Solutions - Mu Alpha Theta
... 26. Let N M 2 . Since the last two digits of N are 25, the units digit of M must be 5. In particular, if M is 5, then x = 0. Now suppose that M has at least two digits. If the tens digit of M is 1, 3, 6, or 8, then x = 2. If the tens digit of M is 2 or 7, then x = 6. If the tens digit is 4, 5, or ...
... 26. Let N M 2 . Since the last two digits of N are 25, the units digit of M must be 5. In particular, if M is 5, then x = 0. Now suppose that M has at least two digits. If the tens digit of M is 1, 3, 6, or 8, then x = 2. If the tens digit of M is 2 or 7, then x = 6. If the tens digit is 4, 5, or ...
solutions - UCI Math
... and S(N = S(N ≥ 45 . Thus 5S(N0 ) < 4N0 and 5S(N0 ) + 5 ≥ 4N0 + 4 so that N0 +1 N0 +1 5S(N0 ) < 4N0 ≤ 5S(N0 ) + 1. Since this is an inequality among integers, it must be that, in fact, 4N0 = 5S(N0 ) + 1. Thus, S(N0 + 1) = S(N0 ) + 1 = 4N50 −1 + 1 = 45 (N0 + 1), and so S(N0 + 1) is exactly 80% of N0 ...
... and S(N = S(N ≥ 45 . Thus 5S(N0 ) < 4N0 and 5S(N0 ) + 5 ≥ 4N0 + 4 so that N0 +1 N0 +1 5S(N0 ) < 4N0 ≤ 5S(N0 ) + 1. Since this is an inequality among integers, it must be that, in fact, 4N0 = 5S(N0 ) + 1. Thus, S(N0 + 1) = S(N0 ) + 1 = 4N50 −1 + 1 = 45 (N0 + 1), and so S(N0 + 1) is exactly 80% of N0 ...
SectionModularArithm..
... Modular Arithmetic To begin, we first review what it means to divide two numbers.. Definition 1.1: We say that a divides b, denoted as a | b , if b ka for some integer a. For instance, we know that 7 | 21 since 21 3 7 . However, we know that 5 | 21 since there is no integer multiple of 5 that ...
... Modular Arithmetic To begin, we first review what it means to divide two numbers.. Definition 1.1: We say that a divides b, denoted as a | b , if b ka for some integer a. For instance, we know that 7 | 21 since 21 3 7 . However, we know that 5 | 21 since there is no integer multiple of 5 that ...
Some explorations about repeated roots
... This led me to consider why the iterates of 2 and 6 produced rational values, and I realized rather quickly (for me) that the sequence of values 2, 6, 12, 20, 30…. etc. would produce exact integers since those were the numbers for which 4a+1 was a perfect square, that is, values in the sequence n(n+ ...
... This led me to consider why the iterates of 2 and 6 produced rational values, and I realized rather quickly (for me) that the sequence of values 2, 6, 12, 20, 30…. etc. would produce exact integers since those were the numbers for which 4a+1 was a perfect square, that is, values in the sequence n(n+ ...
Complex numbers
... If you have not seen much about complex numbers, here are some elementary properties. (If you have any questions, feel free to ask!) The complex numbers C are the set of all numbers of the form a + bi where a, b ∈ R and i2 = −1. We may define addition of complex numbers by (a + bi) + (c + di) = (a + ...
... If you have not seen much about complex numbers, here are some elementary properties. (If you have any questions, feel free to ask!) The complex numbers C are the set of all numbers of the form a + bi where a, b ∈ R and i2 = −1. We may define addition of complex numbers by (a + bi) + (c + di) = (a + ...
SummerLecture15.pdf
... In the example of the airplane, the function f gives the altitude of the plane at time t. Since the plane has an altitude at all times between 9:00 AM and 9:15 AM, the function is continuous on the interval [9:00 AM, 9:15 AM]. Using the designations from the formal statement of the theorem, a = 9 : ...
... In the example of the airplane, the function f gives the altitude of the plane at time t. Since the plane has an altitude at all times between 9:00 AM and 9:15 AM, the function is continuous on the interval [9:00 AM, 9:15 AM]. Using the designations from the formal statement of the theorem, a = 9 : ...
