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 an even integer, and let 
denote the number of (positive) divisors of . does not divide 
.
+1 w
such that 
for all 
a regular hexagon inscribed in a circle 
. Let 
be a point inside and 
its polar reflection wrt . The rays 
meet again at 
. Call 
the polygon formed by the vertices . Similarly construct the polygon 
using instead. and are congruent.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
with an area of 
, with 
. If 
, Hence find the value of 
.
such that for all 
we have:
be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
Let 
be positive reals such that 
.Prove the inequality 
pieces of equal weights. Can Vova prevent Pasha's win?
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
has been partitioned into disjoint pairs 
(
) so that for all 
, 
equals 
or 
. Prove that the sum ![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
or we can just switch 
and 
.
.
is odd. is odd as well. is the sum of 
's or 's and is odd, we know there must be an even number of 's. Let 
be the number of 's.
as desired. QED.
and 
. For 
, we say that 
links onto if if 
or 
. links onto if and only if links onto , there are an even number of ordered pairs of integers 
such that pair 
links onto pair .
, there are an odd number of integers between and , so the number of pairs that link onto is odd. Thus, the total number of ordered pairs of integers such that pair links onto pair is congruent to![\[ \sum_{i=1}^k(\text{number of pairs that link onto pair \textit{i}}) \equiv \sum_{i=1}^k 1\equiv k \bmod 2. \]](http://latex.artofproblemsolving.com/d/3/2/d329c536fdd045d921eb518e92a266b261a917f2.png)
Therefore, since we already know that this number is even, we find that 
is even, and the value of the given sum is![\[ 6k+(999-k) = 999+5k\equiv 9\bmod10. \]](http://latex.artofproblemsolving.com/5/6/7/567b4b4748515f02829e222347cb938debad384f.png)
modulo 
part. Why can we add the numbers like that? (I don't really get the 
part).
implying the desired sum is congruent to 
Combining gives the result.
comes 'even-even' or 'odd-odd'. We claim that the number of 's must be even. If odd then we get greater than odd or even number, this is a contradiction because 
have even and odd. So
and 
For 
note that 
since absolute value doesn't matter in 
So our sum is just 
For 
just notice that its always 
for all of 
so our answer is just 
Using CRT, we get an answer of 
for all integers 
.
and 
, so by CRT we have that the sum is 
, as desired. 
the sum of 999 terms has a remainder of 1 when divided by 2 similarly we see that 
so the sum of 999 terms has a remainder of 4 when divided by 5 so by CRT we see that the sum has a remainder of 9 when divided by 10 which implies that the units digit is 9
. Note that 
. Also note that has the same parity as 
which clearly odd, so 
. Hence the remainder of 
is unique by the Chinese Remainder Theorem, and it is easy to check that it is indeed.
the sum of 999 terms has a remainder of 1 when divided by 2 similarly we see that so the sum of 999 terms has a remainder of 4 when divided by 5 so by CRT we see that the sum has a remainder of 9 when divided by 10 which implies that the units digit is 9
???
and differ by 
, the ending digit is whenever there is an odd number of such that 
and is 
otherwise. There are odds and evens so there can't be an even number of such that , because that would imply an even number of odds and an even number of evens. Therefore, the ending digit must be .
has been partitioned into disjoint pairs () so that for all , equals or . Prove that the sum ends in the digit .
, so it remains to show that it’s 
.
for any integers 
and 
, this means![\[|a_1-b_1|+|a_2-b_2|+\dots+|a_{999}-b_{999}|\equiv 1+2+\dots + 1998\equiv 1\pmod 2\]](http://latex.artofproblemsolving.com/f/4/2/f42a811a8eb834c662a16847d8f14d3074516c22.png)
and we are done. 
for each .![\[\sum a_i + \sum b_i = 999 \cdot 1999,\]](http://latex.artofproblemsolving.com/c/5/c/c5cfa60d865257af1f2074885e56c96b03777da1.png)
is odd. Thus, the number of 's, denote as 
, is even. We have![\[\sum a_i - \sum b_1 = 6 \cdot 2k + (999-2k) = 10k+999 \equiv 9 \pmod{10}. \ \square\]](http://latex.artofproblemsolving.com/e/f/d/efd5f612eb6c209b6e41929f00282ac3ca0737de.png)
or . Say for of these pairs. The sum would then be equal to 
. Note that a pair with difference has same parity and a pair with difference does not. All pairs of difference would give us an even number of even numbers and an even number of odd numbers. If was even, that suggests there are even numbers of odd numbers and even numbers of even numbers. However, since there is a total of even numbers and odd numbers, this cannot be true. So thus, is odd, and the sum would end with digit .
such that 
.
six-pairs 
such that 
. Then, there are 
one-pairs such that . The given expression evaluates to 
For this to be 
, it suffices to show that is even.
, in order to make successful pairings, we must have either exactly one of the six-pairs 
or one of the following sets of three six-pairs 
is even, as desired 
is 
, the sum is clearly . It suffices to show that the sum is odd. Clearly the absolute value when taking the parity, so the sum has the same parity as . Let 
. We have 
is odd, so 
has the opposite parity of 
, implying that is odd, as desired.
pairs that sum to 
and 
pairs that sum to 
Then 
Meanwhile, if WLOG for all 
then our sum equals 
so by CRT it ends in the digit 
so we are done. QED
(mod )
(mod )
(mod )
(mod )
(mod ) (mod 
) as desired.
such that:
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
Prove that
Let 
Prove that
Prove that
.
as the number of pairs where 
,
as the number of pairs where 
.
.
,
and 
are positive integers, where the middle numbers weren't any 
or 
.
and 
from the lemma, we know that
.
.
)
.
.
,
.
,
is odd.
for all 
since 
for all integers 
is odd, or equivalently, 
and 
have different parities. Assume otherwise. This implies 
is even. However,![\[\displaystyle \sum_{i=1}^{999} a_i + \displaystyle \sum_{i=1}^{999} b_i = \displaystyle \sum_{i=1}^{1998} i = 999 \times 1999\]](http://latex.artofproblemsolving.com/b/5/f/b5f1d2f0aec740f64bf91caa0089ff75c0ebeffe.png)
is odd. Since the only possible values 
is even. Let the number of integers 
We wish to show that 
which is true since we have already shown that 
.
.
such that ![\[\sum_{i=1}^{999} |a_i-b_i| \equiv k + (999-k)\cdot 6 \equiv 4-5k\pmod {10}.\]](http://latex.artofproblemsolving.com/b/9/0/b90e9dfcff5c8390e8e5554a9561e2e29d3d399c.png)
However, since 
is even.
,
,
denote the number of even-even, even-odd, and odd-odd pairs. Clearly, there are 
evens, hence 
as desired. 
![\[ S = \sum |a_i-b_i| \equiv \sum (a_i+b_i) = 1 + 2 + \dots + 1998 \equiv 1 \pmod 2. \]](http://latex.artofproblemsolving.com/2/b/2/2b2d29e81ec5446574365666892450113fcce7a5.png)
Modulo ![\[ S = \sum |a_i-b_i| = 1 \cdot 999 \equiv 4 \pmod 5. \]](http://latex.artofproblemsolving.com/3/5/6/3569ffd364e1e1e1b1e457ab0cee1d62330f8982.png)
So 
.
for all integers 
regardless of whether 
.
.
equals 
Notice that 
and we are done. 
and we have 
,
![\[(999-x)+6x = 999+5x,\]](http://latex.artofproblemsolving.com/5/f/4/5f4a6b12990e8b7e3635bd8c5e38bb58cb379b5e.png)
which has units digit either 
,![\[\lvert a_i - b_i \rvert \equiv a_i - b_i \equiv a_i + b_i \mod 2.\]](http://latex.artofproblemsolving.com/d/e/1/de1a1a014ccc2723f786957c0c1b60f128ba099a.png)
we need to prove that this is 
note that 
has the same parity as 
so we need to prove 
so 
hence proved 
![\[ \displaystyle\sum |a_i-b_i| = \displaystyle\sum a_i - b_i \equiv \displaystyle\sum a_i + b_i = \dfrac{1998\cdot 1999}{2} \equiv 1 \pmod{2}. \]](http://latex.artofproblemsolving.com/c/a/c/cac30537422175850423b5e4cc253d9f555352b3.png)
![\[ \displaystyle\sum a_i - b_i \equiv \displaystyle\sum 1 = 999 \equiv 4 \pmod{5}. \]](http://latex.artofproblemsolving.com/6/f/4/6f4bbef6189b53a64a92b4a7057f57f4776631db.png)
.
,
.
.
,
,
Note that 
,
Now from the Chinese Remainder Theorem, 
.
,
,
.
,
.
.
The only thing left is to prove that 
However, 
.
modulo 5. But that is trivial since there are 999 terms in the sum, each of which are 1 modulo 5.
is odd.
,