[/quote]
,
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 
(the number of positive divisors of 
,
.
that satisfies:
satisfy 
and 
Prove that always exists positive integer 
.
is "not congruence"
and 
Prove that
Solution:
and 
then 
and 
.
![$$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$](http://latex.artofproblemsolving.com/b/1/d/b1d283f2e7b8e574b6c0ff72236c52a39d3da9e8.png)
Prove that
Prove that
and 
intersect at points 
intersects 
,
.
,
.
Y by suvamkonar, parmenides51, Adventure10, Mango247
Welcome to the second mock AIME of the year!
Please signup in thread if you are interested.
Note: This is the first test of a sequence of tests deciding who will officially take a mock USAMO. The next test will be an AMC 12, and then the top 50% indices will make USAMO.
Please signup in thread if you are interested.
Note: This is the first test of a sequence of tests deciding who will officially take a mock USAMO. The next test will be an AMC 12, and then the top 50% indices will make USAMO.
a week, so good luck.![\[\left(\dfrac{6^2-1}{6^2+11}\right)\left(\dfrac{7^2-2}{7^2+12}\right)\left(\dfrac{8^2-3}{8^2+13}\right)\cdots\left(\dfrac{2012^2-2007}{2012^2+2017}\right)=\dfrac{m}{n},\]](http://latex.artofproblemsolving.com/1/b/b/1bb17c1bc295194d9f50e1be9517bb7786bae42a.png)
where 
is divided by 
.
.
,![\[\frac{6^2-1}{2012^2+2017}\]](http://latex.artofproblemsolving.com/7/e/d/7ed09fe9861e1db786c9571158e2dcf10cba18df.png)
however 
.
however 
.
,
.
,
and 
to give us 
.
,
be defined as 
where 
,
for 
.
is 
.
.
,
.
such that 
.
before a number 
can be seen to be 
.
.
,
.
and 
satisfy this, however, since 
,
,
.
.
.
terms in it. Before the last group, we have 
terms, and 
is therefore the 
t
,
.
and will take more than 
as the value of 
,
as our sum which is greater than 
since 
.
is 
)
.
.
,
.
or 
.
,
.
,
,
in the exponents. Since 
,
.
,
.
,
.
,
,
be the points where the angle bisectors of 
,
,
,
can be written in the form 
where 
and 
.
,
,
.
,
,
.
to get 
,
.
.
,
so the answer is 
.
sides numbered 
is nonzero and is equivalent to the probability that a sum of 
.
is thus the same as the coefficient of 
.
is the same as the coefficient of 
.
,
is thus 
,
.
and center in the first quadrant is placed so that it is tangent to the 
-
,
where 
.
.
,
for some positive 
.
,
.
.
.
or 
.
.
be the angle between the line 
and the 
-
,
and 
.
.
also equals 
,
,
and 
.
.
are positive real numbers such that 
,
can be expressed as 
,
and 
.
.
.
.
is positive, we can divide by 
is already a perfect square, we begin with that. We have 
.
to get 
.
from our original equation as 
,
.
.
and 
.
.
.
.
,
and 
.
.
,
and 
.
.
.![\[3\left(a-\dfrac{1}{2}\right)^2-\dfrac{3}{4}+\left(2b-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1=0\]](http://latex.artofproblemsolving.com/1/e/a/1ea094233b3a34395f6676970b6672a5124e1c2e.png)
,
and 
.
and 
,
and 
.
with center 
.
first intersect the circle at 
.
,
.
intersect 
at 
.
,
.
.
,
.
.
,
,
.
and 
,
,
,
,
can be expressed in the form 
where 
are all positive integers, and 
.
.
.
because 
in 
.
is the same thing as 
,
.
.
.
.
from adding point 
triangle 
as shown in this diagram (where 
angle, and 
angle):
and 
and 
.
,
and the coordinates of 
.![$[EDF]$](http://latex.artofproblemsolving.com/e/d/1/ed1edc46be5f921c15369bae991d164322f892e1.png)
by using 
.
,
.![$[EDF]=\frac12*b*h$](http://latex.artofproblemsolving.com/c/0/b/c0bbde315a1330176f8fa4581c3a9c05dcdfa0ac.png)
give us 
.
.![$[DEF]=DE\cdot DF\cdot\frac{\sin\angle ADC}{2}=\frac{3}{16}[ABC]$](http://latex.artofproblemsolving.com/4/8/4/48434f533fd3979b2d0cee29208aecb8675adb9d.png)
,![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
and 
with 
such that the product of the elements in 
.
be 
and 
to be 
,
.
and 
,
,
for some value of 
be the sum of all possible values of 
.
,
.
.
.
.
.
,
.
,
.
satisfying the equations.)
.
has three real solutions 
.
,
.
and 
.![\[a^{15x}=b\]](http://latex.artofproblemsolving.com/d/1/f/d1faebbba8451917acc776e916baddab70672794.png)
![\[b^{3x}=ac^4\]](http://latex.artofproblemsolving.com/7/4/b/74bf320f9d78e73fa4fd8f09160dce6b1c0c52a9.png)
![\[c^{5x}=a^2b\]](http://latex.artofproblemsolving.com/4/2/1/421759d76d8ada823e20dedef1ede488550f4399.png)
![\[a^{45x^2}=ac^4\rightarrow a^{\frac{45x^2-1}{4}}=c\]](http://latex.artofproblemsolving.com/6/f/5/6f54e572160a2e35b85802a17a0710c393dff728.png)
.
.
.
,
,
is![\[\frac{8}{225}\cdot 15^3=8\cdot 15=120\]](http://latex.artofproblemsolving.com/8/c/3/8c3781cd3399442b3336b05c805edae742d1965a.png)
.
.
,
.
(such that points 
,
,
is 
.
,
.
divides 
and 
for 
are \textit{near Carmichael numbers}. The proof of this is noting that 
,
,
.
.
.
,
and 
,
.
together because this would imply that 
,
and 
.
or 
.
,
.
gives us no values for 
and this gives us 
to give another \textit{near Carmichael number}: 
or 
or 
.
,
where 
.
.
,
.
.
gives 
which isn’t a two digit prime number, 
gives 
which gives a largest value of 
gives us 
,
or 
which aren’t prime. 
gives us no two digit \textit{near Carmichael numbers}, since 
for 
is greater than 
.
to note that either 
and the second gives us 
,
which isn’t a two digit prime number. Therefore, we look at the first case. We need for 
or 
.
or 
.
,
.
which isn’t a two digit prime number, we must have 
.
therefore 
or 
.
and 
both false, and the second one gives 
and 
,
which gives us a sum of 
.
for 
and 
.
is a polynomial, and 
is an arithmetic sequence, find the smallest positive integer value of 
.
is an arithmetic sequence, difference(let this be h
,
.
.
,
,
.
.
,
term and the 
term, because by method of finite differences, you only have to take 
.
or 
.
,
and 
.
and 
.
or 
.
or 
.
.
,
.
for 
to give us 
.
.
approximately. Note that 
(where 
,
)
,
to give us 
and hence the smallest possible positive integer value for 
.
with common difference 
That is,![\[f(n)=d-(n-1)^3+(n-2)^3\]](http://latex.artofproblemsolving.com/2/a/2/2a2c1d8b4e1f3570c1fc33231188057dd4326a66.png)
for every positive integer 
for every positive integer 
.
so 
.![\[f(n)<-2012\iff -3x^2+9x+2012<0\iff x>\frac{-9\pm\sqrt{81+12\cdot 2012}}{6}.\]](http://latex.artofproblemsolving.com/6/3/9/6398e5665ddd96dea17c6f7fb7d7941c109b5510.png)
.
.
