2018 AIME I Problems/Problem 12

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

Ignore the $0$ 's because we're gonna multiply $2^6$ at the end. Let $a$ be the $1's$ and $b$ be the $2's$ . The key here is that $2 \equiv -1 \pmod{3}$ so the difference between the number of $a$ and $b$ is a multiple of $3$ .

1. Counted twice because $a$ and $b$ can be switched: $6a$ $6a,3b$ $5a,2b$ $4a,b$ $3a$ 2. Counted once: $6a,6b$ ... $0a,0b$

By Vandermonde's identity on the second case, this is $2^6 \left( 2\left(1+20+90+90+20\right) + \binom{12}{6} \right)\implies \boxed{683}$

Solution 1

Rewrite the set after mod 3

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 $1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366$

$P=\frac{1366}{2^{12}}=\frac{683}{2^{11}}$

ANS= $\boxed{683}$

By SpecialBeing2017

Solution 2

Consider the numbers $\{1,4,7,10,13,16\}$ . Each of those are congruent to $1 \pmod 3$ . 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 ${6 \choose 0}+{6 \choose 3}+{6 \choose 6} = 22$ possible subsets congruent to $0\pmod 3$ . There are $2^6=64$ possible subsets of these numbers. By symmetry there are $21$ subsets each for $1 \pmod 3$ and $2 \pmod3$ .

We get the same numbers for the subsets of $\{2,5,8,11,14,17\}$ .

For $\{3,6,9,12,15,18\}$ , all $2^6$ subsets are $0 \pmod3$ .

So the probability is: $\frac{(22\cdot22+2\cdot21\cdot21)\cdot2^6}{2^{18}}=\frac{683}{2^{11}}$

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 $\boxed{683}$ .

Solution 4

We use generating functions. Each element of $U$ has two choices that occur with equal probability--either it is in the set or out of the set. Therefore, given $n\in U$ , the probability generating function is $\frac{1}{2}+\frac{1}{2}x^n.$ Therefore, in the generating function $\frac{1}{2^{18}}(1+x)(1+x^2)(1+x^3)\cdots (1+x^{18}),$ the coefficient of $x^k$ represents the probability of obtaining a sum of $k$ . We wish to find the sum of the coefficients of all terms of the form $x^{3k}$ . If $\omega=e^{2\pi i/3}$ is a cube root of unity, then it is well know that for a polynomial $P(x)$ , $\frac{P(1)+P(\omega)+P(\omega^2)}{3}$ will yield the sum of the coefficients of the terms of the form $x^{3k}$ . Then we find $\begin{align*} \frac{1}{2^{18}}(1+1)(1+1^2)(1+1^3)\cdots (1+1^{18})&=1\\\frac{1}{2^{18}}(1+\omega)(1+\omega^2)(1+\omega^3)\cdots (1+\omega^{18})&=\frac{1}{2^{12}}\\\frac{1}{2^{18}}(1+\omega^2)(1+\omega^4)(1+\omega^6)\cdots (1+\omega^{36})&=\frac{1}{2^{12}}. \end{align*}$ To evaluate the last two products, we utilized the facts that $\omega^3=1$ and $1+\omega+\omega^2=0$ . Therefore, the desired probability is $\frac{1+1/2^{12}+1/2^{12}}{3}=\frac{683}{2^{11}}.$ Thus the answer is $\boxed{683}$ .

Solution 5

Define a set that "works" to be a set for which the sum of the terms is $0$ mod $3$ . The given set mod $3$ is $\{1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0\}.$ Let $w(N)$ be the number of subsets of the first $N$ terms of this set that "work." Consider just the first $3$ terms: $\{1,2,0\}.$ There are $2^3=8$ total subsets, and $w(3)=4$ (the subsets are $\emptyset, \{0\}, \{1,2\},$ and $\{1,2,0\}$ ). Now consider the first $6$ terms: $\{1,2,0,1,2,0\}.$ To find $w(6)$ , we set aside the last $3$ terms as a "reserve" that we can pair with subsets of the first $3$ terms (which we looked at earlier).

First, all $2^3$ subsets of the first $3$ terms can either be paired with a $1$ or a $2$ (or nothing) from the "reserve" terms so that they "work," creating $2^3$ subsets that "work" already. But for each of these, we have the option to add a $0$ from the reserve, so we now have $2\cdot2^3$ subsets that "work." For each of the $w(3)=4$ subsets of the first $3$ terms that "work", we can also add on a $\{1,2\}$ or a $\{1,2,0\}$ from the reserves, creating $2w(3)$ new subsets that "work." And that covers all of them. With all of this information, we can write $w(6)$ as $w(6)=2\cdot2^3+2w(3)=2(2^3+w(3)).$ The same process can be used to calculate $w(9)$ then $w(12)$ etc. so we can generalize it to $w(N)=2(2^{N-3}+w(N-3)).$ Thus, we calculate $w(18)$ with this formula: $w(18)=2( 2^{15} + 2( 2^{12} + 2( 2^9 + 2( 2^6 + 2( 2^3 + 4 ) ) ) ) )=87424.$ Because there are $2^{18}$ total possible subsets, the desired probability is $\frac{w(3)}{2^{18}}=\frac{87424}{2^{18}}=\frac{683}{2048},$ so the answer is $\boxed{683}.$

Solution 6

Try smaller cases and find a pattern. Using similar casework as in Solution 1, we can easily find the desired probability if $U$ is of a small size.

If $U = \{ 1,2,0\} \pmod 3$ , then $4$ out of $8$ subsets work, for a probability of $\frac12$ .

If $U = \{ 1,2,0,1,2,0\} \pmod 3$ , then $24$ out of $64$ subsets work, for a probability of $\frac38$ .

If $U = \{ 1,2,0,1,2,0,1,2,0\} \pmod 3$ , then $176$ out of $512$ subsets work, for a probability of $\frac{11}{32}$ .

Let $a_n$ be the numerator of the desired probability if $U$ is of size $3n$ . Then $a_1 = 1, a_2 = 3,$ and $a_3 = 11$ . Noticing that the denominators are multiplied by $4$ each time, we can conclude that the pattern of the numerators is $a_n = 4a_{n-1} - 1$ , so $a_6 = \boxed{683}$ .

Solution 7 (Quick guesswork for about 2 minutes remaining)

We conjecture that the difference from the probability will be as small as possible from 1/3

(The value approached as n-->infinity, where n is the number of terms in the largest subset.)

We also see that there are 2^18 subsets and know the denominator will be a power of 2 (since the numerator is an integer).

We essentially want to guess that the greatest integer n satisfying (2^n/3)-1 <1000 can be plugged in to get the solution of round(2^n/3)

We see that this occurs at n=11, and get round(2^11/3)=round(682.66...)= 683.

2018 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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