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and 
invertible. Prove that if 
for any positive integer 
then 
be periodical functions such that the function 
has finite limit at 
Prove that 
is constant.
are real numbers such that 
for every 
then 
be a nondegenerate interval and 
a differentiable function. We denote 
Prove that:
is an interval;
and the set 
contains at most two elements;
Using parts and 
deduce that 
has the intermediate value property.
and 
be two sequences of positive numbers. Show that the following statements are equivalent:
of positive numbers such that 
and 
both converge;[/*]
converges.[/*]
, let 
be the number of positive divisors of . Evaluate
dividing by x
be a positive root of the equation 
. Prove that the following sequence converges: 
, 
, 
, 
such that ![\[1 + xy + x^2y^2= a(x)c(y) + b(x)d(y)\]](http://latex.artofproblemsolving.com/4/d/7/4d74d8d44b9484443805d2b08e84c912aaa355a0.png)
holds identically?
![\[
\int_{1}^{2} \frac{x^{3}+x^2}{x^5}dx
\]](http://latex.artofproblemsolving.com/5/a/b/5ab9b8014b74dbfcf4db15873e0348bbdc0171ef.png)
![\[\int_{2025}^{2025^{2025}}\frac{1}{\ln\left(2025\right)\cdot x}dx\]](http://latex.artofproblemsolving.com/9/f/3/9f3fe22d62ad9fc5259d97fcf60eb3c45d240bfa.png)
![\[
\int(\sin^2(x)+\cos^2(x)+\sec^2(x)+\csc^2(x))dx
\]](http://latex.artofproblemsolving.com/8/3/d/83da36cded674b594ce7d585b1c1fee7df4956ae.png)
![\[
\int_{-2025.2025}^{2025.2025}\sin^{2025}(2025x)\cos^{2025}(2025x)dx
\]](http://latex.artofproblemsolving.com/9/1/0/91045d7c16160403507ee5f66a7cd021dfe0a0c3.png)
![\[
\int_{\frac \pi 6}^{\frac \pi 3} \tan(\theta)^2d\theta
\]](http://latex.artofproblemsolving.com/3/4/f/34f4936db8bf49e9d0c77de8539ba6ee144f94d5.png)
![\[
\int \frac{1+\sqrt{t}}{1+t}dt
\]](http://latex.artofproblemsolving.com/b/4/b/b4bead66ef90e5677281987993f82b1e4bb864df.png)
-----![\[\int \frac{x^{2}+2x+1}{x^{3}+3x^{2}+3x+3}dx
\]](http://latex.artofproblemsolving.com/9/2/2/922c2e35f18adffdd453ad926d0ae91fe759dd50.png)
![\[\int_{0}^{\frac{3\pi}{2}}\left(\frac{\pi}{2}-x\right)\sin\left(x\right)dx\]](http://latex.artofproblemsolving.com/3/6/b/36b02db24ba60cb50e0027a1d0d8a3ce1d07e72f.png)
![\[
\int_{-\pi/2}^{\pi/2}x^3e^{-x^2}\cos(x^2)\sin^2(x)dx
\]](http://latex.artofproblemsolving.com/a/c/5/ac50d60b5ff6d26d8ef2079efaaf7e2d293c8798.png)
![\[
\int\frac{1}{\sqrt{12-t^{2}+4t}}dt
\]](http://latex.artofproblemsolving.com/9/0/d/90d94cd0cdb89832dbf6b91dfb2af29169005963.png)
![\[
\int \frac{\sqrt{e^{8x}}}{e^{8x}-1}dx
\]](http://latex.artofproblemsolving.com/1/c/0/1c091415a17d162a745bd79dfb6c1632297d9ca0.png)
-----![\[
\int \frac{e^x}{e^{2x}+1} dx
\]](http://latex.artofproblemsolving.com/c/9/c/c9c93d4a2fc4e8856e0b6d30cffd5bf2581521bf.png)
![\[
\int_{-5}^5\sqrt{25-u^2}du
\]](http://latex.artofproblemsolving.com/7/6/0/760ca61e6e609ada7052eddb31c0b5cc8bdc7d8a.png)
![\[
\int_{-\frac12}^\frac121+x+x^2+x^3\ldots dx
\]](http://latex.artofproblemsolving.com/f/7/a/f7a1aa77b71a36c1800ed7bf5e44fa6744ee9a04.png)
![\[\int \cos(\cos(\cos(\ln \theta)))\sin(\cos(\ln \theta))\sin(\ln \theta)\frac{1}{\theta}d\theta\]](http://latex.artofproblemsolving.com/e/0/3/e030efdb1f1d73326dd09a60c961ac4b7bf987df.png)
![\[\int_{0}^{\frac{1}{6}}\frac{8^{2x}}{64^{2x}-8^{\left(2x+\frac{1}{3}\right)}+2}dx\]](http://latex.artofproblemsolving.com/4/a/2/4a274f4254113198782619e28ac63edcb41664ff.png)
-----![\[\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(x\right)}{\sin\left(x\right)+\cos\left(x\right)}+\frac{\sin\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}dx\]](http://latex.artofproblemsolving.com/2/0/5/2051d51ec5ec892ce1f571385bfc1f74d05bea9f.png)
![\[
\int_0^{\infty}e^{-x}\Bigl(\cos(20x)+\sin(20x)\Bigr) dx
\]](http://latex.artofproblemsolving.com/4/e/4/4e482f5c117a0f3984c822614dd4a4214cc03b86.png)
![\[
\lim_{n\to \infty}\frac{1}{n}\int_{1}^{n}\sin(nt)^2dt
\]](http://latex.artofproblemsolving.com/3/b/3/3b3bf68b93dbe49e9979471ba04b82c3a2eb96c4.png)
![\[
\int_{x=0}^{x=1}\left( \int_{y=-x}^{y=x} \frac{y^2}{x^2+y^2}dy\right)dx
\]](http://latex.artofproblemsolving.com/c/0/3/c0314f295bc45fa87cfffc93ee348f38b44f51b2.png)
![\[
\int_{0}^{13}\left\lceil\log_{10}\left(2^{\lceil x\rceil }x\right)\right\rceil dx
\]](http://latex.artofproblemsolving.com/4/f/0/4f012045fcbc8210ea752729f609149b8a534e4f.png)
-----![\[
\int \frac{6x^2}{x^6+2x^3+2}dx
\]](http://latex.artofproblemsolving.com/0/d/1/0d1131a496881bcf7036de0b2a36191008925840.png)
![\[
\int -\sin(2\theta)\cos(\theta)d\theta
\]](http://latex.artofproblemsolving.com/b/0/0/b006ec3908ed9d9c00fe354578d50b5f91d56ddb.png)
![\[
\int_{0}^{5}\sin(\frac{\pi}2 \lfloor{x}\rfloor x) dx
\]](http://latex.artofproblemsolving.com/a/2/1/a21381f8fa2432165c02e7cd28dfb00ae3ad5475.png)
![\[
\int_{0}^{1} \frac{\sin^{-1}(\sqrt{x})^2}{\sqrt{x-x^2}}dx
\]](http://latex.artofproblemsolving.com/b/c/b/bcbe046b36c5d072bd6e064ee98a94683ccadfcc.png)
![\[
\int\left(\cot(\theta)+\tan(\theta)\right)^2\cot(2\theta)^{100}d\theta
\]](http://latex.artofproblemsolving.com/0/1/2/012796576e56002d2983da33436ddd6208ae3507.png)
-----![\[
\int \ln\left\{\sqrt[7]{x}^\frac1{\ln\left\{\sqrt[5]{x}^\frac1{\ln\left\{\sqrt[3]{x}^\frac1{\ln\left\{\sqrt{x}\right\}}\right\}}\right\}}\right\}dx
\]](http://latex.artofproblemsolving.com/d/6/f/d6fa1f1a12b988fef90f2a39f9b77f6e9822f76c.png)
![\[\int_{1}^{{e}^{\pi}} \cos(\ln(\sqrt{u}))du\]](http://latex.artofproblemsolving.com/e/1/5/e152dacb1601c772f32ab9f3494d45bed43c635f.png)
![\[
\int_e^{\infty}\frac {1-x\ln{x}}{xe^x}dx
\]](http://latex.artofproblemsolving.com/4/e/8/4e8ae310d63fae062122d6ff9ba1570ffaf98a42.png)
![\[\int_{0}^{1}\frac{e^{x}}{\left(x^{2}+3x+2\right)^{\frac{1}{2^{1}}}}\times\frac{e^{-\frac{x^{2}}{2}}}{\left(x^{2}+3x+2\right)^{\frac{1}{2^{2}}}}\times\frac{e^{\frac{x^{3}}{3}}}{\left(x^{2}+3x+2\right)^{\frac{1}{2^{3}}}}\times\frac{e^{-\frac{x^{4}}{4}}}{\left(x^{2}+3x+2\right)^{\frac{1}{2^{4}}}} \ldots \,dx\]](http://latex.artofproblemsolving.com/4/2/f/42feb97860007e1ef6ce36e03c71743b253f2576.png)
on the domain 
it is known that ![\[\displaystyle f(x)=\sin\left(\int_{0}^x \sqrt[3]{\cos\left(\frac{\pi}{2} t\right)^3+26}\ dt\right)\]](http://latex.artofproblemsolving.com/d/0/a/d0af1de2e29589d6205f05bad4cf0412742535b8.png)
is invertible. What is 
?
by![\[f(n) = \sum_{i\ge 0} \bigg\lfloor \frac{n + a^i}{a^{i+1}}\bigg\rfloor=\bigg\lfloor \frac{n + 1}{a} \bigg\rfloor + \bigg\lfloor \frac{n + a}{a^2} \bigg\rfloor + \bigg\lfloor \frac{n + a^2}{a^3} \bigg\rfloor + \cdots\]](http://latex.artofproblemsolving.com/8/f/5/8f5369fb9dfb998ea45bbe1ec16985cc6c2e8325.png)
for some fixed 
. Prove that![\[\lim_{n \rightarrow \infty} \frac{f(n)}{n/(a-1)} = 1.\]](http://latex.artofproblemsolving.com/0/6/c/06c0bf2d4619df935ccc5ad74832be520a796a74.png)
for 
.]
be the set of all triples 
of positive integers for which there exist triangles with side lengths 
Express ![\[\sum_{(a,b,c)\in T}\frac{2^a}{3^b5^c}\]](http://latex.artofproblemsolving.com/9/f/6/9f6e0cee87e5b6958e143ff23b887bcd876f6cba.png)
as a rational number in lowest terms.
be a function such that 
and 
for all 
is continuous. is bounded.![\[A=\begin{bmatrix} 1&1&1&1&1\\1&-1&1&-1&1\\?&?&?&?&?\\?&?&?&?&?\\?&?&?&?&?\end{bmatrix}\in\mathbb R^{5\times 5}\]](http://latex.artofproblemsolving.com/f/c/b/fcbdeb5107deefc1ec6441dfc0b7e93006729b9e.png)
satisfy 
Determine the maximum value of 
. The general setting with arbitrary starting values leads to "essentially" the same problem, but restricting those values to 1 helps to avoid unnecessary variables.
, so that 
will be the biggest prime divisor of 
. In order to gain some familiarity with the situation, the following table provides the relevant values for 
: 
2 2 2/3 4 2 1/2 6 3 3/5 9 3 1/2 12 3 3/7 15 5 5/8 20 5 5/9 25 5 1/2 30 5 5/11 35 7 7/12 42 7 7/13 49 7 1/2 56 7 7/15 63 7 7/16 70 7 7/17 77 11 11/18 88 11 11/19 99 11 11/20 110 11 11/21 121 11 1/2 132 11 11/23 143 13 13/24 156 13 13/25 169 13 1/2 182 13 13/27 195 13 13/28 208 13 13/29 221 17 17/30
-column, to the third column as the 
-column, and to the respective entries as -values and -values.
is 
, and we will now examine that hypothesis.. In lines 12 to 17, for example, we consistently have the value 7 in the -column. We will now investigate these sections a little closer, focussing on the values of and . For that purpose, we define 
to be the 
-th prime number (in their natural increasing order), i.e. 
, 
etc.. We observe that the -value and the -value, that is 6 and 3, can be written as 
and for 
. Plainly speaking, the -value is the product of the corresponding prime in the -column and the previous prime. We will show that this is no coincidence for the first line of such a section.-value 3, we do indeed have and in those two columns. By definition of , the value in the -column in the next line is 
which is the same as 
. If the value in the -column of that line does not change, is added to the -value in the following line once again, resulting in the value 
there, and as long as nothing changes in the -column, the values in the -column will be of the form 
, 
in the following lines.-column will change once we reach the smallest 
such that has a prime factor 
larger than . As two different prime numbers are always relatively prime, this is equivalent to the fact that divides 
. We start with 
, when is smaller than and also smaller than any candidate prime number (which must even be greater than ). Clearly, the first such that has a prime divisor greater than is the one with 
, so that our new prime number will be 
, which then does not only divide , but will be equal to it.
lines (as that was the crucial value of which initiated a change in the -column). The last line of that section will have the value 
in the -column and in the -column, and the new section will therefore begin with a line that has 
in the -column and 
in the -column. In particular, this shows that the values in the -column run through all the prime numbers in a monotonously increasing way.-column in terms of the sequence 
. Now we have to consider them as actual elements of the sequence 
, which, loosely speaking, means that we have to find a relation between those values and the line number.-column, the prime number prevails for exactly lines. Thus the prime number 
, which appears for the first time in line 5, is succeeded by 
in line 
. By induction, the prime number (for some integer ) will appear in the -column for the first time in line
and this telescoping sum is the same as
-value does not change within a section (by definition of a section), we can conclude that, at the end of the section, which comes 
lines later, we have 
are clearly strictly decreasing, as it is only the index which changes. Therefore, to establish the limit of , which is the ultimate objective of this text, it suffices to focus on the values of at the beginning and at the end of each section. If the values of the subsequence at the beginning, i.e.
and at the end, i.e.
converge to the same limit, the entire sequence 
will converge to that same limit by the squeezing theorem. There is the (still open) conjecture that there are infinitely many twin primes. If we assume that the conjecture is true, this would easily show that the only possible limit of our two subsequences from above, and therefore the whole sequence, is indeed . is needed. I am somewhat hesitant to proceed here, however, as I feel that this is beyond what is reasonable for a problem from a Math olympiad. To me, the attraction of this problem lies in uncovering the more elementary results from above. If others can produce an elementary proof of the limit statement, though, I am very interested in hearing about it.
Find 
.
for some 
in domain then also we know second condition given in problem implies 
for all 
.
. Similarly, 
.
for all 
,
is the greatest prime divisor of 
.
and 
?