AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be the set of real numbers. Let 
be a function such that ![\[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]](http://latex.artofproblemsolving.com/5/d/3/5d3476fd9195d085d0030d6036a9fde1cf6756a7.png)
for every 
. Assume that the inequality is strict for some 
.
for every 
or 
for every .
be an integer. Ion draws a regular 
-gon and all its diagonals. On every diagonal and edge, Ion writes a positive integer, such that for any triangle formed with the vertices of the -gon, one of the numbers on its edges is the sum of the two other numbers on its edges. Determine the smallest possible number of distinct values that Ion can write.
be an acute-angled triangle and 
, and 
be the feet of the altitudes from 
, and 
, respectively. On the rays 
, and 
, we have points 
, and 
respectively, lying outside of , such that![\[\frac{A_1A_2}{AA_1} = \frac{B_1B_2}{BB_1} = \frac{C_1C_2}{CC_1}.\]](http://latex.artofproblemsolving.com/d/f/a/dfa94e10638971d53066c20c4589a9c125059911.png)
If the intersections of 
and 
, 
and 
, and 
and 
are 
, and 
respectively, prove that 
, and 
have a common point.
such that
, 
, and 
, let 
denote the greatest common divisor of 
.
is an integer.
satisfying the equation![\[ 4x^2 - 3y^2 = 1 \]](http://latex.artofproblemsolving.com/d/2/8/d284c2c3a7d7fbe12a7988a4534df16e7adc8947.png)
such that 
.
and 
are standing in front of the room in which there is . They know that there is a chessboard in the room and that on every square there is a coin. Every coin is black on one side and white on the other side and is flipped randomly. enters the room and then points at exactly one square on the chessboard. After that, must flip exactly one coin of his choice on the chessboard to the other side and leave. Finally, enters the room ( and haven't met again after entered the room) and he has to guess which square did point at. and have that will make this happen every time?
such that for any real value of 
we have:
be a triangle with incenter 
and circumcircle 
with circumcenter 
. The incircle touches 
at 
respectively. 
is another intersection point of external bisector of 
with , and 
is 
incircle touch point to . Let 
be points lie on . 
intersect 
at 
. Assume that 
. Suppose that 
and 
are concyclic, and 
are concurrent.
are concurrent. 
, tangent line to at and 
are concurrent.
be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:
, where 
and 
are two (not necessarily distinct) numbers on his napkin.
, where and are
, where 
are four (not necessarily distinct) numbers from the first line. such that, regardless of the numbers on Gugu's napkin, every number in the second line is also a number in the third line.
be a positive integer. Show that there exist a polynomial 
with integer coefficient that satisfy the following is at most 
for each 
natural numbers around the perimeter of a circle. She wants the products of adjacent pairs of numbers to be exactly the set 
Can she accomplish this?
is not.
are present twice, making it a perfect square. while if all of these products are in the set, 
must be a perfect square as well. To prove that it's not, in we'd have to take a prime less than (actually <
, this method works, just slightly less motivated and more work). 
, or it'll have an even power. (following numbers denote the power of 43 in their factorials). Notice that:
So the total power of 43 is 
. However, we still need to add 
(to)since 
have a power of 
in them, so the total power is 
, which is odd.
![\[n-s_2(n)=2017\]](http://latex.artofproblemsolving.com/a/6/e/a6eaeffc50c89718da2f6f3c8e206d57c1bcf4b0.png)
![\[ N = 1! \cdot 2! \cdots 2024!\]](http://latex.artofproblemsolving.com/a/8/8/a88205b920dc9303b604db61ff66f7dba5c3e869.png)
is a perfect square. Now we consider the 
of this. For any 
, we have 
, if 
, we have 
, and if 
, we have 
. Hence the parity of the of 
is the number of positive integers with , which is just 
, hence 
is odd, so isn't a perfect square, contradiction.
! Congrats on solving it (me who used 43 
)
is not a square. To show this, consider powers of 
. We can pair 
and 
which cancels (in terms of parity of the exponent of ), leaving us with 
. We then calculate 
which is odd.
. Every integer 
appears 
times in the product, so odd integers appear an even number of times and even integers appear an odd number of times.
, there is always a prime 
such that 
. By Nagura's Theorem, there exists a prime such that 
. In the product, this prime appears only in the terms and 
, since 
and 
. The term appears an even number of times and the term appears an odd number of times in the product, so 
is odd, which is a contradiction to 
being a perfect square.
natural numbers described. Multiplying together the product of every adjacent pair, we see that every number appears twice, so 
must be a perfect square. However, note that is prime, so we can compute
which implies that 
cannot be a perfect square, contradiction.
as a perfect square. After checking, the p-adic value of turns out to be odd. We are done
. Every integer appears times in the product, so odd integers appear an even number of times and even integers appear an odd number of times., there is always a prime such that . By Nagura's Theorem, there exists a prime such that . In the product, this prime appears only in the terms and , since and . The term appears an even number of times and the term appears an odd number of times in the product, so is odd, which is a contradiction to being a perfect square.
are the numbers around the circle is![\[ \left ( \prod_{k=1}^{2024} a_k \right )^2 = \prod_{k=1}^{2024} a_ka_{k+1} = \prod_{i=1}^{2024} k! \]](http://latex.artofproblemsolving.com/3/c/3/3c3c22aeede73fe70341fcaaf1031a37d57ec7b6.png)
![\[ \nu_{47} \left ( \prod_{i=1}^{2024} k! \right ) = 47(1+2+ \cdots +42) + 4 \cdot 43 \]](http://latex.artofproblemsolving.com/e/2/9/e29a5d3788e5f0f612f6fb78a4875d2c632e579e.png)
Which is odd. and thus the product can't be a square. 
is not a perfect square.
. Note is a prime. 
. There are in total 
s in the number. 
, so there is no way the number is a square. Probably AMC level ngl
, Suppose, 
are permutation of 
for 
, now suppose this is possible then, 
where, 
,
This will imply,
Let, 
be the total number of positive divisors of then we have,
We know that total number of divisors of perfect square is always odd.
is always odd.
is prime dividing 
But, 
but, 
is divisible by 
thus must be even but since 
and is odd, so contradiction!!! 
natural numbers around the perimeter of a circle. She wants the products of adjacent pairs of numbers to be exactly the set 
.From wich can she accomplish this?
she can not accomplish this.
is not.
we can just select an odd prime number bigger than 
and we are done.
we want to find an odd prime number belongs to ![$[\frac{n}{3},\frac{n}{2}]$](http://latex.artofproblemsolving.com/2/4/d/24d91d2493f1ca56d7499e4bd9e25deb93470157.png)
and we know that for 
there exist a prime belongs to ![$[a,\frac{6a}{5}]$](http://latex.artofproblemsolving.com/d/4/6/d46dc7c04de19770bd94cec0926a09d9b49525ff.png)
so sinse there exist a prime in ![$[\frac{n}{3},\frac{6n}{15}]=[\frac{n}{3},\frac{2n}{5}]$](http://latex.artofproblemsolving.com/2/e/4/2e4e2c6246085604d68aeadd810a7b380d688229.png)
done.
is odd 
as before.
we can't.
then
we can take 
we can take 
we can get 
we can get 
we can get 
we see 
we see 
we see 
we see 
we can get the numbers 
. Let be such that 
, then is a perfect square.
is a perfect square, then all the prime factors of are elevated to an even number., there are 
factors in , but 
wich is odd therefore Jane cant accomplish what she wants.
. Consider the product of the products:
, where 
.
must be a perfect square; in other words, in the prime factorization of , all the exponents must be even.
, so clearly, 
and by Legendre's, 
.
, meaning that the exponent of is odd, a contradiction. 
, there are 2 "" on the circle, and they are next to each other.
, which is a prime number.
, and for all 
. They indicate that there are 1012 numbers which can all be divided by and they cannot be next to each other. But it is contradictory to the 2 "" next to each other. Thus Jane cannot accomplish this.
such
and 
are adjacent
,
....
must be a square number.
, 
is 
.
(Don't get me wrong for choosing that's what came to my mind) which is a prime.
are all where there are odd number of .
are all 2
is odd which is contradiction 
. they situated on circle in this order .
must be a square.
so, will be even (
for some d ) lefts and from legendres formula we find 
.
. So a contradiction for being perfect square. Answer is no
natural numbers around the perimeter of a circle. She wants the products of adjacent pairs of numbers to be exactly the set Can she accomplish this?
where 
and the products of adjacent pairs are 
. The product of each of these products of adjacent pairs is:
so 
must be a square.
and 
, since 
is prime.
.
we have 
, so 
we have 
, so 
.
we have 
since 
, so 
.
is odd. to be a square, therefore impossible for Jane to write down these natural numbers. This is because if 
for some 
, we would need to have 
be even.
is a prime? anyways, I used 
instead (because 
.) This is a good example of the global idea, imo.
in order. Then
must be a square of some integer.
is odd. Note that is prime, and . We have
which is obviously odd, contradiction.
is a prime? anyways, I used instead (because .) This is a good example of the global idea, imo. in order. Thenmust be a square of some integer. is odd. Note that is prime, and . We have
which is obviously odd, contradiction.
so knowing prime upto 
is enough.Yes it will take some time but not more than 
to 
mins
such that 
coprime, 
and 
.
.
,
However, note that for the prime 
,
which is odd, a contradiction, as 
must be a perfect square. 
for 
and a prime 
,
which is odd, thus our task cannot be accomplished.
,
modulo 
,
is not possible for any integer 
where 
for some integer 
and 
a contradiction
is odd which finishes