2016 Chapter Competition Solutions
... Initially, exactly 5 of the lights are on. Three people enter the room sequentially and independently flip one switch randomly. Find: the probability that after the third person has exited the room, exactly six of the lights are on. In order to end up with exactly 6 lights on, the three peop ...
... Initially, exactly 5 of the lights are on. Three people enter the room sequentially and independently flip one switch randomly. Find: the probability that after the third person has exited the room, exactly six of the lights are on. In order to end up with exactly 6 lights on, the three peop ...
Discrete Mathematics
... “How many ways can you guys be matched with each other?” she wonders. “This is clearly the same problem as seating you on six chairs; it does not matter whether the chairs are around the dinner table of at the three boards. So the answer is 720 as before.” “I think you should not count it as a diff ...
... “How many ways can you guys be matched with each other?” she wonders. “This is clearly the same problem as seating you on six chairs; it does not matter whether the chairs are around the dinner table of at the three boards. So the answer is 720 as before.” “I think you should not count it as a diff ...
Number Theory for Mathematical Contests
... state, are very hard to solve. Some number-theoretic problems that are yet unsolved are: 1. (Goldbach’s Conjecture) Is every even integer greater than 2 the sum of distinct primes? 2. (Twin Prime Problem) Are there infinitely many primes p such that p + 2 is also a prime? 3. Are there infinitely man ...
... state, are very hard to solve. Some number-theoretic problems that are yet unsolved are: 1. (Goldbach’s Conjecture) Is every even integer greater than 2 the sum of distinct primes? 2. (Twin Prime Problem) Are there infinitely many primes p such that p + 2 is also a prime? 3. Are there infinitely man ...
Algebraic Number Theory - School of Mathematics, TIFR
... some (m > 0). Proof: Let G be generated by a. Consider the map f : Z → G taking n ∈ Z to an . This is a surjective homomorphism and ker f is a subgroup mZ of Z generated by m ≥ 0 in Z. If m = 0, G is isomorphic to Z. If m ≥ 0, G is isomorphic to Z/(m). Definition 1.7 Let G be a group and a ∈ G. We s ...
... some (m > 0). Proof: Let G be generated by a. Consider the map f : Z → G taking n ∈ Z to an . This is a surjective homomorphism and ker f is a subgroup mZ of Z generated by m ≥ 0 in Z. If m = 0, G is isomorphic to Z. If m ≥ 0, G is isomorphic to Z/(m). Definition 1.7 Let G be a group and a ∈ G. We s ...
1 - UCSD Mathematics
... We assume that you are familiar with the integers. Let us list the algebra axioms for Z: By an axiom, we mean that it is a basic unproved assumption. We have a long list of axioms (9). Hopefully you won’t go to sleep before the end since the last one (well ordering) is particularly important. Algebr ...
... We assume that you are familiar with the integers. Let us list the algebra axioms for Z: By an axiom, we mean that it is a basic unproved assumption. We have a long list of axioms (9). Hopefully you won’t go to sleep before the end since the last one (well ordering) is particularly important. Algebr ...
PDF only - at www.arxiv.org.
... both have maximum amplitude at time zero. Each time either waveform reaches maximum amplitude, a yellow dot is added. All the yellow dots are at multiples of time 0.1 = (1/10). In general, with coprime frequencies a and b, the peak amplitudes would be reached at times a multiple of (a·b)-1 = (BH)-1, ...
... both have maximum amplitude at time zero. Each time either waveform reaches maximum amplitude, a yellow dot is added. All the yellow dots are at multiples of time 0.1 = (1/10). In general, with coprime frequencies a and b, the peak amplitudes would be reached at times a multiple of (a·b)-1 = (BH)-1, ...
