,
,
such that the sequence 
, for 
, is periodic modulo 
be an integer. Isaac wants to cover an 
board with some tiles, without overlaps and without going off the board. For this purpose, Isaac has many tiles of the form:
squares of the board. Isaac is allowed to choose any tile and, with a straight cut, divide the tile into two new tiles. Each new tile must be able to cover exactly an integer number of squares of the board. For which values of 
is it possible for Isaac to achieve his goal by making at most two cuts?
such that 
for any 
are real numbers. Prove that 
for all real numbers 
be a positive integer. On an 
board, players 
and 
take turns in a game. On each turn, a player selects an edge of the board (not on the board border) and makes a cut along that edge. If after the move the board is split into more than one piece, then that player loses the game.
Y by Adventure10, kamatadu, Mango247
Let 
be the prime numbers listed in increasing order, and let 
be a real number between 0 and 1. For positive integer 
, define
![\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in] {\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}} & \mbox{if} \; x_{k-1} \neq 0, \end{cases} \]](//latex.artofproblemsolving.com/e/1/7/e17b3fc5962fba938dee877dac266c3cd024a6e3.png)
where
denotes the fractional part of 
. (The fractional part of is given by 
where 
is the greatest integer less than or equal to .) Find, with proof, all satisfying 
for which the sequence 
eventually becomes 0.
be the prime numbers listed in increasing order, and let 
be a real number between 0 and 1. For positive integer 
, define![\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in] {\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}} & \mbox{if} \; x_{k-1} \neq 0, \end{cases} \]](http://latex.artofproblemsolving.com/e/1/7/e17b3fc5962fba938dee877dac266c3cd024a6e3.png)
where
denotes the fractional part of 
. (The fractional part of is given by 
where 
is the greatest integer less than or equal to .) Find, with proof, all satisfying 
for which the sequence 
eventually becomes 0.
except that it's integral... 
with 
.
,
again and so on, thus the sequence 
defined by this is (strictly) decreasing as long as 
.
,
is irrational, also 
will be, so for all other starting values it doesn't become 
.
.
direction: suppose that 
.
.
.
.
such that 
.
.
and 
,
,
direction: pretty clear that this is true, but to be totally rigorous, suppose that 
.
for integral 
,
,
satisfy the condition.
for some positive integers 
and 
.
,
.
.
.
,
.
.
where 
and 
.
is strictly greater than that of 
;
is strictly monovariant. Therefore, the numerator of the numbers of the sequence will become 
.
is rational, but this implies that 
is necessary and sufficient. Indeed, for necessity, suppose 
.
,
is rational; finish with induction.
for positive coprime 
,
and all nonnegative integers 
,
and consequently 
must also be irrational (otherwise 
is a rational number equal to the product of a rational number and an irrational one, contradiction) and taking the fractional part only subtracts this result by an integer, so 
is also irrational. Rinse and repeat.
we have that the numerator of 
we can write 
for some integer 
is an integer, so the sequence evaluates to 
will also be irrational etc. Thus, the sequence can never terminate to 
such that 
.
,
such that the numerator of 
.
,
are strictly decreasing, until we reach 
.
could have been any sequence of integers.
in simplest terms. Observe that 
for all 
for some 
by definition. Hence 
,
is decreasing. Therefore, in finitely many terms 
where 
.
when 
.
thus 
.
that are rational tend to 
denote the denominator of 
for coprime integers 
.
.
Now as the numerators of all of these fractions are strictly less than the denominator, we obtain a sequence of decreasing numerators. Therefore the numerator tends to 
.
are irrational, and therefore cannot be 
If 
Then, 
where we choose 
such that 
Hence, the numerators of 
.
denote the set of irrational numbers
,
So it cannot be zero. Assume that 
.
So 
.
for 
.
be a rational number. Then, we have
where 
and 
.
where 
and 
.
and 
will be so small such that 
.
Now since for all 
we have 
can be any rational number between 
needs to be an integer so 
must be rational. For 
must also be rational, then so on until 
.
which must have a denominator of 
,
.
,
is an integer.
is strictly decreasing and that 
for all whole numbers 
.![\[ \left\{ \frac{p_k}{\frac{a}{b}} \right\} = \frac{p_kb \pmod{a}}{a} \]](http://latex.artofproblemsolving.com/f/e/1/fe11acb7164361a84d9e4e40ac1f24223358d87b.png)
Thus, for rationals 
,
where 
,
and 
,
which implies when 
is fully simplified its denominator is at most 
is irrational. Similarly 
is irrational and so on. Thus every element of 
is irrational. Since 
for 
and 
,
.
,
.
work.
Now, if 
,
where 
then 
which is true. However, we can't have that the denominator goes less than 0. Thus, eventually, it must become 0.
