AoPS Academy
and the distribution function is 
then
and find the expression of 
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
be a function whose derivative is continuous on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
. Show that![$$\lim_{n \to \infty} \sum^n_{k = 1}\left[f\left(\frac{k}{n}\right) - f\left(\frac{2k - 1}{2n}\right)\right] = \frac{f(1) - f(0)}{2}.$$](http://latex.artofproblemsolving.com/3/1/4/314a1597acae030381f980e6e26c12432f31e069.png)
be a positive integer greater than 
. Show that
where each of 
is either or 
.
with integer coefficients, such that 
and 
for some distinct primes 
and 
.
be a finite subset of 
such that 
. Let 
be functions satisfying
for all 
.
such that
for all 
,
for all 
.
be an integer and let 
be the set of all 
invertible matrices in which their entries are 
or . Let 
be the number of 's in the matrix 
. Determine the minimum and maximum values of in terms of , as varies over 
.
![$f : [0,1] \longrightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/9/a/4/9a4b7997c35990f839742d29866122791ade8b7c.png)
is differentiable with 
. If 
for all ![$x \in [0,1]$](http://latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
, then show that 
for all 
.
If G has a normal subgroup of order 
then prove that 
is abelian without using Sylow Theorems.
is differentiable and 
for all . Show that for some 
, 
.
such that 
Edit: Proposed by Marian Vasile
where , 
, and are nonnegative integers.
. Note that 
, 
, 
, and 
. Thus 
there exist nonnegative integers ,, and such that 
. Note that 
implies 
but then if 
is congruent to modulo 
it must be congruent to modulo 
as well as we inspect on the 
cases where the set of residues for , , and is 
,
,
, or 
.
with 
. Let 
. These constructions suffice:
Clearly 
is a nonnegative integer (AM-GM, 
is a ring). So it suffices to show that 
is impossible.
. If 
then 
, so 
. So 
. Then 
is divisible by , so 
. But then .
. We claim 
can take on any nonnegative value that isn't 
or 
. Note that plugging in 
gives 
, plugging in 
gives 
, plugging in 
gives 
, and plugging in 
gives . is divisible by , then it is also divisible by . This is true because we can factor as![\[T=(A+B+C)((A+B+C)^2-3(AB+BC+CA)).\]](http://latex.artofproblemsolving.com/e/a/3/ea375bca4fa257115092cf95765c5600547fcf5a.png)
It is evident that if divides one of the factors, it must divide the other, so 
, as desired. This completes the solution.
![$f(x) \in \mathbb{Q}[x]$](http://latex.artofproblemsolving.com/e/d/a/eda31c06f8b59911d0cedd1128bbd5499094fc2c.png)
and 
is an element of ![$R := \mathbb{Q}[x]/(f(x))$](http://latex.artofproblemsolving.com/4/9/d/49d336ecc1bf5fe0cc84202407461abf69c1b5ff.png)
, then the Norm of is: taken over all 
roots of 
-linear map 
given by multiplication by .
is the Norm of 
when we take 
. Since has two factors over and three factors over 
, the same will also be true for 
. The fact that ![$\mathbb{N}[x]/(x^3-1)$](http://latex.artofproblemsolving.com/e/f/4/ef492e47d0fdfd9a66b9ddf7b160a90d395d432d.png)
is closed under multiplication implies that its set of Norms (the set of values in question A1) is also closed under multiplication. This may be a good way to construct similar exercises.
![\[a^3+b^3+c^3-3abc=\frac 12\cdot (a+b+c)\sum_{\mbox{cyc}}(a-b)^2.\]](http://latex.artofproblemsolving.com/9/8/c/98cdc7d9bb55c6ed37113ec2fc45ac992d8e279c.png)
So by taking 
we can achieve any 
and by taking 
we can achieve any 
. It remains to figure out which multiples of we can achieve. We can achieve multiples of by setting 
to get 
.. Then exactly one of 
and 
is a multiple of . If the latter is the case: if some two of 
are the same mod , then all are, but this cannot occur so are all distinct modulo . This is also a contradiction. So then the first case holds. Then are all congruent modulo or all distinct modulo which again ends up causing a contradiction. and multiples of are achievable.
and Introduction to Number Theory level Modular Arithmetic? If so, can someone provide a basic solution like that? Just wondering.
as residues 
such that everything else works?
residues. is a multiple of . We claim that 
and 
, which together imply that 
.
by FLT, so .
and permutations then . These are the only possibilities since . can only be 
. We can show that all of these are achievable:
, is possible.
, 
(except for ) is possible.
, 
is possible.
, 
is possible.
, 
is possible.
, 
is possible.
, 
is possible.
, 
is possible.
.
..
in , and then show 
.
(indeed, this covers 
mod ).
works.
(indeed, this covers 
mod ).
works.
, set 
. This works. For 
, set 
.
or 
.![\[f(A,B,C)=(A+B+C)(A^2+B^2+C^2-AB-BC-AC)=(A+B+C)((A+B+C)^2-3(AB+BC+AC))\]](http://latex.artofproblemsolving.com/c/7/d/c7d7c28b8d40b487a1ccd24d5bed182a19e99cb1.png)
must divide and 
. If divides 
, then divides , a contradiction. If 
, then , a contradiction.
.![\[f(A,B,C)=(A+B+C)(A^2+B^2+C^2-AB-BC-AC)\]](http://latex.artofproblemsolving.com/3/d/5/3d5cec918d9d55efa51b68e1fca0cde90998a0cd.png)
, because 
.
.
, which implies is nonnegative. 
that are either or 
mod 3 or mod 9.
, then 
is a construction
, then 
is a construction
, then 
is a construction is divisible by , it must be divisible by . This can be seen by factoring and looking over all residue combinations of 
módulo that allow the expression to be divisible by . Thus, these are all the cases.
with 
.
, we get .
, we get 
. Similarly, by taking 
, we get 
.
, we get 
. 
and 
.
to get the result. 
. while . Finally, . We now show that it is impossible for 
.
and 
is possible only if:
,
.
, then 
and 
which means 
. Hence, 
.
.
then 
. If 
then 
, while 
. Hence, . then , as claimed.
work for every 
.
work. Let's see the constructs for these values:
(constitutes for 
) for all non negative integers .
(constitutes for 
) for all positive integers .
(constitutes for 
) for all non negative integers .
Thus 
are not achievable. 
works.
First we prove if 
then 
, notice 
, now 
, now that forces 
, now that implies 
for some 
and hence 
, so clearly 
are not possible. 
gives 
under , which gives as residues under .
we get 
which gives residues under , and finally for 
we get , which finally accounts all possible values of for non negative integers . 
is a multiple of 3, denote 
must be a multiple of .
, denote 
, where the integers in the parenthesis are the remainders of divided by . Test three cases, 
, 
yielding 
as desired.
gives 
,
gives 
, and
gives 
. Letting 
vary gives all solutions.
, 
follows by inspection.
Thus, we can obtain all integers 
such that 
.
Thus, we cannot obtain numbers that are or 
, and we are done.
or 
mod . Indeed, by setting 
we get 
which covers all multiples of , and by setting 
we get 
which covers all non-multiples of . or mod integers fail. This just stems from the observation that![\[a^2+b^2+c^2-ab-bc-ca = (a+b+c)^2 - 3(ab+bc+ca) \equiv (a+b+c)^2 \pmod 3\]](http://latex.artofproblemsolving.com/8/1/0/8104b784cf4fd2f5fd7f86d0e8ce0826200ccb74.png)
so either 
is the product of two multiples of or two numbers relatively prime to .
such that 
. To construct all possible values, we have
to get 
,
to get 
,
to get 
,
to get 
.
covers all numbers that are not 
.
, we have
is impossible.
2. 
3. 
Bracket 2
2. 
3. 
Bracket 3![$$\int_0^{1}\frac{dx}{1-\sqrt[3]{1-\sqrt{x}}}$$](http://latex.artofproblemsolving.com/d/5/e/d5e8db00985944b2eca341e37a94e34661cb5c86.png)
2. 
3. 
Bracket 4
2. 
3. 
Bracket 5
2. 
3. 
Bracket 6
2. 
3. 
Bracket 7
2. 
3. 
Bracket 8
2. 
3. 
![$$\int_0^{2\pi}\sqrt[4]{\cos^4(x)-\cos(2x)}dx$$](http://latex.artofproblemsolving.com/7/e/e/7eebb4a488fec702e0130a253ef87f0f6dfedb2d.png)
2. ![$$\int_0^1 \frac{x^5}{\sqrt[3]{1-x^3}}dx$$](http://latex.artofproblemsolving.com/6/0/5/605bb9d81d0ccff54a276c9f7bf1c1726c136398.png)
3. 
Bracket 2
2. 
3. 
Bracket 3
2. ![$$\int_0^1 \frac{dx}{(\sqrt{x}+\sqrt[3]{x})^2}$$](http://latex.artofproblemsolving.com/6/9/8/69869c0ed754c0dfacf412ac6411742ffa164a6d.png)
3. 
Bracket 4
2. 
3. 
2. 
3. 
Bracket 2
2. 
3. 
2. 
3. 
2. 
3. 
4. 
5. 
and 
are achievable, we will show that 
,
implies 
.
iff 
,
iff 
,
iff 
.
,
,
,
are 0, -1, 1 
(in some order). Using the above observation, it can be checked that cases i) and ii) imply 
.
equivalent to 
.
.
gives 
and 
gives 
,
gives 
.
work. To construct you just take 
,
,
.
.
