AoPS Academy
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 
.]
![\[
\int_{1}^{2} \frac{x^{3}+x^2}{x^5}dx
\]](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 
?
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.
weights, each weighing 
and another 
weights with 
each. Write a algorithm that determines the ways in which a scale can be balanced with a weight 
on the left pan, and display the number of possible solutions. (The weights can be placed on both pans and the program starts with the numbers 
. What will be displayed after three successive runs: 5,2,5,1,4 | 5,2,5,1,11 | 5,2,5,1,20?
from the domain 
and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the -th derivative of at 
) is divisible by 
is 
For 
whereas for 
Note that the value of that at any integer is divisible by 
which is the product of eight consecutive positive integers. Any 
consecutive positive integers contains a multiple of and a different number which is a multiple of 
hence the product of those numbers is divisible by 
Any consecutive positive integers contains two multiples of 
and is divisible by 
Any consecutive positive integers contains at least one multiple of 
and is divisible by 
Hence, the product 
is divisible by 
This this applies to every nonzero term, we have that 
divides 
for any integer 
then we may set 
and note that 
But 
is not divisible by 
and nether are 
Hence, no number smaller than will suffice as 
. We can easily show 
as 
doesn't work (just take the case 
odd).
and the first observation incinerate the case 
.
is .
let 
be a polynomial with integer coefficients of degree 
. WLOG, let 
. We get 
for all integers . As 
we are done. So, the least such satisfies 
.
and note that the -th derivative is . Thus, 
establishing our claim.
doesn't work but not smaller. Is this a one point deduction or more?
, etc, worked then would also work. So once you've shown that doesn't work but does work the rest seems trivial to me. Whether the graders will think so too I have no idea.
where 
.
Hence we find minimum such that 
. We note that 
. For 
, we consider 
. Then 
which is not divisible by . Hence 
.
derivative of all such polynomials in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
is divisible by we pick a 
degree polynomial with integer coefficient such that is the minimal integer for which the problem statement holds true.
forall such 
, 
which implies 
so such a minimal that works is since 
, so is the smallest such positive integer .
for all such and ![$p \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/2/6/0267b98c16b400dda1f081877e9b84fe672ce0e7.png)
for 
for every such 
, we have 
, which is indeed divisible by since and 
for all 
, hence we are done. 
is the desired value.
, choosing gives 
, and 
for all 
. So these values of do not work.
. Then 
, and the result follows since 
.
. The question is equivalent to: the product of consecutive integers is a multiple of 
. Thus, 
. When 
which isn't a multiple of . There must exist 
even numbers and one of them is a multiple of , so the power of 
must exceed 
. Moreover, there exists at least two multiples of , and at least one multiple of . We are done.
. Then for 
every coefficient has a product of j consecutive integers multiplied with some integer.
for all
Therefore the whole rhs is divisible by 
observe that 
but unfortunately does not divide 
therefore the minimum possible value is 
8 ,
it still satisfies , also the minimum possible degree of the polynomial should be 
, otherwise it cant be differentiated times , also in general for a polynomial to be the -th derivative, the the degree of the polynomial must be atleast 
Find 
.
and 
.
![$ a,b,c\in [0,1] $](http://latex.artofproblemsolving.com/2/1/b/21b329d0a92c18c6a2861742177969d5977d4558.png)
. Prove that
.
.
.
.
.
.
and 
we have 
.
from which we obtain 
.
for all 
forms a complete residue class modulo 7 and thus 
.
forms a complete residue class modulo 6 and 
forms a complete residue class modulo 8, thus 9 and 32 divide the product as well. Hence 
as desired and 
.
is necessary, for 
note that 
has 
,
works, write 
and note ![\[ P^{(8)}(x) = \sum_{n = 8}^d n(n-1) \dots (n-7) a_n x^{n-8} = 40320 \sum_{n=8}^d \binom n8 a_n x^{n-8} \]](http://latex.artofproblemsolving.com/0/f/5/0f5c8dd6c1b81a3bead4e3e3402c4e3dd5b737f1.png)
which is clearly divisible by 