Difference between revisions of "2018 AIME I Problems/Problem 12"

Revision as of 21:54, 8 March 2018

Problem

For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ .

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 ${6 \choose 0}=1$ way to choose zero numbers ${6 \choose 1}=6$ ways to choose 1 and so on. There ends up being ${6 \choose 0}+{6 \choose 3}+{6 \choose 6}$ possible subsets congruent to 0mod 3. There are $2^6=64$ possible subsets of these numbers.

Solution 3

Notice that six numbers are $0\pmod3$ , six are $1\pmod3$ , and six are $2\pmod3$ . Having numbers $0\pmod3$ will not change the remainder when $s(T)$ is divided by $3$ , so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are $1\pmod3$ , minus the number of numbers that are $2\pmod3$ , must be a multiple of $3$ , possibly zero or negative. We can now split into cases based on how many numbers that are $1\pmod3$ are in the set.

Case 1- $0$ , $3$ , or $6$ integers: There can be $0$ , $3$ , or $6$ integers that are $2\pmod3$ . We can choose these in $\left(\binom60+\binom63+\binom66\right)\cdot\left(\binom60+\binom63+\binom66\right)=(1+20+1)^2=484$ ways.

Case 2- $1$ or $4$ integers: There can be $2$ or $5$ integers that are $2\pmod3$ . We can choose these in $\left(\binom61+\binom64\right)\cdot\left(\binom62+\binom65\right)=(6+15)^2=441$ ways.

Case 3- $2$ or $5$ integers: There can be $1$ or $4$ integers that are $2\pmod3$ . We can choose these in $\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441$ ways.

Adding these up, we get that there are $1366$ 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 $3$ in our set, we have that there are $1366\cdot\left(\binom60+\binom61+\binom62+\binom63+\binom64+\binom65+\binom66+\right)=1366\cdot2^6$ subsets $T$ with a sum that is a multiple of $3$ . Since there are $2^{18}$ total subsets, the probability is $\frac{1366\cdot2^6}{2^{18}}=\frac{683}{2^{11}}$ , so the answer is $\boxed{683}$ .

@@ Line 75: / Line 75: @@
 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}}

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