Math 445 Homework 5 Solutions Due Wednesday, October 6 21. If
... b d then n can be expressed as the sum of the squares of two integers. From (*), clearing denomenators, we have that nb2 d2 = a2 d2 + c2 b2 = (ad)2 + (bc)2 is a sum of two squares. So for every prime p with p ≡ 3 (mod 4) , pk ||nb2 d2 = n(bd)2 with k even. But since (bd)2 is a perfect square, pm ||( ...
... b d then n can be expressed as the sum of the squares of two integers. From (*), clearing denomenators, we have that nb2 d2 = a2 d2 + c2 b2 = (ad)2 + (bc)2 is a sum of two squares. So for every prime p with p ≡ 3 (mod 4) , pk ||nb2 d2 = n(bd)2 with k even. But since (bd)2 is a perfect square, pm ||( ...
MTH2125: Discrete Mathematics Tutorial Questions 3 Dr. John O
... A parking lot has 31 visitor spaces, numbered from 0 to 30. Visitors are assigned parking spaces using the hashing function: h(k) = k mod m, where k is the number formed from the first three digits on a visitor’s license plates, and m is the number of available parking spaces. Which spaces are assig ...
... A parking lot has 31 visitor spaces, numbered from 0 to 30. Visitors are assigned parking spaces using the hashing function: h(k) = k mod m, where k is the number formed from the first three digits on a visitor’s license plates, and m is the number of available parking spaces. Which spaces are assig ...
Solutions
... If a = 4, then since a2 < 2(b – a), we have 2b > a2 + 2a = 24, so b > 12 and then only possibility is that b = 13, c = 14 and d = 15. This contradicts equations (1) and (2) so there is no solution with a = 4. If a = 2, then equation (3) gives b + c – d = 4. Substituting a = 2 and d = b + c – 4 back ...
... If a = 4, then since a2 < 2(b – a), we have 2b > a2 + 2a = 24, so b > 12 and then only possibility is that b = 13, c = 14 and d = 15. This contradicts equations (1) and (2) so there is no solution with a = 4. If a = 2, then equation (3) gives b + c – d = 4. Substituting a = 2 and d = b + c – 4 back ...
