Math Circle Beginners Group March 6, 2016 Euclid and Prime
... For example, 5 × 4 + 1 = 21 and 5 × 2 − 1 = 9, which are both composite. This is a converse to what we found in part a, and this shows that converses are not always true if the statements are true. ...
... For example, 5 × 4 + 1 = 21 and 5 × 2 − 1 = 9, which are both composite. This is a converse to what we found in part a, and this shows that converses are not always true if the statements are true. ...
Lecture slides (full content)
... “There are infinitely many even natural numbers.” Proof: assume there are a finite number of even numbers # assuming ¬Q then there exists a largest even number, m then let m’ = 2m then m’ is an even number, and larger than m # contradiction, since m is the largest even number then there are infinit ...
... “There are infinitely many even natural numbers.” Proof: assume there are a finite number of even numbers # assuming ¬Q then there exists a largest even number, m then let m’ = 2m then m’ is an even number, and larger than m # contradiction, since m is the largest even number then there are infinit ...
How to find prime numbers 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1
... • A prime number can be divided only by itself and 1 with nothing left over. • It has exactly two factors. A factor is a whole number b which hi h di divides id iinto a target number b with ih no remainder. For example, the factors of 12 are 1, 2, 3, 4, 6, 12 because I can divide 12 by all of them w ...
... • A prime number can be divided only by itself and 1 with nothing left over. • It has exactly two factors. A factor is a whole number b which hi h di divides id iinto a target number b with ih no remainder. For example, the factors of 12 are 1, 2, 3, 4, 6, 12 because I can divide 12 by all of them w ...
Elementary Number Theory Solutions
... 3) The prime factorization of 225 is 32 52 = 9 × 25 = 225. For a number n to be divisible by 9 the sum of n’s digits must be divisible by 9. Thus there must be 9 1’s making up the number. For a number to be divisible by 25 it must end with the last two digits 25, 50, 75, or 00. The only one that is ...
... 3) The prime factorization of 225 is 32 52 = 9 × 25 = 225. For a number n to be divisible by 9 the sum of n’s digits must be divisible by 9. Thus there must be 9 1’s making up the number. For a number to be divisible by 25 it must end with the last two digits 25, 50, 75, or 00. The only one that is ...
lecture notes 4
... |mπ − n| < . Outline. Fix a positive integer N such that N1 < . The fractional part of any positive real lies in precisely one of the sets [0, 1/N ), [1/N, 2/N ), . . . , [(N − 1)/N, 1). Some two of elements of {π, 2π, . . . , (N + 1)π}, have a fractional part in the same interval; their differenc ...
... |mπ − n| < . Outline. Fix a positive integer N such that N1 < . The fractional part of any positive real lies in precisely one of the sets [0, 1/N ), [1/N, 2/N ), . . . , [(N − 1)/N, 1). Some two of elements of {π, 2π, . . . , (N + 1)π}, have a fractional part in the same interval; their differenc ...
