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.![\[{x^3}{y^2}\left( {2y - x} \right) = {x^2}{y^4} - 36\]](http://latex.artofproblemsolving.com/d/5/6/d5668b2a486f5264e1cf2faeb363ac69bf91e3f0.png)
, where 
is a positive integer.
with exactly 
elements such that the sum of its elements is a prime number.
+1 w
satisfies the following condition.
, ![$\sum_{k=1}^{n}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n}{k}\right]}\right)a_k=1$](http://latex.artofproblemsolving.com/3/1/0/3109e3f024dc59c2cdad60efa270494d6095c330.png)
holds.
, compute 
.
such that 
for all 
.
. 
amount of people stand on coordinates 
where 
. Every person got a water cup and two people are considered to be neighbour if the distance between them is 
. At the first minute, the person standing on coordinates 
got litres of water, and the other 
people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups. will not have more than 
litres of water.
be positive real numbers such that 
. Prove that 
has centroid 
. A line parallel to 
passing through intersects the circumcircle of at a point 
. Let lines 
and intersect at 
. Suppose a point 
is chosen on such that the tangent of the circumcircle of 
at , the tangent of the circumcircle of at 
and concur. Prove that 
. is called beautiful if, for every integer 
, the base-
representation of contains the consecutive digits 
, 
, , 
(in this order, from left to right). Determine whether the set of all beautiful integers is finite.
such that for all positive integers 
,![\[
f(mn) = \mathrm{lcm}(m, n) \cdot \gcd(f(m), f(n)),
\]](http://latex.artofproblemsolving.com/6/0/7/6071692e2716901e88cc1cfc19ed60388c10701d.png)
where 
and 
denote the least common multiple and the greatest common divisor of 
and 
, respectively. be a function satisfying the following conditions for all 
:![\[
\begin{cases}
f(n+1) > f(n) \\
f(f(n)) = 2n + 2
\end{cases}
\]](http://latex.artofproblemsolving.com/0/4/7/047e28c9d899d7de443ae40ac2c756fda851fafb.png)
Find the value of 
.
differents subjects, each consist of 
students. Given that for any different subjects, there exists a student compete in both subjects. Prove that there exists a student who compete in at least 
different subjects.
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 
coprime, 
and 
.
.
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.
,