Difference between revisions of "2018 AIME I Problems/Problem 12"
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Case 3- <math>2</math> or <math>5</math> integers: There can be <math>1</math> or <math>4</math> integers that are <math>2\pmod3</math>. We can choose these in <math>\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441</math> ways. | Case 3- <math>2</math> or <math>5</math> integers: There can be <math>1</math> or <math>4</math> integers that are <math>2\pmod3</math>. We can choose these in <math>\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441</math> ways. | ||
− | Adding these up, we get that there are <math>1366</math> ways to choose the numbers such that their sum is a multiple of three. Putting back in the possibility that there can be multiples of <math>3</math> in our set, we have that there are <math>1366\cdot\left(\binom60+\binom61+\binom62+\binom63+\binom64+\binom65+\binom66 | + | Adding these up, we get that there are <math>1366</math> ways to choose the numbers such that their sum is a multiple of three. Putting back in the possibility that there can be multiples of <math>3</math> in our set, we have that there are <math>1366\cdot\left(\binom60+\binom61+\binom62+\binom63+\binom64+\binom65+\binom66\right)=1366\cdot2^6</math> subsets <math>T</math> with a sum that is a multiple of <math>3</math>. Since there are <math>2^{18}</math> total subsets, the probability is <math>\frac{1366\cdot2^6}{2^{18}}=\frac{683}{2^{11}}</math>, so the answer is <math>\boxed{683}</math>. |
==Solution 4== | ==Solution 4== |
Revision as of 18:29, 28 January 2019
Problem
For every subset of
, let
be the sum of the elements of
, with
defined to be
. If
is chosen at random among all subsets of
, the probability that
is divisible by
is
, where
and
are relatively prime positive integers. Find
.
Solution 1
Rewrite the set after mod3
1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0
All 0s can be omitted
Case 1
No 1 No 2
Case 2
Case 3
Case 4
Case 5
Case 6
Case 7
Case 8
Case 9
Case 10
Case 11
Case 12
Case 13
Case 14
Case 15
Case 16
Case 17
Total
ANS=
By S.B.
Solution 2
Consider the numbers . Each of those are congruent to
. There is
way to choose zero numbers
ways to choose
and so on. There ends up being
possible subsets congruent to
. There are
possible subsets of these numbers. By symmetry there are
subsets each for
and
.
We get the same numbers for the subsets of .
For , all
subsets are
.
So the probability is:
Solution 3
Notice that six numbers are , six are
, and six are
. Having numbers
will not change the remainder when
is divided by
, so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are
, minus the number of numbers that are
, must be a multiple of
, possibly zero or negative. We can now split into cases based on how many numbers that are
are in the set.
Case 1- ,
, or
integers: There can be
,
, or
integers that are
. We can choose these in
ways.
Case 2- or
integers: There can be
or
integers that are
. We can choose these in
ways.
Case 3- or
integers: There can be
or
integers that are
. We can choose these in
ways.
Adding these up, we get that there are ways to choose the numbers such that their sum is a multiple of three. Putting back in the possibility that there can be multiples of
in our set, we have that there are
subsets
with a sum that is a multiple of
. Since there are
total subsets, the probability is
, so the answer is
.
Solution 4
We use generating functions. Each element of has two choices that occur with equal probability--either it is in the set or out of the set. Therefore, given
, the probability generating function is
Therefore, in the generating function
the coefficient of
represents the probability of obtaining a sum of
. We wish to find the sum of the coefficients of all terms of the form
. If
is a cube root of unity, then it is well know that for a polynomial
,
will yield the sum of the coefficients of the terms of the form
. Then we find
To evaluate the last two products, we utilized the facts that
and
. Therefore, the desired probability is
Thus the answer is
.
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.