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 an acute triangle with 
and let 
and 
be its orthocenter and circumcenter, respectively. Let 
be the circle 
. The line 
and the circle of radius centered at 
cross at 
and 
, respectively. Prove that , the circle on diameter 
and circle 
are concurrent.
of functions from the set of real numbers to itself that satisfy ![\[g(f(x+y)) = f(x) + (2x + y)g(y)\]](http://latex.artofproblemsolving.com/0/a/3/0a3bc25299a0bcbd3f3602a0e6881d68fce750c5.png)
for all real numbers 
and 
.
such that for all 
be given pairwise coprime positive integers where 
. Let 
be positive integers. We call 
to be a grandson of 
if and only if, for all possible piles of stones whose total mass adds up to and consist of stones with masses , it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of . Find the greatest possible number that doesn't have any grandsons.
+1 w
of primes (greater than 5) such that : 
of positive integers such that:
.Prove that: 
has incentre 
and incircle touching 
at 
. Let 
be the antipode of in the circumcircle of . Point 
is chosen on the internal angle bisector of 
such that 
. Let 
be the midpoint of arc 
, and let 
be the midpoint of 
. Prove that 
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly positive integer factors 
. Assume that no prime larger than 
divides any of the 
's. Show that there must be some perfect cube among the 's.
and 
intersect at points and 
. A line through intersects and at points 
and , respectively. Line 
intersects at point 
, and line 
intersects at point 
. If is the circumcenter of 
, prove that 
.
be such that![\[u_{1} = 1; \quad u_{n+1} = \dfrac{u_{1} + u_{2} +...+u_{n}}{n}+n-1 \quad \forall n \in \mathbb{N^{*}}\]](http://latex.artofproblemsolving.com/1/5/e/15e5bf0cc14794e7028dbfe4a8754f6803e850ea.png)
and 
. Find 
.
denote the set of positive real numbers. Find all functions 
such that for each 
, there is exactly one 
satisfying 
and primes 
such that 
.
, one can check the only solutions are: 
Now we prove there are no solutions for 
.
, since 
and by inspection one notes that no three elements have vanishing sum modulo .). First, we analyze the cases where 
:
: No solutions for parity reasons.
: since 
, we find 
even and 
odd (hence 
). Now looking modulo 
gives that 
,
: From 
, we find is odd and is even. Now looking modulo gives a contradiction, even if 
, since 
but 
.
. Next, by taking modulo 
we have 
, which forces both and to be odd (in particular, 
). We now have 
The first equation implies 
is even, but the second equation requires to be odd, contradiction. Hence no solutions with .
solution here)-
. One can easily check that all of these work.
, there are no solutions. First, take the equation mod 
to get 
, or 
, implying is odd. Taking the equation mod gives 
, implying is odd as well. Taking the equation mod 
gives 
, implying is even since is odd. Finally, taking the equation mod 
gives 
(from the parities above). One can easily check that no combination of these residues sums to 
mod .
, a finite case check gives the solutions listed above. From now, we assume 
. First, assume 
, or 
. Then, 
. If , then 
, which is impossible since 
is never divisible by . Thus, 
. Taking the equation mod gives 
, so 
. However, mod gives 
, contradiction. and , then we do casework on the value of . If 
, we get 
is even, which is impossible. If , we get 
. Taking mod gives 
, so is even. Taking mod gives 
, so is odd. However, taking mod gives 
, so must be odd, contradiction. If , then 
. As before, mod gives odd, so mod gives 
, which is impossible. Thus, we have exhausted all cases and the above solutions are the only ones.
. Remark that 
, so 
. Additionally, remark 
, so 
. Let 
.
.. Rearranging yields 
. Take modulo to get 
, hence is even. In the case 
, we could take modulo to get 
, which can't occur if is even. Hence, we have 
and 
. This has no solution if 
, so we may assume 
. Then, 
, which is absurd.
. Take modulo to get 
, so is odd. Take modulo to get 
. Since is odd, 
, implying that 
and is odd as well. In particular, this implies 
. Take modulo to get 
, which implies is odd. Take modulo to get 
, which is false. Hence, there are no solutions for 
, and we are done. 
.
. Then, we have 
. This implies , so 
. This implies 
. Taking gives 
and taking gives 
. These constitute all possibilities in this scenario.
. Then, we have 
. This implies 
, so 
.. Rearrange to get 
. This implies 
. Taking 
gives no solution, taking 
gives no solution, and taking gives no solution, hence there are no solutions in this case.
. Rearrange to get 
. This implies . Taking gives no solution, taking gives 
, and taking gives no solution, hence we have the solution 
.
.
We show 
has no solutions.
so 
Thus by mod 
as well. By mod 
implying 
or 
But by mod 
implying that 
which is a contradiction.
, we can manually check everything, finding that the only solutions are 
..
. We see that 
, so and are opposite parity.
gives the following: then the LHS is odd, but for , 
is even: contradiction, then the LHS becomes 
and the RHS is , so we get even. However, we also get that is even from the opposite parity condition. But then, taking 
gets that the LHS is , while the RHS is because : contradiction.
. then the LHS becomes 
and the RHS is , so is odd.., then is odd, so 
. Since we know is odd, we see that 
, so the LHS is equal to but the RHS must be : contradiction.
. Then we get 
. Since the RHS is , we have , which implies that is odd and is even. is even, is odd, and is odd. Let 
. Then the condition rewrites as:
Now, taking 
, we see that 
. However, we can easily verify that this has no solutions. Thus any fails. 
, we see that the solutions are 
won't have any solutions. Notice how 
with 
, we see that 
which means and have different parities.
, we have two cases.
. Next, for 
, we must have 
and 
, which means 
. Because as 
, we see that and have the same parity, a contradiction.
and 
in this case, which is equivalent to 
. Because , we have and must be different parities, so is even. Finally, if we take 
, we see that 
, while 
, a because 
, we have a contradiction.
, it can be confirmed that the solutions are 
. Henceforth assume .
. Then mod 4 gives 
, so is odd; let 
for some 
. If is even, then 
, so taking mod 5 gives 
. This is impossible since the LHS is even and absolutely bounded by 4. Hence is odd; let 
for some 
. Now 
, so taking mod 3 gives 
. Hence is even; let 
for some 
. Now the equation becomes 
. Then mod 8 gives 
, but this is impossible both when 
and 
.. The equation is . Then mod 4 gives 
, so is even; let 
for some . Taking instead mod 3 gives 
, so is even; let for some . The equation is 
. Taking mod 8 gives 
, contradiction.. The equation is 
. Now mod 4 gives 
, contradiction., and we conclude.
. 
clearly yield no solutions since then the left hand side is less than 3. we have 
, so 
. means 
which can't hold. So . We want to solve 
and we can check that the solutions are 
and 
by inspection. we have 
, so . then 
, so 
, but we can check there are no solutions by hand. then 
. Thus . We can check by inspection that the only solution is 
. and : then 
. The left hand side is odd and the right hand side is even, contradiction. and : then 
. We have already done the case . then we want to solve 
, but the left hand side is not 
and the right hand side is, contradiction.
then we have 
which cannot hold. and : we have already done the cases when or , so 
. Now 
so is odd. then we want to solve 
. The left hand side is not divisible by 
but the right hand side is, contradiction. then since is odd, . Then we have 
which is not possible.
and : then 
, so 
have opposite parity. then by the above, is even. Now 
, so 
, so is odd. But 
. But is odd and is even so the left hand side is 
, contradiction. then by the above, is odd. Now 
so is even. Then 
. But is even and is odd so the left hand side is 
, contradiction. then 
, so 
, so is odd. Then 
. This means 
, so is odd. Therefore is even. Since , 
so 
. But is odd and is even, so the left hand side is 
, contradiction.
, we can find that the only solutions are 
, 
., we have 
. If , we find that there are no solutions by testing out values for . If , then we find no solutions unless . Therefore, 
is the only solution when .
. We can take powers of 
.
.
so we are done.
. Mod 5, we we have 
which implies that 
must be the same parity. Mod 3, we have 
implying that and must be opposite parity. Now assume -- then mod 4, 
implies that and are the same parity mod 5. Combined with the previous two facts, this is an obvious contradiction., then we have 
Modulo 120, 
. But since 
and 
, no combination of the two can attain 118, contradiction. Thus we only need to consider ., 
has no solutions. For , 
solves to get 
; for any smaller , there are also no solutions. Thus we have exhausted this case.
If 
, and if we obtain works. If , then , which has no solutions. For , we also have the solution 
. Thus the answers are 
, 
and 
.
These clearly work. Now we analyse 
. We get 
, 
and 
. Now, clearly, we have 
is not congruent to mod 120, so there exist no non-negative integer which satisfy 
. But we have 
divides for all 
Hence no interger 
satisfies the equation. We are done. 
, since and by inspection one notes that no three elements have vanishing sum modulo .
the solutions are 
. We undergo pain in order to show there are no solutions when . then taking gives odd, and then taking gives odd, so we're solving 
. Taking gives even, and then gives 
: this is impossible. then gives odd, and then (noting that 
, otherwise gives a contradiction) gives 
: this is impossible. then gives even, and then (noting that , otherwise gives a contradiction) gives 
: this is impossible. then 
is odd.
, just case bash to get some (a,b,c,n). Please search up for those. Real fun is for showing no solutions for ..
(*)
.
, where 
for some natural . and hence a contradiction to the assumption that is even.
.
is odd,
, which is not possible. is even, or 
.
.
which is not odd. Contradiction on the initial assumption that . So done!
noting 
We see 
and 
Since 
we know 
so 
which is absurd. Hence, 
Note 
so we have no solutions.
Note 
so we have no solutions.
We proceed by casework on 
If 
we have no solutions. If 
works. If 
works.
Again, we use caswork on If 
we have no solutions. If 
or 
works.
and 
, we have:
Let 
and 
with 
, then 
, so 
. Then:
Taking
gives:
so is odd. Let 
for some 
, then:
Finally,
gives:
a contradiction (since this is impossible for any and ).
and and 
By
, we have 
, so no solutions exist in this case. and 
By, 
and so is even. Let 
., 
and so is even. Let 
.
By, 
which is absurd. Hence no solutions. and 
By, 
and so is odd. Let 
., 
which is impossible.
, so 
. This means that 
. If then which has no solutions. If then , so 
. If then , whereas if then , which gives that 
and are solutions
, so . then , so 
. For each of these cases there is no corresponding value of that works. then , so . If or no solutions exist, but if we have that 
is a solution.
had such nice modulos for powers for 
, and ., take to get the following powers:
). Now finally some more case bashing for 
and 
(the ones below these don't work due to size reasons) show that 
, 
and 
are the only ones that actually work for the tuples of 
respectively (which is again left for the reader to prove ) and we are done.
, are the ones stated in the claim, so from now on, 
assume 
however this in only possible when 
, thus is odd.
furthermore, since is odd, we can rewrite the expression as 
thus since is odd, 
, which forces 
which implies that 
, or is odd.
thus and are pairwise distinct, however this forces 
or to be even, from the fact that is odd.
, thus
however this is only possible when , which contradicts the previously statement that is odd. Thus we have reached a contradiction, so there exist no solution for 
are the only solutions manually caseworked for 
, so assume henceforth that n is at least 5. We see 
which can be checked to see that it doesn't work, so our only solutions are indeed the ones listed above.
Assume 
Then if 
we have 
and 
and 
Then if we look at the expression in modulo 3 we see 
and, 
Say 
Then in modulo 5 we have 
Since 
and 
we have a contradiction and must be and 
.
Looking at the expression in modulo 9 gives us 
and 
but since 
means 
we have a contradiction. b must be equal to 0. Now we have 
but for 
this means 
Contradiction. Our claim is right.
:
we have 
. In this case only 
is possible. we have 
. Only 
meaning either 
or 
is possible in this case.
we have 
but 
a contradiction. The only possible solutions are 
.
. (Feeling kind of lazy) for 
, then![\[2^a + 3^b + 5^c \equiv 0\pmod{12}\]](http://latex.artofproblemsolving.com/6/8/f/68f7b20ca7b169aa5caefb5b6db8b4ac560a89bc.png)
![\[2^a \equiv 2, 4, 8, 16, 32, 64\pmod{120}\]](http://latex.artofproblemsolving.com/c/a/6/ca661228a62068a8ed96392940e23745cde69287.png)
![\[3^b \equiv 3, 9, 27, 81\pmod{120}\]](http://latex.artofproblemsolving.com/c/3/9/c39d30b3671348a072b7c0c04ea2c4cde7e45d7e.png)
![\[5^c \equiv 5, 25\pmod{120}\]](http://latex.artofproblemsolving.com/0/d/3/0d31f1bdff059dbae51e43d218b63239d1511213.png)
., so it is impossible for ..
, we can manually check each case and find each of the above solutions. We will prove that there are no more solutions for . For each , note that![\[120 \mid n! = 2^a+3^b+5^c.\]](http://latex.artofproblemsolving.com/d/8/b/d8b81f119f1dc9949fba2692e20ba660c7a0dba9.png)
. Thus, there are no solutions when .
.. modulo 
is 
, modulo is 
, and modulo is 
. Note that the max sum of residues is 
which means that our residue sum must be exactly . Note that if the second residue is 
then the max sum of residues is 
so our second residue must be 
. Then, the sum of the remaining residues must be 
. Note that this isn't possible (just check that none of 
appear in the first set), so there's no solutions for . It remains to find the solutions for . For 
clearly a contradiction. So, only cases remaining are 
and 
. For we have 
which gives solution 
For 
we have 
whic gives solution 
So, the only solutions are 
, 
, and 
, which clearly work. One can manually check that these are the only solutions for . Now we prove that there are no other solutions for .
. Note that 
Note that if you choose one element from each of these sets, they cannot sum to 
. However, implies 
done.
.
.
.
.
,
,
.
.
.
.
,
,
,
,
,
to the equation 
where 
.
.
to get 
.
to get 
for 
which is obviously impossible. Hence 
can't all hold true simultaneously. Now it's just a straightforward case bash to get solutions as 
don't work. If 
,
Thus, 
have opposite parity, 
are odd 
has no solutions. If 
.
.
If 
,
is odd
have opposite parity
,
for odd 
,
are odd
for odd 
,
are the only solutions for 