AoPS Academy
with 
with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers 
such that all unit squares on the board can be made into all black after a finite number of moves.
such that![\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]](http://latex.artofproblemsolving.com/6/2/4/624afa78b68f810aaaecf0e1e2251047beeddc36.png)
for all rational numbers 
.
such that![\[2^{n!}+1\mid 2^{m!}+19\]](http://latex.artofproblemsolving.com/8/7/e/87ede5616cffcb33aa187f5d8a13827470f741b4.png)
with 
. Points 
and 
are on 
and 
respectively such that 
. Lines 
and 
meet at 
. 
is the incenter of 
and 
is the orthocenter of 
. 
and 
are circles with diameter 
and 
, respectively, intersecting each other at points 
and 
. Prove that and lie on 
.
be a natural number. We write out the fractions 
, 
, 
, 
such that they are all in their simplest form. Let the sum of the numerators be 
. For what 
is one of and 
odd, but the other is even?
be a natural number. We write out the fractions , , , such that they are all in their simplest form. Let the sum of the numerators be . For what is one of and odd, but the other is even?
be the sum of the positive integers less than or equal to and co-prime to . Since 
, by pairing, 
. Then note that, 
and for any 
, 
. Therefore, 
. Since 
and 
are multiplicative, so is 
.
All numbers below are supposed to be nonnegative integers.
has the same parity for 
thus it's suffice to prove, that the sum 
is odd. Note that 
Hence there are exactly 
suitable values of 
which is odd. The conclusion follows.
in term of a function 
such that 
,
will have an even numerator if and only if its 2-adic valuation is strictly greater than the 2-adic valuation of n.
.
an
with 
odd.
.
a
will always have different parities. Therefore all n are solutions.
if 
, where 
is even and 
is odd.