Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$

Solution: First, we know that there are $6$ ways to assign a color to vertex $A$ . Then, there are $6$ ways to assign a color to vertex $B$ , because it doesn't depend on vertex $A$ . There are also $5$ ways to assign a color to vertex $C$ , because it can't be the same color as vertex $A$ . Notice that both vertices $E$ and $D$ have 5 or 4 choices, depending on whether $A$ $B$ , and $C$ are the same color. This means that the we know that the amount of choices must be bigger than $6 \times 6 \times 5 \times 4 \times 4 = 2880$ , which eliminates options $A$ and $B$
Then, notice that the amount of choices must be smaller than $6 \times 6 \times 5 \times 4.5 \times 4.5 = 3645$ , because the expected amount of choices $E$ and $D$ have are less than 4.5. (There is a greater probability that $A$ $B$ , and $B$ $C$ are both different). This eliminates option $E$ , which leaves us with options $C$ and $D$
We know that the amount of choices must be a multiple of $6$ , so $D$ is eliminated, leaving us with option $\boxed{3120}$

Answer: C 3120
Problem: Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$

Solution: First look at the numbers Alice says. $1, 3, 4, 6, 7, 9 \cdots$ skipping every number that is congruent to $2 \pmod 3$ . Thus, Barbara says those numbers EXCEPT every second - being $2 + 3^1 \equiv 5 \pmod{3^2=9}$ . So Barbara skips every number congruent to $5 \pmod 9$ . We continue and see:
Alice skips $2 \pmod 3$ , Barbara skips $5 \pmod 9$ , Candice skips $14 \pmod {27}$ , Debbie skips $41 \pmod {81}$ , Eliza skips $122 \pmod {243}$ , and Fatima skips $365 \pmod {729}$
Since the only number congruent to $365 \pmod {729}$ and less than $1,000$ is $365$ , the correct answer is $\boxed{365}$

Answer: C 365
Problem: Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$

Solution: After Alice says all her numbers, the numbers not mentioned yet are \[\text{Alice: } 2,5,8,11,14,17,\cdots,998.\] After Barbara says all her numbers, the numbers that haven't been said yet are \[\text{Barbara: } 5,14,23,32,41,50,\cdots,995.\] After Candice, the list is \[\text{Candice: } 14,41,68,\cdots,986.\] Notice how each list is an arithmetic sequence where the common differene is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: \[\text{Debbie: } 41,122,203,\cdots,959\] \[\text{Eliza: } 122,365,608,878\] \[\text{Fatima: } 365\]
Thus, George says $\boxed{365}$

Answer: C 365
Problem: Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$

Solution: Notice that Alice has skipped the numbers $3n-1$ for $n=1,2,3,...,333$ .
Namely, \[3\cdot1-1,3\cdot2-1,3\cdot3-1,...,3\cdot333-1\] Thus the numbers that Barbara skips are \[3\cdot2-1,3\cdot5-1,3\cdot8-1,...\] or in a more general expression, $3(3n-1)-1$ for $n=1,2,3,...$ .
Namely, \[3\cdot(3\cdot1-1)-1,3\cdot(3\cdot2-1)-1,3\cdot(3\cdot3-1)-1,...\] Repeating the pattern until George, we have the first number he says, \[3(3(3(3(3(3\cdot1-1)-1)-1)-1)-1)-1= \boxed{365}\] In addition, note that the second number George says exceeds $1000$

Answer: C 365
Problem: Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$

Solution: Similar to Solution 1 we find that the only number that George can say must leave a remainder of $2$ when divided by $3$ , and that it must also leave a remainder of $5$ when divided by $9$ . Since we as human beings are usually lazy, and that MAA provides answer choices, we check all the possible numbers and find that our answer is $\boxed{365}$

Answer: C 365
Problem: Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$

Solution: Every integer from 1 to 1000 can be written in the form $t_n+1$ in base 10, where $t_n$ is a trinary integer with no more than 7 significant digits. The important insight is that person $1\le{k}\le6$ will not say $n$ if and only if the $k$ th digit from the right of $t_n$ is 1. Therefore, the last 6 digits of $t_n$ must be 1, as the first 6 people never said the number. The only options for $t_n$ are thus $2111111$ $1111111$ , and $0111111$ . But, since George only said one number, the first two must have been too big for it to be $\le1000$ . Our answer is therefore $111111_3+1=\frac{3^6-1}{3-1}+1=364+1=\boxed{365}$

Answer: C 365
Problem: Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textbf{(B)}\ 1560 \qquad \textbf{(C)}\ 2320 \qquad \textbf{(D)}\ 2480 \qquad \textbf{(E)}\ 2500$

Solution: There must be four rays emanating from $X$ that intersect the four corners of the square region. Depending on the location of $X$ , the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is $100$ -ray partitional (let this point be the bottom-left-most point).
We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining $96$ rays are divided among the other two triangular sectors, each sector with $48$ rays, thus dividing these two sectors into $49$ triangles of equal areas.
Let the distance from this corner point to the closest side be $a$ and the side of the square be $s$ . From this, we get the equation $\frac{a\times s}{2}=\frac{(s-a)\times s}{2}\times\frac1{49}$ . Solve for $a$ to get $a=\frac s{50}$ . Therefore, point $X$ is $\frac1{50}$ of the side length away from the two sides it is closest to. By moving $X$ $\frac s{50}$ to the right, we also move one ray from the right sector to the left sector, which determines another $100$ -ray partitional point. We can continue moving $X$ right and up to derive the set of points that are $100$ -ray partitional.
In the end, we get a square grid of points each $\frac s{50}$ apart from one another. Since this grid ranges from a distance of $\frac s{50}$ from one side to $\frac{49s}{50}$ from the same side, we have a $49\times49$ grid, a total of $2401$ $100$ -ray partitional points. To find the overlap from the $60$ -ray partitional, we must find the distance from the corner-most $60$ -ray partitional point to the sides closest to it. Since the $100$ -ray partitional points form a $49\times49$ grid, each point $\frac s{50}$ apart from each other, we can deduce that the $60$ -ray partitional points form a $29\times29$ grid, each point $\frac s{30}$ apart from each other. To find the overlap points, we must find the common divisors of $30$ and $50$ which are $1, 2, 5,$ and $10$ . Therefore, the overlapping points will form grids with points $s$ $\frac s{2}$ $\frac s{5}$ , and $\frac s{10}$ away from each other respectively. Since the grid with points $\frac s{10}$ away from each other includes the other points, we can disregard the other grids. The total overlapping set of points is a $9\times9$ grid, which has $81$ points. Subtract $81$ from $2401$ to get $2401-81=\boxed{2320}$

Answer: C 2320
Problem: Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textbf{(B)}\ 1560 \qquad \textbf{(C)}\ 2320 \qquad \textbf{(D)}\ 2480 \qquad \textbf{(E)}\ 2500$

Solution: Position the square region $R$ so that the bottom-left corner of the square is at the origin. Then define $s$ to be the sidelength of $R$ and $X$ to be the point $(rs, qs)$ , where $0<r,q<1$
There must be four rays emanating from $X$ that intersect the four corners of $R$ . The areas of the four triangles formed by these rays are then $A_1=\frac{qs\times s}{2}$ $A_2=\frac{(s-rs)\times s}{2}$ $A_3=\frac{(s-qs)\times s}{2}$ , and $A_4=\frac{rs\times s}{2}$
If a point is $n$ -ray partitional, then there exist positive integers $a, b, c, d$ such that $a+b+c+d=n$ and $\frac{A_1}{a}=\frac{A_2}{b}=\frac{A_3}{c}=\frac{A_4}{d}$ . Substituting in our formulas for $A_1$ $A_2$ $A_3$ , and $A_4$ and canceling equal terms, we get \[\frac{q}{a}=\frac{1-r}{b}=\frac{1-q}{c}=\frac{r}{d}.\]
Taking $\frac{q}{a}=\frac{1-q}{c}$ and solving for $q$ , we get $q=\frac{a}{a+c}$ , and taking $\frac{1-r}{b}=\frac{r}{d}$ and solving for $r$ , we get $r=\frac{d}{b+d}$ . Finally, from $\frac{q}{a}=\frac{r}{d}$ , we have $qd=ar$ $\Leftrightarrow$ $\frac{ad}{a+c}=\frac{ad}{b+d}$ $\Leftrightarrow$ $a+c=b+d$
So for a point $X$ to be $100$ -ray partitional, $a+b+c+d=100$ , so $a+c=b+d=50$ $X$ must then be of the form $\left(\frac{d}{50}s, \frac{a}{50}s\right)$ . Since $X$ is in the interior of $R$ $a$ and $d$ can be any positive integer from $1$ to $49$ (with $b$ and $c$ just equaling $50-d$ and $50-a$ , respectively). Thus, there are $49\times 49=2401$ points that are $100$ -ray partitional.
However, the problem asks for points that are not only $100$ -ray partitional but also not $60$ -ray partitional. Points that are $60$ -ray partitional are of the form $\left(\frac{m}{30}s, \frac{n}{30}s\right)$ , where $m$ and $n$ are also positive integers. We count the number of points $\left(\frac{d}{50}s, \frac{a}{50}s\right)$ that can also be written in this form. For a given $d$ $\frac{d}{50}=\frac{m}{30}$ if and only if $m=\frac{3}{5}d$ , and likewise with $a$ and $n$ . We can then see that a point is both $100$ -ray partitional and $60$ -ray partitional if and only if $a$ and $d$ are both divisible by $5$ . There are $9$ integers between $1$ and $49$ that are divisible by $5$ , so out of our $2401$ points that are $100$ -ray partitional, $9\times 9=81$ are also $60$ -ray partitional.
Our answer then is just $2401-81=\boxed{2320}$

Answer: C 2320
Problem: Josanna's test scores to date are $90, 80, 70, 60,$ and $85$ . Her goal is to raise here test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 82 \qquad\textbf{(C)}\ 85 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 95$

Solution: The average of her current scores is $77$ . To raise it $3$ points, she needs an average of $80$ , and so after her $6$ tests, a sum of $480$ . Her current sum is $385$ , so she needs a $480 - 385 = \boxed{95}$

Answer: E 95
Problem: At a store, when a length or a width is reported as $x$ inches that means it is at least $x - 0.5$ inches and at most $x + 0.5$ inches. Suppose the dimensions of a rectangular tile are reported as $2$ inches by $3$ inches. In square inches, what is the minimum area for the rectangle?
$\textbf{(A)}\ 3.75 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8.75$

Solution: The minimum dimensions of the rectangle are $1.5$ inches by $2.5$ inches. The minimum area is $1.5\times2.5=\boxed{3.75}$ square inches.

Answer: A 3.75
Problem: LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A < B$ . How many dollars must LeRoy give to Bernardo so that they share the costs equally?
$\textbf{(A)}\ \frac{A + B}{2} \qquad\textbf{(B)}\ \dfrac{A - B}{2}\qquad\textbf{(C)}\ \dfrac{B - A}{2}\qquad\textbf{(D)}\ B - A \qquad\textbf{(E)}\ A + B$

Solution: The difference in how much LeRoy and Bernardo paid is $B-A$ . To share the costs equally, LeRoy must give Bernardo half of the difference, which is $\boxed{2}$

Answer: C 2
Problem: LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A < B$ . How many dollars must LeRoy give to Bernardo so that they share the costs equally?
$\textbf{(A)}\ \frac{A + B}{2} \qquad\textbf{(B)}\ \dfrac{A - B}{2}\qquad\textbf{(C)}\ \dfrac{B - A}{2}\qquad\textbf{(D)}\ B - A \qquad\textbf{(E)}\ A + B$

Solution: Since there are no restrictions on cost paid besides $A<B$ , we can use an example where $A = 40$ and $B = 50$ . Quickly, we realize the only way they could pay the same amount of money is if they both pay 45 dollars. This means LeRoy must give Bernardo $50 - 45 = 5$ . Looking at the answer choices we see only $\frac{50-40}{2} = 5$ works. $\boxed{2}$

Answer: C 2
Problem: In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161$ . What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad\textbf{(B)}\  161 \qquad\textbf{(C)}\  204 \qquad\textbf{(D)}\ 214 \qquad\textbf{(E)}\ 224$

Solution: We have $161 = 7 \cdot 23.$ Since $a$ has two digits, the factors must be $23$ and $7,$ so $a = 32$ and $b = 7.$ Then, $ab = 7 \times 32 = \boxed{224}.$

Answer: E 224
Problem: On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66$

Solution: Let $x$ represent the amount of candies Casper had at the beginning.
\begin{align*} \frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &= 0\\ \frac{2}{3} x - 2 &= 18\\ \frac{2}{3} x &= 20\\ x &= \boxed{30}

Answer: A 30
Problem: On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66$

Solution: We work backwards. If he had 8 candies at the end, then before he gave candies to his sister he had 12 candies. This means that at the end of Halloween he had 18 candies, so before he gave candies to his brother he had 20 candies. Therefore, at the start he had $\boxed{30}$

Answer: A 30
Problem: On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66$

Solution: A solve by algebra is more secure and safe (and usually faster), but you can also test the answer choices. We are lucky and option A works $\Longrightarrow \boxed{30}$

Answer: A 30
Problem: The sum of two angles of a triangle is $\frac{6}{5}$ of a right angle, and one of these two angles is $30^{\circ}$ larger than the other. What is the degree measure of the largest angle in the triangle?
$\textbf{(A)}\ 69 \qquad\textbf{(B)}\ 72 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 102 \qquad\textbf{(E)}\ 108$

Solution: The sum of two angles in a triangle is $\frac{6}{5}$ of a right angle $\longrightarrow \frac{6}{5} \times 90 = 108$
If $x$ is the measure of the first angle, then the measure of the second angle is $x+30$ \[x + x + 30 = 108 \longrightarrow 2x = 78 \longrightarrow x = 39\]
Now we know the measure of two angles are $39^{\circ}$ and $69^{\circ}$ . By the Triangle Sum Theorem, the sum of all angles in a triangle is $180^{\circ},$ so the final angle is $72^{\circ}$ . Therefore, the largest angle in the triangle is $\boxed{72}$

Answer: B 72
Problem: Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$

Solution: The requested ratio is \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Using the formula for a geometric series, we have \[10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},\] which is very close to $\dfrac{10^{10}}{9},$ so the ratio is very close to $\boxed{9}.$

Answer: B 9
Problem: Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$

Solution: The problem asks for the value of \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Written in base 10, we can find the value of $10^9 + 10^8 + \ldots + 10 + 1$ to be $1111111111.$ Long division gives us the answer to be $\boxed{9}.$

Answer: B 9
Problem: Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$

Solution: Let $f(n)=\dfrac{10^n}{1+10+10^2+10^3+\cdots+10^{n-1}}$ . We are approximating $f(10)$ . Trying several small values of $n$ gives answers very close to $9$ , so our answer is $\boxed{9}\approx9.09.$ ~Technodoggo

Answer: B 9
Problem: There are $52$ people in a room. what is the largest value of $n$ such that the statement "At least $n$ people in this room have birthdays falling in the same month" is always true?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12$

Solution: Pretend you have $52$ people you want to place  in $12$ boxes, because there are $12$ months in a year. By the Pigeonhole Principle , one box must have at least $\left\lceil \frac{52}{12} \right\rceil$ people $\longrightarrow \boxed{5}$

Answer: D 5
Problem: Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?
$\textbf{(A)}\ \frac{\pi}{3} \qquad\textbf{(B)}\ \frac{2\pi}{3} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}$

Solution: Let $s$ be Keiko's speed in meters per second, $a$ be the length of the straight parts of the track, $b$ be the radius of the smaller circles, and $b+6$ be the radius of the larger circles. The length of the inner edge will be $2a+2b \pi$ and the length of the outer edge will be $2a+2\pi (b+6).$ Since it takes $36$ seconds longer for Keiko to walk on the outer edge,
\begin{align*} \frac{2a+2b \pi}{s} + 36 &= \frac{2a+2\pi (b+6)}{s}\\ 2a+2b\pi +36s &= 2a+2b\pi +12\pi\\ 36s&=12\pi\\ s&=\boxed{3}

Answer: A 3