2018 AIME I Problems/Problem 12
Contents
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
Consider the elements of modulo
Ignore the 's because we're gonna multiply
at the end. Let
be the
and
be the
. The key here is that
so the difference between the number of
and
is a multiple of
.
1. Counted twice because and
can be switched:
2. Counted once:
...
By Vandermonde's Identity on the second case, this is
Solution 2 (Simple Casework)
The question asks us for the probability that a randomly chosen subset of the set of the fist 18 positive integers has the property that the sum of its elements is divisible by 3. To simplify the problem, let’s convert the set to mod 3:
Note that there are six elements congruent to 0 mod 3, 6 congruent to 1 mod 3, and 6 congruent to 2 mod 3. After conversion to mod three, the problem is the same but we’re dealing with much simpler numbers. Let’s apply casework on this converted set based on the sum of the subset’s elements.
The sum is 0.
In this case, we can restrict the subsets to subsets that only contain 0. There are six 0’s and each one can be in our out of the subset, for a total of subsets. In fact, for every case we will have, we can always add a 0 and the total sum will not change, so we will have to multiply by 64 for each case. Therefore, let’s just calculate the total number of ways we can have each case and remember to multiply it in after summing up the cases. Therefore this case gives us
way.
The sum is 3.
In this case and each of the subsequent cases, we denote the number of 1’s in the case and the number of two’s in the case as respectively. Then in this case we have two subcases;
x, y = 3, 0: This case has ways.
x,
Solution 3
Rewrite the set after mod 3 as above
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 SpecialBeing2017
Solution 4
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 5
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
Case 2- or
integers: There can be
or
integers that are
. We can choose these in
Case 3- or
integers: There can be
or
integers that are
. We can choose these in
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 6
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
.
Solution 7
Define a set that "works" to be a set for which the sum of the terms is mod
. The given set mod
is
Let
be the number of subsets of the first
terms of this set that "work."
Consider just the first
terms:
There are
total subsets, and
(the subsets are
and
). Now consider the first
terms:
To find
, we set aside the last
terms as a "reserve" that we can pair with subsets of the first
terms (which we looked at earlier).
First, all subsets of the first
terms can either be paired with a
or a
(or nothing) from the "reserve" terms so that they "work," creating
subsets that "work" already. But for each of these, we have the option to add a
from the reserve, so we now have
subsets that "work." For each of the
subsets of the first
terms that "work", we can also add on a
or a
from the reserves, creating
new subsets that "work." And that covers all of them. With all of this information, we can write
as
The same process can be used to calculate
then
etc. so we can generalize it to
Thus, we calculate
with this formula:
Because there are
total possible subsets, the desired probability is
so the answer is
Solution 8
Try smaller cases and find a pattern. Using similar casework as in Solution 1, we can easily find the desired probability if is of a small size.
If , then
out of
subsets work, for a probability of
.
If , then
out of
subsets work, for a probability of
.
If , then
out of
subsets work, for a probability of
.
Let be the numerator of the desired probability if
is of size
. Then
and
. Noticing that the denominators are multiplied by
each time, we can conclude that the pattern of the numerators is
, so
.
Solution 9 (Quick guesswork for about 2 minutes remaining)
We conjecture that the difference from the probability will be as small as possible from (The value approached as
, where
is the number of terms in the largest subset).
We also see that there are subsets and know the denominator will be a power of
(since the numerator is an integer).
We essentially want to guess that the greatest integer satisfying
can be plugged in to get the solution of round
.
We see that this occurs at , and get round
round(
)=
.
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 |
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