Difference between revisions of "2018 AIME I Problems/Problem 12"
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<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \frac{1}{2^{18}}(1+ | + | \frac{1}{2^{18}}(1+1)(1+1^2)(1+1^3)\cdots (1+1^{18})&=1\\\frac{1}{2^{18}}(1+\omega)(1+\omega^2)(1+\omega^3)\cdots (1+\omega^{18})&=\frac{1}{2^{12}}\\\frac{1}{2^{18}}(1+\omega^2)(1+\omega^4)(1+\omega^6)\cdots (1+\omega^{36})&=\frac{1}{2^{12}}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
To evaluate the last two products, we utilized the facts that <math>\omega^3=1</math> and <math>1+\omega+\omega^2=0</math>. Therefore, the desired probability is | To evaluate the last two products, we utilized the facts that <math>\omega^3=1</math> and <math>1+\omega+\omega^2=0</math>. Therefore, the desired probability is |
Revision as of 20:05, 13 March 2018
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 1
Case 2 222 20
Case 3 222222 1
Case 4 12 6*6=36
Case 5 12222 6*15=90
Case 6 1122 15*15=225
Case 7 1122222 15*6=90
Case 8 111 20
Case 9 111222 20*20=400
Case 10 111222222 20
Case 11 11112 15*6=90
Case 12 11112222 15*15=225
Case 13 1111122 6*15=90
Case 14 1111122222 6*6=36
Case 15 111111 1
Case 16 111111222 20
Case 17 111111222222 1
Total 1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366
P=1362/2^12=683/2^11
ANS=683
By S.B.
Solution 2
Consider the numbers {1,4,7,10,13,16}. Each of those are congruent to 1mod3. There is way to choose zero numbers
ways to choose 1 and so on. There ends up being
possible subsets congruent to 0mod 3. There are
possible subsets of these numbers.
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.