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.
be a cyclic quadrilateral. Let 
, 
, 
be the feet of the perpendiculars from 
to the lines 
, 
, 
, respectively. Show that 
if and only if the bisectors of 
and 
are concurrent with 
.
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 positive integers. Prove that it is impossible to have all of the three numbers 
to be perfect squares.
such that for all real numbers 
and 
,
be a quadrilateral inscribed in a circle of center 
. Let M and N be the midpoints of diagonals 
and 
, respectively and let 
be the intersection point of the diagonals and of the given quadrilateral .It is known that the points 
are distinct. Prove that the points 
are concyclic if and only if the points 
are concyclic.
such that there exists a pair 
of positive integers, such that 
is not divisible by the cube of any prime, and 
such that has exactly 2000 prime divisors and divides 
?
be triangle with 
and 
on the segment such that 
. Consider point 
on 
such that 
. Point 
is the intersection of tangent line through point to circle 
respectively . If the tangent through point to 
intersect at 
. Line 
intersect at . Prove that 
bisect 
.
grid is made. 
unit squares are created by separating the side with length 
in portions and separating the side with length 
in portions. Find, with proof, a formula for how many rectangles with sides parallel to the grid can be created in terms of and . For example, when 
and 
, it can be found by casework that 
rectangles can be created.
such that 
.
and 
intersect at the point 
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
families, there are 
children respectively. If a random child from any of the families is chosen, what is the probability that the child has 
siblings? Express your answer as a common fraction.[/quote]
,
people, exactly 
.
lies on diagonal 
.
.
has 
,
,
.
.
to 
have the digit 
and are divisible by 
now, we have covered all the numbers in the 7 ones place. (27, if I counted right) now, lets find numbers with tens digit 7 that do NOT have 7 in its one place, since we already counted the _77 in the first case.
this totals up to 25 numbers, so the total is
so 
?
without a calculator.
for 
:
:
.
and 
where the rest has been omitted because of the floor function. Now 
and multiplying by 
yields 
grid, each unit square is created by dividing the sides of lengths 
.
horizontal lines and 
vertical lines:
that can be formed, we multiply the number of ways to choose the horizontal lines and the ways to choose the vertical lines:
satisfy 
,
and 
are positive integers, what is the probability that 
is divisible by 
,
have?
not 
has the most factors?
?
.
,
represents a distinct digit from 
to 
.
is our answer. (By python, 
does not work and surprisingly, 
is the only permutation number that is divisible by 
8907