/sci/ - Science & Math

[Anonymous No.4695030]

I want to get into solving putnam problems, but most of the times I don't even know how to start.
Are there any easy starting problems and explanations.
Help appreciated.

[Anonymous No.4695039]

>>4695030

inb4 spivak

[Anonymous No.4695043]

spivak

[Anonymous No.4695045]

Yes there are some very easy problems, generally the problems with lowest numbers are easy. I'll try and find some for you.

[EK !EKFnQG0UBo No.4695046]

>>4695039
you win all my fucckin internets for the day!

[Anonymous No.4695047]

"The Art and Craft of Problem Solving" is supposed to be a very good book regarding this kind of competitional problems.

[Anonymous No.4695054]

Here's 2010 problem set. A1 and A5 are both pretty easy. Try them, let us know if you need a hint before you give up.

[Anonymous No.4695069]

>>4695054

[Anonymous No.4695073]

>>4695069

[Anonymous No.4695089]

Look at A1. The sum of all sets is the nth pyramid number. 1 + 2 + 3 + 4 + ... + n = n(n+1)/2.
k is the number of equal sets, it has to be a factor of n(n+1)/2. Each set sums to n(n+1)/(2k) which has to be greater than or equal to n. (Because one set has to contain n) Just break it down into when n is even and n is odd and it's easy.

[Anonymous No.4695092]

>>4695089
i have nothing to look at
couldnt find it
was hoping you would post it

[Anonymous No.4695095]

>>4695054
He meant: http://www.artofproblemsolving.com/Forum/resources.php?c=2&cid=23&year=2010&sid=da9754df
25e818832abefedbc43235ac

But to be honest, you should try your hand at AMC problems and work your way up to USAMO/IMO problems and then try to do Putnam stuff.

Problems from math competitions like the IMO and Putnam require breadth of knowledge, which you'll be able to fill in as you do AMC/AIME/USAMO problems; of course, they also require depth and creativity, which will come as you practice.

[Anonymous No.4695117]

for A5 i don't know how to start either

[Anonymous No.4695120]

can we go through this very slow?

[Anonymous No.4695128]

>>4695117
A5 is easy. Just keep in mind that a cross product takes two vectors, and returns a vector perpendicular to both. A group is a set and a function that takes to elements of the group and follows a few simple rules.
http://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition
See how the cross product matches these definitions.

[Anonymous No.4695148]

>>4695128
can you do it step by step?

[Anonymous No.4695183]

>>4695128
i don't really understand what the problem is asking for, although i maybe could do it.

[Anonymous No.4695214]

Sure. I've also got A1 solved if you're curious about it. Because G is a group, there must be a vector e in G, such that a * e = a, for every element a.
Suppose a is a non-zero element in G.
This would mean that either
<div class="math"> a \times e = a </div>
or
<div class="math"> a * e = 0 = a</div>
The first option is not possible because <span class="math"> a \times e [/spoiler] is perpendicular to a. The second isn't possible either because a is nonzero. This leads to a contradiction, and the assumption that G contained a nonzero vector must be wrong. The product of two vectors in G has to 0 because 0 is the only vector in G.

[Anonymous No.4695243]

>>4695214
That's called "reductio ad absurdum"
You prove something ISN'T there by assuming it is there then getting a contradiction.

[Anonymous No.4695263]

>>4695214
I have asked this a bazzillion times, but nobody ever answers: how do I include LaTeX?

[Anonymous No.4695344]

>>4695263
see what I mean?