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# Tag Info

## Hot answers tagged arithmetic

109

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 ...

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69

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)

38

A puts in her age plus a random addition, B, C and D add the same.. Then A removes her addition, then B, C and D do the same. In equation form where A is the actual age and a is the random addition chosen by A: (A + a) + (B + b) + (C + c) + (D + d) - a - b - c - d = A + B + C + D which is sum of all their ages Then divide by 4.

35

Making an assumption:- The Summation becomes: Process:

35

Ten solutions.

26

Well, first I have to solve the math problem, I guess.

25

Using these provided operations, let's try to define primitives that can be used to build up the numbers we want. This is a constructive answer: the end result is a sequence of keys that should work. To get 1 from 0, press $x!$, $e^x$ or $\cos$. To get 1/2 from 1, we can use e.g. your solution ($\arctan$, $\sin$, $x^2$). To get $1/x$, press $\arctan$ then $\... 23 Easy math: 22 Your question with the XOR operator mathematically reduces to: In the$\mathbb{F}_{2^m}$field (field of characteristic 2 whose elements can be represented as sequences of$m$bits, and where addition is bitwise XOR), is the following matrix: $$M = \left( \array{1\ 1\ 0\\1\ 0\ 1\\0\ 1\ 1}\right)$$ invertible ? Indeed, for a dimension-3 vector$V = (x, y, ...

22

Ah, just scooped by Gareth! I'll post anyway since I've got a picture: The place to start is In the same way, you can solve C, D, F, G, and H. Then, looking at this picture Then we have Then things get tricky, because it seems to me that this The top two circles provide The bottom three circles, when So in all, this moderate visual puzzle gives us

22

I have a solution using 448 consecutive integers. The integers are centered on a number which I will call $c$, which is: 342807324437386669890245640930601418907623281362205720311372555412334319102271973072975564592463524267335118648596000454654370063914007920553665901460149824033735950242152959820 $c$ is approximately $3.4*10^{176}$. The integers are the ...

21

I wrote a program to solve all possible conditions including everything. The code is running for some days now and I have found lots of close results. According to the benchmark, it will take a couple of days to go and as a result I would have checked every single possibility and share the result with you guys. For $1,2,3,4,5,6,7,8,9$, I am going to update ...

21

The ages are: In order to see this, EDIT: The above implicitly assumes that all ages are non-zero. If we drop this assumption, we get some more solutions (Although I do not think this is intended):

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(edit) I will post one of the 10 solutions as an example:

19

The strategy suggested by Jiminion might work fairly well in practice, but it does risk revealing some extra information about the ladies' ages if the distribution of the random number(s) is known (or can be guessed). For example, let's say Ana picks her random number uniformly between 0 and 100; that's surely random enough to totally obscure her age, right?...

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Here is another solution. Your daughters are ages This works, because

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I got: The steps I took to get there were:

18

No, since, A ⊕ B = B ⊕ C = A ⊕ C = 0 can be either A = B = C = 0 or A = B = C = 1

18

Here's one solution (not sure if it's unique): How I found it: by following the logic used to answer this similar question. How I found this specific solution:

17

My try: And another one (same digits with same shape):

17

Not sure if it's unique, but this works: The common factors: How I found it:

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16

How about 624 Before you read the huge number below, realize my strategy: I am creating a number so huge that it cannot be written out, BUT, at the end it is multiplied by zero, and then 624 is added. I am sure that someone could use this same strategy to get a much larger (longer) number than this. Also note that the numbers twenty-one through ninety-...

16

1,720 $\Rightarrow$ 4,242 $\Rightarrow$ 24,749 This can be expanded infinitely so long as we can come up with new names for large numbers. In fact, notice that it starts with a new prefix after $10^{3,000}$. You could do the whole list again with the Milli- prefix and then Billi-, Trilli-, etc. A Million Billion Trillion Quadrillion Quintillion Sextillion ...

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There are a few nodes that can be linked together immediately, giving us a good starting point: Most of those starting links are on the bottom half of the triangle, so that's where I started. Specifically, I was quickly able to fill in the bottom-right: Followed by the rest of the bottom: The next bit threw me briefly, but I managed to figure out the ...

16

I think the answer is Proof: It's a bit like: See the following table: So we can make In theory, we want Concatenating all the pairs we get the answer:

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If then If then because then Hence,

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