[/quote]
,
,
is chosen on the side 
.
and 
,
and 
are drawn, respectively. The point 
is the circumcenter of 
.
and 
lies on the line 
.
be a point in the plane of triangle 
such that the segments 
,
,
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to 
is acute.
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray 
and ray 
.
be a point on segment 
.
be the intersection of lines 
and 
be the intersection of line 
is tangent to 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of 
,
denote the incenter and 
.
perpendicular to 
intersects the tangent from 
.
and 
.
.
such that for any positive integer number 
and any arrangement of the numbers from 1 to 
on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than 
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
.
![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be the binary representation of 
.
,
,
,
equals 
.
such that![\[
4k^3 + 4k + 1
\]](http://latex.artofproblemsolving.com/f/3/5/f355e456cd47ecfdb13597d52fec3d848b67c594.png)
numbers, none of which appears more than 
times. Prove that they can be divided into 
groups of 
be a cyclic quadrilateral. 
be the projections of 
,
is a kite.
.
and 
,
,
.
such moves the blackboard will contain a single number. What is its maximum possible value? (Reminder that 
is the maximum integer not exceeding 
Y by Hydrogen-Helium, Davi-8191, tenplusten, anantmudgal09, opptoinfinity, itslumi, HWenslawski, Adventure10, Mango247, Tastymooncake2
Let 
be a given integer. Prove that infinitely many terms of the sequence 
, defined by ![\[a_k=\left\lfloor\frac{n^k}{k}\right\rfloor,\]](//latex.artofproblemsolving.com/8/3/2/832113b1ef1fe211e1eaead25129873aead1dab4.png)
are odd. (For a real number 
, 
denotes the largest integer not exceeding .)
Proposed by Hong Kong
be a given integer. Prove that infinitely many terms of the sequence 
, defined by ![\[a_k=\left\lfloor\frac{n^k}{k}\right\rfloor,\]](http://latex.artofproblemsolving.com/8/3/2/832113b1ef1fe211e1eaead25129873aead1dab4.png)
are odd. (For a real number 
, 
denotes the largest integer not exceeding .)Proposed by Hong Kong
This post has been edited 3 times. Last edited by djmathman, Jul 24, 2015, 8:15 PM
dividing 
for sufficiently large integers ![\[ n^{n^r-r} - 1. \]](http://latex.artofproblemsolving.com/9/0/5/9057001be3b00a20d7393eacb736e3f3b681b3bc.png)
for some integer 
.
be such a prime, with corresponding integer ![\[ n^{n^rp} \equiv n^r \pmod{n^rp} \]](http://latex.artofproblemsolving.com/e/b/1/eb1a237f657877daf1d01ad6576dc381a052d632.png)
since this is clearly correct mod 
,
,
,![\[ \left\lfloor \frac{n^k}{k} \right\rfloor = \frac{n^k - n^r}{n^rp} \]](http://latex.artofproblemsolving.com/6/c/b/6cbb8bdbc3610f156abfc527cf6dc6aac43d7ff3.png)
which is odd.
for some prime 
and 
.
that does not divide 
.
is the order of 
.
.
.
which is clearly odd.
for 
.
is odd and 
,
,
,![\[\ \lfloor\frac{n^{p^l}}{p^l}\rfloor = \frac{n^{p^l}-1}{p^l} \]](http://latex.artofproblemsolving.com/9/6/9/96931d0b10e107922eed3d3d564d10c861c2d5e0.png)
by LTE, and it is obviously odd.
.
,
works. Indeed![\[\ \lfloor\frac{2^{3(2^{2j})}}{3(2^{2j})} \rfloor = \lfloor \frac{2^{3(2^{2j})-2j}}{3} \rfloor \]](http://latex.artofproblemsolving.com/8/a/3/8a3e64118b523c7ed081019bc8dc2a8a04cb53e4.png)
.
is always even, so then this quotient becomes![\[ \frac{2^{3(2^{2j})-2j}-1}{3} \]](http://latex.artofproblemsolving.com/1/e/b/1eb76f0f8b97384c4d6fdb1d70b57ae8afbb88e6.png)
, which is clearly odd, so we are done.
for some integer 
.
for some positive integer 
.
,
which is odd.
for some positive integer 
for some positive integer 
and then I guess 
.
mod 
.
.
then it's easy to see that 
which is odd.
and 
from zigmondy's theorem 
has an odd prime divisor 
so 
(
for every 
now notice that for large value of 
which is odd.
then if 
then 
so we can find infinitely odd 
such that 
which is odd.
for any positive integer 
is divisble by 
,
is odd.
,
If 
so that 
,
is even, choose a prime divisor 
From the equality ![\[n^{2^{\alpha}} - 2^{\alpha} = 2^{\alpha}\left(2^{2^{\alpha} - \alpha}m^{2^{\alpha}} - 1\right),\]](http://latex.artofproblemsolving.com/b/4/f/b4f19a0c9aac68e9d8efef97f724002a63f8ad5a.png)
it follows that 
Moreover, 
,
would force 
,
,
By considering 
,
gives rise to a distinct 
since 
Hence, we can write ![\[0 < n^k - \left\lfloor \frac{n^k}{k} \right\rfloor k < k.\]](http://latex.artofproblemsolving.com/5/f/9/5f9d242f516d860281e5019044e3d86197f1cad9.png)
If 
is odd, we're done. Thus, suppose that 
divides 
because ![\[2^{\alpha + 1} \le n^k - \left\lfloor \frac{n^k}{k} \right\rfloor k < k \implies 2 \le \frac{n^k}{2^{\alpha}} - \left\lfloor \frac{n^k}{k} \right\rfloor p < p. \quad (\star)\]](http://latex.artofproblemsolving.com/c/c/2/cc23d4853871fc6a0307eda57aa666707c90106d.png)
But Fermat's Little Theorem implies that ![\[\frac{n^k}{2^{\alpha}} = \frac{\left(n^p\right)^{2^{\alpha}}}{2^{\alpha}} \equiv \frac{n^{2^{\alpha}}}{2^{\alpha}} \equiv 1 \pmod{p}.\]](http://latex.artofproblemsolving.com/c/9/7/c97c5d40f0b396e2a06070db96fe6225f2161566.png)
Thus, when reducing 
modulo 
:
.
:
.
.
.
,
,
is even and 
,
Thanks!
.
.
works for all 
.
.
.
which is obviously odd, since 
works in a similar fashion.
to get ![\[\left\lfloor\frac{n^{n^{x}}}{n^x}\right\rfloor=n^{n^x-x}\]](http://latex.artofproblemsolving.com/e/d/9/ed99ebaa0e8764d1ea1cc693f6edf2a92894fab0.png)
which is odd. If 
to get ![\[\left\lfloor\frac{n^{n^{2m}(n+1)-2n}}{n+1}\right\rfloor\]](http://latex.artofproblemsolving.com/6/e/9/6e91b8eaf889205ab2a7c0405adc4748f233dfcc.png)
Since 
is even, 
.![\[\left\lfloor\frac{n^{n^{2m}(n+1)-2n}}{n+1}\right\rfloor=\frac{n^{n^{2m}(n+1)-2n}-1}{n+1}\]](http://latex.artofproblemsolving.com/5/7/7/577d9f6096a4228afff9c543d7f1bab7e4931779.png)
which is odd.
(which exists by Zsigmondy) and take 
for integers 
so that
.
.
for prime 
We can construct infinitely many 
and 
;
with 
,
As there are infinitely many 
for any 
where 
for any 
.
.
,![\[\left\lfloor \frac{n^k}{k} \right\rfloor = \frac{n^k-2^v}{k}\]](http://latex.artofproblemsolving.com/5/8/f/58f6d38d5cc4003e8e6c29da3e56003b9e2d3a58.png)
Note that 
since 
since 
for any positive integer 
for some prime odd 
such that![\[n^{v \cdot p} \equiv 2^v \pmod p \iff n^v \equiv 2^v \pmod p\]](http://latex.artofproblemsolving.com/6/0/6/6060453868177d54ee36435ae7f398d398bff678.png)
Thus to finish just take any odd prime 
.
by the following reason. If 
,
,
factors into 
times an odd number greater than 
,
,
(for 
)
,![\[
a_k = \left\lfloor \frac{n^{n^x}}{{n^x}} \right\rfloor = n^{n^x - x}
\]](http://latex.artofproblemsolving.com/a/9/a/a9a37df2bd740cbffafb343934215b2eecf5cc2b.png)
is clearly odd since 
,
since 
,
,
must be odd, 
,
.
,
.![\[
a_k = \left\lfloor \frac{2^{3\cdot2^{2x}}}{3\cdot2^{2x}} \right\rfloor = \frac{2^{3\cdot2^{2x}} - 2^{2x}}{3\cdot2^{2x}} = \frac{2^{3\cdot2^{2x} - 2x} - 1}{3}
\]](http://latex.artofproblemsolving.com/4/5/b/45bedeaecb0245ead74f64e73a4aaf48868fc855.png)
Since the numerator is odd, and the denominator is odd, we are done.
where 
with 
,
,
,
thus 
and moreover take the desired 
,
and 
are 
so now we only vary 
and we find infinitely many 
for a positive integer ![\[n^k \equiv \begin{cases} 0 & \pmod{2^{(p-1)a}} \\ 1 & \pmod p \end{cases} \implies n^k \equiv 2^{(p-1)a} \pmod k\]](http://latex.artofproblemsolving.com/7/d/5/7d52ca002ec9e66820408d79c91e479ef2533c7c.png)
![\[\left\lfloor\frac{n^k}{k}\right\rfloor = \frac{n^k - 2^{(p-1)a}}{2^{(p-1)a} \cdot p} = \frac{\frac{n^k}{2^{(p-1)a}}-1}{p} = \frac{\text{odd}}{\text{odd}} = \text{odd}.\]](http://latex.artofproblemsolving.com/2/d/1/2d18821b342d7dfb3f51bf1eec605d34f5e3093a.png)
.
for any 
Indeed this works because,
is an even integer, we take 
for an arbitrary 
By division algorithm, we write 
,
Now is sufficient to show that ![\[\min(\nu_2(q), \nu_2(r)) \geq\nu_2((n-1)^mq + r) = \nu_2(n^{(n-1)^m}) \geq (n-1)^m.\]](http://latex.artofproblemsolving.com/7/7/a/77a25ec1bf65f63efe70a3c91a3e7acabc1082f6.png)
But this means 
,
.
,
,
.
,
are odd.
for all 
,
.
,
.
.
.
(mod 
)
,
where p is a prime factor of n and m is a suitably good number. then we are done.
now take 
here the second thing is due to the order condition that zigmondy brings.
for each positive integer ![\[ \left \lfloor \frac{n^{n^r}}{n^r} \right \rfloor = n^{n^r-r}\]](http://latex.artofproblemsolving.com/6/5/3/653d4a29b5cb3551bd69595602aace56cfec1914.png)
which is clearly an odd positive integer.
,
.
for each positive integer ![\[\nu_p(n^{p^r}-1) = \nu_p(n-1)+\nu_p(p^r) > r \]](http://latex.artofproblemsolving.com/4/b/8/4b8b41fea534429ff1c80d1d83da75e52736e8bc.png)
So 
and 
.![\[n^{p^r} = p^r \left \lfloor \frac{n^{p^r}}{p^r} \right \rfloor +1\]](http://latex.artofproblemsolving.com/f/4/4/f44c45333c2d1948dac4d2a9ad83526ef0a98b17.png)
which since 
is odd as desired.
for each odd positive integer 
so 
.![\[\nu_3(2^{4\cdot 3^r}-1)= \nu_3(2^4-1) + \nu_3(3^r)> r\]](http://latex.artofproblemsolving.com/c/d/4/cd4a608c4f6b20851f6d115e0c5ea218ef72863a.png)
so 
and 
.
.![\[2^{4\cdot 3^r} = 4\cdot 3^r \left \lfloor \frac{2^{4\cdot 3^r}}{4\cdot 3^r} \right \rfloor + (1+3^r)\]](http://latex.artofproblemsolving.com/7/9/6/7965f40735acb25dfdb5eb5fd15341d2f9f9d77a.png)
To finish note that by Lifting the Exponent we also have 
(since 
must indeed be odd, and we are done.
,
and we get 
is odd.
and the rest follows by LTE as 
which is odd.
works and 
makes the expression an integer, it makes sense to consider 
.
and that 
,
the floor is 
and is odd for even 
.
if and only if 
is even and in that case the ratio 
is odd, since the numerator and denominator are both odd.
![\[ n^{n^r-1} - 1. \]](http://latex.artofproblemsolving.com/1/9/6/196d22d042884e7ad63e109857ffc6008dd70dad.png)
for some integer 
.
for ![\[\left\lfloor\frac{n^{n^{2t}(n+1)}}{n^{2t}(n+1)}\right\rfloor=\left\lfloor\frac{n^{n^{2t}(n+1)-2t}}{n+1}\right\rfloor=\frac{n^{n^{2t}(n+1)-2t}-1}{n+1}\]](http://latex.artofproblemsolving.com/8/9/9/89964e47ebd8356d6f73370ce705e26590417913.png)
which is an odd integer, as desired.
