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 of integers with the following properties:
.
and 
, then 
.
, 
is composite, then all positive divisors of are in . contains all positive integers.
+1 w
and 
be positive integers satisfying the equation ![$(a, b) + [a, b]=2021^c$](http://latex.artofproblemsolving.com/2/4/6/246ccbb22acecb7adc72219afb8afb34c9b05101.png)
. If 
is a prime number, prove that the number 
is composite.
be the set of real numbers. Determine all functions 
such that ![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
and 
.
into 
or 
can sometimes be quite complex. That's why I’ve written a program to assist with this process.
be real numbers such that 
Prove that 
Where 
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even.
, prove that there exists a positive integer 
, such that for any integer 
, 
can be expressed by a sum of positive integers 
's as![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
, 
, 
, 
and 
.
equally spaced points on a circular track 
of circumference . The points are labeled 
in some order, each label used once. Initially, Bunbun the Bunny begins at 
. She hops along from to 
, then from to 
, until she reaches 
, after which she hops back to . When hopping from 
to 
, she always hops along the shorter of the two arcs 
of ; if 
is a diameter of , she moves along either semicircle. arcs which Bunbun traveled, over all possible labellings of the points.
satisfying 
and![\[ a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6{,}000{,}000.\]](http://latex.artofproblemsolving.com/0/4/4/0446861fb2e63782803c4f182a529b79e213fdfd.png)
sadge
to get 
, so ![\[100(ab + bc + ca) = 2{,}000{,}000 + abc.\]](http://latex.artofproblemsolving.com/f/5/8/f5875bf64f0a266440a09bec2652d5a5c87fc823.png)
Rearrange this to 
. Ok let's try to factor this again. Add 
to both sides to get ![\[(a - 100)(b - 100)(c - 100) = 0.\]](http://latex.artofproblemsolving.com/f/d/7/fd7a08361cd4d39a4e0a4636deb54e5ba7a2d5b9.png)
. Then we just need 
, so there are 
options. Multiply this by 
, but 
is overcounted twice, so the answer is 
.
, achieved by 
and permutations for nonnegative integers at least 
. One can verify that these work. To show that this gives exactly 
solutions, note that if 
, then there are 
possibilities for and three possibilities to place (the possibilities won't overlap as ), so we have 
. Now if 
, then we have to add in one additional solution 
, giving in total. Now we prove must be a permutation of 
(in other words, one of the variables is ). is a solution, so now assume there exists at least two of 
that doesn't equal . WLOG 
. Note that 
.![\[a^2 b + a^2 c + b^2 a + b^2 c + c^2a + c^2 b = ab(a+b) + c^2(a+b) + c(a^2 + b^2)\]](http://latex.artofproblemsolving.com/0/4/f/04fb03668c244156ac5ad71f93c7807c9625f477.png)
, which equals![\[ ab(300 - c) + c^2 (300- c) + c((300-c)^2 - 2ab) \]](http://latex.artofproblemsolving.com/b/d/3/bd36f7ada933440517fdec53a591788feea0ded5.png)
This can be further simplified to 
. Dividing by , we get![\[ ab(100 - c) + 100 c(300 - c) = 2,000,000\]](http://latex.artofproblemsolving.com/5/4/b/54bf18cf29701c9474238ea236831f395e7e36e9.png)
This implies 
. However, we see that 
, so since 
, 
must equal 
. Now we find 
and 
(using the fact that 
). We have that 
are the roots of the quadratic![\[ X^2 - (300 - c) X + (100c - 20,000) = 0\]](http://latex.artofproblemsolving.com/7/a/0/7a0ee8888e1ae800028ce583472996eac4626461.png)
By inspection, 
is a root of this equation, and the other root is 
. Therefore one of the variables is . This means must be a permutation of 
for some .
, so ![\[ 100(ab+bc+ac)-abc = 2 \cdot 10^6. \]](http://latex.artofproblemsolving.com/5/b/b/5bb55b52bccc4e535b372771bdf84902c622269d.png)
Let 
be the monic polynomial with roots , , and . Then it is easy to verify given the prior equation and 
that 
. WLOG 
. Plugging into the original equation, we have that and only need to satisfy 
. If neither and are , then we obtain triples. Adding back the triple, there are 
triples.
and 
.
.
. It follows that either or 
. The second condition cannot be cyclically true for all 
(simply plot 
on a number line: the absolute difference between the middle two cannot equal the absolute difference between the outer two unless all are equal.) It follows that WLOG . Thus from the condition, we get 
for all integers 
cyclically are the only valid solutions, giving .
are the roots of the equation 
which 
. Now, we have that
, then we have 
, which shows that any ordered pairs 
will satisfy. Therefore, we need at least one of to be equal to , and the rest follows.
.
Unless one of is 100, exactly one of 
is a different sign than the others, contradicting Fermat's Last Theorem upon rearranging the equation. One of being 100 is also sufficient for equality. We can then extract .
.
and 
.
.
:
is 0. We can proceed with casework.
so we split into cases:
such that 
. Since 
.
, the equation cancels out and we get 
which is true, so as long as one variable is 100, we are fine.
, we get 
, this means that one is 0 and the other is 100.
obviously gives 
and permutations, which doesn't work.
and 
so 
. This once again implies that 
and the only way to still satisfy our case 2 is to have 
from which we get 
implying once again that 
. If none of these are divisible by 10, 
, so we get a contradiction. If one of them is divisible by 10, then one of 
.
but we have some numbers in the form 
and others in the form 
for positive integers 
. However, if we have done that, then WLOG let them be 
, but then 
, a contradiction to 
, and the same if we had 2 numbers with factors of 2, we would end up with the same contradiction that none of these will satisfy.
to both sides of the second equation, we obtain 
so , , and are roots of the equation 
Since is a root of this equation, we have 
work.
for some 
. We verify the second equation: 
as desired. such that , , and 
.
: There is clearly one case.
: If , then , so there are 
total ways here.
or something similar. The answer is .
And substituite 
to make that 
Solve this as a quadratic in 
to get or . i.e. our condition is equivalent to one of the variables being . Trivial counting gives 
(i dont remember exactly if it was 8032038 or some other number, so correct me on this)
(i posted this at 12:29) lol
?
Prove that 
From the second condition we find 
Now 
However 
is a root of this, so as long as one of 
the second condition will hold. Thus simple counting gives 
so 
,
,
.
.
.
for 
.
so they can go to Jane Street and then make 
dollars.
work
Thus we must have one of 
![\[\sum_{sym} a^2b = \sum_{cyc} ab(a + b) = \sum_{cyc} ab(300 - c) = 300 \left( \sum_{cyc} ab \right) - 3abc\]](http://latex.artofproblemsolving.com/0/1/c/01c9fdd1c151521a926b725c7efd03d13051fc05.png)
![\[= 3000000 - 30000 (a + b + c) + 300 \left( \sum_{cyc} ab \right) - 3abc - 3000000 + 9000000\]](http://latex.artofproblemsolving.com/2/6/c/26c5208457abf64d413063a02c5474a10369847a.png)
if and only if,![\[3000000 - 30000 (a + b + c) + 300 \left( \sum_{cyc} ab \right) - 3abc = 0\]](http://latex.artofproblemsolving.com/2/a/d/2ad4eeb19fa3f0a944d74e2a111922e008d4030f.png)
![\[\iff 3(100 - a)(100 - b)(100 - c) = 0\]](http://latex.artofproblemsolving.com/a/3/a/a3a0d914a216c3510c259e26990b23ee30c27f3e.png)
cases. Answer is 
by alg manipulation and Jensen's Inequality. Thus, equality cases can only occur when 
triples along with 
,