# Final_practice_solutions.pdf - Math 240 u2013 Autumn 2020...

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Math 240 – Autumn 2020 – Addario–Berry–Paquette Final preparation solutions 1. Find gcd(5180 , 3404) and express gcd(5180 , 3404) as a linear combina- tion of 5180 and 3404 . Solution. We use Euler’s algorithm: b 5180 / 3404 c = 1 = 5180 - 1 · 3404 = 1776 , b 3404 / 1776 c = 1 = 3404 - 1 · 1776 = 1628 , b 1776 / 1628 c = 1 = 1776 - 1 · 1628 = 148 , b 1628 / 148 c = 11 = 1628 - 11 · 148 = 0 , So 148 is the greatest common divisor. To find the coefficiets in the linear combinations, we back-substitute to find: 148 = 1 · 1776 + - 1 · 1628 = 1776 - (3404 - 1776) , 148 = - 1 · 3404 + 2 · 1776 = - 3404 + 2(5180 - 3404) , 148 = 2 · 5180 + - 3 · 3404 . 2. The least common multiple of integers a, b N is the number lcm( a, b ) := min { m N : a | m and b | m } . Prove that if gcd( a, b ) > 1 then lcm( a, b ) < ab. Solution. Write g = gcd( a, b ). Then we can write a = gx and b = gy for positive integers x, y with gcd( x, y ) = 1. Then gxy is a multiple of both a and b , so if g > 1 then lcm( a, b ) gxy < gxgy = ab. 3. Let G be a simple planar graph with v 5 vertices and 3 v - 8 edges. Prove that G is connected. Solution. Suppose that G is disconnected and that G can be de- composed as G 1 = ( V 1 , E 1 ) and G 2 = ( V 2 , E 2 ) which are disconnected. If both have size 3 or larger then | E 1 | ≤ 3 | V 1 | - 6 . and | E 2 | ≤ 3 | V 2 | - 6 . So the number of edges E in G can be estimated by | E | ≤ 3 | V 1 | - 6 + 3 | V 2 | - 6 = 3 v - 12 . Otherwise, without loss of generality, either | V 1 | = 1 or | V 1 | = 2 . In the first case | E | = | E 2 | ≤ 3 | V 2 | - 6 = 3 v - 9 , which is excluded by the number of edges. In the second case | E | ≤ 1 + | E 2 | ≤ 3 | V 2 | - 6 = 3 v - 11 , which is also excluded by the number of edges. 1
Math 240 – Autumn 2020 – Addario–Berry–Paquette Final preparation solutions 4. Suppose that G is a planar graph with n vertices in which every face has 4 sides. Show that the average degree 1 n x ∈V ( G ) deg( x ) < 4 . Solution. We let v, e, f be the number of vertices, edges and faces respectively. So v = n. On the one hand, we have that 1 n X x ∈V ( G ) = 2 e v . By the assumption, every face of G has 4 sides. So 4 f = 2 e. We also know that since G is planar, so that v - e + f = 2 . Substituting f = e/ 2 , 2 v - 4 = 2( e - f ) = e. Therefore 2 e v = 4 v - 4 v < 4 . 5. Show that a connected graph with n vertices has exactly 1 cycle if and only if it has exactly n edges.