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12 / 3 / 11 / ?

File: question18.png ( 11.57 KB , 693x61 , 1690377094530.png )

Image
  No. 456
>>459 >>457 >>482
Prove you are not a Midwit.
>>
  No. 457 >>458
>>456
sure
ill start with the case where n is between 1 and 6
in this case, ceil(n(n-6)/19) <= 0.
at least 0 elements being unrepresentable is always true.
>>
  No. 458 >>533
>>457
Pre-School level proof.

Let
m:=n(n−1)2m:=\frac{n(n-1)}{2}
,
M:={1,…,m}M:=\{1,\ldots,m\}
and
D:={ai−aj:ai>aj,1≤i,j≤n}={d1,…,dm}D:=\{a_i-a_j: a_i>a_j, 1\leq i,j\leq n\}=\{d_1,\ldots,d_m\}
be the multiset of differences.
Then let
A:=M∩DA:=M\cap D
,
B:=M\A={b1,…,bt}B:=M\backslash A=\{b_1,\ldots,b_t\}
,
C:=D\A={c1,…,ct}C:=D\backslash A=\{c_1,\ldots,c_t\}
. Here
tt
is the number wanted to be larger than
⌈n(n−6)19⌉\lceil \frac{n(n-6)}{19} \rceil
.
Consider the generating function
F(z):=za1+⋯+zanF(z):=z^{a_1}+\cdots+z^{a_n}
,
G(z):=F(z)F(1z)=n+∑(zdk+z−dk)G(z):=F(z)F(\frac{1}{z})=n+\sum (z^{d_k}+z^{-d_k})
,
H(z):=z−m+z−m+1+⋯+1+z+⋯+zmH(z):=z^{-m}+z^{-m+1}+\cdots+1+z+\cdots+z^m
. We have
G(z)=n−1+H(z)−∑(zbl+z−bl)+∑(zcl+z−cl)G(z)=n-1+H(z)-\sum (z^{b_l}+z^{-b_l}) + \sum (z^{c_l}+z^{-c_l})
.
Take
∣z∣=1|z|=1
, then
G(z)=∣F(z)∣2≥0G(z)=|F(z)|^2\geq 0
and
∣G(z)−(n−1)−H(z)∣≤4t|G(z)-(n-1)-H(z)|\leq 4t
. Write
zz
as
e2iθe^{2i\theta}
then
H(z)=sin⁡(2m+1)θsin⁡θH(z)=\frac{\sin(2m+1)\theta}{\sin\theta}
. Take
(2m+1)θ=32π(2m+1)\theta=\frac{3}{2}\pi
, we have
H(z)=−1sin⁡θ<−n2−n+132πH(z)=-\frac{1}{\sin\theta}<-\frac{n^2-n+1}{\frac{3}{2}\pi}
. Hence
∣G(z)−(n−1)−H(z)∣≥23π(n2−(1+3π2)n)|G(z)-(n-1)-H(z)|\geq \frac{2}{3\pi}(n^2-(1+\frac{3\pi}{2})n)
. So we can get
t≥16πn(n−6)t\geq \frac{1}{6\pi}n(n-6)
.

Great problem.
>>
  No. 459
>>456
Isn't the captha enough?
>>
  No. 460 >>514

File: IMG_20230803_204442.jpg ( 434.51 KB , 975x937 , 1691085059640.jpg )

Image
No one in this site can do these two 1/2
>>
  No. 461

File: IMG_20230803_204731.jpg ( 217.67 KB , 939x486 , 1691085174123.jpg )

Image
2/2
>>
  No. 482 >>487
>>456 Where did you get these questions from? Just curious. I could solve the first one with ease, the other ones are a little harder for me.
>>
  No. 487
>>482
All these questions are from China Team Selection Test for IMO.
And also if you actually can solve first one you're pretty above average.
>>
  No. 514 >>533
>>460
Can be done in a couple of ways; summing
⌊akm⌋⌊bkm⌋\lfloor \frac{ak}{m} \rfloor \lfloor \frac{bk}{m} \rfloor
or computing
∫01{ax}{bx}\int_0^1 \{ax\}\{bx\}
works.

For(1),We will show that
λ=∫01{ax}{bx}dx\lambda = \int_{0}^{1}\left \{ ax \right \} \left \{ bx \right \}dx
satisfied. In fact,we would show that
[ab(k−1)m,abkm]∩Z=∅\left [ \frac{ab\left ( k-1 \right )}{m} ,\frac{abk}{m}\right ]\cap\mathbb{Z}= \emptyset

then we are done. This
⟺{akm}{bkm}−m∫k−1mkm{ax}{bx}dx=O(1m)\Longleftrightarrow\left \{ \frac{ak}{m} \right \} \left \{ \frac{bk}{m} \right \} -m\int_{\frac{k-1}{m} }^{\frac{k}{m} }\left \{ ax \right \}\left \{ bx \right \} dx=\mathrm {O}\left ( \frac{1}{m} \right )
and
⟺sup⁡x∈[k−1m,km]{ax}{bx}−inf⁡x∈[k−1m,km]{ax}{bx}=O(1m)\Longleftrightarrow \sup_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} -\inf_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} =\mathrm {O} \left ( \frac{1}{m} \right )
It's obviously right.

■\blacksquare
>>
  No. 533 >>534 >>536
>>458
>>514
>mathchan
>can't even render weblatex
this is sad
>>
  No. 534 >>547
>>533
You have to use 
\[ ... \]
instead of dollars.
>>
  No. 536
>>533
Exactly just what I was thinking lmao
>>
  No. 547
>>534
>he did not read the sticky

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