/sci/ - Science & Math

[Anonymous No.4280688]

Putnam/IMO problem of the day from
http://math.harvard.edu/putnam/

Let <span class="math">S[/spoiler] be a set of n distinct real numbers. Let <span class="math">A_S[/spoiler] be the set of numbers that occur as averages of two distinct elements of <span class="math">S[/spoiler]. For a given <span class="math">n \geq 2[/spoiler], what is the smallest possible number of elements in <span class="math">A_S[/spoiler]?

[Anonymous No.4280698]

1

[Anonymous No.4280697]

2n - 3

[Anonymous No.4280717]

1, 10, 100, 1000, 10000, etc...doesn't have averages

the answer is 0

[Anonymous No.4280732]

>>4280717
(1+10)/2 = 5.5
(1+100)/2 = 50.5

Those are averages.

[Anonymous No.4280740]

>>4280697
This would be my geussem too. We can show that it's an upper bound as follows: If S is the integers 0 through n-1, then <span class="math">A_S[/spoiler] is the set of half-integers 1/2 through n-(3/2), of which there are 2n-3.

[Anonymous No.4280797]

>>4280740
Nice, all that's left to show is that this is the minimum, which is easy enough.

Suppose you have n distinct real numbers
<span class="math">x_1 < x_2 < ... < x_n[/spoiler]

Then <span class="math"> x_1 + x_2 < x_1 + x_3 < ... < x_1 + x_n[/spoiler]
There are clearly n-1 of these.
And
<span class="math"> x_1 + x_n < x_2 + x_n < x_3 + x_n < ... < x_{n-1} + x_n [/spoiler]
Of which there are clearly n-2 of these.

So there are at least n-1 + n-2 = 2n-3 averages.

[Anonymous No.4280900]

>>4280797
Well that concludes the /sci/ math business for the day. Back to Philosophy and Religion!

[ponymath !JD241DkHoE No.4280950]

You should be able to solve this.

Let <span class="math">a[/spoiler] be a positive integer and let <span class="math">X[/spoiler] be a random variable with the Poisson distribution and mean <span class="math">a[/spoiler]. Prove that for any prime <span class="math">p[/spoiler], the <span class="math">p[/spoiler]th moment of <span class="math">X[/spoiler] is <span class="math">\equiv 2a \pmod p[/spoiler].

[Anonymous No.4281092]

There is no way this was an actual Putnam question, shit is trivial as hell.

[Anonymous No.4281298]

/sci/, really quickly here, why is it that d(u(t))/dt / (u(t)) when integrated gives ln|u(t)| I mean this would make total sense if it was

d (1/u(t))/dt but it makes no sense to me as it is above.

[Anonymous No.4281310]

>>4281298

because chain rule, when you differentiate ln(u) you do

d(ln u)/du * du/dt=u'(t)/u(t)

[Anonymous No.4281317]

>>4281310

Right, that was pretty retarded of me, thanks for the quick response!

[Anonymous No.4281436]

>>4281092

B1 from 1992... it's easy on purpose

[Anonymous No.4281442]

>>4281092
>trivial as hell
>no proof

[Anonymous No.4281451]

>>4281442
>result already shown
>herpin and derpin

[Anonymous No.4281465]

>>4281442
There seems to usually be someone in each thread who shows up just to complain about how trivial the problem is while offering either no answer or a completely wrong, <10 word hint at an "answer".

Ofc if you show up after it's been done already (or just google it) there's no risk of looking like an idiot.

[Anonymous No.4281519]

n^2-n

This is just my guess though; I'm probably misunderstanding the question.

[Anonymous No.4281675]

>>4281092
It was probably a USAMO problem, and it was probably supposed to be one of the easier ones.

[Anonymous No.4281692]

>>4281675
1992 B1

http://amc.maa.org/a-activities/a7-problems/putnam/-pdf/1992.pdf

Why didn't you just google it like >>4280797 did?

[Anonymous No.4281713]

>>4281692
Okay, so right answers are always Googled. The question was so easy I'd expect any of the people here who like math and have a moment to spare to be able to solve it.

Also, that's not how I roll.

[Anonymous No.4281780]

>>4281675
the easiest putnam problem are actually usually easier than the easiest usamo problems

[Anonymous No.4282643]

>>4281780
First question on this test isn't too hard:

http://amc.maa.org/a-activities/a7-problems/USAMO-IMO/q-usamo/-pdf/usamo2008.pdf

Didn't take me any more time than the one in this thread.

[Anonymous No.4282673]

>>4282643
Plus, USAMO gives you 3x more time per problem.

[Anonymous No.4282949]

>>4281519
I see where you're coming from, and that was my first thought too (well, averaging is commutative, so it'd actually be (n^2-n)/2, because the average of a and b is the same as the average of b and a).

But remember that the question is asking for the lowest possible number of elements in <span class="math">A_S[/spoiler]. The reason that's lower than our first thought is because once you have four or more elements, some of the averages are going to be the same. Just look at S={1,2,3,4}. The average of 1 and 4 is the same as the average of 2 and 3.