Točno
13. ožujka 2018. 22:02 (6 godine, 4 mjeseci)
Let
![A = (a_1, a_2, \ldots, a_{2001})](/media/m/d/e/a/dea1bb6c045d836086dac9a2662ba5c6.png)
be a sequence of positive integers. Let
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
be the number of 3-element subsequences
![(a_i,a_j,a_k)](/media/m/e/2/c/e2c5aee66748fb5f5ae6f6de58c3e546.png)
with
![1 \leq i < j < k \leq 2001](/media/m/7/8/a/78aee77a7aa45f72d01c1e0c4f1829a9.png)
, such that
![a_j = a_i + 1](/media/m/8/d/b/8db3887b5379b05667605bcab808066f.png)
and
![a_k = a_j + 1](/media/m/5/8/5/585dc667a841b5401156af6eebd83025.png)
. Considering all such sequences
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
, find the greatest value of
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
.
%V0
Let $A = (a_1, a_2, \ldots, a_{2001})$ be a sequence of positive integers. Let $m$ be the number of 3-element subsequences $(a_i,a_j,a_k)$ with $1 \leq i < j < k \leq 2001$, such that $a_j = a_i + 1$ and $a_k = a_j + 1$. Considering all such sequences $A$, find the greatest value of $m$.
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Prvo izaberemo prvih
![667](/media/m/2/a/f/2af78f34973b2d2a5e5b17771714304b.png)
članova kao
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
sljedećih toliko sa
![a+1](/media/m/7/1/5/7153c0612039acb72638f6dc8f9eb833.png)
i zadnjih
![667](/media/m/2/a/f/2af78f34973b2d2a5e5b17771714304b.png)
sa
![a+2](/media/m/c/6/c/c6c99b89231dd906701c20fad0d6d43e.png)
, ovom konstrukcijom uviđamo da je
![m \geq 667^3](/media/m/c/d/2/cd21da725653f5fccf2f586a3cf41fbe.png)
Uočimo da su u tročlanom podskupu svi elementi različiti modulo
![3](/media/m/b/8/2/b82f544df38f2ea97fa029fc3f9644e0.png)
to nas motivira da posmatramo neki skup
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
sa
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
brojeva kongruentni
![0](/media/m/7/b/8/7b8b0b058cf5852d38ded7a42d6292f5.png)
modulo
![3](/media/m/b/8/2/b82f544df38f2ea97fa029fc3f9644e0.png)
,
![y](/media/m/c/c/0/cc082a07a517ebbe9b72fd580832a939.png)
brojeva kongruentni
![1](/media/m/a/9/1/a913f49384c0227c8ea296a725bfc987.png)
modulo
![3](/media/m/b/8/2/b82f544df38f2ea97fa029fc3f9644e0.png)
i
![z](/media/m/d/2/4/d241a79f1fdd0ce9a8f3f91570ba5d62.png)
brojeva kongruentno
![2](/media/m/e/e/e/eeef773d19a3b3f7bdf4c64f501e0291.png)
. Uočimo da vrijedi slijedeća tvrdnja
![-\enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace\enspace \enspace \enspace \enspace m \leq \#(](/media/m/0/a/e/0ae35eb0d5cdafbca24aa972218a5f26.png)
podskupova
![(x,y,z)](/media/m/5/2/e/52e5ec762ff5263770c6e6c12cb9838e.png)
gdje su
![x,y,z](/media/m/b/7/2/b72c022e9d438802d328d34eb61bb4ba.png)
u parovima različti
![mod \enspace 3 \enspace) \leq xyz \leq (\frac{x+y+z}{3})^3\leq 667^3](/media/m/3/7/c/37c5f1293a41ffeb582a36cd1feac737.png)
Kombiniranjem konstrukcije i gornje ograde na
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
zaključujemo da
%V0
Prvo izaberemo prvih $667$ članova kao $a$ sljedećih toliko sa $a+1$ i zadnjih $667$ sa $a+2$, ovom konstrukcijom uviđamo da je $m \geq 667^3$
Uočimo da su u tročlanom podskupu svi elementi različiti modulo $3$ to nas motivira da posmatramo neki skup $A$ sa $x$ brojeva kongruentni $0$ modulo $3$, $y$ brojeva kongruentni $1$ modulo $3$ i $z$ brojeva kongruentno $2$. Uočimo da vrijedi slijedeća tvrdnja
$-\enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace\enspace \enspace \enspace \enspace m \leq \#($ podskupova $(x,y,z)$ gdje su $x,y,z$ u parovima različti $mod \enspace 3 \enspace) \leq xyz \leq (\frac{x+y+z}{3})^3\leq 667^3$
Kombiniranjem konstrukcije i gornje ograde na $m$ zaključujemo da $$m=667^3$$