Puzzle Thread

>>7782039
paul erdos

there's two people to each hand shake to the total number of handshakes must be even. the sum of the handshakes of people with an even number must be even. the difference between the total handshake and the total handshakes of people with even number of handshake must be even since both are even. therefore the number of people with odd handshake which is equal to the difference must also be even.

now gtfo you pleb, this isn't science or maths.

>>7782039
Total handshakes H is even because each handshake involves two people. The number of handshakes by people with an even number of handshakes is even since it is the sum of even numbers. The number of handshakes by people with an odd number of handshakes O must also be even since

H = E + O

Since O is even yet it is the sum of odd numbers, it must be the sum of an even number of odd numbers.

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>>7782058

nice!

here is the next one. the question is poorly worded but i wasnt sure how to explain it, so ask if its not clear

File: Super Hard coutnerfeit extension.png (131KB, 1051x326px) Image search: [Google]

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oh and this one is effectively impossible, so good fucking luck.
if anyone solves it i will be REALLY impressed!

this comes from an old Soviet maths book I found in a library - written in the 60s I think. its fiendish!

>>7782067
makes no sense, if you count a number itself as a product then the solution is trivial.

if not then a counterexample is 4 which is the smallest element of s

if you meant s as a product of elements in k, then by the fundamental theorem of arithmetic it must be true since k is the set of natural numbers and all elements of S are integers>2

>>7782081

oh dear, i have made a bit of a hash of my question i think.

So, what im asking is kind of like "What if we had a different set of prime numbers?"

So for set S, the smallest numbers that exist are 4,7,10 etc, and so the "prime factorisation" of 28 is not 2x2x7, but 4x7, because in this scenario 2 doesnt exist.

So 4 is a faux-prime, kind of.

Does that make sense?

>>7782080
trivial again

it's binary search but with an extra layer so that you get 3 equal, one different pile and then that will reveal whether the coin is heavier after which it is a straight forward binary search excluding 1 coin when an odd number is encountered.

>>7782088
already answered that here
>if not then a counterexample is 4 which is the smallest element of s
if you discount the smallest element then 7 and 10 are also counter examples

you're really shit at this stuff, you should pick up a maths book sometime. or if you actually think these are anything but trivial then maybe not haha

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>>7782090
>>7782093

wow! you are solving these faster than i can write them! I actually don't understand your solution to the coutnerfeit one, but I've saved it and I will try and examine it later to figure it out.

while I write the next one, do you think you could tell me of your education? I would very much like to be like you one day!

>>7782096
I'm an EE student at one of the best unis in the world

I wouldn't bother with writing anymore until you've got a grasp for what makes a challenging logic puzzle. the CLRS intro to algorithms book would do the trick.

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>>7782106

okay, next one is done!

Thats cool, which uni? I'm going to Cambridge year after next (got a deferred offer), but im going for Physics, not Maths - as you can tell my maths isn't good enough for that!

what kind of work do/did you do in and out of uni to get better at maths? i want to learn as well, if thats okay

>>7782109
Second is Lagrange's four-square theorem, first is a specific example of it.

>>7782118

So you already know the answer? thats no fun...

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>>7782067
(sorry for the edit)

Looking beyond the fact that you didn't take the identity into account, nor that some numbers have no factorization into two smaller numbers at all,
I see the idea you had behind the question and the answer is still no.

The script in pic related tells me that a counter example is
112 = (3 · 37 + 1)
which can be written as both
(3 · 1 + 1) · (3 · 9 + 1) = 4 · 28
(3 · 2 + 1) · (3 · 5 + 1) = 7 · 16

Come on everyone knows the handshaking theorem from graph theory. And yes, people here know who Paul Erdos is.

>>7782080
7 weighings?

>>7782109
not going to say, but I rejected oxford for it

maths is just a hobby that I've always been decent at. I self-taught further maths when I was 17 and just stuck with it out of habit, getting more and more difficult books as I went along.

since you have the potential not be a popsci meme master I'll recommend maths methods for physics and engineering by riley, hobson and bence. if you finish it properly you'll know your basics really thoroughly. after that you'll probably have your own opinions but I liked baby rudin and CLRS (more of a CS book but still).

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NB: Disclaimer: i havent actually solved this one before, and I have NO idea how to do it. But I do know it is possible, somehow!

>>7782129

Ah, well, I suppose thats one way to do it.

But you're lucky that the sequence just so happened to have a false result relatively early on - for all you knew there could have been no false results for the first billion natural numbers.

still, good job!

>>7782138

It is! fantastic work! That one took me a solid few months - you must be real smart

>>7782140

So which university is it?

>>7782039
>Shake 2 hands at once
yeah fuck you and your false proof OP

>>7782145

it doesnt say "shaken hands with a certain number of hand"

it says "shaken hands with a certain number of people"

so that wouldnt count as shaking hands with anyone! sorry!

>>7782140
>>7782106
>>7782093
>>7782090

OP ignore him. He clearly just googled the answers and thats why hes being so retardedly evasive about his studies and trying to end the conversation at every opportunity.

My bullshit meter is off the charts right now.

>>7782080
Separate coins into three piles, two with 333 and one with 334, weigh the two with 333 against each other, then:

1) If equal weight, then coin is in the 334 pile, discard the rest
2) If unequal weight, then coin is in one of the 333 pile, discard the rest

If (1), then separate into three piles again, two with 111 and one with 112, repeat first step.
If (2), then separate into three piles of 111 and repeat first step.

[math]...[/math]

I assume that if you constantly keep getting the coin to be in the pile with the highest number (like 334, 112), it's going to be the highest amount weighings, so eventually you're gonna get two piles of 4 and one of 5 after 5 weighings. Measure the two 4's, if same, it's in the pile of 5, if not measure the two single coins in the heavier pile. If in the pile of 5, measure again two 2's, etc. So if my arithmetic is correct, it's 8 weighings at most, if you always happen to get the heavier coin in the pile you don't weigh.

>>7782158

hm, you are just 1 off the answer. you must have made some small oversight - but very close!

You can guarantee finding it in 7 moves, if you look hard enough

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>>7782140

>>7782090
This isn't optimal

>>7782144
Am I retarded or does that fraction simply have no denominator? I mean, if you add the fractions and make it into a reduced form you're gonna get the same series back

[eqn]\sum_{n=1}^{p-1}\frac{1}{n^2}=\frac{(p-1)!^2+\frac{(p-1)!^2}{2^2}+...+\frac{(p-1)!^2}{(p-1)^2}}{(p-1)!^2}=1+\frac{1}{2^2}+...+\frac{1}{(p-1)^2}
[/eqn]

>>7782180

On a phone and can't read your equation, but I assume youve tried plugging in p = 5 to see what happens?

>>7782182
no i just wrote down the series, summed the terms and made a big fraction, had (p-1)!^2 all over the place, reduced the fraction and got back the original series

>>7782080
>>7782160
That's actually a very interesting problem. If my understanding of it is correct then you can have a pile of 2187 coins and still be able to do it in 7.

>>7782193
No, you can't know the solution if the sides are even every time.

>>7782067
You seem to be asking if that is closed under multiplication.

>>7782144
Your numerator mod p looks like

[math] ((p-1)!)^2[1^{-2}+\dots+(p-1)^{-2}] [/math]

but since p is prime, this is a field, so that reordering the sum it is congruent to

[math] ((p-1)!)^2[1^2+\dots+(p-1)^2] [/math]

but now, there is a closed formula (also easy to derive) for the sum of the first n squares, which will make you see is equal to

[math] 1^2+\dots+(p-1)^2=\frac{(p-1)p(2(p-1)+1)}{6} [/math]

But now, since p>3, p-1=0 mod 2, and either p-1=0 mod 3 or 2(p-1)+1=0 mod 3 (maybe you have to check p=5 separately, but it also works),

so your numerator mod p is some number times p. Since p is surely not in the denominator, p divides the numerator as you wanted.

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>>7782080