[/quote]
,
.
-
,
be the result.
,
is the Euler's totient theorem, which calculates the number of integers less than 
.
for 
.
be the remainder 
leaves when you divide it by 
.
be your predicted USA(J)MO score.![$\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$](http://latex.artofproblemsolving.com/c/d/3/cd3f7d917b0ec944ec3017e1f737c538e1edd10e.png)
.
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 
,
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.
.
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 
and 
.
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
intersect 
at 
and 
and 
.
Where 
such that :
.
be reals such that 
Prove that 
Let 
Prove that 
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
consecutive beads every color appears at least once. Find the largest value of 
possible.
This easy via the binomial theorem for the quantity is just 
,
satisfies 
.
,
.
Y by cheese1000, Diehard, Quadratic64, mathman523, lucylai, Tan, AkshajK, droid347, megarnie, parmenides51, Adventure10, Mango247
The title says it all: four 2013-14 WOOTers, all in the Northeastern USA area, namely codyj, djmathman, yugrey, and myself have decided to write a Mock AIME in prep for the AIME I coming up next month. You must signup in this thread to participate, since submissions will be via form and to keep identification security, we will send you passwords before the contest is released.
I will be running the tournament, grading submissions, etc. I thought of this yesterday so we aren't done writing problems, but we expect them to be released next week. Please copy the signup list already existent and add your name.
Note: We may do a Mock AIME II depending on the interest in this one.
I will be running the tournament, grading submissions, etc. I thought of this yesterday so we aren't done writing problems, but we expect them to be released next week. Please copy the signup list already existent and add your name.
Note: We may do a Mock AIME II depending on the interest in this one.
This post has been edited 6 times. Last edited by ahaanomegas, Mar 7, 2014, 3:10 AM
![[asy]
unitsize(1 cm);
real x, y;
real[] hist;
x = 0.6;
y = 0.1;
hist[1] = 16;
hist[2] = 20;
hist[3] = 11;
hist[4] = 10;
hist[5] = 12;
hist[6] = 14;
hist[7] = 4;
hist[8] = 13;
hist[9] = 8;
hist[10] = 7;
hist[11] = 6;
hist[12] = 5;
hist[13] = 5;
hist[14] = 7;
hist[15] = 4;
draw((x,0)--(16*x,0));
for(int i = 1; i <= 15; ++i) {
draw((i*x,0)--(i*x,hist[i]*y)--((i + 1)*x,hist[i]*y)--((i + 1)*x,0));
label(string(i),((i + 0.5)*x,-3*y),fontsize(10));
label(string(hist[i]),((i + 0.5)*x,(hist[i] + 3)*y),fontsize(10));
}[/asy]](http://latex.artofproblemsolving.com/2/a/2/2a2de093b97a5d584dc2d7e3b8cdbcca3bb870c4.png)
countingkg and AkshajK with a score of 12.
mssmath with a score of 10.
