Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23$

Solution: The favorable responses are either they both show a green bean or they both show a red bean. The probability that both show a green bean is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$ . The probability that both show a red bean is $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$ . Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\boxed{38}$

Answer: C 38
Problem: Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23$

Solution: We can list out all the combinations and we get this: $GG, GY, GR_1, GR_2, RG, RY, RR_1, RR_2$ . There is a total of 8 combinations and 3 that are the same. Hence, we yield the answer $\boxed{38}$

Answer: C 38
Problem: A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$

Solution: We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:
$5:3 = 5(8):3(8) = 40:24$
$8:5 = 8(5):5(5) = 40:25$
Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40:25:24$ . Since the ratio is in lowest terms, the smallest number of students participating in the project is $40+25+24 = \boxed{89}$

Answer: E 89
Problem: A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$

Solution: The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are $40*\frac{3}{5}=24$ 6th graders and $40*\frac{5}{8}=25$ 7th graders. The numbers of students is $40+24+25=\boxed{89}$

Answer: E 89
Problem: The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

Solution: The arithmetic mean of these numbers is $\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5$ . Therefore the numbers are $333$ $334$ $335$ $336$ $337$ $338$ , so the answer is $\boxed{338}$

Answer: B 338
Problem: The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

Solution: Let the $4^{\text{th}}$ number be $x$ . Then our desired number is $x+2$
Our integers are $x-3,x-2,x-1,x,x+1,x+2$ , so we have that $6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{338}$

Answer: B 338
Problem: The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

Solution: Let the first term be $x$ . Our integers are $x,x+1,x+2,x+3,x+4,x+5$ . We have, $6x+15=2013\implies x=333\implies x+5=\boxed{338}$

Answer: B 338
Problem: The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

Solution: Since there are $6$ numbers, we divide $2013$ by $6$ to find the mean of the numbers. $\frac{2013}{6} = 335 \frac{1}{2}$ .
Then, $335 \frac{1}{2} + \frac{1}{2} = 336$ (the fourth number). Fifth: $337$ ; Sixth: $\boxed{338}$

Answer:  338
Problem: The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

Solution: Let the $6th$ number be $x$ . Then our list is: $x-6+x-5+x-4+x-3+x-x-1=2013$ . Simplifying this gets you $6x-21=2013\implies 6x=2034$ , which means that $x = \boxed{338}$

Answer: B 338
Problem: Isabella uses one-foot cubical blocks to build a rectangular fort that is $12$ feet long, $10$ feet wide, and $5$ feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle);  label("10",(A--B),SE); label("12",(C--B),SW); label("5",(F--B),W);[/asy]
$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$

Solution: There are $10 \cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11 + 9 + 11 = 40$ cubes. Hence, the answer is $120 + 4 \cdot 40 = \boxed{280}$

Answer: B 280
Problem: Isabella uses one-foot cubical blocks to build a rectangular fort that is $12$ feet long, $10$ feet wide, and $5$ feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle);  label("10",(A--B),SE); label("12",(C--B),SW); label("5",(F--B),W);[/asy]
$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$

Solution: We can just calculate the volume of the prism that was cut out of the original $12\times 10\times 5$ box. Each interior side of the fort will be $2$ feet shorter than each side of the outside. Since the floor is $1$ foot, the height will be $4$ feet. So the volume of the interior box is $10\times 8\times 4=320\text{ ft}^3$
The volume of the original box is $12\times 10\times 5=600\text{ ft}^3$ . Therefore, the number of blocks contained in the fort is $600-320=\boxed{280}$

Answer: B 280
Problem: Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighbourhood picnic?
$\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf{(E)}\hspace{.05in}9$

Solution: Since Rachelle uses $3$ pounds of meat to make $8$ hamburgers, she uses $\frac{3}{8}$ pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or $\frac{3}{8} \cdot 24 = \boxed{9}$

Answer: E 9
Problem: In the country of East Westmore, statisticians estimate there is a baby born every $8$ hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?
$\textbf{(A)}\hspace{.05in}600\qquad\textbf{(B)}\hspace{.05in}700\qquad\textbf{(C)}\hspace{.05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000$

Solution: There are $24\text{ hours}\div8\text{ hours} = 3$ births and one death everyday in East Westmore. Therefore, the population increases by $3$ $1$ $2$ people everyday. Thus, there are $2 \times 365 = 730$ people added to the population every year. Rounding, we find the answer is $\boxed{700}$

Answer: B 700
Problem: In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , $X$ in centimeters?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=(4,7); M=(4,6); N=(0,6); O=(0,5); P=(2,5); Q=(2,3); R=(4,3); draw(A--B--C--D--E--F--G--H--I--J--K--L--M--N--O--P--Q--R--cycle); label("$X$",(3.4,1.5)); label("6",(7.6,1.5)); label("1",(7.6,3.5)); label("1",(8.4,4.6)); label("2",(9.4,4.6)); label("2",(10.4,6)); label("3",(8.4,7.4)); label("1",(7.5,7.8)); label("2",(6,8.5)); label("1",(4.7,7.8)); label("1",(4.3,7.5)); label("1",(3.5,6.5)); label("4",(1.8,6.5)); label("1",(-0.5,5.5)); label("2",(0.8,4.5)); label("2",(1.5,3.8)); label("2",(2.8,2.6));[/asy]
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}2\qquad\textbf{(C)}\hspace{.05in}3\qquad\textbf{(D)}\hspace{.05in}4\qquad\textbf{(E)}\hspace{.05in}5$

Solution: 2012amc85.png
$1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\ 5 + X = 10\\ X = 5$
Thus, the answer is $\boxed{5}$

Answer: E 5
Problem: In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , $X$ in centimeters?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=(4,7); M=(4,6); N=(0,6); O=(0,5); P=(2,5); Q=(2,3); R=(4,3); draw(A--B--C--D--E--F--G--H--I--J--K--L--M--N--O--P--Q--R--cycle); label("$X$",(3.4,1.5)); label("6",(7.6,1.5)); label("1",(7.6,3.5)); label("1",(8.4,4.6)); label("2",(9.4,4.6)); label("2",(10.4,6)); label("3",(8.4,7.4)); label("1",(7.5,7.8)); label("2",(6,8.5)); label("1",(4.7,7.8)); label("1",(4.3,7.5)); label("1",(3.5,6.5)); label("4",(1.8,6.5)); label("1",(-0.5,5.5)); label("2",(0.8,4.5)); label("2",(1.5,3.8)); label("2",(2.8,2.6));[/asy]
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}2\qquad\textbf{(C)}\hspace{.05in}3\qquad\textbf{(D)}\hspace{.05in}4\qquad\textbf{(E)}\hspace{.05in}5$

Solution: Note that we only need to consider the value below the marked red line, so we have the equation: \[X + 2 = 6 + 1\] \[X = 5\]
Hence, the answer is $\boxed{5}$

Answer: E 5
Problem: A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures $8$ inches high and $10$ inches wide. What is the area of the border, in square inches?
$\textbf{(A)}\hspace{.05in}36\qquad\textbf{(B)}\hspace{.05in}40\qquad\textbf{(C)}\hspace{.05in}64\qquad\textbf{(D)}\hspace{.05in}72\qquad\textbf{(E)}\hspace{.05in}88$

Solution: In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is $8 \times 10 = 80$ square inches. The height of the whole frame (including the photograph) would be $8+2+2 = 12$ , and the width of the whole frame, $10+2+2 = 14$ . Therefore, the area of the whole figure would be $12 \times 14 = 168$ square inches. Subtracting the area of the photograph from the area of both the frame and photograph, we find the answer to be $168-80 = \boxed{88}$

Answer: E 88
Problem: Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?
$\textbf{(A)}\hspace{.05in}90\qquad\textbf{(B)}\hspace{.05in}92\qquad\textbf{(C)}\hspace{.05in}95\qquad\textbf{(D)}\hspace{.05in}96\qquad\textbf{(E)}\hspace{.05in}97$

Solution: Isabella wants an average grade of $95$ on her 4 tests; this also means that she wants the sum of her test scores to be at least $95 \times 4 = 380$ (if she goes over this number, she'll be over her goal!). She's already taken two tests, which sum to $97+91 = 188$ , which means she needs $192$ more points to achieve her desired average. In order to minimize the score on the third test, we assume that Isabella will receive all $100$ points on the fourth test. Therefore, the lowest Isabella could have scored on the third test would be $192-100 = \boxed{92}$

Answer: B 92
Problem: Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?
$\textbf{(A)}\hspace{.05in}90\qquad\textbf{(B)}\hspace{.05in}92\qquad\textbf{(C)}\hspace{.05in}95\qquad\textbf{(D)}\hspace{.05in}96\qquad\textbf{(E)}\hspace{.05in}97$

Solution: Isabella needs to lose a max of $20$ . When she got $97$ and $91$ , the max goes down to $20 - 3 - 9 = 8$ . If we get $100$ on her last test, then she will get $\boxed{92}$ on the third test.

Answer: B 92
Problem: A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$

Solution: Let the original price of an item be $x$
First, everything is half-off, so the price is now $\frac{x}{2} = 0.5x$
Next, the extra coupon applies 20% off on the sale price , so the price after this discount will be $100\% - 20\% = 80\%$ of what it was before. (Notice how this is not applied to the original price; if it were, the solution would be applying 50% + 20 % = 70% off the original price.)
$80\% \cdot 0.5 x = \frac{4}{5} \cdot 0.5 x = 0.4x$
The price of the item after all discounts have been applied is $0.4x = 40\% \cdot x$ . However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract $40\% x$ from $100\% x$ (the original price of the item), to find the answer, $\boxed{60}$

Answer: D 60
Problem: A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$

Solution: Since the problem implies that the percentage off the original price will be the same for every item in the store, fakesolving is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a total of $\frac{10-4}{10} = \frac{6}{10} = \frac{60}{100} = \boxed{60} \%$

Answer: D 60
Problem: A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$

Solution: The price is $50\%$ off, which means the price would be $100\% - 50\% = 50\%$ of what it was before. Then, you get $20\%$ off. Sidenote: A common mistake is finding $20\%$ of the price, but forgetting to subtract it. $20\%$ off the price is $100\% - 20\% = 80\%.$ Now, we can multiply the two discounts as decimals. $0.5 \times 0.8 = 0.4 = 40\%.$ Our final answer is $100\% - 40\% = \boxed{60}$

Answer: D 60
Problem: The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
$\textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161$

Solution: Let the number of two-legged birds be $x$ and the number of four-legged mammals be $y$ . We can now use systems of equations to solve this problem.
Write two equations:
$2x + 4y = 522$
$x + y = 200$
Now multiply the latter equation by $2$
$2x + 4y = 522$
$2x + 2y = 400$
By subtracting the second equation from the first equation, we find that $2y = 122 \implies y = 61$ . Since there were $200$ heads, meaning that there were $200$ animals, there were $200 - 61 =  \boxed{139}$ two-legged birds.

Answer: C 139
Problem: The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
$\textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161$

Solution: Let's add 2 legs for each bird and assume each bird has four-legged as each mammal. So the total legs of these birds and mammals would be $4*200=800$ . Actually, there were only $522$ legs. The difference between these two numbers exactly gives us the number of all the legs we added for all birds: $800 - 522 = 278$ . Because each bird was added by 2 legs, so the total number of birds would be $278/2 = \boxed{139}$ two-legged birds. ---LarryFlora

Answer: C 139
Problem: How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$

Solution: For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$ , since all of the valid 4-digit number will always be greater than $1000$ . The best way to solve this problem is by using casework.
There can be only two leading digits, namely $1$ or $2$
When the leading digit is $1$ , you can make $\frac{3!}{2!1!} \implies 3$ such numbers.
When the leading digit is $2$ , you can make $3! \implies 6$ such numbers.
Summing the amounts of numbers, we find that there are $\boxed{9}$ such numbers.

Answer: D 9
Problem: How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$

Solution: Notice that the first digit cannot be $0$ , as the number is greater than $1000$ . Therefore, there are three digits that can be in the thousands.
The rest three digits of the number have no restrictions, and therefore there are $3! \implies 6$ for each leading digit.
Since the two $2$ 's are indistinguishable, there are $\frac {3\cdot6}{2}$ such numbers $\implies \boxed{9}$

Answer: D 9
Problem: The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$

Solution: We can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$ , because of the above statement. Now we need to test the remaining answer choices.
Case 1: $x = 6$
Mode: $6$
Median: $6$
Mean: $\frac{37}{7}$
Since the mean does not equal the median or mode, ${\textbf{(B)}\ 6}$ can also be eliminated.
Case 2: $x = 11$
Mode: $6$
Median: $6$
Mean: $6$
We are done with this problem, because we have found when $x = 11$ , the condition is satisfied. Therefore, the answer is $\boxed{11}$

Answer: D 11
Problem: The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$

Solution: Notice that the mean of this set of numbers, in terms of $x$ , is:
$\frac{3+4+5+6+6+7+x}{7} = \frac{31+x}{7}$
Because we know that the mode must be $6$ (it can't be any of the numbers already listed, as shown above, and no matter what $x$ is, either $6$ or a new number,  it will not affect $6$ being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and $6$ equal:
\[\dfrac{31+x}{7}=6\] \[31+x=42\] \[x=\boxed{11}\]

Answer: D 11