Probability AdditionPrinciple AIME Intermediate
2018


Problem - 4112

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 $m$.


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Answer     683

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

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


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

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