Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: We can see that $3^n-1 = 143^2*x$ , which means that $v_{11}(3^n-1) \geq 2$ $v_{13}(3^n-1) \geq 2$ $v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})$ $v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})$ by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as $v_{13}(3^n-1) \geq 2$ . Therefore the minimum possible value of n is $3\times5\times13=\boxed{195}$

Answer:  195
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: Note that the problem is basically asking for the least positive integer $n$ such that $11^2 \cdot 13^2 | 3^n - 1.$ It is easy to see that $n = \text{lcm } (a, b),$ where $a$ is the least positive integer satisfying $11^2 | 3^a - 1$ and $b$ the least positive integer satisfying $13^2 | 3^b - 1$ . Luckily, finding $a$ is a relatively trivial task, as one can simply notice that $3^5 = 243 \equiv 1 \mod 121$ . However, finding $b$ is slightly more nontrivial. The order of $3^k$ modulo $13$ (which is $3$ ) is trivial to find, as one can either bash out a pattern of remainders upon dividing powers of $3$ by $13$ , or one can notice that $3^3 = 27 \equiv 1 \mod 13$ (the latter which is the definition of period/orders by FLT). We can thus rewrite $3^3$ as $(2 \cdot 13 + 1) \mod 13^2$ . Now suppose that \[3^{3k} \equiv (13n + 1) \mod 13^2.\] I claim that $3^{3(k+1)} \equiv (13(n+2) + 1) \mod 13^2.$
Proof:
To find $3^{3(k+1)},$ we can simply multiply $3^{3k}$ by $3^3,$ which is congruent to $2 \cdot 13 + 1$ modulo $13^2$ .
By expanding the product out, we obtain \[3^{3(k+1)} \equiv (13n + 1)(2 \cdot 13 + 1) = 13^2 \cdot 2n + 13n + 2 \cdot 13 + 1 \mod 13^2,\] and since the $13^2$ on the LHS cancels out, we're left with \[13n + 2 \cdot 13 + 1 \mod 13^2 \implies 13(n+2) + 1 \mod 13^2\] . Thus, our claim is proven.
Let $f(n)$ be the second to last digit when $3^{3k}$ is written in base $13^2$ . 
Using our proof, it is easy to see that $f(n)$ satisfies the recurrence $f(1) = 2$ and $f(n+1) = f(n) + 2$ . Since this implies $f(n) = 2n,$ we just have to find the least positive integer $n$ such that $2n$ is a multiple of $13$ , which is trivially obtained as $13$ .
The least integer $n$ such that $3^n - 1$ is divisible by $13^2$ is $3 \cdot 13 = 39,$ so our final answer is $\text{lcm } (5, 39) = \boxed{195}.$

Answer:  195
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: The requested positive integer is the least value of $n>0$ such that $3^n\equiv 1\pmod{143^2}.$ Note that $143=11\cdot 13.$ The least power of $3$ that is congruent to $1$ modulo $11^2$ is $3^5=243=2\cdot 11^2+1.$ It follows that. $3^n\equiv 1\pmod {11^2}$ if and only if $n=5j$ for some positive integer $j$
The least power of $3$ that is congruent to $1$ modulo $13$ is $3^3=27=2\cdot 13+1.$ It follows that $3^n\equiv 1\pmod{13}$ if and only if $n=3k$ for some positive integer $k$ . Additionally, for some positive integer $k$ , the Binomial Theorem shows that $3^{3k}=(26+1)^k=26\cdot k+1 \pmod{13^2}$ . In particular, $3^n=3^{3k}\equiv 1\pmod {13^2}$ if and only if $k=13m$ for some positive integer $m$ , that is, if and only if $n=39m.$
Because $11^2$ and $13^2$ are relatively prime, $3^n\equiv 1\pmod {143^2}$ if and only if $3^n\equiv 1\pmod{11^2}$ and $3^n \equiv 1\pmod {13^2}$ . This occurs if and only if $n$ is a multiple of both of the relatively prime integers $5$ and $39$ , so the least possible value of $n$ is $5\cdot 39=\boxed{195}.$

Answer:  195
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: We first note that we wish to find $3^n \equiv 1 \pmod{11^2}$ and $3^n \equiv 1 \pmod{13^2}.$ Not thinking of anything else, we try a few numbers for the first condition to get that $5 \mid n.$ For the second condition, upon testing up to 729, we find that it doesn't yield a solution readily, so we use Lifting the Exponent from our toolkit to get that \[v_{13}(27^m-1^m)=v_{13}(26)+v(m)=1+v(m), 3m = n\] which clearly implies $m=13$ and $39 | n.$ Our answer is then obviously $39 \cdot 5 = \boxed{195}.$

Answer:  195
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: The question asks us for the probability that a randomly chosen subset of the set of the first 18 positive integers has the property that the sum of its elements is divisible by 3. Note that the total number of subsets is $2^{18}$ because each element can either be in or not in the subset. To find the probability, we will find the total numbers of ways the problem can occur and divide by $2^{18}$
To simplify the problem, let’s convert the set to mod 3:
\[U' = \{1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0\}\]
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 $S = s(T')$ , the sum of the elements of a subset $T'$ of $U'$
$\textbf{Case 1: }S=0$
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 or out of the subset, for a total of $2^{6} = 64$ subsets. In fact, for every case we will have, we can always add a sequence of 0’s 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. This is equivalent to finding the number of ways you can choose such subsets without including the 0's. Therefore this case gives us $\boxed{1}$ way.

Answer:  1
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: Consider the elements of $U$ modulo $3.$
Ignore the $0$ 's because we're gonna multiply $\binom{6}{0}+..+\binom{6}{6}=2^6$ 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 $2^6 \left( 2\left(1+20+90+90+20\right) + \binom{12}{6} \right)\implies \boxed{683}$

Answer:  683
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: Elements $0 \pmod{3}$ can either be included or excluded, for a total of $2^6$ . Then, like the previous solution, let $a$ be the number of elements $1 \pmod{3}$ and $b$ be the number of elements $2 \pmod{3}$ . Then, $a + 2b \equiv 0 \pmod{3} \implies a - b \equiv 0 \pmod{3}$ . Since $0 \le a, b \le 6$ , there are 3 cases:
$\>\>\>\> 1. \> a - b = 0$
Then, we have, \[\binom{6}{0}\binom{6}{0} + \binom{6}{1}\binom{6}{1} + \cdots + \binom{6}{6} \binom{6}{6}\]
Using the substitution $\binom{6}{k} = \binom{6}{6-k}$ on every second binomial coefficient, it is clear that the bottom numbers sum to $6$ , and the ones above sum to $12$ . Apply Vandermonde's, we obtain $\binom{12}{6}$
$\>\>\>\> 2. \> a - b = \pm 3$
For $a - b = 3$ , using the same substitution and applying Vandermonde's, we get: \[\binom{6}{3} \binom{6}{0} + \binom{6}{4} \binom{6}{1} + \cdots + \binom{6}{6} \binom{6}{3} = \binom{12}{3}\]
Then, for $a - b = -3$ , it is completely symmetric with $a - b = 3$ . We get $\binom{12}{3}$
$\>\>\>\> 3. \> a - b = \pm 6$
We have $\binom{6}{6} \binom{6}{0} + \binom{6}{0} \binom{6}{6} = 2$
We get $\frac{2^6 (\binom{12}{6} + 2 \binom{12}{3} + 2)}{2^{18}} = \frac{683}{2^{11}} \implies m = \boxed{683}$

Answer:  683
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: Note that in general, the answer will be around $1/3$ of $2^{18}$ . We will use this to our advantage. Partition $U$ into disjoint subsets $\{1,2,3\}, \{4,5,6\}, \cdots, \{16,17,18\}$ . Within each of these subsets, there are 3 possible remainders $\mod 3$ depending on what elements we choose to include into $T$ : (Using set $\{1,2,3\}$ for demonstration purposes)
Remainder of zero: Include $\{3\}, \{1,2\}, \{\}, \{1,2,3\}$
Remainder of one: Include $\{1\}, \{1,3\}$
Remainder of two: Include $\{2\}, \{2,3\}$
Suppose all subsets are of the form $\{\}, \{1,2,3\}$ . Then, since both choices are $0 \pmod{3}$ , we can choose either one for all six subsets, giving us $2^6$ combinations.
On the other hand, suppose there exists a subset $V$ that isn't of the form $\{\}, \{1,2,3\}$ . Then, for $V$ , it is equally likely that a remainder of zero, one, or two is chosen, since each remainder has $2$ ways to be achieved, i.e. there is a $1/3$ chance for each. Consider the sum of the other subsets $T \setminus V$ . The sum is either $0, 1, 2 \pmod{3}$ . Whatever that remainder $r$ might be, we can always choose $3-r$ as the remainder for our set $V$ , giving us a total of $r + 3 - r \equiv 0 \pmod{3}$ . The probability we choose remainder $3-r$ is $1/3$ as previously mentioned. So, we get $\frac{1}{3}(2^{18} - 2^6)$ total combinations.
Therefore, we get $\frac{2^6 + \frac{1}{3}(2^{18} - 2^6)}{2^{18}} = \frac{683}{2^{11}} \implies m = \boxed{683}$

Answer:  683
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: 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 $1$
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= $\boxed{683}$

Answer:  683
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: 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 \[\left(\binom60+\binom63+\binom66\right)\cdot\left(\binom60+\binom63+\binom66\right)=(1+20+1)^2=484.\]
CASE 2- $1$ or $4$ integers:  There can be $2$ or $5$ integers that are $2\pmod3$ .  We can choose these in \[\left(\binom61+\binom64\right)\cdot\left(\binom62+\binom65\right)=(6+15)^2=441.\]
CASE 3- $2$ or $5$ integers:  There can be $1$ or $4$ integers that are $2\pmod3$ .  We can choose these in \[\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441.\]
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 $\boxed{683}$

Answer:  683
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: 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 $x^{3k}$ (where $k \in \mathbb{Z}_{\geq0}$ ). 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 $x^{3k}$ . (This is known as the third Roots of Unity Filter.) 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 $\boxed{683}$

Answer:  683
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: 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 \[\frac{w(18)}{2^{18}}=\frac{87424}{2^{18}}=\frac{683}{2048},\] so the answer is $\boxed{683}.$

Answer:  683
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: 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 $\tfrac12$
If $U = \{ 1,2,0,1,2,0\} \pmod 3$ , then $24$ out of $64$ subsets work, for a probability of $\tfrac38$
If $U = \{ 1,2,0,1,2,0,1,2,0\} \pmod 3$ , then $176$ out of $512$ subsets work, for a probability of $\tfrac{11}{32}$
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 $a_6 = \boxed{683}$

Answer:  683
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: The total number of subsets is simply $2^{18}.$ Now we need to find the number of subsets that have a sum divisible by $3.$
Ignore the 6 numbers in the list that are divisible by 3. We look only at the number of subsets of $\{1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17\}$ then multiply by $2^6$ at the end. This is because adding a multiple of $3$ to the sum will not change whether it is divisible by $3$ or not and for each of the $6$ multiples, we have two options of whether to add it to the subset or not.
Now, let $f(3k)$ be the number of subsets of $\{1, 2, 4, 5, \dots, 3k-2, 3k-1\}$ that have a sum divisible by $3$ . There are $2k$ numbers in the set. Let us look at the first $2k - 2$ numbers: all subsets of the first $2k - 2$ numbers will have a residue $0, 1,$ or $2$ mod $3.$ If it is $0,$ add both $3k-2$ and $3k-1$ to the subset. If it is $1$ add just $3k-1$ to the set, and if it is $2$ add just $3k-2.$ We don't have to compute all three cases separately; since there is a 1 to 1 correspondence, we only need the sum of the three cases (which we know is $2^{2k-2}$ ).
Now, this counts all the subsets except for those that include neither $3k-1$ nor $3k-2.$ However, this is just $f(3k - 3).$ Thus, $f(3k) = 2^{2k-2} + f(3k-3)$ and the base case is $f(0) = 1.$ We have,
$f(18) = 2^{10} + f(15) = 2^{10} + 2^{8} + f(12) = \dots = 2^{10} + 2^8 + 2^6 + 2^4 + 2^2 + 2^0 + f(0).$ $=2^{10} + 2^8 + \dots + 2^2 + 2.$
Multiplying this by $2^6$ we have, \[\frac{2^7(1 + 2^1 + 2^3 + 2^5 + 2^7 + 2^9)}{2^{18}} .\] The $2^7$ cancels out with the denominator. However, the second term in the product is obviously odd and so does not cancel further with the denominator, which is just a power of $2.$ Thus, we want to find $1 + 2 + 2^3 + 2^5 + 2^7 + 2^9,$ which equals $\boxed{683}.$

Answer:  683
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: The total number of subsets is $\sum_{i=0}^{18}\tbinom{18}{i}=2^{18}$ . If $s(T)\equiv 0\bmod{3}$ , the sum of the elements divides 3. We can rewrite the set as 6 0s, 6 1s, and 6 2s. We can ignore the zeros for now, since they won't influence the sum so we focus on each configuration of the 6 1s and 6 2s such that the sum is divisible by 3 and then multiply by $\sum_{i=0}^{6}\tbinom{6}{i}=2^6$ at the end. We proceed with casework on the number of values that are equivalent to $2\pmod{3}$ as follows:
Case 1: There are zero elements:
Then, there are only configurations of the values congruent to 1 mod 3. There are $\tbinom{6}{0}+\tbinom{6}{3}+\tbinom{6}{6}$ ways to assign values in the set that are congruent to 1 mod 3 such that the sum of those values divides 3. Therefore there are $22$ configurations for this case.
Case 2: There is one element:
There are $6$ ways to choose which element is included in the subset and $\tbinom{6}{1}+\tbinom{6}{4}$ ways to assign values to the numbers congruent to 1 mod 3 in the set. Therefore there are $6\cdot21=126$ configurations for this case.
Case 3: There are two elements:
There are $\tbinom{6}{2}=12$ ways to choose the 2 elements in the set that are congruent to 2 mod 3 and $\tbinom{6}{2}+\tbinom{6}{5}$ possible ways to assign values to the numbers congruent to 1 mod 3. Therefore there are $15\cdot21=315$ configurations for this case.
Case 4: There are three elements:
There are $\tbinom{6}{3}=20$ ways to choose three elements in the set that are congruent to 2 mod 3. There are $\tbinom{6}{0}+\tbinom{6}{3}+\tbinom{6}{6}$ ways to assign values to the numbers congruent to 1 mod 3. Therefore, there are $20\cdot22=440$ configurations for this case.
Case 5, Case 6, and Case 7 are symmetric to Case 3, Case 2, and Case 1 respectively so we can multiply by 2 for those cases.
Putting all the cases together we obtain $2(22+126+315) + 440=1366$ . Multiplying by $2^6$ gives $2^6\cdot1366=2^7\cdot683$ . Since $n=2^{18}$ $\frac{m}{n}=\frac{683}{2^{11}}$ . Therefore, $m=\boxed{683}$

Answer:  683