>>7911345
lol u dont need math irl bro

>>7911357
you are the shittiest poster in the board. useless fuck

>On A5 I'm literally visualizing riding the Dyck

Math is cancer

>>7911357
fucking idiot

>>7911400
A dyk is what they use to keep out water you inept dolt

I have once and I will again this year.

>>7911501
isn't dyk some kind of sexual orientation?

>>7911501
Dyck is a surname

solutions anyone?

>>7911740
??
Isn't it trivial?

>>7911750

is it? I'm poorly educated.

>>7911719
It's an archaic spelling of dyk = dyke

>>7911740
For A1, all the terms in the sum are either [math]a_1[/math] or [math]a_1+1[/math], with the constraint that at least one of the terms must be [math]a_1[/math]. If [math]j[/math] of the [math]k[/math] terms in the sum are [math]a_1+1[/math], then the sum is [math](k-j) a_1 + j (a_1+1) = k a_1 + j[/math]. Since [math]0 \leq j < k[/math], [math]a_1[/math] and [math]j[/math] are the quotient and remainder, respectively, when the value of the sum [math]n[/math] is divided by [math]k[/math]; these exist and are uniquely defined for any positive [math]n[/math] and [math]k[/math]. We satisfy the criteria that [math]a_1[/math] be positive if and only if [math]0 < k \leq n[/math]. That gives [math]n[/math] possible sums.

yeah i got reemed lol it's pretty hard and i don't do well under pressure

>>7911501
No, that's a dyke.
Dyk isn't an English word.

>>7911740
Partial solution for A4. If anyone has an idea for how to conclude, I would be glad:
Let [math]p(x) = ax^2 + bx + c[/math] and [math]P(x) = Ax^2 + Bx + C[/math] such that [math]\forall x \in \mathbb R |p(x)| \le |P(x)|[/math]. Without loss of generality, we may assume a, A > 0.
. If P has two real roots, then p has the same roots, therefore there is a constant [math]\alpha \ne 0[/math] such that [math] p = \alpha P[/math]. The hypothesis imposes [math]|\alpha| \le 1[/math], which implies [math]|b^2 -4ac| = \alpha^2 |B^2 -4AC| \le |B^2-4AC|[/math].

. If P has a double root x, then x is also a double root of p. Indeed, p vanishes at x and, if p' did not vanish at x, then we would have [math]p(x+h) = h(p'(x)+\eta(h))[/math] with [math]\eta[/math] a function that goes to 0 when h goes to 0 (this is Taylor's theorem). But we also have [math]P(x+h) = Ah^2[/math].
Then, for h sufficiently small, we have the inequalities [math]|P(x+h)| = Ah^2 < \frac{|p'(x)|}{2}|h| \le |p(x+h)|[/math], which is a contradiction.
Therefore x is a double root of p, therefore [math]|b^2 -4ac| = |B^2-4AC|=0[/math].

. We may assume from now on that P has no real roots (ie. [math]B^2-4AC < 0[/math]). Then, we have [math]\forall x \in \mathbb R Ax^2 + Bx + C \ge ax^2 + bx + c[/math] and [math]\forall x \in \mathbb R Ax^2 + Bx + C \ge -ax^2 -bx -c[/math].
This implies [math](B-b)^2 - 4(A-a)(C-c) \le 0[/math] and [math](B+b)^2 -4(A+a)(C+c) \le 0[/math].
Summing the two inequalities, we get [math]b^2-4ac \le 4AC-B^2 = |B^2-4AC|[/math], which is half of what we want. But I don't know how to get the other inequality...
We can notice that since [math]\displaystyle \lim_{x\to\infty} \frac{|P(x)|}{|p(x)|} \ge 1[/math], we have [math]A \ge a[/math]. Since [math]4(A-a)(C-c) \ge (B-b)^2[/math], this implies [math]C \ge c[/math] (unless A = a, in which case B=b and C = +/-c, in which case the problem is also solved), maybe this could be used

>>7911345
>Supposedly the median score is zero.
Only because too many brainlets participate. The problems aren't that hard.

>>7914118
Mathematicians confirmed for being brainlets.

>>7912326
Eesh, am I the only one that found this smart?

>>7912326
>these exist and are uniquely defined for any positive nn and kk.

Is this okay to say like that?

Is there some proof for the fact that for kk equals a particular number (and nn is something) there is a unique (only 1) a1a1 and jj to give you nn?

In other words, is there anyway to show (or rather, could you elaborate on how one can see that) each different value for kk only has one "way" from which the sum of nn can be reached? Since the question asks for number of ways.

>>7911345
I took it once in sophomore year. I got two problems out of the twelve.

They're insanely difficult, sure, but I found them really boring.
An interesting solution to an aimless, contrived problem isn't that interesting.

I felt the same way about high school math competitions.

>>7911345
I've been working on B6 for a while and it's kicking my ass

>>7911345
Honestly it's not so much that the questions are unsolvable as it is solving them in the allocated time.

How does one begin to study for such a test? Looking at this makes my head hurt. Where is square 1?

>>7911345
1 is purely deterministic, anyoen could do it if they sat down for long enough

2 is just algebra, but tricky

3 is easy

4 is interesting, because the second quadratic muct be both steeper and further shifted up you would think its discriminant must reach zero *before* the first one! Tough

For B6 btw, differentiate with respect to y, d|x|/dy = 0, so RHS becomes 0, our integrated modulus on LHS is clearly greater than 0 or equal to zero in the case f(y) = -f(x)

>>7911345
A6 is easy peasy, do a subtraction comparison between sets and allocate an ordered list as appropriate (although there is a fancy way of saying that set theoretically aswell)

>>7911357
Fuck off, honestly

>>7915713
Have faith in yourself :)

>>7911345
Almost certain the answer to B1 is no

>>7911345
I hope the golden years of sci arent over....

>>7914173
Except most people taking the test aren't mathematicians, and most mathematicians would do decently on the putnam.

>>7914556
I feel the same way about competitions. Interesting problems with interesting solutions are only interesting insofar as they help you understand the mathematics better. This idea of mathematicians all just loving the thrill of cleverly solving hard problems applies to only a small handful of geniuses. For most, it's about wanting the best way to explore the subject, and that involves figuring out some challenging problems.

>>7915713
Doing these competitions involves doing lots and lots of problems. It's very, very rarely about creativity. I'd say 100% of Putnam problems can be solved with techniques that are a part of SOME mathematicians' toolkits. The idea is to do so many problems that you have a very large toolkit, and to do enough problems such that you start getting good intuition for when to apply what tools.

>>7915866
Where would I start to do problems? What about for someone with little math background?

I'm >>7914587
B6:

The Stone-Weierstrass Theorem provides that [math] f [/math] is the limit of a uniformly convergent sequence of polynomials [math] \{ p_n \} [/math]. Observe then that
[math] \int\limits_{0}^{1} \int\limits_{0}^{1} |f(x)+f(y)| dx dy = \int\limits_{0}^{1} \int\limits_{0}^{1} | \lim_{n\to\infty} (p_{n}(x)+p_{n}(y)) | dx dy = \lim_{n\to\infty} \int\limits_{0}^{1} \int\limits_{0}^{1} | (p_{n}(x)+p_{n}(y)) | dx dy [/math]
and
[math] \int\limits_{0}^{1} |f(x)|dx = \int\limits_{0}^{1} | \lim_{n\to\infty} (p_{n}(x) |dx = \lim_{n\to\infty} \int\limits_{0}^{1} | (p_{n}(x) |dx = \lim_{n\to\infty} \int\limits_{0}^{1} \int\limits_{0}^{1} | (p_{n}(x) |dx dy [/math]
where the interchange of operations is justified by uniform convergence of [math] \{ p_n \} [/math]. Let [math] p_{n}(x)= \sum\limits_{i=1}^{n} a_{i}x^{i} [/math].

Clearly for all nonnegative integers [math] i [/math] and [math] 0< x,y<1 [/math], [math] |x^i+y^i| \geq |x^i| [/math] which implies [math] \int\limits_{0}^{1} \int\limits_{0}^{1} |x^i+y^i| dx dy \geq \int\limits_{0}^{1} \int\limits_{0}^{1} |x^i| dx dy [/math]. This gives for all [math] n [/math] that [math] \sum\limits_{i=1}^{n} |a_i| \int\limits_{0}^{1} \int\limits_{0}^{1} |x^i+y^i| dx dy \geq \sum\limits_{i=1}^{n} |a_i| \int\limits_{0}^{1} \int\limits_{0}^{1} |x^i| dx [/math], which is equivalent to [math] \int\limits_{0}^{1} \int\limits_{0}^{1} | (p_{n}(x)+p_{n}(y)) | dx dy \geq \int\limits_{0}^{1} \int\limits_{0}^{1} | (p_{n}(x) |dx dy [/math] by the same uniform convergence argument.

Because [math] n [/math] is arbitrary, [math] \lim_{n\to\infty} \int\limits_{0}^{1} \int\limits_{0}^{1} | (p_{n}(x)+p_{n}(y)) | dx dy \geq \lim_{n\to\infty} \int\limits_{0}^{1} \int\limits_{0}^{1} | (p_{n}(x) |dx dy [/math], establishing the result.