MathDB
2021 BmMT Individual Round - Berkley mini Math Tournament

Source:

November 6, 2023
bmmtalgebrageometrycombinatoricsnumber theory

Problem Statement

p1. What is the largest number of five dollar footlongs Jimmy can buy with 88 dollars?
p2. Austin, Derwin, and Sylvia are deciding on roles for BMT 20212021. There must be a single Tournament Director and a single Head Problem Writer, but one person cannot take on both roles. In how many ways can the roles be assigned to Austin, Derwin, and Sylvia?
p3. Sofia has7 7 unique shirts. How many ways can she place 22 shirts into a suitcase, where the order in which Sofia places the shirts into the suitcase does not matter?
p4. Compute the sum of the prime factors of 20212021.
p5. A sphere has volume 36π36\pi cubic feet. If its radius increases by 100%100\%, then its volume increases by aπa\pi cubic feet. Compute aa.
p6. The full price of a movie ticket is $10\$10, but a matinee ticket to the same movie costs only 70%70\% of the full price. If 30%30\% of the tickets sold for the movie are matinee tickets, and the total revenue from movie tickets is $1001\$1001, compute the total number of tickets sold.
p7. Anisa rolls a fair six-sided die twice. The probability that the value Anisa rolls the second time is greater than or equal to the value Anisa rolls the first time can be expressed as mn\frac{m}{n} , where mm and nn are relatively prime positive integers. Compute m+nm + n.
p8. Square ABCDABCD has side length AB=6AB = 6. Let point EE be the midpoint of BC\overline{BC}. Line segments AC\overline{AC} and DE\overline{DE} intersect at point FF. Compute the area of quadrilateral ABEF.
p9. Justine has a large bag of candy. She splits the candy equally between herself and her 44 friends, but she needs to discard three candies before dividing so that everyone gets an equal number of candies. Justine then splits her share of the candy between herself and her two siblings, but she needs to discard one candy before dividing so that she and her siblings get an equal number of candies. If Justine had instead split all of the candy that was originally in the large bag between herself and 1414 of her classmates, what is the fewest number of candies that she would need to discard before dividing so that Justine and her 1414 classmates get an equal number of candies?
p10. For some positive integers aa and bb, a2b2=400a^2 - b^2 = 400. If aa is even, compute aa.
p11. Let ABCDEFGHIJKLABCDEFGHIJKL be the equilateral dodecagon shown below, and each angle is either 90o90^o or 270o270^o. Let MM be the midpoint of CD\overline{CD}, and suppose HM\overline{HM} splits the dodecagon into two regions. The ratio of the area of the larger region to the area of the smaller region can be expressed as mn\frac{m}{n} , where mm and nn are relatively prime positive integers. Compute m+nm + n. https://cdn.artofproblemsolving.com/attachments/3/e/387bcdf2a6c39fcada4f21f24ceebd18a7f887.png
p12. Nelson, who never studies for tests, takes several tests in his math class. Each test has a passing score of 60/10060/100. Since Nelson's test average is at least 60/10060/100, he manages to pass the class. If only nonnegative integer scores are attainable on each test, and Nelson gets a di erent score on every test, compute the largest possible ratio of tests failed to tests passed. Assume that for each test, Nelson either passes it or fails it, and the maximum possible score for each test is 100.
p13. For each positive integer nn, let f(n)=nn+1+n+1nf(n) = \frac{n}{n+1} + \frac{n+1}{n} . Then f(1)+f(2)+f(3)+...+f(10)f(1)+f(2)+f(3)+...+f(10) can be expressed as mn\frac{m}{n} where mm and nn are relatively prime positive integers. Compute m+nm + n.
p14. Triangle ABC\vartriangle ABC has point DD lying on line segment BC\overline{BC} between BB and CC such that triangle ABD\vartriangle ABD is equilateral. If the area of triangle ADC\vartriangle ADC is 14\frac14 the area of triangle ABC\vartriangle ABC, then (ACAB)2\left( \frac{AC}{AB}\right)^2 can be expressed as mn\frac{m}{n} , where mm and nn are relatively prime positive integers. Compute m+nm + n.
p15. In hexagon ABCDEFABCDEF, AB=60AB = 60, AF=40AF = 40, EF=20EF = 20, DE=20DE = 20, and each pair of adjacent edges are perpendicular to each other, as shown in the below diagram. The probability that a random point inside hexagon ABCDEFABCDEF is at least 20220\sqrt2 units away from point DD can be expressed in the form abπc\frac{a-b\pi}{c} , where aa, bb, cc are positive integers such that gcd(a,b,c)=1(a, b, c) = 1. Compute a+b+ca + b + c. https://cdn.artofproblemsolving.com/attachments/3/c/1b45470265d10a73de7b83eff1d3e3087d6456.png
p16. The equation x+20x=20+20xx2\sqrt{x} +\sqrt{20-x} =\sqrt{20 + 20x - x^2} has 44 distinct real solutions, x1x_1, x2x_2, x3x_3, and x4x_4. Compute x1+x2+x3+x4x_1 + x_2 + x_3 + x_4.
p17. How many distinct words with letters chosen from B,M,TB, M, T have exactly 1212 distinct permutations, given that the words can be of any length, and not all the letters need to be used? For example, the word BMMTBMMT has 1212 permutations. Two words are still distinct even if one is a permutation of the other. For example, BMMTBMMT is distinct from TMMBTMMB.
p18. We call a positive integer binary-okay if at least half of the digits in its binary (base 22) representation are 11's, but no two 11s are consecutive. For example, 1010=1010210_{10} = 1010_2 and 510=10125_{10} = 101_2 are both binary-okay, but 1610=10000216_{10} = 10000_2 and 1110=1011211_{10} = 1011_2 are not. Compute the number of binary-okay positive integers less than or equal to 20202020 (in base 1010).
p19. A regular octahedron (a polyhedron with 88 equilateral triangles) has side length 22. An ant starts on the center of one face, and walks on the surface of the octahedron to the center of the opposite face in as short a path as possible. The square of the distance the ant travels can be expressed as mn\frac{m}{n} , where mm and nn are relatively prime positive integers. Compute m+nm + n. https://cdn.artofproblemsolving.com/attachments/f/8/3aa6abe02e813095e6991f63fbcf22f2e0431a.png
p20. The sum of 1a\frac{1}{a} over all positive factors aa of the number 360360 can be expressed in the form mn\frac{m}{n} ,where mm and nn are relatively prime positive integers. Compute m+nm + n.
PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.