AoPS Academy
be 
. The tangents to at 
meet at point 
. For a point 
on line 
which is not on the segment , let the midpoint of 
be 
. Lines 
meet again at points 
respectively. Let 
be the midpoint of . Prove that the points 
lie on a circle.
red cards, 
white cards, and 
blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
and a set 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
, for 
in at least 
different ways.
, point 
lies on side and point 
lies on side 
. Let 
and 
be points on segments 
and 
, respectively, such that 
is parallel to 
. Let 
be a point on line 
, such that lies strictly between and , and 
. Similarly, let 
be the point on line 
, such that lies strictly between and , and 
.
, and are concyclic.
with 
. The points 
lie on 
respectively, such that 
. Extend 
to meet the circumcircle of at 
respectively. Let the circumcircles of 
meet at points 
. Extend 
to at , given 
, find ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
with 
, denote the incenter of the triangle as 
. Extend 
to meet the circumcircle of 
at 
, find the length of 
.
and 
. Prove that
Equality holds when 
Equality holds when 
Equality holds when 
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/6/1/a/61ad92b49bd9b4b8839e6f6c4e81c758a11835a3.png)
are all powers of 
). Propose a positive 
performing 
. By the presentation of , we perceive that 
proportionately partitions all parts of the progression. proportionately parts 
.
power of phive. Our party personally presents:
as preferred.
,
purporting 
. Provoking LTE, since 
, 
The partisanship proves 
, proving 
, a preposterous predicament, proving the proposition. 
for some 
. My claim is that do not devide any 
. devides some of .Also assume 
.
Is not a perfect power of which is a contradiction !!
.![\[\Rightarrow a_1^{2018^2} +a_3 \equiv a_{2}^{2018}+a_3 \pmod{5}\]](http://latex.artofproblemsolving.com/6/7/8/678562f143317850a039fbd656121e5fa851bfc2.png)
![\[ \Rightarrow a_1^{2018^{2018}} +a_1 = a_{2018}^{2018}+a_1\equiv 0 \pmod{5}\]](http://latex.artofproblemsolving.com/5/7/2/572ac277302871d81c04af696fd20e9b3e48c66d.png)
do not devide 
hence ![\[ 5\mid a_1^{2018^{2018}-1} +1\]](http://latex.artofproblemsolving.com/c/8/b/c8bc717ca63b4562c0d2024dbd6390ef4eb42e76.png)
which forces ![\[5\mid a_1^{2.2018^{2018} -2} -1\]](http://latex.artofproblemsolving.com/2/d/0/2d04ef0a624adef4cf4bb1c93ff938a2a1ee72d0.png)
it means 
. Clearly It is a contradiction.
. Note that if 
for any 
, we have an immediate contradiction, as then 
will be divisible by a prime other than . Hence, 
for all . Iterating this yields ![\[ \nu_5(a_1) = 2018\nu_5(a_{2018}) = 2018^2\nu_5(a_{2017}) = \cdots = 2018^{2018}\nu_5(a_1), \]](http://latex.artofproblemsolving.com/d/a/e/dae690c335bde48a9f33113dfbff67a08c771804.png)
so 
, implying that 
for all .
. Iterating this yields
Since 
, we have 
, so 
. This is enough to imply that 
for all .
is minimal. Then divides all of the other terms, so
Iterating this yields
where dividing by makes sense because 
(if 
, then would be divisible by a prime other than , as 
). Recalling that 
, we have from LTE that
However, 
, so 
. Therefore, since is a power of , we have
Hence, we must have 
, but 
and 
(since 
). This is a contradiction, so we are done. 
.
is maximal and positive, then 
, so cannot be a power of . 
denote the smallest of the powers of , so all elements are divisible by . Evidently, 
, and for every index , ![\[ a_{i+2} \equiv - a_i^{2018^2} \pmod{5^e} \]](http://latex.artofproblemsolving.com/7/7/8/77850a6652a997353c584a317c45684b35e5fed3.png)
for all .
. But if we iterate this 
times we obtain ![\[ x \equiv x^{2018^{2018}} \pmod{5^e} \]](http://latex.artofproblemsolving.com/d/2/f/d2fd0de1f66c3570c2bcbb885a26eb314fcdbf75.png)
Thus 
has order dividing 
. But since 
, we find 
, So this forces 
.
, and the same is true for any other term. But 
and we have a size contradiction.
for all .
be maximal. It is evident that , a contradiction.
and as is not divisible by , we must have 
for all . Now suppose 
is minimal, so 
Now we are almost done; note that 
so 
which is obviously false.
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/b/a/4/ba4ac7fb2a62792bf76b5914c85994e6111069ff.png)
are all powers of does not divide for any divides one of it divides all.Now consider the number 
where 
is chosen so that 
.
cant be a power of since 
which concludes the claim.
Substituting repeatedly we get 
.Thus 
.Thus 
.But this is a contradiction.
for all .
and 
, if 
is a power of , then 
. Therefore, since 
is a power of for each , we have 
. Iterating this, we have 
since the sequence cycles back, which means 
, and hence 
for all . Therefore, 
for all , as claimed. 
be the smallest power of in the list, WLOG it is 
. (We can WLOG since the variables are cyclically symmetric.) Then all of the elements of the list are multiples of . We have for all (indices cycle) that 
, so![\[-a_{i+1}\equiv (-a_i)^{2018} \pmod{5^m}.\]](http://latex.artofproblemsolving.com/9/9/2/9926d23e3d013808b5cef25106db0a5362fb62b8.png)
Iterating this, we have 
. Since 
, we have![\[(-a_1)^{2018^{2018}-1} \equiv 1 \pmod{5^m}. \qquad (\clubsuit) \]](http://latex.artofproblemsolving.com/c/f/9/cf93d8cf864be5045c475ea053bf882758aa5f95.png)
In particular, said equivalence also holds in 
. Since 
, this implies 
, so 
. Now, 
implies![\[ m\le \nu_5\left( a_1^{2018^{2018}-1}+1\right) = \nu_5(a_1+1) + \nu_5(2018^{2018}-1) = \nu_5(a_1+1)\]](http://latex.artofproblemsolving.com/2/9/8/298c001fa7c58fbe6501024b54e212427307119a.png)
by LTE. Therefore, 
, but 
, a size contradiction.
such that the numbers are all powers of 
we have 
then all the terms of sequence 
is divisible by . Then we will Prove that this is never possible. For it let's start with 
then observe as 
Similarly we can get 
and so on. But then 
will be the clear contradiction.
is the smallest possible power of in this then we will have 
and as 
So we must have 
now this is enough to prove the same for all 
and Hence we will get Size Contradiction by LTE. So we are done there doesn't exist such sequence of natural numbers. 
Since , we have , so . This is enough to imply that for all .
means ?
Since , we have , so . This is enough to imply that for all . means ?
we get 
.
. Let 
be the index such that 
.![\[ a_{i_0}^{2018} + a_{i_0+1} = 5^{k_{i_0}}, \]](http://latex.artofproblemsolving.com/d/f/4/df496af200a5a24bdb1361e4fa05dc3a1356e72d.png)
we know 
, thus 
, but because of size reasons this is a clear contradiction.
. Perform a change of variables 
.
such that 
then we are done because of size reasons. So, 
, which means all of them are equal, making another change of variables 
as otherwise we are done because of size reasons. Thus, ![\[c_2 \equiv -2c_1, c_3 \equiv 4c_1, \cdots, c_{2018} \equiv (-2)^{2017}c_1 \implies c_1 \equiv 2^{2018}c_1 \pmod{5}.\]](http://latex.artofproblemsolving.com/7/9/b/79b1b5c65b21ce878954ce2d099e4a6588223ded.png)
From which we get 
, a contradiction.
. Now if one of the is divisible by , then so are the rest, and if WLOG 
we have a contradiction with 
being a power of 
none of the 
's is divisible by .
is the smallest out of all of those powers of ,so it divides all of the other numbers. We get that 
thus 
. Continuing in this fashion by induction we can get that 
and thus 
so 
. If 
, then 
, so either way 
. This however is impossible if 
because 
, so we must have 
. But then 
, which isn't possible, so the system has indeed no solutions!
and now note that by FLT we have 
hence since 
is 
on mod 5 we get that all 
are on mod 5.
multiple of . be the one with 
maximun compared with the others hence let 
and now using 
we have that 
which is not possible since 
. such that 
and note that its a recursion and in some moment we will get 
which is not possible are 
in mod 5.
where 
is the smallest power of between 
. It suffices to show that 
and now note that by the conditions.
Hence proved!.
Setting 
we get that 
but that means 
is a power of which is a contradiction!! exists! (thus we are done 
)
should be divisible by , otherwise we 
is not a power of where 
. Then using FLT we get
So each is modulo . Write 
. Let 
, with 
. Write each 
. Then 
. By definition not all 
are divisible by . Now note each power of is at least 
(as 
). Hence,
Using 
repeatedly gives
But this means divides every , contradiction. 
, we have two cases:
so 
then 
this obvious is contradiction because 
so 
isn't a power of 
with 
. We rewrite 
is a power of 5 then by LTE 
so 
(otherwise the contradiction is the same like the case 1) but: 
If we divide all original expressions by 
we have that the new expressions are multiples of , but: 
with 
, this implies the problem. 
such that the numbers are all powers of we see 
for all (Just assuming randomly,you can see that choosing any will work).Since we get 
,so repeating this process on other terms continuously gives 
, which is impossible.
Practicing this we get 
,then 
and so on.....finally we derive 
which immediately implies 
so we get that 
for all . as the minnimum of all possible .So 
for all .
Thus we get 
But 
,thus there doesnt exist such numbers 
For someone interested in how this works,
such that the numbers are all powers of does not divide for any divides one of it divides all.Now consider the number where is chosen so that . cant be a power of since which concludes the claim.Substituting repeatedly we get .Thus .Thus .But this is a contradiction..
such that the numbers are all powers of does not divide for any divides one of it divides all.Now consider the number where is chosen so that . cant be a power of since which concludes the claim.Substituting repeatedly we get .Thus .Thus .But this is a contradiction..
is not a power of if there exists 
such that 
. Suppose otherwise; we first narrow down the value of 
.
for any . for any then for all . Thus pick 
such that 
is minimal. We have
but then 
cannot be a power of . is even we have 
for all , which means that we must have 
. But then this means 
, so 
for all .
be the least power of present in the sequence, and WLOG let . We have
so 
as 
. Since 
, by Lifting the Exponent
but since we must have 
(otherwise which is impossible), it follows that is not a power of —contradiction. 
.
. Since it is the minimum, we have 
![$5\nmid a_i\forall i\in [2018]$](http://latex.artofproblemsolving.com/e/9/0/e90f074e19568f67968692cb796eacc35ec4db6a.png)
be the lowest. Then 
Which is not possible.![$$a_1((a_1)^{2018^{2018}-1}+1)\equiv 0\mod 5^{\alpha}\implies a_1\equiv 4\mod 5\implies a_i\equiv 4\mod 5\forall i\in [2018].$$](http://latex.artofproblemsolving.com/b/b/9/bb941ada4dd337e634a86c162fa1d7851f84a24b.png)
is a power of , we get that 
But 
as 
.
such that the numbers are all powers of 
for all 
.
be the lowest power of . We have
Let 
. Then 
and 
imply 
or 
.
for all , which is a contradiction since 
.
claims. are divisible by . then all should be. Hence say 
is the highest then 
which is a size contradiction. 
.
in 
is either 
; 
; 
; 
and only the last one works as it is the only one which is cyclic. 
is the smallest number in the sequence then see that in 
we have ![\[a_1^{2018}= -a_2 \implies a_1^{2018^2} = a_2^{2018} = -a_3 \implies \text{ } \dots \text{ } \implies a_1^{2018^{2018}} = a_{2018}^{2018} = -a_1\]](http://latex.artofproblemsolving.com/6/1/e/61eb12506a259355b236f650e50f5d12da1523cb.png)
And so we have ![\[t \le \nu_5 \left(a_1^{2018^{2018}-1}+1 \right) \overset{\text{LTE}}= \nu_5(a_1+1)+\nu_5(2018^{2018}-1)=\nu_5(a_1+1)\]](http://latex.artofproblemsolving.com/2/d/b/2dbcd72661f8bbacabeeac56e39670f127511c56.png)
And so we have ![\[a_1^{2018}+a_2=5^t \le 5^{\nu_5(a_1+1)} \le a_1+1\]](http://latex.artofproblemsolving.com/6/f/9/6f95df4cd4a7827c1d62a5b73a9db544132786c8.png)
which is a size contradiction (as 
).
integers with these as powers of ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2019})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/1/8/f/18f27b4c93ed5abbe6c0f9f96fc8021059bc3411.png)
. Tumon2001 found it too low lavil for his taste so the typo was fixed
denote the highest exponent of 
for all 
.
Then 
and hence 
,
.
for 
.
Let 
Then since all 
'
for all 
Note that 
.
Proceeding similarly, 
for 
;
divides 
.
,
.
denote the smallest natural number such that 
.
Also, since 
it follows that 
.
,
and therefore 
divides 
,
,
be positive integers such that 
is the perfect power of prime 
.
or(analogously) 
, you immediately get a contradiction.
and so 
.
,
.
,
.
and so 
.
and using the lemma we have 
which thanks to lifting the exponent, reduces to 
,
is a constant.
for some naturals 
(note that since 
,
)
,
and so 
,
.
and so we have 
which implies 
.
.
we must have 
,
Thus there exist no such numbers, as required.
,
If 
then it will divide all 
. Since 
,
,
to be of the least power of 
we get 
.
,
which is not possible as 
and we are done!