Fun questions :)

[humph]

Unrelated question, but I was wondering if anyone had a neat solution (mine was split into a couple of cases, then induction, liberally applying a convenient inequality - really lengthy though).

Let be positive integers with . Show that

Here we set if .

[kamil9876]

equivalent to:



Now first term on LHS, is less than first term on RHS. Hence we are done if we can show that a similair thing happens to the second terms, that is:



equivalent to:



Now to prove this you can pair up the factors as follows:

with
with
with
.
.
.
with

And for the term on the right is always greater than the term on the left.

While for we can check the original inequality manually. (edit:it doesn't actually hold for )

edit: btw how did this pop up? I thought you were more of an analysis junkie  

[humph]

It was a differential geometry question: if is a finite-dimensional vector space, show that

when . Then show that this does not hold when . I did this by showing that

which is precisely the inequality


Here is the (real) vector space of tensors over of type , is the subspace of alternating tensors, and is the subspace of symmetric tensors.

[TrueTears]

hey humph, are you researching prime number theory ? or have you moved on to another area?

[humph]

Yup, I've done a couple of Advanced Studies Courses on them, one on (analytic techniques on) prime numbers in arithmetic progressions, and one on the elementary proof of the prime number theorem. Why?

[kamil9876]

1.) A fly starts off at (0,0) and flies to (2010,2010). His only movements are that he can move from (x,y) to (x+1,y+1), or (x+1,y-1) or (x-1,y+1) or (x-1,y-1) in one second. In how many ways can he do this if his journey lasts for 4020 seconds?

2.) You have a 16x16 grid such that each row and column has at most 4 distinct entries, what is the maximum number of distinct entries in the whole grid?

[dcc]

Heres my try - bit of handwaving, but meh.

Let be the set of all -tuples with strictly increasing elements chosen from .  Similarly, let be the set of all -tuples with monotonic increasing elements chosen from and let be the set of all -tuples with elements chosen from .

Now consider the set .  Create a new set by "reversing" each element of .  We note that . If , then but .  Therefore .

Since , (we can order each distinct -multiset of to become a monotonic increasing -tuple) and , we find that



edit: I feel like you easily strengthen this to  (count the number of strictly decreasing sequences of then add an on the end to make an -tuple)

[kamil9876]

Nice I was trying to find a "subset argument" as well but unfortunately I couldn't so instead I made up the crap you can find above. The fact that there is such a proof probably means that humph's original problem could've been solved like that directly (finding a subset in one of your spaces) rather than converting it into an inequality (and then converting it back into sets).

[dude]

if 3^x=4^y=12^z, show that z=xy/(x+y)

[googoo]

3^x=4^y=12^z=a, x=loga/log3, y=loga/log4, z=loga/log12

log12=log3+log4, log12/loga=log3/loga+log4/loga, 1/z=1/x+1/y, z=xy/(x+y)

[kamil9876]

A nice one: Consider a network of (finitely many) cities such that between every city there exists a road, and each road is strictly one way (you can travel in one direction ONLY). Show that there exists a path that visits each city exactly once. (A directed Hamiltonian path)

[/0]

Do you have to start on the right city? e.g. for a 2-city network if you have a path going from city A to city B but you start on city B...

[kamil9876]

Yeah you can start anywhere, ie to show that there exists a sequence of cities such that there exists a road from to for   (each city appears once and only once in that sequence)

[pi]

Should Q42 be worded differently... As 0 is a solution, and zero is an integer...

[enpassant]

a, b, c and d are natural numbers such that a^2 + b^2 + c^2 = d^2 and a<b<c<d. Find b, c and d if a = 2011.