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.
a function satifying 
for all 
.
, then what is 
, the answer is in terms of 
.
. The segment 
measures 40 units. Let 
be the orthocenter of triangle and 
its circumcenter. Let 
be the foot of the altitude from 
and 
the foot of the altitude from 
. Additionally, point divides the segment such that 
. If the perpendicular bisector of segment 
passes through point , calculate the area of quadrilateral 
.
be real number such that 
Prove that
Let 
be real number . Prove that
be a triangle. The points 
and 
lie on the segments 
and 
respectively, such that the lines 
and 
intersect in a common point. Prove that it is possible to choose two of the triangles 
and 
whose inradii sum up to at least the inradius of the triangle .
and 
be two positive integer numbers such that 
.Prove that 
divides 
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
be a point interior to triangle (with 
). The lines 
, 
and 
meet again its circumcircle 
at 
, 
, respectively . The tangent line at 
to meets the line 
at 
. Show that from 
follows 
.
. Let 
be the midpoint of the arc 
not containing 
and define 
similarly. Show that the orthocenter of 
is the incenter 
of .
of positive integers such that 
is an integer.
+1 w
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
, the digits in the base-
representation of 
are all greater than .
, , 
, 
, 
, 
represent different digits. Suppose 
is the largest number that satisfies the equation 
What is the value of 
?
, so 
. However, 
results in 
, which is not allowed, thus 
. In fact, 
results in 
which is allowed. Our final answer is 
.
)
can be is 
so let equal to since can't be greater than 
otherwise it would become a 
digit number, which wouldn't satisfy the expression to the right of the equation. digit number we note that can either be 
or as the digits must be distinct since 
. hence .
, , , , , represent different digits. Suppose is the largest number that satisfies the equation What is the value of ?
which is C
works. Then, we get C.
always equals 
then there will be no carryovers before that when doing the multiplication, because if there were carryovers, then 
which is not equal to 
So the problem just condenses into 
because if it were 
then 
then 
and if 
and if 
because 
or 
because 
We can multiply 124 by 8 to get 992, which isn't what we want, because 
in this case. So 
and when we do 
will become 
which satisfies the fact that all of the letters must be distinct.
results in 
,
Then 
.
,
,
gives 
,
and we are done.
