AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a set consisting of non-negative integers such that:
,
, both 
and 
..
such that![\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]](http://latex.artofproblemsolving.com/8/0/0/8009a4709ffd6708023d67db2b544e605d1fbe68.png)
for all real numbers 
satisfying 
.
and 
such that![\[ {a^2+b\over b^2-a}\quad\mbox{and}\quad{b^2+a\over a^2-b} \]](http://latex.artofproblemsolving.com/e/3/b/e3bc96af9f635e1c10e21dc91894c8afbf113a6d.png)
. Points 
and 
lie on segments 
and 
, respectively. Lines 
and 
intersect at point 
. The circumcircles of triangles 
and 
intersect at a second point 
. The circumcircles of triangles and 
intersect at a second point 
. Point 
lies on segment 
such that 
. Prove that triangles 
and 
are similar.
| 
with 
are positive integers and 
is prime (
), we get: 
, both greater than 1, such that the following holds: of degree 
with integer coefficients and for any prime 
, there exists a set 
of at most 
integers, such that![\[\bigcup_{s \in S} \{s,\; Q(s),\; Q(Q(s)),\; Q(Q(Q(s))),\; \dots\}\]](http://latex.artofproblemsolving.com/f/9/1/f919f2391121c2b0551d29d4f09ea6cde631a8ba.png)
contains a complete residue system modulo (i.e., intersects with every residue class modulo ).
such that there exists a prime number , such that 
is a power of 
. is a number of the form 
where is a positive integer.
from the non-zero reals to the non-zero reals, such that ![\[f(x+y) \left(f(x) + f(y)\right) = f(xy)\]](http://latex.artofproblemsolving.com/2/f/e/2fe8a09e628f60faedaa07ceb154891375763a83.png)
for all non-zero reals 
such that 
.
+1 w
and 
such that 
.
players. Every player played every other player once, with no draws. In addition, each player had a numerical rating before the tournament began, with no two players having equal ratings. It turns out there were exactly 
games in which the lower-rated player beat the higher-rated player. Prove that there is some player who won no less than 
and no more than 
games.
be positive integers such that is a permutation of 
, and
is a permutation of 
.
.
2021 IMOC Problems
for all integers . But now 
, so we must have 
.
is attainable; consider 
. 
on both sides, we see that the RHS is 
. Since perfect cubes are only 
, we see that 
.
, we have the following construction: 
. Our answer is 
. 
. This is achievable when 
Now, to show that this is optimal, note that taking 
gives that 
, but cubes are 
. Therefore, it takes at least numbers to get to , so we are done.
for all integers . Note that 
.
( As if 
, then 
, then 
. But we want 
which is not possible)
is mentioned by many posts above
gives 
but since each 
clearly 
Now we construct a solution for 
showing that this is the answer. Note that we can just do 
so this finishes off the problem.
is the minimum. The construction is:
Now I prove minimality.
By taking our original equation modulo 9, we get that 
Thus, 
.
such that there exist integers 
with ![\[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\]](http://latex.artofproblemsolving.com/e/f/a/efaa98e77ac0273284ee3446e4335d8cc4f681cb.png)
is congruent to 4 (mod9)., we need at least 4 terms.
so 
.
.
are coprime, 
,
by Euler's Theorem. Therefore 
or 
.
,
.
.
,
is not satisfies. We have ![\[ 2002^{2002}\equiv 4^{2002}\equiv 4\pmod 9.\]](http://latex.artofproblemsolving.com/5/e/6/5e672ff0642825fc556062006af9ebb4f5642969.png)
Since 
,
for 
and 
,
from Euler's Theorem. Since the set of cubic residues mod 
,
.
so 
.
works.
where 
.
works.
.
.
,
.
.
is indeed a really big number, so it would be quite nice if we could somehow reduce it to something feasible, at least something to the order of a thousand... wait this is a no-brainer, let the 
of our numbers be 
,
to get:
.
must satisfy this equation- aha so we have found that 
(aka 
and 
)
will be quite useless here, so let's start with 
.
:
-
,
,
we get that![\[x_1^3 + x_2^3 + \cdots + x_t^3 = 2002^{2002} \equiv 4 \pmod{9}\]](http://latex.artofproblemsolving.com/0/3/0/030592aae3cd928965b8458eb339b21232320616.png)
and since the only cubic residues modulo 
,
.
:![\[(2002^{667})^3 + (2002^{667})^3 + (10\cdot 2002^{667})^3 + (10\cdot 2002^{667})^3 = 2002^{2002}\]](http://latex.artofproblemsolving.com/4/c/3/4c3ba5157d0469fa46b02bd79f8aed1a3cde2646.png)
Using 
we get 
so we need at least 
to end the problem.
and cubes are 
in 
.
.
,
Taking the equation 
and 
.
and 
:
.
.