Discrete Mathematics—Introduction
... There are infinitely many prime numbers. Lemma: Given an integer a and p a prime number, if p divides a then p does not divide a+1. Proof: (by contradiction) Suppose p divides a and p divides a+1. then a=np and a+1=mp. 1=a+1-a=mp-np=(m-n)p. Therefore p is a factor of 1 so p is 1 or –1. Therefore p i ...
... There are infinitely many prime numbers. Lemma: Given an integer a and p a prime number, if p divides a then p does not divide a+1. Proof: (by contradiction) Suppose p divides a and p divides a+1. then a=np and a+1=mp. 1=a+1-a=mp-np=(m-n)p. Therefore p is a factor of 1 so p is 1 or –1. Therefore p i ...
Exam
... 15. Isosceles ΔABC has base AB = 4 and altitude CP = 6. Choose point D with AD ⊥ AB, AD = AB, and BD intersecting AC. Choose point E so that ΔADE ≅ ΔABC and AE intersects BC . Find the area common to the two triangles. ...
... 15. Isosceles ΔABC has base AB = 4 and altitude CP = 6. Choose point D with AD ⊥ AB, AD = AB, and BD intersecting AC. Choose point E so that ΔADE ≅ ΔABC and AE intersects BC . Find the area common to the two triangles. ...
ON ABUNDANT-LIKE NUMBERS
... Problem 188, [3], stated: Apart from finitely many primesp show that if n, is the smallest abundant number for whichp is the smallest prime divisor of n,, then n, is not squarefree. Let 2=pl
... Problem 188, [3], stated: Apart from finitely many primesp show that if n, is the smallest abundant number for whichp is the smallest prime divisor of n,, then n, is not squarefree. Let 2=pl
MTH6128 Number Theory 9 Sums of squares
... of a and b is a square. For any prime divisor of ab occurs to an even power, and must occur in one of a and b and not the other; so each of a and b is a product of even powers of primes, and so is a square. More generally, if the product of any number of pairwise coprime factors is a square, then ea ...
... of a and b is a square. For any prime divisor of ab occurs to an even power, and must occur in one of a and b and not the other; so each of a and b is a product of even powers of primes, and so is a square. More generally, if the product of any number of pairwise coprime factors is a square, then ea ...
Team Round Solutions
... most 29 + 98 + 98 = 225 to GOT . This means T is either 3 or 4 or 5. But then, T O is at most 59, and so in fact, we add at most 59 + 59 + 29 = 147 to GOT . So in fact, T is either 3 or 4. So GOT is something of the form 2 ∗ 3 or 2 ∗ 4. Some calculator bashing now yields (G, O, T, P ) = (2, 6, 3, 1) ...
... most 29 + 98 + 98 = 225 to GOT . This means T is either 3 or 4 or 5. But then, T O is at most 59, and so in fact, we add at most 59 + 59 + 29 = 147 to GOT . So in fact, T is either 3 or 4. So GOT is something of the form 2 ∗ 3 or 2 ∗ 4. Some calculator bashing now yields (G, O, T, P ) = (2, 6, 3, 1) ...
The Yellowstone permutation
... Hypothesis A is only a conjecture, since we cannot rule out the possibility that this behavior breaks down at some much later point in the sequence. It is theoretically possible, for example, that a term that is twice a prime is not followed two steps later by the prime itself (as happens after a(8) ...
... Hypothesis A is only a conjecture, since we cannot rule out the possibility that this behavior breaks down at some much later point in the sequence. It is theoretically possible, for example, that a term that is twice a prime is not followed two steps later by the prime itself (as happens after a(8) ...
SOLUTION 7 1. Solution Problem 1 From the program on the web
... Each d has a one-one correspondence to d0 ≤ n where dd0 = n. ...
... Each d has a one-one correspondence to d0 ≤ n where dd0 = n. ...
1 A little probability of error goes a long way
... algorithm is not used much for string matching, it is widely used when one wants to check for more than one string, i.e. when we have substrings Y1 , Y2 , . . . , Yk and we would like to know if any Yi occurs as a substring of X. Karp-Rabin algorithm. Choose a prime p ∈ {2, 3, . . . , T } uniformly ...
... algorithm is not used much for string matching, it is widely used when one wants to check for more than one string, i.e. when we have substrings Y1 , Y2 , . . . , Yk and we would like to know if any Yi occurs as a substring of X. Karp-Rabin algorithm. Choose a prime p ∈ {2, 3, . . . , T } uniformly ...
