(a) Find an inverse for 73 modulo 660. First use the extended Euclidean algorithm to find the greatest common divisor of 660 and 73 and express it as a linear combination of 660 and 73. Step 1: Find q, and r, so that 660 - 73 -4, + where Osr, <73. Then r,- 660 - 73 - 4,- Step 2: Find q, and r, so that 73 -9+ where 0sr Then r,-73- Step 3: Find and so that 1-24+ where osr <- Then, 4, -0 Step 4: Select the comrect statement from the options below. O Becauser-0, god (660, 73) - 73 -r,4,-1. O Because r,- 0, god (660, 73) -- - O Becauser- 1, god (660, 73) - 73 - 2-0. O Because r - 1, god (660, 73)--73-4-0. O Becauser- 1, god (660, 73) -- -0. Next, substitute numerical values backward through the preceding steps, simplifying the results for each step, to find god (660, 73) -1- 660s + 73r, where s- and t- It follows that 73 - (0 =1 (mod 660), and so |is an inverse for 73 modulo 660. (b) Find the least positive solution for the congruence: 73x = 125 (mod 660). Step 1: By part (a), 73 0) =1 (mad 660). Multiply both sides by 125 to obtain 73 - 125 =1- 125 (mad 660) = 125 (mod 660). Let x- 125- Then 73x = 125 (mod 660). Nate that x is a solution to the given congruence, but it may not be the least positive solution. Step 2 Let v be the remainder abtained when x is divided by 660. Then v- Jand, by Theorem 8.4.1 and Theorem 8.4.3, 73x = 73v (mod 660). Hence, v is also a solution to the congruence. In other words, 73v = 125 (mod 660). Suppose u is any positive solution to the congruence that is less than or equal to v. Then 73- u= 73 - J(mad 660). Since 73 is prime, gdc(73, 660) - 1. Thus, Selec- can be applied to conclude that u- Therefore, is the least positive solution to the congruence.