Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution: Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sum of $90$ and an integer between $1$ and $10$ , we rewrite the problem into receiving scores between $1$ and $10$ . Later, we can add $90$ to her score to obtain the real answer.
From this point of view, the problem states that Isabella's score on the seventh test was $5$ . We note that Isabella received $7$ integer scores out of $1$ to $10$ . Since $5$ is already given as the seventh test score, the possible scores for Isabella on the other six tests are $S={1,2,3,4,6,7,8,9,10}$
The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set $S$ above, and their sum with $5$ must be a multiple of $7$ . The interval containing the possible sums of the six numbers in S are from $1 +2+3+4+6+7=23$ to $4+6+7+8+9+10=44$ . We must now find multiples of $7$ from the interval $23+5 = 28$ to $44+5=49$ . There are four possibilities: $28$ $35$ $42$ $49$ .
However, we also note that the sum of the six numbers (besides $5$ ) must be a multiple of $6$ as well. Thus, $35$ is the only valid choice.(The six numbers sum to $30$ .)
Thus the sum of the six numbers equals to $30$ . We apply the logic above in a similar way for the sum of the scores from the first test to the fifth test. The sum must be a multiple of $5$ . The possible interval is from $1+2+3+4+6=16$ to $6+7+8+9+10=40$ . Since the sum of the five scores must be less than $30$ , the only possibilities are $20$ and $25$ . However, we notice that $25$ does not work because the seventh score turns out to be $5$ from the calculation. Therefore, the sum of Isabella's scores from test $1$ to $5$ is $20$ . Therefore, her score on the sixth test is $10$ .
Our final answer is $10+90=  \boxed{100}$

Answer: E 100
Problem: Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution: Let $n$ be Isabella's average after $6$ tests. $6n+95 \equiv 0 \pmod{7}$ , so $n \equiv 4 \pmod{7}$ . The only integer between $90$ and $100$ that satisfies this condition is $95$ . Let $m$ be Isabella's average after $5$ tests, and let $a$ be her sixth test score. $5m+a= 6\cdot95 \equiv 0 \pmod{5}$ , so $a$ is a multiple of $5$ . Since $100$ is the only choice that is a multiple of $5$ , the answer is $\boxed{100}$

Answer: E 100
Problem: Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution: Let $T$ be the total sum of Isabella's first five test scores, and let $S$ be her score on the sixth test. It follows that $T\equiv 0\pmod {5}$ $T+S\equiv 0\pmod {6}$ , and $T+S+95\equiv 0\pmod {7}$ , since at each step, her average score was an integer. Using the last equivalence, $T+S\equiv -95\equiv 3\pmod{7}$ , so we have a system of equivalences for $T+S$ . Solving this using the Chinese Remainder Theorem, we get $T+S\equiv (0)(7)(1)+(3)(6)(-1) \equiv -18 \equiv 24 \pmod{42}$
Now let's put a bound on $T$ . Using the given information that each test score was a distinct integer from $91$ to $100$ inclusive and that the seventh score was 95, we get $91+92+93+94+96\leq T\leq 100+99+98+97+96$ . Since $T\equiv 0 \pmod{5}$ , we get $T=470,475,480,485,490$ . Therefore, $T\equiv 8,13,18,23,28\pmod{42}$
The last preparation step will involve calculating all the possible test scores $\pmod {42}$ . Here they are: $91\equiv 7\pmod{42},92\equiv 8\pmod{42},93\equiv 9\pmod{42},94\equiv 10\pmod{42},95\equiv 11\pmod{42},96\equiv 12\pmod{42},97\equiv 13\pmod{42},98\equiv 14\pmod{42},99\equiv 15\pmod{42},100\equiv 16\pmod{42}$ . This means that $S\equiv 7,8,9,10,12,13,14,15,16\pmod{42}$ . Note that $11$ is not in the previous list because it corresponds to a score of $95$ , which we cannot have.
We must have $T+S\equiv 24\pmod{42}$ , and using the possible values we found for $T$ and $S$ , the only two that sum to $24$ are $T\equiv 8\pmod{42}$ and $S\equiv 16\pmod{42}$ . This corresponds to an $S$ value of $100$ , so the answer is $\boxed{100}$

Answer: E 100
Problem: Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution: We know that Isabella's score on the 7th test is 95, and we work backwards. Let $a_n$ denote the average of the first $n$ scores. We are given that all $a_n$ are integers. If the $a_6$ is different from the $a_7$ , then the last score must be a multiple of 7 away from $a_6$ . However, $95-7=88$ and $95+7=102$ clearly cannot be $a_6$ since they are outside the range of 91 to 100. Thus, $a_6$ must be 95. Additionally, $a_5$ cannot be 95 since that would mean the sixth score is 95, which isn't possible since scores can't repeat (also, it's not an answer choice). Intuitively, $a_5$ should be close to 95 due to the constraint on the range of scores. We can try 94 and 96 for $a_5$ . Again, the sixth score must be a multiple of 6 (preferably only one multiple) away from $a_5$ . The only one that works is $94+6=100$ , and we can check that the rest don't work since they exceed the range. The answer is $\boxed{100}$

Answer: E 100
Problem: Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution: Say the sum of the first 6 scores is $n$ . Then, we know that $n + 95 \equiv 0 \pmod{7}$ . Thus, $n \equiv 3 \pmod{7}$ . Additionally, since the average of these 6 scores was an integer, we know that $n \equiv 0 \pmod{6}$ . We find that the smallest $n$ to satisfy both of these inequalities is 24 (series below). \[3, 10, 17, 24, ...\] To get to the next one, we have to add $6 \cdot 7 = 42$ so as not to ruin the moduli.
Next, we know that $n$ is between $91 \cdot 6 = 546$ and $100 \cdot 6 = 600$ . It can't take on either value as all the scores have to be different. Now, we calculate the first $n$ that satisfies both moduli equations:
\[546 < 42 + 24x\] \[91 < 7 + 4x\] \[84 < 4x\] \[21 < x\]
$x = 22$ works while $x = 23$ does not (as it hits the upper limit of 600), and we find that $n = 42 + 24 \cdot 22 = 570$ . So, the sum of the first 6 scores is 570. We also know that the sum of the first 5 scores is a multiple of 5. Since $5~\mid~570$ , the 6th score must be a multiple of 5. 95 is taken, so the only possibility $\boxed{100}$

Answer: E 100
Problem: A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$ -player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n$ $9\leq n\leq 2017$ , can be the number of participants?
$\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562$

Solution: Let there be $T$ teams. For each team, there are ${n-5\choose 4}$ different subsets of $9$ players that includes a given full team, so the total number of team-(group of 9) pairs is
\[T{n-5\choose 4}.\]
Thus, the expected value of the number of full teams in a random set of $9$ players is
\[\frac{T{n-5\choose 4}}{{n\choose 9}}.\]
Similarly, the expected value of the number of full teams in a random set of $8$ players is
\[\frac{T{n-5\choose 3}}{{n\choose 8}}.\]
The condition is thus equivalent to the existence of a positive integer $T$ such that
\[\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.\]
\[T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}\]
\[T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}\]
Note that this is always less than ${n\choose 5}$ , so as long as $T$ is integral, $n$ is a possibility. Thus, we have that this is equivalent to
\[2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).\]
It is obvious that $5$ divides the RHS, and that $7$ does iff $n\equiv 0,1,2,3,4\mod 7$ . Also, $3^2$ divides it iff $n\not\equiv 5,8\mod 9$ . One can also bash out that $2^5$ divides it in $16$ out of the $32$ possible residues $\mod 32$
Note that $2016 = 7*9*32$ so by using all numbers from $2$ to $2017$ , inclusive, it is clear that each possible residue $\mod 7,9,32$ is reached an equal number of times, so the total number of working $n$ in that range is $5\cdot 7\cdot 16 = 560$ . However, we must subtract the number of "working" $2\leq n\leq 8$ , which is $3$ . Thus, the answer is $\boxed{557}$

Answer: D 557
Problem: The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] real x=.369; draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray); filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray); filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray); filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray); filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray); [/asy]
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution: Let $s$ be the side length of the small squares.
The diagonal of the big square can be written in two ways: $\sqrt{2}$ and $s \sqrt{2} + s + s \sqrt{2}$
Solving for $s$ , we get $s = \frac{4 - \sqrt{2}}{7}$ , so our answer is $4 + 7 \Rightarrow \boxed{11}$

Answer: E 11
Problem: The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] real x=.369; draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray); filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray); filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray); filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray); filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray); [/asy]
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution: The diagonal of the small square can be written in two ways: $s \sqrt(2)$ and $2*(1-2s).$ Equating and simplifying gives $s = \frac{4 - \sqrt{2}}{7}$ . Hence our answer is $4 + 7 \Rightarrow \boxed{11}.$

Answer: E 11
Problem: Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64$

Solution: Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act.
Let $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act.
From the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$
Adding these equations together, we get $2(a + b + c) + ab + bc + ac = 136$
Since there are a total of $100$ students, $a + b + c + ab + bc + ac = 100$
Subtracting these equations, we get $a + b + c = 36$
Our answer is $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{64}$

Answer: E 64
Problem: Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64$

Solution: An easier way to solve the problem:
Since $42$ students cannot sing, there are $100-42=58$ students who can.
Similarly $65$ students cannot dance, there are $100-65=35$ students who can.
And $29$ students cannot act, there are $100-29=71$ students who can.
Therefore, there are $58+35+71=164$ students in all ignoring the overlaps between $2$ of $3$ talent categories.
There are no students who have all $3$ talents, nor those who have none $(0)$ , so only $1$ or $2$ talents are viable.
Thus, there are $164-100=\boxed{64}$ students who have $2$ of $3$ talents.

Answer: E 64
Problem: The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution: Setting the first two equations equal to each other, $\log_3 x = \log_x 3$
Solving this, we get $\left(3, 1\right)$ and $\left(\frac{1}{3}, -1\right)$
Similarly with the last two equations, we get $\left(3, -1\right)$ and $\left(\frac{1}{3}, 1\right)$
Now, by setting the first and third equations equal to each other, we get $\left(1, 0\right)$
Pairing the first and fourth or second and third equations won't work because then $\log x \leq 0$
Pairing the second and fourth equations will yield $x = 1$ , but since you can't divide by $\log 1 = 0$ , it doesn't work.
After trying all pairs, we have a total of $5$ solutions $\rightarrow \boxed{5}$

Answer: D 5
Problem: The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution: Note that $\log_b a =\log_c a / \log_c b$
Then $\log_b a = \log_a a / \log_a b = 1/ \log_a b$
$\log_\frac{1}{a} b = \log_a \frac{1}{a} / \log_a b = -1/ \log_a b$
$\log_\frac{1}{b} a = -\log_a b$
Therefore, the system of equations can be simplified to:
$y = t$
$y = -t$
$y = \frac{1}{t}$
$y = -\frac{1}{t}$
where $t = \log_3 x$ . Note that all values of $t$ correspond to exactly one positive $x$ value, so all $(t,y)$ intersections will correspond to exactly one $(x,y)$ intersection in the positive-x area.
Graphing this system of functions will generate a total of $5$ solutions $\rightarrow \boxed{5}$

Answer: D 5
Problem: Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$
$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$

Solution: For $6$ to $8$ heads, we are guaranteed to hit $4$ heads, so the sum here is $\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37$
For $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$
For $5$ heads, we either start off with $4$ heads, which gives us $_4\text{C}_1=4$ ways to arrange the other flips, or we start off with five heads and one tail, which has $6$ ways minus the $2$ overlapping cases, $\text{HHHHHTTT}$ and $\text{HHHHTHTT}$ . Total ways: $8$
Then we sum to get $46$ . There are a total of $2^8=256$ possible sequences of $8$ coin flips, so the probability is $\frac{46}{256}=\frac{23}{128}$ . Summing, we get $23+128=\boxed{151}$

Answer: B 151
Problem: Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$
$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$

Solution: Reaching 4 will require either 4, 6, or 8 flips. Therefore we can split into 3 cases:
(Case 1): The first four flips are heads. Then, the last four flips can be anything so $2^4=16$ possibilities work.
(Case 2): It takes 6 flips to reach 4. There must be one tail in the first four flips so we don't repeat case 1. The tail can be in one of 4 positions. The next two flips must be heads. The last two flips can be anything so $2^2=4$ flips work. $4*4=16$
(Case 3): It takes 8 flips to reach 4. We can split this case into 2 sub-cases. There can either be 1 or 2 tails in the first 4 flips.
(1 tail in first four flips). In this case, the first tail can be in 4 positions. The second tail can be in either the 5th or 6th position so we don't repeat case 2. Thus, there are $4*2=8$ possibilities.
(2 tails in first four flips). In this case, the tails can be in $\binom{4}{2}=6$ positions.
Adding these cases up and taking the total out of $2^8=256$ yields $\frac{16+16+8+6}{256}=\frac{46}{256}=\frac{23}{128}$ . This means the answer is $23+128=\boxed{151}$

Answer: B 151
Problem: Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$
$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$

Solution: Notice every $2$ flips, there is a $\dfrac{1}{4}$ chance to go $2$ steps left $(L),$ $\dfrac{1}{2}$ chance to stay put $(P),$ and $\dfrac{1}{4}$ chance to move $2$ steps right $(R).$ We have $4$ choices for how to get to $4:$ \[RR-1\text{ arrangement}-\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}\] \[PRR-2\text{ arrangements}-2\cdot\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}\] \[LRRR-2\text{ arrangements}-2\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{128}\] \[PPRR-3\text{ arrangements}-3\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{3}{64}\] Finally, our answer will be $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{128}+\dfrac{3}{64}=\dfrac{23}{128}\implies a+b=\boxed{151}.$

Answer: B 151
Problem: Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$

Solution: Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be $x,y,$ and $a$ . WLOG assume $a$ is the largest. Then, $x+y>a$ , meaning the solution is $\boxed{12}$ , as shown in the graph below. [asy] pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair E = (0,0);  draw(A--B--C--D--cycle); draw(B--D,dashed); fill(B--D--C--cycle,gray);  label("$0$",A,SW); label("$a$",B,S); label("$a$",D,W); label("$y$",(0,.5),W); label("$x$",(.5,0),S); label("$x+y>a$",(5/7,5/7)); [/asy]

Answer: C 12
Problem: Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$

Solution: WLOG, let the largest of the three numbers drawn be $a>0$ .  Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ .  The probability that their sum is greater than $a$ is $\boxed{12}$

Answer: C 12