Difference between revisions of "2018 AIME I Problems/Problem 12"
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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>. | 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2018|n=I|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Revision as of 22:54, 8 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
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
.
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.