This is a famous physical puzzle that can be tied to the fibonacci series. To answer the question as posed, the issue is that the two slopes are different ($\frac25$ vs $\frac38$). Note that all those numbers are in the fibonacci series ($1,1,2,3,5,8,13,21,\ldots$). Successive fractions are closer approximations to $\varphi$, alternating between above and ...


The diagram is misleading, as it hides a gap in the middle of the second configuration. This is what we actually get if we rearrange the shapes in question. Notice that the diagonal “bows” slightly, leaving some extra space between the shapes – this is where the extra unit of area creeps in. But you shouldn’t trust me any more than the person who drew ...


Does this count?


If you're given just a⊕b and b⊕c, then you can calculate (a⊕b) ⊕ (b⊕c) = a ⊕ (b ⊕ b) ⊕ c (since ⊕ is associative) = a ⊕ 0 ⊕ c (since X⊕X=0 for any X) = a ⊕ c (since X⊕0=X for any X) so in effect when you're given a⊕b, b⊕c, a⊕c you've only been given two numbers (because the last one is redundant). So (assuming a,b,c are 8 bits as ...


If you are allowed to use decimals, then


Note: This answer only applies prior to the edit that clarifies that the expression on the left must evaluate to 100, rather than simply the equation being true. If you allow exponents, you can get away with just two:



Answer: How?


Here's an answer which


Yes. Guess 1st set (1,3,5,7,9,11,13,15) -> If the number is in the set, write down 1 Guess 2nd set (2,3,6,7,10,11,14,15) -> If the number is in the set, write down 2 Guess 3rd set (4,5,6,7,12,13,14,15) -> If the number is in the set, write down 4 Guess 4th set (8,9,10,11,12,13,14,15) -> If the number is in the set, write down 8 After all 4 guesses, add the ...


This is not possible. Consider the two cases where a, b and c are all true or all false. Now in both cases we have a⊕b = b⊕c = a⊕c = false And more generally, $(¬a)⊕(¬b)=a⊕b$, so if $a,b,c$ is a solution, then so is $¬a,¬b,¬c$. (from Klaus Draeger)


Since the puzzle oddly and specifically mentions the symbol for the square root, I used this: but rotated and reflected it giving:


I believe that this is the smallest:


Another answer could be


I'm gonna say: Explanation:


If you consider the numbers as Then


4 ops = 1.9934200404 points: Off by 0.00108199. 5 ops = 2.2864604146 points: Off by 0.0000340537. 6 ops = 2.7136051067 points: Off by only 0.000000266764(!) Now we can keep taking square roots of 1 in this expression to get a lower score bound for $n$ operations where $n \geq 6$, namely: $$s_n = -\ln \left( \frac{ 355/113 } \pi - 1 \right) / n$$ Which ...


What about this where


Remove one egg. Now the number of eggs is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10. Therefore it is divisible by their least common multiple. There are algorithms to compute the LCM; for example you can write the prime factor decompositions and take the maximum for each factor. Here the result is 7*5*9*8 = 2520. So the original number of eggs is of the form ...


Let's define $J$=Jennifer and $D$=Douglas. The problem can be rewritten as: $J=D+21$ $J+6=5(D+6)$ According to my math, Douglas is $D=(-\frac{3}{4})$ years old, which means $-9$ months. Pregnancy lasts for nine months, so Jennifer's husband is in the bed with her right now.


Also, you can use any operation. Ok then. $\begin{array}{c|c} 0 & \log_{\frac1 2} \left( \log_{4!!-3} 5 \right) \\ 1 & \log_{\frac1 2} \left( \log_{4!!-3} \sqrt 5 \right) \\ 2 & \log_{\frac1 2} \left( \log_{4!!-3} \sqrt{\sqrt 5\,} \right) \\ 3 & \log_{\frac1 2} \left( \log_{4!!-3} \sqrt{\sqrt{\sqrt 5\,}\,} \right) \\ 4 & \log_{\frac1 2} ...


A simple way is to pick the number $X$ which is half-way through the range and ask Is it less than $X$? From the answer you can discard either the lower half or the upper half of the range. Repeat until you have 1 number left. This method is known as a binary search or binary chop. In general, If you are allowed $q$ queries, you can get an answer from ...


Another solution:


No rules? Looks like 88 to me if I squint.


As Jo has already shown, this can be accomplished in To help visualize this problem, we can imagine: Proving minimality:



Such an old chestnut! Each of them eats 5/3 loaves. The first traveler contributes 3 loaves, eats 5/3 himself, and gives 4/3 to the extra traveler. The second traveler contributes 2 loaves, eats 5/3 himself, and gives 1/3 to the extra traveler. Since the first traveler gives 4 times as much bread to the extra guy, he should also receive 4 times as much ...


Build two straight runways side to side, 1 mile long each. Then, put a portal-conducting surface at the end of the first runway and another one at the beginning of the second one. Finally, use your portal gun to place a yellow portal on one of those surfaces and a blue portal in the other one.


Here's one way: For the image-impaired:


And three more à la Perry

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