,
.
lie on a moving circle 
.
at 
and 
at 
.
intersect at 
.
intersect at 
.
is the perpendicular image of 
.
and the orthocenter 
are collinear.
be a positive integer. Let 
be the number of pairs of integers 
satisfying 
for odd values of 
.
be the number of pairs of integers 
.
and 
be positive integers and let 
be a set of 
of 
![\[ k < \frac {1}{2} + \sqrt {2 \cdot n}
\]](http://latex.artofproblemsolving.com/d/a/5/da56146bf1ad38f257e203e4a0231bd604060a7a.png)
Y by
For each integer 
we consider the last digit different from zero in the decimal expansion of 
. The infinite sequence of these digits starts with 
. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
we consider the last digit different from zero in the decimal expansion of 
. The infinite sequence of these digits starts with 
. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
analysis, and thus it suffices to look at the last digit mod t. Then we can verify the sequence repeats, and this gives (I think?) All even numbers.
for every integer 
to reduce it down to the digits 2, 4, 6, 8. Now, using Legendre and the fact that 
one can show that if 
for some positive integer 
,![\[\frac{n!}{10^{\nu_5(n!)}} \equiv 2^{\ell} \pmod 5\]](http://latex.artofproblemsolving.com/e/f/2/ef24c9784a5588939d2d7899dac76317e3893b20.png)
which is enough, as 
all have different and non-zero residues modulo 5.
be the given sequences. Use 
as above to show that all members of the sequence must be even.
,
,
all even digits appear once (you can find this if you simply start to calculate the first couple terms of the sequence). When 
,
.
is not so easy. In the example above we have 
as the ending 
does not change anything.
up to 
where we have 
and 
by the same argument as before. Here, we can compute all 
,
,
,
,
,
if it works for 
.
,
all appear infinitely often.
.
where 
is an odd number
,
,
,
denote the nonzero part of 
This works perfectly until 
.
,
(Keep in note that 
,
is odd, it can't appear). Using these, with enough patience , one can find a loop
