Difference between revisions of "G285 2021 Summer Problem Set"
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Geometry285 is playing the game "Guess And Choose". In this game, Geometry285 selects a subset of not necessarily distinct integers <math>P=\{a,b,c \cdots \}</math> from the set <math>S=\{1,2,3,4 \cdots k-1,k \}</math> such that the sum of all elements in <math>P</math> is <math>k</math>. Each distinct is selected chronologically and placed in <math>P</math>, such that <math>1 \le a \le k</math>, <math>1 \le b \le a</math>, <math>1 \le c \le b</math>, and so on. Then, the elements are randomly arranged. Suppose <math>S_{p,k}</math> represents the total number of outcomes that a subset <math>P</math> containing <math>p</math> integers sums to <math>k</math>. If distinct permutations of the same set <math>P</math> are considered unique, find the remainder when <cmath>\sum_{p=1}^{1000}\sum_{k=1}^{1000} S_{a,k}</cmath> is divided by <math>100</math>. | Geometry285 is playing the game "Guess And Choose". In this game, Geometry285 selects a subset of not necessarily distinct integers <math>P=\{a,b,c \cdots \}</math> from the set <math>S=\{1,2,3,4 \cdots k-1,k \}</math> such that the sum of all elements in <math>P</math> is <math>k</math>. Each distinct is selected chronologically and placed in <math>P</math>, such that <math>1 \le a \le k</math>, <math>1 \le b \le a</math>, <math>1 \le c \le b</math>, and so on. Then, the elements are randomly arranged. Suppose <math>S_{p,k}</math> represents the total number of outcomes that a subset <math>P</math> containing <math>p</math> integers sums to <math>k</math>. If distinct permutations of the same set <math>P</math> are considered unique, find the remainder when <cmath>\sum_{p=1}^{1000}\sum_{k=1}^{1000} S_{a,k}</cmath> is divided by <math>100</math>. | ||
− | <math>\textbf{(A)} \0 \qquad\textbf{(B)} \1 \qquad\textbf{(C)} \50 \qquad\textbf{(D)} \51 \qquad\textbf{(E)} \124</math> | + | <math>\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 50 \qquad\textbf{(D)}\ 51 \qquad\textbf{(E)}\ 124</math> |
[[G285 Summer Problem Set Problem 7|Solution]] | [[G285 Summer Problem Set Problem 7|Solution]] | ||
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<math>\textbf{(A)}\ 67 \qquad\textbf{(B)}\ 69 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 71 \qquad\textbf{(E)}\ 72</math> | <math>\textbf{(A)}\ 67 \qquad\textbf{(B)}\ 69 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 71 \qquad\textbf{(E)}\ 72</math> | ||
+ | |||
+ | [[G285 Summer Problem Set Problem 8|Solution]] | ||
Solution: The first summation is simply <math>9 \cdot 9 = 81</math> by Vieta's. The second summation is <math>-9+r_{12}</math>. The minimum possible value is <math>72+r_{12}</math>, so we need to minimize <math>r_{12}</math>. If we do bounding, when <math>x=0</math> we have <math>p(x)=1</math>, and when <math>x=-1</math> we have <math>p(x)=-230</math>. The shift implies there is a root <math>r_k</math> where <math>0 \le k \le 12</math> such that <math>-1 < r_k < 0</math>. However, <math>x=-1.5</math> seems very close to <math>0</math>, and <math>x<-1.5</math> approaches infinity, so there is another root <math>-2 < r_k < -1</math>. Therefore, we have the smallest root <math>r_{12}</math> must be <math>72+(-2+\{r_{12} \})</math>, where <math>\{x \}</math> is the fractional part. The answer is <math>\boxed{70}</math> | Solution: The first summation is simply <math>9 \cdot 9 = 81</math> by Vieta's. The second summation is <math>-9+r_{12}</math>. The minimum possible value is <math>72+r_{12}</math>, so we need to minimize <math>r_{12}</math>. If we do bounding, when <math>x=0</math> we have <math>p(x)=1</math>, and when <math>x=-1</math> we have <math>p(x)=-230</math>. The shift implies there is a root <math>r_k</math> where <math>0 \le k \le 12</math> such that <math>-1 < r_k < 0</math>. However, <math>x=-1.5</math> seems very close to <math>0</math>, and <math>x<-1.5</math> approaches infinity, so there is another root <math>-2 < r_k < -1</math>. Therefore, we have the smallest root <math>r_{12}</math> must be <math>72+(-2+\{r_{12} \})</math>, where <math>\{x \}</math> is the fractional part. The answer is <math>\boxed{70}</math> |
Revision as of 12:49, 24 June 2021
Welcome to the Birthday Problem Set! In this set, there are multiple choice AND free-response questions. Feel free to look at the solutions if you are stuck:
Contents
Problem 1
Find
Problem 2
Let If
is a positive integer, find the sum of all values of
such that
for some constant
.
Problem 3
Let circles and
with centers
and
concur at points
and
such that
,
. Suppose a point
on the extension of
is formed such that
and lines
and
intersect
and
at
and
respectively. If
, the value of
can be represented as
, where
and
are relatively prime positive integers, and
is square free. Find
Problem 4
Let be a rectangle with
and
. Let points
and
lie on
such that
is the midpoint of
and
lies on
. Let point
be the center of the circumcircle of quadrilateral
such that
and
lie on the circumcircle of
and
respectively, along with
and
. If the shortest distance between
and
is
,
and
are degenerate, and
, find
Problem 5
Suppose is an equilateral triangle. Let points
and
lie on the extensions of
and
respectively such that
and
. If there exists a point
outside of
such that
, and there exists a point
outside outside of
such that
, the area
can be represented as
, where
and
are squarefree,. Find
Problem 6
people are attending a hotel conference,
of which are executives, and
of which are speakers. Each person is designated a seat at one of
round tables, each containing
seats. If executives must sit at least one speaker and executive, there are
ways the people can be seated. Find
. Assume seats, people, and table rotations are distinguishable.
Problem 7
Geometry285 is playing the game "Guess And Choose". In this game, Geometry285 selects a subset of not necessarily distinct integers from the set
such that the sum of all elements in
is
. Each distinct is selected chronologically and placed in
, such that
,
,
, and so on. Then, the elements are randomly arranged. Suppose
represents the total number of outcomes that a subset
containing
integers sums to
. If distinct permutations of the same set
are considered unique, find the remainder when
is divided by
.
Problem 8
Let , Let
be the twelve roots that satisfies
, find the least possible value of
Solution: The first summation is simply by Vieta's. The second summation is
. The minimum possible value is
, so we need to minimize
. If we do bounding, when
we have
, and when
we have
. The shift implies there is a root
where
such that
. However,
seems very close to
, and
approaches infinity, so there is another root
. Therefore, we have the smallest root
must be
, where
is the fractional part. The answer is