[/quote]
,
board, show that there is a 
block on the board in which at least 3 of the squares are colored red.
be a quadrilateral such that 
and 
are acute and 
.
,
.
and let 
be a point on the shorter arc 
of 
and 
and 
have equal areas.
towns. For every pair of towns, there is a narrow road going from one town to the other. One day, all the roads are declared to be “one way” only. Alice has no information on the direction of the roads, but the King of Hearts has offered to help her. She is allowed to ask him a number of questions. For each question in turn, Alice chooses a pair of towns and the King of Hearts tells her the direction of the road connecting those two towns.
questions.
be any point on side 
of 
.Let 
and 
be points on 
such that 
and 
respectively. Prove that the locus of circumcenter of 
is a line.
and 
pebbles, respectively. In the first round Barry chooses a positive integer 
and Andrew takes away 
pebbles from one of the two heaps (if 
be a triangle with 
.
lies inside the triangle in such a way that 
.
be the midpoint of 
.
,
at 
.
similarly.Prove that 
are collinear.
grid with 
the there exist two identical rows (meaning same numbers in the same order).
![$$\left(a^{3}+b^{3}+c^{3}-3abc\right)^{2}-\left[a^{3}+b^{3}+c^{3}+3abc-ab(a+b)-bc(b+c)-ca(c+a)\right]^{2}\ge 2k\cdot(a-b)^{2}(b-c)^{2}(c-a)^{2}$$](http://latex.artofproblemsolving.com/d/2/4/d24b46e65a587fbd977467b6e4e31838a0eb57aa.png)
forall 
such that p is a prime, 
and 
is a divisor of 
.
Y by rightways, navi_09220114, tenplusten, megarnie, Adventure10, Mango247, rudransh61
Let 
denote the number of positive integers less than 
that are relatively prime to . Prove that there exists a positive integer 
for which the equation 
has at least 
solutions in .
Proposed by Iurie Boreico
denote the number of positive integers less than 
that are relatively prime to . Prove that there exists a positive integer 
for which the equation 
has at least 
solutions in .Proposed by Iurie Boreico
This post has been edited 2 times. Last edited by v_Enhance, Aug 23, 2016, 12:47 AM
,
,
as we like, which can be made far greater than 2015 (If you want to be rigorous, you can use several well known bounds on 
.
(to avoid boundary cases) that satisfies 
.
,
prime factors (the maximal case is when 
.
be the set of primes less than 
,
.
,
.
,
.
,
above 
,
,
.
,
,
![$S = [a_1,a_2,..a_n]$](http://latex.artofproblemsolving.com/1/6/a/16ae4dc3727a6f16120ee4491a7166d14b9d28ab.png)
be a set and let ![$S' = [b_1,b_2,..b_m]$](http://latex.artofproblemsolving.com/8/3/0/830959438c6fd8dafec6a52d6f93aaa0f7fffafa.png)
be a set. Suppose 
and 
.
,
,
divides all the 
is fixed for all 
.
and use the fact that 
is multiplicative over and over again. Basically we want to show 
but then write 
and then use algebra.
and for any 
,![$A_i = [(p_i -1)^2, p_i (p_i -1)]$](http://latex.artofproblemsolving.com/1/5/8/158e32e1b8063d3542b4ae1ac2e51cfab064b566.png)
,
.
to consist of the productts of all numbers in 
with all numbers in 
.
'
for all 
(this is easy with Dirichlet + CRT). Then by lemma 2, we can basically let 
to get 
and we have over 2015 numbers with the same phi value.
such that 
distinct prime divisors is 
,
.
;
integers less than 
integers 
and 
,
can take on at most 
different values. Thus some value 
times.
that there is exactly 2015 ?
.
,
,
.
,![\[ M = \phi\left( \frac{N}{\prod_{p\in T} (p-1)} \prod_{p \in T} p \right). \]](http://latex.artofproblemsolving.com/e/e/9/ee9a1e1b49327d846bfb51082c7744c81faf9a0e.png)
Since 
we're done. 
![\[ (2^2-2)(3^2-3) = (7-1)(2-1)(3-1) \implies \phi(2^2\cdot3^2) = \phi(2\cdot3\cdot7). \]](http://latex.artofproblemsolving.com/6/8/0/680928b9f2cc7a2fa2e50fdff219f11a78b65175.png)
denote the number of integers 
that can be represented (not counting multiplicity) as the Euler Totient function of some other integer, then 
for any 
Now, if we let 
denote the same function, but counting multiplicity 
for any 
would be more than sufficient for our needs though.) In particular, if this problem where not true, then we would get 
which would lead to contradiction.
if 
if 
if 
if 
if 
if 
if 
if 
if 
if 
intergers positive such that 
for all 
the we get:
is constant 
has at least 
we can find a positive integer 
has at least 
. The base case is obvious (we can choose 
for 
as 
)
with 
and 
for any 
be all the odd prime divisors of the 
.
and 
.
and the only possible common divisor of 
and 
is 
.
.
has at least 
solutions (and they respect the ratio hypothesis). Note that 
.
,
.
and 
for 
as the numbers are distinct and 
,
.
for 
from the hypothesis of induction. We also have that 
,
which is in contradiction with 
.
if 
otherwise).
(
if 
otherwise) and respect the ratio hypothesis.
or 
,
and 
,
,
(and in particular 
things in the hypothesis (which, as you can imagine, led me to writing and rewriting and adding and deleting conditions in the hypothesis due to silly mistakes) .
and wanted to find 
numbers with the same Euler totient. Knowing that the problem was true, I thought that I could find 
very big with 
and 
for any 
and 
and I'd be done. I tried to construct explicitly 
and 
,
,
could generate me anything. And the easiest thing you could wish for (and be in the same time 
) is a 
.
part from the solution. I wanted now these two numbers to be coprime with the 
.
,
?
,
where 
are integers large enough. Let 
be a set of primes. Noted that for every 
and 
.
, herein 
. Then there are at least 
solutions in 
solutions.
can be arbitrarily large
.
where 
is not divisible by any element in 
or 
and each is independent since they are relatively prime. Clearly, 
.
be an increasing sequence of primes and fix a positive integer 
and 
satisfies 
since 
,
works. We can let 
equal 
or 
for every prime in 
and adjust the number of factors of 
in 
works.
.
.
.
,
.
.
.![\[\frac{\varphi\left(\prod_{x\in S}x\right)}{\varphi\left(\prod_{x\in T}x\right)}=\varphi\left(\prod_{x\in S, x\not\in T}x\right)\]](http://latex.artofproblemsolving.com/c/5/c/c5cab39c6dfc2aed881ab8548c7ec6c758260cf5.png)
Note that this value can be written in the form 
,
.
gives 
works.
ways to choose 
different values of 
![\[m=312 \cdot 408 \cdot 576 \cdot 660 \cdot 936 \cdot 1152 \cdot 1800 \cdot 1320 \cdot 1008 \cdot 1200=149967695669069120274432000000\]](http://latex.artofproblemsolving.com/5/5/f/55fc9964984fe313161c34f37cf7f0b00b97baa7.png)
works.
,
Since there are 
choices for 
works.
Each of these primes has the property that the number 1 less than it contains no prime factors other than possibly 2 or 3. We will generate 2048 different 
in the following manner:
.
for 
Then, since 
which gives us a solution for each of the 2048 values of 
such that 
for all 
of 
be a big integer. Take numbers 
,
in the following way: noticing 
for all 
,
,
.
has at least 
has at least 
has at least 
solutions as long as each of the 
and the 
,
are coprime, then 
.
has more than one solution will do the trick. For this,, just pick 
pairs of paiwise distinct primes 
such that 
is prime, and which are each also pairwise distinct from those constructed numbers 
(just do small cases, sorry 
). As then 
,
by Dirichlet, and let 
for these primes multiplied by 
.
for each of these primes 
.
,
.
besides 
values of 
,
is the product of all the prime factors of 
or 
,
pairs of primes 
.
numbers generated from picking one prime from each pair and taking the product have Totient functions differing by powers of 
, such that 
does not divide 
, such that 
for all 
. Such primes can be selected by using Dirichlet.
for all 
be the first ![\[m=\phi(p_1p_2 \cdots p_N) = (p_1-1)(p_2-1)\cdots(p_N-1).\]](http://latex.artofproblemsolving.com/6/6/4/6648238f43cda8378a8e3cdcab8f3ed3dec0dc82.png)
The claim is that for all 
,
,
for all 
,
.
be the maximal prime powers that divide 
.
such that 
.![\[n = (p_k - 1)\prod_{i \neq k} p_i\]](http://latex.artofproblemsolving.com/c/e/2/ce249b18ef92c99dcd80c63b58b4f4463ecb0b50.png)
works for each 
works, where 
where 
since 
.
,
for all 
regardless of all previous 
,
and by Dirichlet we can obtain 
has at least 
be a sequence of primes such that 
satisfies. Provided that 
,
also satisfies and clearly 
,
as the 
