Supplementary Notes
... has a solution. Any two solutions x1 , x2 are congruent mod n1 n2 . . . nk . Proof. Let mi denote the product of all elements of the set {n1 , n2 , . . . , nk } other than ni . Note that gcd(mi , ni ) = 1 so Lemma P 7 implies that there is a number ri such that mi ri ≡ 1 (mod ni ). Now let x = ki=1 ...
... has a solution. Any two solutions x1 , x2 are congruent mod n1 n2 . . . nk . Proof. Let mi denote the product of all elements of the set {n1 , n2 , . . . , nk } other than ni . Note that gcd(mi , ni ) = 1 so Lemma P 7 implies that there is a number ri such that mi ri ≡ 1 (mod ni ). Now let x = ki=1 ...
Subtracting negative numbers 28OCT
... Starter Questions 1. Calculate 85% of 160 2. Calculate a) -2 + 6 ...
... Starter Questions 1. Calculate 85% of 160 2. Calculate a) -2 + 6 ...
Textbook Section 4.2
... What are the numbers you multiply to get 24? Can you arrange the beans into a different rectangle? What product does this represent? ...
... What are the numbers you multiply to get 24? Can you arrange the beans into a different rectangle? What product does this represent? ...
Solutions - Mu Alpha Theta
... Now, each divisor of n is composed of the same prime factors, where the ith exponent can range from 0 to ai. Hence there are a1 + 1 choices for the first exponent, a2 + 1 choices for the second, and so on. Therefore the number of positive divisors of n is (a1 + 1)(a2 + 1) ... (ar + 1). The unique pr ...
... Now, each divisor of n is composed of the same prime factors, where the ith exponent can range from 0 to ai. Hence there are a1 + 1 choices for the first exponent, a2 + 1 choices for the second, and so on. Therefore the number of positive divisors of n is (a1 + 1)(a2 + 1) ... (ar + 1). The unique pr ...
Lecture 6 (powerpoint): finding a gigantic prime number
... Key Lemma doesn’t apply if n is a prime-power. However, it doesn’t matter since it cannot pass the test of step (3), i.e., we are sure that a(n-1)/2 <> 1,-1 mod n for all a. Proof (assume all operations are mod n): ...
... Key Lemma doesn’t apply if n is a prime-power. However, it doesn’t matter since it cannot pass the test of step (3), i.e., we are sure that a(n-1)/2 <> 1,-1 mod n for all a. Proof (assume all operations are mod n): ...
H6
... 2. In this problem we want to try and understand the “sums of two squares” problem for polynomials over R. The question is: When can a polynomial n(x) ∈ R[x] be written as the sum of two squares, n(x) = (f (x))2 + (g(x))2 , with f (x), g(x) ∈ R[x]? We will use one fact not proved in this class: Eve ...
... 2. In this problem we want to try and understand the “sums of two squares” problem for polynomials over R. The question is: When can a polynomial n(x) ∈ R[x] be written as the sum of two squares, n(x) = (f (x))2 + (g(x))2 , with f (x), g(x) ∈ R[x]? We will use one fact not proved in this class: Eve ...
2008 - Maths
... Part of the graph y = f(x) is shown below, where the dotted lines indicate asymptotes. Sketch the graph y = –f(x + 1) showing its asymptotes. Write down the equations of the asymptotes. ...
... Part of the graph y = f(x) is shown below, where the dotted lines indicate asymptotes. Sketch the graph y = –f(x + 1) showing its asymptotes. Write down the equations of the asymptotes. ...