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.
, inscribed in the circle 
, satisfies 
and 
, and 
is the intersection point of the diagonals 
and 
. The circle with center 
and radius 
intersects in two points 
and 
. Prove that the line 
is tangent to the circles with diameters 
and 
.
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
intersect at the point 
. Let 
be a point on the arc (not containing ) such that 
and 
. Let the lines 
and 
intersect at point 
. Let 
be the reflection of with respect to the line . The lines 
and 
intersect at point 
, and intersects the circumcircle of again at point 
. lies on the circumcircle of 
.
be non-negative real numbers such that 
Prove that
such that 
and 
for any real numbers 
and 
.
is written on the blackboard. Every time, Carmela can erase the number 
on the black board and replace it with a new number 
, if and only if 
is a perfect square. Prove or disprove that all positive integers 
can be written exactly once on the blackboard.
and 
as ![\[\sqrt{2}=1.a_1a_2a_3\dots\quad\quad\sqrt{3}=1.b_1b_2b_3\dots\]](http://latex.artofproblemsolving.com/4/d/5/4d5049d5c3b26757a25c5dd34462c5462228b83d.png)
where each 
or 
is a digit between 0 and 9. Prove that there exist at least 1000 values of 
between 
and 
such that 
.
of positive integers with 
is called good if we can colour each integer red, blue, green or purple, in such a way that of each 
consecutive integers at least one is coloured red;
of each 
consecutive integers at least one is coloured blue;
of each 
consecutive integers at least one is coloured green;
of each 
consecutive integers at least one is coloured purple.
. From the fourth term of the sequence onwards, each term of the sequence is the average of the previous three. Given that the first six terms of the sequence are natural numbers and that the sixth number written was 
, what is the fifth term of the sequence?
points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 
scalene triangles with vertices of that color.
such that :
for all prime numbers 
.
for all positive integers 
such that 
, draw exterior equilateral triangles on sides 
and to obtain 
and 
, respectively. Let 
be the intersection of the altitude through and the median through . Let 
be the intersection of the altitude through and line 
. Let 
be the intersection of the median through and the line 
. Prove that , , and lie on a common line. be an even positive integer. A sequence of real numbers is called complete if for every integer with 
either the sum of the first terms of the sum or the sum of the last terms is integral. Determine numbers.
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.
satisfying the equation![\[
(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/2/5/e/25eed911f303ed7cc4e03765a2f943ad4641c451.png)
.
,
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