Amusing Math Problem, Which Even Lawyers Know Enough Math To Solve:
From Misha Lemeshko, via my friend Haym Hirsh. I'm not sure I rate it as highly as Lemeshko did, but it's worth checking out.
8809 = 6
7111 = 0
2172 = 0
6666 = 4
1111 = 0
3213 = 0
7662 = 2
9312 = 1
0000 = 4
2222 = 0
3333 = 0
5555 = 0
8193 = 3
8096 = 5
7777 = 0
9999 = 4
7756 = 1
6855 = 3
9881 = 5
5531 = 0
2581 = ?
The source page is apparently in Byelorussian, or so I'd guess: The text is not in Russian, the text is roughly intelligible by a Russian speaker, and the page is on a .by domain. I'm told the source page is in Russian, though with a heavy dose of slang.
The answer is 2.
Any chance I am first?
It's the number of "0"s in the number. So for each 8, add two, for each 6 or 9, add 1, etc.
Or perhaps that all these analytical methods that you professors are teaching us can't help us solve the most fundamental or simple of problems? :)
Who said math couldn't be fun...
I actually tried to solve it algebraically. Looks like my undergraduate degree in physics was for naught.
Depends whether your four has a closed loop in it. Depending on the font used, the answer could change.
5 is 0
8 is 2
1 is 0
0+0+2+0 = 2
i was able to figure out that a 0, a 6 or a 9 were worth 1 while an 8 was worth 2 and the rest (besides 4) were worth 0....so i could solve the puzzle..but didn't get the loop part.
I had the same initial reaction. But here's one possible statement of the "math problem" implicit in the problem as presented:
Let T be the set of integer representations 0, ..., 9 viewed as topological objects and let f be a mapping from T into the integers in the usual sense (actually, onto the integers 0, 1, and 2). Then given the examples of f(i)+f(j)+f(k)+f(l), what is a possible value of f(2)+f(5)+f(8)+f(l)? Ie, what is a candidate for f?
Since - as the first couple of comments in the linked post suggest - being immersed in "math thinking" may be a disadvantage, this post might better have been titled "Amusing problem which lawyers know little enough math to solve".
0000=4
1111=0
2222=0
3333=0
5555=0
6666=4
7777=0
9999=4
Looking at that set alone, the "how many circles are there" jumps out at you. If you removed these, it would make the puzzle much harder. Especially if you add some trick misdirection patterns like:
4321=0
9753=1
A better question:
1, 2, 3, 5, ?
List three different possibilities for rule by which these numbers are being generated, and for each, give the next number in the sequence.
2. Fibonacci - 8
3. Add 1 twice, then 2 twice - 7
I only make this observation because none of the examples include a numeral 4, and without such an example, we have no idea whether 4 = 1 or 4 = 0.
3. should have been "add n twice, n = 1, 2, 3, ..."
And I forgot
4. random positive integer - any
I also assume that 4 is excluded because of the ambiguities that different fonts have on whether or not it has an enclosed area.
It's not Byelorussian. It's a slangified Russian that's considered cool by some Russian net users. There are various dialects of this Russ-nettese, depending on the type of site. Note, that the site is called "Brainf**k" in Cyrillic transcription.
If the question is about the number of closed loops in the figure, then it's certainly a math question, just not an arithmetic question. Geometry (or topology) is certainly an area of mathematics.
Although, again, pedantically, there should be some sort of an operator symbol around the figure, such as a "norm," "measure" or function.
Perhaps along the lines of:
Q(8809) = 6, etc., then find Q(2581).
And if it's closed loops, then a Q(4) = 1, and yes, it's definitely font-dependent.
Q(ABCD)=4 would give the game away, I suppose.
Eh, I'm pretty sure the intended domain and image of this mystery function are both finite sets.
Yeah, I'm pretty sure you're right and I was wrong about the case in point. What I was trying to get at, and doing it badly, was that this: given any set of n-tuples, finding a relation that binds all the n-tuples does not admit of a unique solution, etc. Kinda trivial, now that I think of it.
"to be pedantic, 1 is not a prime number"
What can I say - Lebesgue and I were contemporaries, although for only a year (see wikipedia's "Prime Number" entry). Perhaps Mr. Dege is also ancient and made the same mistake.
It wasn't clear whether the next part of your comment was directed at me, but if so see my comment at 1.12.2009 4:53pm.
- Charles
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