[/quote]
,
,
and the function 
defined by![\[f(n)=\left\lfloor \frac{an+b}{cn+d}\right\rfloor\text{ for all } n\in\mathbb{Z}_{\ge 0}.\]](http://latex.artofproblemsolving.com/8/f/b/8fb372df4b44c3d7c0690b310b12c684134fee74.png)
Prove that the following are equivalent:
is surjective;
,
and 
.
and 
be on sides 
and 
respectively in 
such that 
.
be the circumcenter of 
and let 
.
,
is the set of real numbers, such that
be the n-th prime number in increasing order for 
.
satisfying 
for all 
be an acute triangle. Let 
be the midpoint of side 
,
be the feet of the altitudes from 
and 
,
and 
intersect at a point 
,
is perpendicular to line 
such that ![\[f(f(x)) + xf(xy) = x + f(y)\]](http://latex.artofproblemsolving.com/a/6/f/a6f84dd53eec985b90bad3df6580eeaf29d73195.png)
for all positive real numbers 
and 
.
,
with center 
and radius 
intersects the circle 
and 
.
.
such that for any real value of 
we have:
and 
are points, such that 
,
= 
and 
and 
passes through the fix point.
Consider an 
-
in the plane. An eligible set is a non-empty set of the form 
where 
determine the smallest possible number of eligible subsets 
such that :
Y by Davi-8191, Amir Hossein, megarnie, Adventure10, Mango247
A positive integer is called fancy if it can be expressed in the form 
where 
are non-negative integers that are not necessarily distinct. Find the smallest positive integer 
such that no multiple of is a fancy number.
Senior Problems Committee of the Australian Mathematical Olympiad Committee
where 
are non-negative integers that are not necessarily distinct. Find the smallest positive integer 
such that no multiple of is a fancy number.Senior Problems Committee of the Australian Mathematical Olympiad Committee
This post has been edited 1 time. Last edited by MellowMelon, May 18, 2017, 3:30 AM
Reason: add proposer
Reason: add proposer
is the one. Indeed, consider any number 
then 
with 
and 
.
for large enough 
to get 
and using a simple separation 
to separate each 
until we reach exactly 
numbers. Hence, 
is a fancy number. This means 
.
that 
is not a fancy number and 
is not a fancy number. It's obviously true for 
.
,
are not fancy numbers.
or (plus 
)
with 
.
.
so 
can be written as 
for any 
.
so the number of digits of 
.
as a fancy number. This gives a contradiction to the induction hypothesis since 
.
is not a fancy number.
,
with 
implies 
is also fancy, a contradiction to the induction hypothesis. Thus, 
cannot be a fancy number.
![$[2^{100}-1,2^{101}-2]$](http://latex.artofproblemsolving.com/0/7/f/07f0918ccc8329b4c2f163be8c0fc3c76160758d.png)
is a fancy number
where 
,
,
and so on
where 
can take on any number in ![$[0,2^{100}-1]$](http://latex.artofproblemsolving.com/6/a/7/6a7aa17b0e3ab0653cd422582d31184313a44dbb.png)
such that
suppose 
that 
and 
have the same modulo i.e 
to 
which is still divided by 
terms, a contradiction
where 
are different number in 
cannot be divided by 
,
is fancy?
is not fancy, 
is not fancy,.....hey, that's a hell lot of work! I wish I could be lazy and sweep out all those cases with one stroke...and yes, we can! Let's reduce all those multiples to 
,
.
modulo 
for some 
'
are all between 
and 
,
.
,
,
is not fancy).
![\[\begin{array}{ccccc}
2=2^0+2^0&3=2^1+2^0&4=2^1+2^1&5=2^2+2^0&6=2^2+2^1
\end{array}\]](http://latex.artofproblemsolving.com/a/6/a/a6a788d606723c791dea73e8a250d4cec58f93c3.png)
So I find that 
is not fancy when 
Analogously we analyze the case that 
and we can find that 
is not fancy. So I conjecture that the answer is 
for general 
We consider module 
Hence it suffices to prove that![\[2^{a_1}+2^{a_2}+\dots+2^{a_n}\neq 0 \pmod{2^{n+1}-1}.\]](http://latex.artofproblemsolving.com/f/1/7/f17feb4abd4258e83151396e77da27b12643a2b2.png)
Where 
for each 
Let ![\[s_2(n)\stackrel{\text{def}}{=}\text{the sum of the digits of $n$ when it is written in base $2$.}\]](http://latex.artofproblemsolving.com/4/4/1/44104dc89d5119822b8a912fb03783f563e30d38.png)
Clearly 
but 
Contradiction.
.
?
?
?
we do the following
if 
merge 
into 
,
that satisfies 
has at least 
.
.
than:
digits a contradiction.Hence the lemma is proved.
with 
if needed.
is fancy using 
.
can be written as a sum of 
appropriately. For proving that 
after merging same powers of 
and WLOG 
.
is fancy. If 
then we have 
,
,
.
if there exists a multiple of 
.
s
representation of 
less than 
is 
isn't 
.
representation of 
,
.
distinct residues whose sum is congruent to zero modulo 
,
.
digits in binary representation, then just take 
and use the fact that 
can be split into 
to obtain the 
is fancy. This is just equivalent to the multiple having 
mod 
by 
.
where 
times; 
is at least 51 but at most 100, so we can separate exactly 
of the 1's from 
.
for all 
where 
are non-negative integers that are not necessarily distinct. Find the smallest positive integer 
such that no multiple of 
works.Clearly this is fancy re-write this as 
i.e 
i.e 
yet
If 
is even then 
,
the 
,
denote the sum of the digits in binary. The smallest number of powers of 2 required to make 
,
,
,
which means that the stuff above 
.
and freely split powers of 
.
can be written as 
.
,
for all 
since otherwise we can subtract 
to reduce 
we have 
since otherwise 
can be combined into 
to reduce 
we have 
,
.
.
.
as distinct positive integers so that 
and 
(which is possible by expressing 
be a number so that 
and 
(this has to exist because 
)
,![\[2^{x-y} + 2^{x-y } + 2^{x-y + 1} + 2^{x-y + 2} + \cdots + 2^{x - y + (y - 1) } \ \ \ \ \square\]](http://latex.artofproblemsolving.com/2/8/8/288d7fa5a7e1469c7400b4bac3b82231758413c9.png)
as the sum of 
powers of ![\[ \left( \sum_{i=1}^{k - 1} 2^{b_i} \right) + 2^{b_{k + 1}}\]](http://latex.artofproblemsolving.com/e/c/6/ec65d53c467e5e3dc04fe242f7538541b5c7fbd0.png)
is both the sum of 
,
if 
and 
if 
.
.
giving the minimum value 
for some integer 
,
,
,
in binary, so the expanded sum looks something like 
,
,
where 
.
