.
-
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
,
are all greater than 
.
,
.
and 
to be the remainder when 
is divided by 
.
and 
is odd for 
.
or an equivalent formulation. To prove this construction works, we need to show that
is always an integer between 
and 
inclusive.
is divided by 
.
,
since it telescopes.
,![\[\frac{r_{i+1} n - r_i}{2^i} > \frac{r_{i+1} n}{2^i} - 1 \geq \frac{n}{2^i} - 1 \geq \frac{n}{2^k} - 1.\]](http://latex.artofproblemsolving.com/4/2/0/420d26b6099ca264846885e7eef23b2c9c7b2dd0.png)
Thus, each digit is lower bounded by 
,
becomes arbitrarily large.
fails.
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
,
such that 
has no nonreal roots. We can also assume 
case is trivial as the constant term must be nonzero, so fix 
.
.
.
which has the 
for all integers 
.
be the coefficient of the 
term of 
for all 
,
for which 
.
and 
.
and 
.
is the coefficient of the 
term in 
and 
.
,
.
,
)
has 
distinct real roots as well, along with two consecutive nonleading coefficients equal to zero, but one degree lower, by the Power Rule. By doing this repeatedly, we eventually end up with a polynomial where the two consecutive coefficients have degree 
,
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
distinct from 
and 
,
such[/center]
is directly similar to either 
or 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
,
to 
,
to 
.
to be the set of points strictly outside the circle with diameter 
.
is always acute. Suppose two roads 
and 
cross. Then, 
is a convex quadrilateral. Since 
.
is not acute but there does not exist an 
such that 
are connected by some path. This is trivially true for 
,
are also connected.
is obtuse, we have 
.
,
,
is not mentioned. Do not reward points if the induction argument is incomplete or incorrect.
be the orthocenter of acute triangle 
,
be the foot of the altitude from 
.
intersects line 
.
be the center of 
.
be the intersection of 
and the line through 
,
.
be the foot of the altitude from 
and 
,
is a midline and thus 
.
is a midline of 
,
and thus 
.
(by definition), by either HL congruence or the Pythagorean theorem, we must have 
.
.
and 
to attempt to identify 
.
or 
,
.
.![\[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]](http://latex.artofproblemsolving.com/3/c/5/3c5984d0003c1ca2df2e68841dd9e93ed6c2d27c.png)
is an integer for every positive integer 
is divisible by 
,
be a prime power dividing 
.
.
,
if 
and 
otherwise, so for each term, either both the numerator are divisible by 
,
,
,
.
has an inverse modulo 
.![\[\binom ni \equiv \pm 1 \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor} \pmod{p^a}.\]](http://latex.artofproblemsolving.com/0/1/c/01c00beb6efc07ed028fd0273f262b8bfb965d33.png)
modulo 
.
,![\[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (\pm 1)^k \equiv p \equiv 0 \pmod p.\]](http://latex.artofproblemsolving.com/2/e/c/2ececb6afa8a10e116056cdc57a6c948298d0c29.png)
where 
,
satisfies the induction hypothesis for the prime 
.![\[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (\pm 1)^k \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}^k \equiv p\sum_{i=0}^d \binom di^k \pmod {p^a}\]](http://latex.artofproblemsolving.com/6/c/9/6c9a4b6aac62b1befef1511451c9fa30993d2cf7.png)
By the induction hypothesis, 
divides the inner binomial sum, so since we are multiplying it by 
.
.
is incorrect, reward up to 
points total.
and 
.
between the set of people except Evan 
if the total score of the cupcakes in that bubble with respect to 
.
denotes the set of neighbors of 
people.
.
from the graph, noting that no person in 
Y by bharatputra, Royalreter1, hamup1, sicilianfan, MathematicsOfPi, Darn, jh235, yimingz89, abishek99, 15Pandabears, xantho, amburger66, AkshajK, TheMaskedMagician, MSTang, IsabeltheCat, spartan168, SHARKYBOY, Tan, DVA6102, pinetree1, flamefoxx99, thkim1011, droid347, hesa57, m-c-geometry, ahaanomegas, Binomial-theorem, jhoer, dantx5, DrMath, TheCrafter, rjiang16, bobthesmartypants, tigerat, csmath, 62861, librian2000, Eugenis, mathisawesome2169, inequality3206, PersonPsychopath, MathAwesome123, blep, joey8189681, thinkinavi, atmchallenge, Carrots101, yojan_sushi, Gmud, bearytasty, sampai7, yrnsmurf, shiningsunnyday, mathguy5041, ohmcfifth, hwl0304, elitechicken, epiclucario, Shaddoll, Wave-Particle, DeathLlama9, champion999, gradysocool, PiDude314, bestwillcui1, DivideBy0, m1234567, brianapa, niyu, oceanair, phi_ftw1618, SW1, hockey10, weilunsun28, mathfun5, BIGBUBBLE, Marzi, kk108, always_correct, Chessking345, Aspen, itised, ItzVineeth, jeffisepic, DaniyalQazi2, mathwiz0803, Ultroid999OCPN, jangkun217, Liopleurodon, AFK16, Heisenberg09, Mathisfun04, green_dog_7983, reckoner, ccx09, TheMagician, Math-Ninja, coolak, AIME12345, andyxpandy99, vsathiam, mathlogician, dchenmathcounts, mathleticguyyy, Vietjung, Williamgolly, Susanssluk, Mathscienceclass, csunllc, FIREDRAGONMATH16, smartninja2000, Imayormaynotknowcalculus, middletonkids, Smile2020, skyscraper, mathbw225, SlimTune, RedFireTruck, IceWolf10, superagh, v4913, centslordm, SuperJJ, fidgetboss_4000, sotpidot, Natsuki, vsamc, OliverA, suvamkonar, aie8920, fuzimiao2013, ihatemath123, GCA, hi.., kankan, LOGYC, RubixMaster21, jhu08, mahaler, megarnie, exp-ipi-1, math31415926535, eagles2018, YuyangZ, sugar_rush, son7, Bradygho, asimov, StopSine, Jndd, Lamboreghini, nikitadas, EpicBird08, s12d34, Adventure10, Paul10, ehuseyinyigit
I know many of us are looking for last-minute AIME practice around this time, since the AIME I is less than two weeks away. With that in mind, here is a collection of AIME level problems that can be used as practice. The problems are sorted into four categories (Algebra, Combinatorics, Geometry, Number Theory) with room for overlap, and there are ten problems in each category arranged (hopefully) from easiest to hardest. Depending on my strengths and weaknesses, some of the subjects have harder problems than others, and not all the problems are properly placed or of high quality, but hopefully this can be used to sharpen your skills before the actual AIME. Enjoy!
EDIT: Oh, and feel free to post solutions to any of the problems!
Edit (December 25, 2019): The solutions packet to this can be found on my website here.
EDIT: Oh, and feel free to post solutions to any of the problems!
Edit (December 25, 2019): The solutions packet to this can be found on my website here.
This post has been edited 2 times. Last edited by djmathman, Dec 26, 2019, 12:15 AM
.
but it asked for "the smallest integer greater than *insert expression here*", making it zero.
.
as the number of ways to label the points if there are 
.
has length 
.
to 
.
.
is prime, so it is very easy to apply the lemma.
.
we have![\[a^2+b^2=\frac{10^6}{a^2}+\frac{10^6}{b^2}=2008.\]](http://latex.artofproblemsolving.com/2/5/0/25010e1afa66cdf66683ca31949a661eea396517.png)
Then because 
![\[\frac{10^6}{2008-b^2}+\frac{10^6}{b^2}=2008.\]](http://latex.artofproblemsolving.com/7/8/e/78ee79c7ce3f77bc217fb5c8b7afc3b4268f1d0a.png)
![\[\rightarrow 10^6=b^2(2008-b^2)\]](http://latex.artofproblemsolving.com/d/c/9/dc9e016af42b0f2827c6d64c14f350a50ca88479.png)
Notice that 
as well, so the polynomial![\[x^2-2008x+10^6=0\]](http://latex.artofproblemsolving.com/b/9/d/b9d68deb110097d62bfde0dabe9bde67ea1d4864.png)
has roots 
.
and 
![\[a+b=\sqrt{4008},\ a-b=\sqrt{8}.\]](http://latex.artofproblemsolving.com/f/0/3/f0326d443e071af49d167c4446b6942c666f2d75.png)
which then means 
(because 
)
.
,
.![\[t^2=(\sqrt3+\sqrt2)^2=5+\sqrt{24}\implies t^2-5=\sqrt{24}.\]](http://latex.artofproblemsolving.com/3/8/a/38a23358440d1849b157e9e54ff4731d6d1efbc7.png)
Squaring again gives 
,
and 
.
and the requested answer is 
.![\[a_{n+1}-1=\sqrt{a_n^2-2a_n+3}=\sqrt{(a_n-1)^2+2}\implies (a_{n+1}-1)^2=(a_n-1)^2+2.\]](http://latex.artofproblemsolving.com/1/c/b/1cbfd071d2d6b5c642c70a969d55625e517c400c.png)
Let 
.
and 
,
for all positive integers 
,![\[(a_{513}-1)^2=1024\implies a_{513}-1=32\implies a_{513}=\boxed{033}.\]](http://latex.artofproblemsolving.com/9/e/3/9e34ac2e7225d292a372c153ed110f9bdb0e6269.png)
(Source: OMO Fall 2012, Ray Li)
gives ![\[x^2-\lfloor x^2\rfloor=(x-\lfloor x\rfloor)^2\implies \lfloor x^2\rfloor=2x\lfloor x\rfloor-\lfloor x\rfloor^2.\]](http://latex.artofproblemsolving.com/5/c/1/5c1cd57de32c54fc83336bdb211a7eddf67a9aca.png)
Let 
.
to be integral, so 
for some positive integer 
.![\[\{x^2\}=\left\{\left(k+\dfrac a{2k}\right)\right\}^2=\left\{k^2+a+\left(\dfrac a{2k}\right)^2\right\}=\left(\dfrac a{2k}\right)^2=\{x\}^2.\]](http://latex.artofproblemsolving.com/a/a/6/aa6df1d694faa33ad09ad1d10a448d70f7c98565.png)
Now it remains to find their sum. This is calculation; if we are careful, we should get that the sum equals ![\begin{align*}&99+\sum_{k=1}^{98}\left[k+\left(k+\dfrac1{2k}\right)+\left(k+\dfrac2{2k}\right)+\cdots+\left(k+\dfrac{2k-1}{2k}\right)\right]\\&=99+\sum_{k=1}^{98}\left[k(2k)+\dfrac1{2k}\sum_{a=0}^{2k-1}a\right]\\&=99+\sum_{k=1}^{98}\left[2k^2+\dfrac{(2k-1)(2k)}{2(2k)}\right]\\&=99+\sum_{k=1}^{98}\left[2k^2+k-\dfrac12\right]\\&=99+2\sum_{k=1}^{98}k^2+\sum_{k=1}^{98}k-\sum_{k=1}^{98}\dfrac12\\&=99+2\left[\dfrac{98(99)(197)}6\right]+\dfrac{98(99)}2-48\\&=641999.\end{align*}](http://latex.artofproblemsolving.com/c/d/a/cdab9cca97aa2896937b245a0872690016daa951.png)
Finding 
as opposed to 
.
,
.![\[(ac)^2+(ad)^2+(bc)^2+(bd)^2=2008^2.\]](http://latex.artofproblemsolving.com/c/2/7/c27ae0ba79a799eb81fc9ded4598b6a2c9c24822.png)
Next note that if 
are two equal numbers then 
.
and 
gives 
.![\[(ac)^2+(ad)^2+(bc)^2+(bd)^2=(ad)^2+2abcd+(bc)^2=(ad+bc)^2=2008^2,\]](http://latex.artofproblemsolving.com/0/7/7/07730baaf25be15b58c8cdd3069432cf61e5ebbb.png)
so 
,![\[a^2+b^2+c^2+d^2-2(ad+bc)=0\implies (a-d)^2+(b-c)^2=0,\]](http://latex.artofproblemsolving.com/0/2/c/02c6f38acb3cbd9127355344bf0d038b94630bfe.png)
implying that 
and 
.![\[(a+b)(c+d)=ac+bd+(ad+bc)=4008.\]](http://latex.artofproblemsolving.com/4/a/1/4a1fb02382e6d4ddfe2a4f45d3e064f6cff250bc.png)
From the equalities established above, we have 
,
.
?
or am I being dumb? Also how did you bound a? Thank you!
is an integer, so the fractional part of 
for some integer 
in the desired range. Then 
.
,
.
answer extraction as it should.
