, 2, 5]: $9+13=22$, $22\times 4=88$, $88:8=11$, $11+6=17$, $17+12=29$, $29\times 7=203$, $203-3=200$, $200-11=189$, $189+1=190$, $190\times 2=380$, $380:5=76$, $76-10=66$. [9, 7, 2, 8, 4, 3, 6, ...

3

Mark the bags $1$ to $10$. Pick $2^1$ balls from the first bag, $2^2$ balls from the second bag and in general pick $2^k$ balls from the bag marked $k$. If all bags had balls of equal weight, the sum would be $20460$g. If bag $i$ has $9$g balls and bag $j$ has $11$g balls, the sum of weight of all balls would be $$20460-2^i + 2^j = x$$ Any integer of the ...

3

Here is an $\mathcal{O}(n)$ solution, once you have sorted the set, which typically costs $\mathcal{O}(n \log n)$. Let $S = \{a_1,a_2,\ldots,a_n\}$, where $a_1 \leq a_2 \ldots \leq a_n$. We have $$\sum_{s \subset S} \max(s) = a_n \cdot 2^{n-1} + a_{n-1} \cdot 2^{n-2} + \cdots + a_2\cdot 2 + a_1 \cdot 1$$ \sum_{s \subset S} \min(s) = a_n \cdot 1 + a_{n-1} ... 2 how do you express “b is a power of 10”. Since Rory already covered the problems with your approach, I'll tackle the question of finding a different solution. In my first attempt to do so, I have made a mistake, so I'm completely rewriting my answer. It is now inspired by this post by Anders Kaseorg, although the wording is mine. All powers at once ... 2 P1-P2-P3-P4-P5-P6 Soccer |AG|BH|CI|DL|EK|FJ| Football |BC|JG|DE|HI|FA|LK| Kick-ball |KJ|CD|GH|EF|IL|AB| Volley Ball |EL|FK|AJ|BG|CH|DI| Hockey |FI|AL|BK|CJ|DG|EH| Water Polo |DH|EI|FL|AK|BJ|CG| @Vince Kroon is incorrect. Here, I have reformatted your schedule using the information from this answer ... 2 Given the premises of the puzzle, you don't need to go beyond A's statements. If his first statement were false -- i.e., if he did do it -- then his second statement would also be false, in violation of the premise that only one statement is false. So his first statement must be true and the second one false. But that means B didn't do it either, which ... 2 A's statement is either (t,f) or (f,t). if it is (f,t) it means both him and B are guilty. Assuming there is a solution to this puzzle A is (t,f) and so C is guilty. Further B is not guilty so his second statement is false. This does not mean C is not guilty, its just that B does not know. Note also C is (f,t) so B does not know who did it anyway. t=true ... 2 This is famously called by George Boolos the "Hardest logic puzzle ever" (1996). He, that, as far as I know, was the first to present and solve it, states the puzzle as follows: Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or ... 2 Always forecast according to the majority prediction (or if there is a tie for sunny and cloudy predictions, forecast randomly). Anytime the majority is correct, you haven't used one of your five failures. Anytime the majority is wrong, you get to fire at least half the remaining meteorologists. So in five wrong forecasts, in the worst case, the number of ... 2 The sum of all digits of each row is always 20. Second row: 1+1+3+3+6+6=20 Third row: 8+3+2+7=20 Then: First row: 2 + 9 +a +b = 11+a+b = 20 \Rightarrow a+b = 9 where 10a + b is the number to be placed in the position (1,3). This is only possible if the number to be placed is 45; in fact a=4, b=5 and a+b=9. 2 There is 1 equation and 9 unknowns, the problem is we have an underdetermined system: http://en.wikipedia.org/wiki/Underdetermined_system. The only relationship we know (in the typical phrasing of the question) is that they all belong to the set [1...9] and are mutually exclusive. While all solutions can be found using brute force (with a program), I don't ... 2 \begin{align} & ? + ? + ? = 30 \\ \implies& 3? = 30 \\ \implies& ? = 10 \end{align} 2 The key word is perpendicularity. Given a parametrised line \cases{x(t)=at+b\\y(t)=ct+d},  the point on that line where it's closest to the origin is is given by $t_0$, which has the property that the position vector $(x(t_0),y(t_0))$ is perpendicular to the direction vector $(a,c)$. Do that for the line $B-A$, and you get your answer.

2

The numbers defined by $x_{n} + y_{n} \, \sqrt{2} = (1+\sqrt{2})^{n}$ are called Pell and Pell-Lucas numbers. As stated by the proposer question B should be $y_{2n+1} = y_{n+1}^{2} + y_{n}^{2}$ as will be shown. 1) From $x_{n} + y_{n} \sqrt{2} = (1+\sqrt{2})^{n}$ it is seen that \begin{align} x_{n+1} + y_{n+1} \sqrt{2} &= (1+\sqrt{2})^{n+1} = ...

2

I'm going to assume that the allowable operations are: empty a bottle completely; and empty one bottle into another until the former is empty or the latter is full, whichever happens first. Then this graph shows the shortest path to reach any reachable state (click here for a full-size version with readable labels): No state with a bottle containing ...

1

I think according to the picture, the "front cover of volume I" is actually between the books I and II, while the back cover of volume 3 is between II and III. I think the wording is just trying to mislead the reader to think the worm eats through all the books, when it really just eats through volume II (which is the only thing between the outside of the ...

1

Think about where the FRONT cover of volume I is and the BACK cover of volume III. It's a trick question.

1

start with: 16 - 0 - 0 fill 16 liter bottle pour into the 6 liter bottle clear 6 liter bottle (10 liter left) pour into the 6 liter bottle clear 6 liter bottle (4 liter left) pour into 3 liter bottle clear 3 liter bottle (1 liter left) pour the remaining liter into the 3 liter bottle fill up the 16 liter bottle pour into the 6 liter bottle now the ...

1

A state like $(x, y, z)$ where $x = 16+3k$ for some $k \in \mathbb{Z}$, $y \in \{0, 3, 6\}$, $z \in \{0, 3\}$ and $x+y+z \geq 11$ can only be transformed into a state satisfying the same criteria. If the problem had a solution, we should have at some point $16+3k = 11$ for some $k \in \mathbb{Z}$, which is impossible.

1

Assuming worst case scenario (tower perfectly focus fires your archers and tower wastes no dps) You will need 10 archers and at least 3 of your archers will survive. If you have 9 archers, you will be wiped out if tower does a perfect job. If tower does not do a perfect job, 9 archers will likely be enough.

1

You already have the answer, so a non-technical proof might be simply checking the answer. First write down every possible starting condition on a piece of paper. Then for each starting position write down every possible result of the first operation on another piece of paper. Then for each result on that piece of paper, write down every possible result of ...

1

Once you write an equation for $d=\sqrt{(x_A(t)-x_B(t))^2+(y_A(t)-y_B(t))^2}$, then consider $d^2$. $d^2=20t^2-60t+50$ $0=20t^2-60t+50-d^2$ $0=2t^2-6t+5-\frac{d^2}{10}$ The axis of symmetry of a parabola is $t=-\frac{b}{2a}$. The minimum or maximum must be at the vertex. We know it is a minimum because ...? (If this isn't enough let me know.)

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