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.
, 
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
of the acute-angled triangle 
is tangent to its side 
at a point 
. Let 
be an altitude of triangle , and let 
be the midpoint of the segment . If 
is the common point of the circle and the line 
(distinct from ), then prove that the incircle and the circumcircle of triangle 
are tangent to each other at the point .
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide . (Note: it is possible that has other prime divisors not among .) Prove that for every prime number 
, there exists a lucky number such that 
.
, such that for any 
, 
be a non-isosceles triangle with incenter 
Let 
be the smaller circle through 
tangent to 
and 
(the addition of indices being mod 3). Let 
be the second point of intersection of 
and 
Prove that the circumcentres of the triangles 
are collinear.
such that for every 
is satisfied? Explain the answer.
be a scalene oblique triangle, and 
a point on the orthocentroidal circle of (
)., trilinear polar of the polar conjugate (
-isoconjugate) of , Droz-Farny axis of are concurrent. with respect to is as follows:
, there exists a pair of orthogonal lines 
, 
through such that the midpoints of the 3 segments cut off by , from the sidelines of are collinear. The line through these 3 midpoints is the Droz-Farny axis of wrt .
such that 
is the square of an integer.
be an acute triangle with 
. Let 
be the centroid of triangle 
and let 
be the foot of the perpendicular from 
to side 
. The median 
intersects the circumcircle 
of triangle at a second point . Let 
be the point where intersects . The line 
intersects the circle at a point 
, such that lies between and . Prove that the points 
and lie on a circle.
such that there exist positive integers 
and 
with 
satifying the condition that,
is an integer.
.
is a nonzero polynomial with integer coefficients. Prove that there exists infinitely many prime numbers 
such that for some natural number , 
.
that is divisible by 
divisible by 
to get 
is even
-digit decimal representations of the numbers 
[i.e., if the number has less than digits, then add appropriate number of zeroes at the beginning]. Define a map 
as follows:
, where 
if 
is even and 
if is odd. If 
where 
then 
for some 
. If any of the 
are nonzero, let 
be the smallest one. Then, the highest power of 
dividing 
is 
, which is a contradiction. Therefore, 
, which implies that 
is injective. Since 
, we get that is surjective as well.
, we are done.
be our constructed number with digits divisible by with all odd digits. For the base case 
take 
Now suppose we have our number 
that is divisible by 
Then we will add a digit to the beginning of to make 
To add the digit at the beginning, we need to add one of 
or 
Adding any of these will not change whether or not the expression is divisible by Moreover, if we let 
the possible numbers we can add become 
and 
Note that 
so taking mod 
gives the list 
in some order. Then if 
we just add the digit which gives the residue 
which will give The induction is complete, so we have solved the problem.
) is trivial, consider 5.
, all of whose digits are odd and which is a multiple of 
, i. e. 
. such that 
. Clearly there is a unique residue 
satisfying this, and so two such numbers in 
; one odd and one even. Pick the which is odd. Then the number 
works.
The number works.
with 
digits, each digit odd that is a multiple of 
digit numbers are 
.
is the number 
are equivalent to a different residue modulo 5, so one of them must be a multiple of . to create a digit number that is a multiple of with all digits odd.
.
-digit number which is divisible by .
, we will show it works for 
.
is a -digit number whose digits are all odd. Then, consider the following numbers:
Factoring out from these numbers yields
Suppose 
and are unique values selected from this list. Then, 
for some 
. Notice cannot be divisible by , thus and are unique values modulo . Therefore, these five values are all unique modulo . So, one of them must be divisible by , as desired.
.
, assume there exits an integer such that 
, and is -digit number with all digits being odd.
, consider the numbers 
, 
.
, which can be written as 
.
give five different remainders when divided by , so one of them must be divisibly by . is divisibly by for one 
, and has odd digits, as required. 
:![\[
n = 1: 5,
\]](http://latex.artofproblemsolving.com/f/e/8/fe8ca6c2e19e69b639d48bc04d3fc44962947152.png)
![\[
n = 2: 75,
\]](http://latex.artofproblemsolving.com/6/a/6/6a60e969829dfb52fcecd121497c56f3b2edaa45.png)
![\[
n = 3: 375,
\]](http://latex.artofproblemsolving.com/8/1/f/81fcae08e6d5c032ffbfe53c9155a9e4297f3313.png)
![\[
n = 4: 9375,
\]](http://latex.artofproblemsolving.com/f/0/a/f0a93d44a57f75c59dc20838ddc72b798d452930.png)
![\[
n = 5: 59375
\]](http://latex.artofproblemsolving.com/4/1/2/412135ae89dc1e11d17c588abfa07e64438e693b.png)
we can assume that This problem could be solved by induction for the first digit from the left.
we have a 
digit number divisible by 
, we call the 
digit number 
.
such that 
and 
is an odd number
to such that 
digit number: 
such that 
we can simply take the number to be . Let us assume that there exists a positive integer with all odd digits divisible by 
for some , then consider the case of . Let the integer we had before be expressed as 
, then we claim that we can always add an odd number to the front of it to make it divisible by . Indeed we want 
. If it is 
we take 
, if it is 
we take 
, if it is 
we take 
. Thus we get a number of the desired qualities.
, 
satisfies the condition, for the inductive step we prove that we can add an extra digit while still making it divisible by 
.
is an odd number, and divides 
and 
from induction hypothesis., we still want the remaining number to be divisible by , and notice that has odd options which is all the residue class mod , so depending on , such an must exist.
such that 
coprime, 
and 
.
.![\[P=\dfrac{a}{\sqrt{2bc+1}}+\dfrac{b}{\sqrt{2ca+1}}+\dfrac{c}{\sqrt{2ab+1}}.\]](http://latex.artofproblemsolving.com/1/f/3/1f39a54e4cf378bfd4e66d97949350b9b37dc1ec.png)
.
,
digits with all digits odd, from the inductive hypothesis. Also from the inductive hypothesis, they are all divisible by 
between consecutive numbers, i.e. they would all have different residues mod 
were m_l+1 digits 
are odd.
,
is even. We construct
with property: digits 
are odd for 
.
such that 
.
and 
.
.
and 
,
and we shall show that you can always pick some 
.
is 
with an odd digit augmented to beginning of the number. Thus, ![\[A_{k+1}=5^{k}(a\cdot 2^k+b_k)\]](http://latex.artofproblemsolving.com/c/2/3/c23b6eed000458d708c6a85f9c05e7bc01fb065e.png)
is divisible by 
can take on any possible residue modulo 
,
are the digits in the decimal representation, i.e. 
with all 
,
;
;
.
.
forms a complete set of residue classes 
as it runs across the 
be an 
and 
.
.
.
where there are 
.
can be 
but 
.
be 2 distinct 
.
be the smallest integer s.t. 
and 
differs on the 
digit (from the right). I.e.
Hence 
.
of 
and both 
are odd, so 
.
.
is divisible by 
,
,
,
or 
will be divisible by 
.
,
.
.
is divisible by 
.
each produce a different residue mod 
varies which is obvious. Thus, at least one of these values for 
,
.
are distinct 
,
.
.
,
d
such that 
.
:
.
can be added to 
,
-
.
,
where 
.
which obviously has a solution which finishes.
,
works. Now that we are done with the base case, by induction hypothesis we assume that 
,
is clearly divisible by