ITT: We post puzzles and complete proofs/explanations of them. I'll kick us off with an old favorite.
Find all positive integers [math]x,y[/math] such that [math]x^y=y^x[/math].
>>7650828
If you looked at this picture and thought this was a Pedo and not a father taking his daughter to swim in a lake or river in a park and teach her the beauty of nature, you are a sick fuck and part of the problem in society.
ylogx = xlogy
y/x = logy/logx
something something
z
I'm a complete amateur so I'm just gonna try to set the thing up. Please correct me if I fuck it up.
P(x) = x^y=y^x
Base case: P(0) = 0^y = y^0
0 = 1;
It fails the base case. Proof by induction?
>>7650828
2 and 4 are the only solution
I can't prove why that is though
>>7650844
You're not trying to prove it for all integers; clearly [math]3^2\neq 2^3[/math]. You're just trying to find the ones for which it works.
>>7650849
x = y = 1
>>7650850
Heh, I've only just started proofs by looking at mit youtube vids.
But, isn't it a logical step to try to prove it for all integers, since that would give a complete and immediate answer? Even if we know it can't be proven, isn't it simply good practice?
>>7650857
>Even if we know it can't be proven, isn't it simply good practice?
No. You should practice by proving things that are true.
>>7650857
No, a better approach for this problem would be to do some calculations and find out when it works. Then you'll have some data when you go to prove it.
>>7650843
You're on a right track. Separate and get [math]\log(y)/y = \log(x)/x [/math] and consider the function [math]f(t)=\log(t)/t[/math]. How can calculus help you here?
[math]{x=y|x,y\in{\mathbb{Z}^+}}[/math]
>>7650872
Ah, I should have said distinct integers...
>>7650872
>[math]{x=y|x,y\in{\mathbb{Z}^+}}[/math]
huh well it worked in the tex editor
ok does this work
[math]\{x=y\mid x,y\in{\mathbb{Z}^+}\} [/math]
>>7650881
Yes, but imagine I said distinct integers.
>>7650882
i meant the tex dummy (i had to escape the outer brackets)
>>7650885
Now that you've gotten the tex to work, perhaps you could try solving the problem?
>>7650886
nah im busy procrastinating on my lab paper
>>7650849
You could try a calculus-based approach. Or a number theory approach ('prime factor' [math]x[/math] and [math]y[/math]; what must be true for their powers to be equal?)
>>7650843
closet Pedo detected
>>7650851
any positive x = y
Since you aspies are jumping on the fact that the original question doesn't say distinct, I'm going to pose a new problem. First reply decides if it's a similar algebraic one, or something different.
>>7650916
No algebra please.
If you look at the graph of the function log(t)/t you see pretty clearly that 2 and 4 are the only possible answers with some monoty argument. I dont feel like taking deravitives right no, so just give another problem.
>>7650918
Yep! The function has a maximum at [math]e[/math] and it's easy to see that if [math]f(y)=f(x)[/math] for integer [math]y>4[/math], then [math]1\leq x <2 [/math], so it can't be an integer.
Here's a great problem. One dark and stormy night, seven robbers break into the king's castle and steal a chest of coins. Back at their lair, they divvy up the loot in equal shares, but there's one coin leftover. They fight and a robber is killed. They pool the coins and try to divide evenly again; again there's an extra, again they fight, and again a robber is killed. They try again and the coins divide evenly. How many coins were in the chest?
>>7650923
x = 1 mod 7
x = 1 mod 6
x = 0 mod 5
It's ANOTHER shitty algebra problem. Chinese Remainder Theorem says x = 85 + 210k with k any non-negative integer.
>>7650923
211 coins
xmod7 = xmod6 = xmod5 = 1
x = 7*6*5 + 1
>>7650932
...y u no like number theory? Correct, though.
Here's a geometry one. A surfer lives on an island in the shape of an equilateral triangle. She wants to build a hut so that the total distance to all three shores is minimal. Where on the island does she build it?
>>7650932
I had x=210k+125, but I did it per hand so maybe i made some errors.
>>7650936
in the middle of one shore. this way avg distance to a shore is 1/2(triangle side) * 2 + 0 / 3 = 1/3.
as opposed to the middle which is 2.
>>7650947
nope
>>7650960
nope
>>7650975
anywhere
>>7650995
prove it
>>7651005
take some random point p in equilateral triangle ABC. draw lines from p to each side of ABC. then take another equilateral triangle A'B'C', such that the triangles overlap where p overlaps side B'C'. draw line parallel to p, horizontally. take altitude A' to this parallel line, z.
A' to z is equal to dist(p to side A'B') because of similar triangles.
notice that the lines drawn to sides A, B, C are thus equal to the altitude of the triangle ABC. so for any p in ABC, this is true
please stop geometry qs i dont want to draw in paint thx
>>7651023
or just:
ABC = PAB + PBC + PCA
.5 * b * h = .5 * PA * b + .5 * PB * b + .5 * PC * b
h = PA + PB + PC
altitude is sum of dists
>>7651023
Okai, no more geometry, and no more algebra by request.
How can you simulate rolling a seven-sided die with a six-sided one?
>>7651036
Or, if you don't like that one:
A positive integer is trapezoidal if it is the sum of at least two consecutive integers. Find all non-trapezoidal positive integers.
>>7651036
Keep rolling the die and call the results of the i-th throw [math]a_i [/math].
Consider the number
[math]x = \sum_{k=1}^\infty (a_i - 1) \left(\frac{1}{6} \right)^k [/math]
>If the number x is between 0 and 1/7 the results is 1.
>If the number x is between 1/7 and 2/7 the results is 2.
>If the number x is between 2/7 and 3/7 the results is 3.
>If the number x is between 3/7 and 4/7 the results is 4.
>If the number x is between 4/7 and 5/7 the results is 5.
>If the number x is between 5/7 and 6/7 the results is 6.
>If the number x is between 6/7 and 1 the results is 7.
With probability 1 you only need to throw the die finitely many times to find out in which interval the number x is.
>>7651037
m*(2n+1) = (m-n) + (m-n+1) + ... + (m+n-1) + (m+1).
That leaves all perfect powers of 2.
>>7651059
Last term should be ... + (m+n).
>>7651037
t is trapezoidal
t = (n+1)+(n+2) +...(n+m)
= (n+m)(n+m+1)/2 - n(n+1)/2
= m(2n+m+1)/2
either m or 2n + m + 1 is even, so it is odd, therefore no trapezoidal number is a power of 2.
e.g. 2, 4, 8, 16, 32, etc. are non-trapezoidal positive integers.
>>7650881
How did you do that?
{x=y∣x,y∈Z+}
>>7651076
Did you read the sticky?
>>7651079
Well, it hasn't changed since I read it last. But the last time I tried [math] wat [/math] it didn't work.
Your post office sells 7c stamps and 12c stamps. What is the largest amount of postage you cannot buy with exact change? Generalize.
>>7650828
0 and 1
2 and 4
ITT: we trick people into doing homework
Here's one for you.
Max, Sally, and Roth want to buy apples. Max has twice as much money as Sally. Roth has enough money to buy 6mil apples. An apple costs $0.29. If Max can buy 183 apples then who did 9/11?
>>7651289
Nice shitpost.
>>7651284
>ITT: we trick people into doing homework
Seriously. And these faggots aren't even trying to make it sound like not homework.
Given a 5*5 chessboard and 3 identical Rooks, how many ways can you place them so that they are never attacking each other
How do I learn to do these sort of problems? Where does one start?
>>7650828
x = y
which is an infinite amount of integers
There are 100 rocks of varying weights. You have a scale which can measure two piles of rocks, and the sum of both quantities of rocks must equal 100.
Determine whether or not one can easily find a solution without checking all combinations of rock piles.
>>7651351
(5*4*3)^2 / 3! ? (Each rook removes a row and a column from the board so I have 25 possibilities for the first one, 16 for the second and 9 for the third but each configuration is counted 3! times since the rooks are identical)
>>7650828
Well, first I'd note that this is valid when x=y. Then I'd come up with a base case. 2^4=4^2, so I'd try to prove that this is concave down.
First we have to see if there's critical points, if there aren't, than this is bounded and can be solved easily.
y'(x) = (y(yx^y-xy^xlog(y)))/(x(xy^x-yx^ylog(x)))
gotta set it equal to 0
0 = yx^y-xy^xlog(y)
yx^y=xy^xlog(y)
x^(y-1)=y^(x-1)log(y)
x^y=y^x so x^(y-1)=y^x/x
y^x/x=y^(x-1)log(y)
1=log(y)/y
y=log(y)
y never equals log(y)
therefore y' never equals 0
y^x=x^y is bounded, and monotonic
2^4=4^2, pair (2,4) (4,2), function is decreasing for higher values of x
as x-> infinity, y goes to some value below 2
we only care about positive integer solutions, so any value less than or equal to 0 doesn't matter. (1,1) is the only integer solution below (2,4). Function reflects over y=x, therefore all solutions are (x,x) for all integer x and (2,4) and (4,2)
>>7651079
sticky is a shit
>>7650828
All positive integers when x=y
>>7651289
max can buy 183 apples
183x$0.29=$53.07
since max has twice as much money as sally that means sally has about $26.54
$53.07/2=$$26.535 ~$26.54
now roth has enough to buy 6mil apples or 6000000
6000000x$0.29=$1740000 or $1.74mil
now the main question hasn't been answered yet, since there are 3 children lets sum up the total money they had and the amount of apples they can buy
name #apples $money
Roth 6mil 1.74mil
Max 183 53.07
Sally 91 26.54
_______________________
6000274 1740079.61
now notice something strange, sally can get 91 apples and there is 1 sally
91 1
911
sally did 911
>>7650843
>father and daughter
If that was your first thought you are a shitty breeder who probably has a mother who got pregnant at 14.
Its obviously a brother and sister.
>>7651283
this
44 responses before getting a correct answer.
>>7650843
Obviously I know what the picture is of. I just wanted it to be oh so much more.
So where can you find problems like these? Can somebody recommend a book or a site or something
>>7652637
>>7652689
Google brought me here :D
>>7652395
Also 1 and 1.
>>7652637
artofproblemsolving
For every ε > 0, there exist only finitely many triples (a, b, c) of positive coprime integers, with a + b = c, such that
c>rad(abc)^(1+E)
Prove this.
quick question guys about notation. What's the difference between the symbol that looks like 7, and ~? Are they the same, do they both stand for not?
>>7652849
Lel.
>>7652858
kek. they're identical. they're both the negation
>>7652724
Google is a wonderful place, is it not?
It can bring you to Heaven and Hell, if it so wishes.
In the 10 cells shown here, you are to inscribe a 10-digit number such that the digit in the first cell tells how many zeros are in the entire 10-digit number, the digit in the cell marked "1" tells how many 1's are in the number, and so on. Thus, if the "4" cell has a 2 in it, then there are two 4's somewhere in the 10-digit number.
Proof is more important than solution.
>>7653137
6210001000
>>7653178
>PROOF IS MORE IMPORTANT THAN SOLUTION
The solution is easy.
>>7653186
The proof is trivial
All variables are nonnegative integers less than 10
b+2c+3d+4e+5f+6g+7h+8i+9j = 10
a+b+c+d+e+f+g+h+i+j = 10
c+2d+3e+4f+5g+6h+7i+8j = a
if a is 9 then j must be 1 but then the equation above does not hold
if a is 8 then i must be 1 and b must be 1, but then b does not properly count 1s.
if a is 7 then h must be 1 and b must be 2, but then the equation above does not hold
if a is 6 then g must be 1 and b must be at least 2. If b is 2 then c is 1 and all equations hold.
>>7653233
Formalize it
0 and 1
>>7653245
Why? it's not an interesting problem.
>>7653137
smells like a project euler problem, homie.
All real numbers. All of them. Even including infinitesmals because the solution can be x=y. Hell, im sure even imaginary numbers would work.
I've been working through some problem sets I found online. With this one I can't make heads or tails of the notation provided in the hint, because obviously 3(2i)^2 != (2i)^2. I know I'm not answering the question as asked and would be marked wrong if this was a real class, but I could intuit a simple deductive proof. Is this at least correct?
Proposition: Z is even if and only if w&x&y are even.
L1: The sum of two odds is even.
L2: The sum of two evens is even.
C1: Squaring preserves parity so it can be discounted.
C2: Order does not matter, so equivalent cases need only be tried once.
PF: Contradiction
Because L1, and L2, it is possible for a pair from (x,y,z) to be both individually odd, and the sum of all the terms to be even. This violates the if and only if clause of the proposition that Z is even if and only if w&x&y are even. Therefore the proposition is proven false by contradiction.
>>7653765
Good attempt, but you'll never be able to find an explicit case, and it's precisely because of the squaring.
L3: Any even square is a multiple of 4.
L4: Any odd square leaves a remainder of 1 when divided by 4.
Suppose exactly two of (w,x,y) are odd. Then w^2+x^2+y^2 is 2 greater than a multiple of 4, and so cannot be a square (z^2). So there is no contradiction.
>>7655132
Thank! I thought this thread was a goner. I was thinking something along similar lines but with powers of two, no multiple or power of two ends in 1, but I couldn't figure out how to use it. But thinking in terms of four makes perfect sense.
When it comes to the "all cases of x,y,z being odd or even," I infer that the summed inversions must either produce 1,2, or 3 greater than a multiple of four, making the proof complete once revised:
I.H. The proposition can be proven if there are no possible contradictions which result in an even Z.
By L3 and L4, we can see four does not divide any combination of sums of w^2+x^2+y^2 when any term is odd. P(z) is true by the I.H. of having no contradictions.
