Prime numbers - Scholastic Shop
... player to decide whether it is a prime number or not, and record how long it took to answer the question. If challenged, a player must prove why their answer is yes or no, with a sensible explanation. ...
... player to decide whether it is a prime number or not, and record how long it took to answer the question. If challenged, a player must prove why their answer is yes or no, with a sensible explanation. ...
Diophantine approximation with primes and powers of two
... primes and a bounded number of powers of two. Under certain conditions, we are able to demonstrate that these values can be made arbitrarily close to any real number by taking sufficiently many powers of two. ...
... primes and a bounded number of powers of two. Under certain conditions, we are able to demonstrate that these values can be made arbitrarily close to any real number by taking sufficiently many powers of two. ...
1 Introduction: Historical Background
... Proof. First remember that each member of C is square-free. If a prime p belongs to C , then by Theorem 2.1 we have p , 1 j p, which implies p = 2. Hence, 2 is the only prime belonging to C . If p < q are two primes and pq 2 C , then we get by Theorem 2.3 that p , 1 j q and q , 1 j p. By p , 1 j q w ...
... Proof. First remember that each member of C is square-free. If a prime p belongs to C , then by Theorem 2.1 we have p , 1 j p, which implies p = 2. Hence, 2 is the only prime belonging to C . If p < q are two primes and pq 2 C , then we get by Theorem 2.3 that p , 1 j q and q , 1 j p. By p , 1 j q w ...
Baltic Way 2016 5 November 2016, Oulu, Finland Working time: 41
... Working time: 4 12 hours. Questions may be asked during the first 30 minutes. Tools for writing and drawing are the only ones allowed. 1. Find all pairs of primes (p, q) such that p3 − q 5 = (p + q)2 . 2. Prove or disprove the following hypotheses. a) For all k ≥ 2, each sequence of k consecutive po ...
... Working time: 4 12 hours. Questions may be asked during the first 30 minutes. Tools for writing and drawing are the only ones allowed. 1. Find all pairs of primes (p, q) such that p3 − q 5 = (p + q)2 . 2. Prove or disprove the following hypotheses. a) For all k ≥ 2, each sequence of k consecutive po ...
Problem of the Month
... unknowns. What to do? Let’s try solving for b, which gives b = r 2 + 2r − 120. At this point, it might occur to try to factor the right side to obtain b = (r + 12)(r − 10). How does this help? Since b is a prime number, then it can’t be factored in many ways! Aha – that is probably useful. If b is a ...
... unknowns. What to do? Let’s try solving for b, which gives b = r 2 + 2r − 120. At this point, it might occur to try to factor the right side to obtain b = (r + 12)(r − 10). How does this help? Since b is a prime number, then it can’t be factored in many ways! Aha – that is probably useful. If b is a ...
Peculiar Primes
... Remember: a positive number is prime if it is only divisible by two numbers—itself and 1. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, . . .. Every integer greater than 1 can be written in a unique way as a product of prime numbers. For instance, 100 = 2 · 2 · 5 · 5, 23 = 23 (since it’s a ...
... Remember: a positive number is prime if it is only divisible by two numbers—itself and 1. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, . . .. Every integer greater than 1 can be written in a unique way as a product of prime numbers. For instance, 100 = 2 · 2 · 5 · 5, 23 = 23 (since it’s a ...
ENGG 2440A: Discrete Mathematics for Engineers Homework 2 The
... • Case 1: a is friends with at least 4 other people in the collection. • Case 2: a is a stranger to at least 4 other people in the collection. One of these two cases must hold. Let’s discuss Case 1. If all the people who are friends with a are strangers among themselves, this is a group of 4 strange ...
... • Case 1: a is friends with at least 4 other people in the collection. • Case 2: a is a stranger to at least 4 other people in the collection. One of these two cases must hold. Let’s discuss Case 1. If all the people who are friends with a are strangers among themselves, this is a group of 4 strange ...
Suggested Solutions of Sharif Internet Contest 09
... want to calculate a!^v mod m. Firstly, calculate u=a! mod m. So, we want to calculate u^v mod m. Notice, that value of ui mod m will be in range [0, m - 1] for any nonnegative integer i. Imagine we are iterating i starting from 0. We'll receive following values: u0 mod m, u1 mod m, and so on. Accord ...
... want to calculate a!^v mod m. Firstly, calculate u=a! mod m. So, we want to calculate u^v mod m. Notice, that value of ui mod m will be in range [0, m - 1] for any nonnegative integer i. Imagine we are iterating i starting from 0. We'll receive following values: u0 mod m, u1 mod m, and so on. Accord ...