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.
, suppose 
. Let 
be the midpoint of 
, and let 
be a point inside the triangle such that 
and 
.
, Find 
.
are written on the board.In each step,two random numbers 
and 
are chosen and deleted.Then,the number 
is written instead.What will be the number remained on the board after the last step.
+1 w
such that for all 
the following equality holds ![\[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \]](http://latex.artofproblemsolving.com/f/3/a/f3ac860e90d671f9fb592a6bd55c102a3000daec.png)
where 
is greatest integer not greater than 
of odd degree with real coefficients such that 
.
be an acute triangle with 
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
be positive integers such that ![$\sqrt[n]{a}+\sqrt[n]{b}$](http://latex.artofproblemsolving.com/d/a/a/daaf95cd3381060559529aa4a318e97b727eac31.png)
is an integer. Prove that 
are both the 
th power of 
positive integers.
such that 
is a prime number and 
. Prove that 
is irreducible over ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
such that![\[ \frac {2^n + 1}{n^2}
\]](http://latex.artofproblemsolving.com/9/2/2/922444e7669c326efcad5fde3ea00f8c2d9fa5d8.png)
is an integer.
+1 w
called ‘nice’ if and only if 
and 
are both perfect squares. How many positive integers 
such that 
are composites for all nice numbers ?
satisfies the conditions![\[\begin{cases}f(x+24)\le f(x)+24\\ f(x+77)\ge f(x)+77\end{cases}\quad\text{for all}\ x\in\mathbb{R}\]](http://latex.artofproblemsolving.com/5/5/6/556f545540d4941d058a1c4099b4b86eb35caf0d.png)
for all real 
.
such that
for all 
.
, there exists an arithmetic sequence 
of rational numbers, where 
are relatively prime positive integers for each 
such that the positive integers 
are all distinct.
![\[A = \left\{1+\frac{1}{k} : k=1,2,3,4,\cdots \right\}.\]](http://latex.artofproblemsolving.com/3/d/f/3dfb1d52835627a2977f0074f8df7e7ed677f2a5.png)
can be written as the product of one or more elements of 
, which are not necessarily different. let denote the minimum integer such that can be written as the elements of , which are not necessarily different.
of integers with 
, 
, and ![\[f(xy)<f(x)+f(y).\]](http://latex.artofproblemsolving.com/c/f/f/cff5554b8e72d99f8989448c2621794242d784d8.png)
(Pairs 
and 
are different if 
or 
).
-digit number be 
.
works for 
. works. Then we can prove that 
works.
for some 
.
are distinct modulo 
, then there must be some that makes 
since both 
and are divisible by 
, so we are done.
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 
and 
.
,
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. 
differs on the 
digit (from the right). I.e.
Hence 
.
of 
is either 
and both 
are odd, so 
.
.
is divisible by 
with all odd digits.
,
,
,
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
.
:
.
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