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1. Use the Euclidian algorithm to find the greatest common divisors, and write the gcd(a, b) as a linear combination of a and b. (a) (513,187) (b) (84.54) (c) (6540,1206) 2. Let a,b,c e Z and prove that if a b and a c then a?|bc. 3. Which of the integers 0,1,2, .... 10 can be expressed in the form 15m +21n, where m, n are integers? 4. Use the prime factorizations of 252 and 180 to find the ged(252,180) and Icm[ 252,180). 5. For positive integers a, b,c, prove that if ged(a,b) = 1 and cb, then ged(a.c) = 1. 6. Let a, b, c be nonzero integers. Prove that if bſa and cla and ged b,c) = d, then be ad. 7. Solve the congruence 42x = 12 mod 90) 8. Solve the congruence 250 = 45mod 60) 9. Prove that 100+1 +4·10" + 4 is divisible by 9, for all positive integers n. 10. Prove that for any integer n, the number n + 5n is divisible by 6. 11. Find the multiplicative inverse of each nonzero element of 211. 12. Find the multiplicative order of each element of Z>. (use ex. 1.4.10 in the text for a small shortcut) 13. Find (9150if possible , in Z301 14. In 224, find all the units (list the multiplicative inverse of each). 15. Show that Z, and Z are cyclic. 16. Make multiplication tables for Z, and 21- (you can use the fact that these are cyclic groups) 17. Show that the function f:R? +R? defined by f(x,y) = (2+3 + y, y) for all (2,4) € R² is a bijection 18. Define f : 212 + Zg by f[r]12) = [2x]g for all (v]12 € Z12 and define g : 212 + Zx by 9(+)12) = (3x]s for all [r]12 € Z12. One of these is a function and one is not. Which one is which? Prove it. 19. Consider the function f : Z10 + Z10 defined by f[2]10 = (3x + 4)10. Show that f is one-to-one and onto by computing all values of the function. Then find a formula of the type 9([r].0) = [mx + b)20 that gives the inverse of f.
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Surname 1

Name
Supervisor
Course
Date
1. a) let 𝑎 = 513 , 𝑏 = 187 then the (513, 187) is calculated as follows:
513 = 2.187 + 139
187 = 1.139 + 48
139 = 2.48 + 43
48 = 1.43 + 5
43 = 8.5 + 3
5 = 1.3 + 2
3 = 1.2 + 1
2 = 1.1 + 1 Thus ( 513,187) = 1
For us to find the linear combination we proceed as follows:
𝑎 = 2𝑏 + 139 ⟹ 𝑎 − 2𝑏 = 139
𝑏 = 𝑎 − 2𝑏 + 48 ⟹ 3𝑏 − 𝑎 = 48
𝑎 − 2𝑏 = 2(3𝑏 − 𝑎) + 43 ⟹ 3𝑎 − 8𝑏 = 43
3𝑏 − 𝑎 = 3𝑎 − 8𝑏 + 5 ⟹ 11𝑏 − 4𝑎 = 5
3𝑎 − 8𝑏 = 8(11𝑏 − 4𝑎) + 3 ⟹ 35𝑎 − 96𝑏 = 3
11𝑏 − 4𝑎 = 35𝑎 − 96𝑏 + 2 ⟹ 107𝑏 − 39𝑎 = 2
35𝑎 − 96𝑏 = 107𝑏 − 39𝑎 + 1 ⟹ 74𝑎 − 203𝑏 = 1
Thus the linear combination is:74.513 − 203.187 = 1

Surname 2

1.b) let 𝑎 = 84 , 𝑏 = 54 then gcd(84,54) is
84 = 1.54 + 30
54 = 1.30 + 24
30 = 1. 24 + 6
24 = 5.6 + 0 ; Thus the gcd(84,54) = 6
We can write as a linear combination as follows:
𝑎 = 𝑏 + 30 ⟹ 𝑎 − 𝑏 = 30
𝑏 = 𝑎 − 𝑏 + 24 ⟹ 2𝑏 − 𝑎 = 24
𝑎 − 𝑏 = 2𝑏 − 𝑎 + 6 ⟹ 2𝑎 − 3𝑏 = 6
Hence the linear combination is: 2.84 − 3.54 = 6
1.c) let 𝑎 = 6540, 𝑏 = 1206 then (6540,1206) is:
6540 = 5. 1206 + 510
1206 = 2.510 + 186
510 = 2.186 + 138
186 = 1. 138 + 48
138 = 2.48 + 42
48 = 1.42 + 6
42 = 7.6 + 0 ℎ𝑒𝑛𝑐𝑒 (6540,1206) = 6
Thus the linear combination of (6540,1206) is written as
𝑎 = 5𝑏 + 510 ⟹ 𝑎 − 5𝑏 = 510
𝑏 = 2(𝑎 − 5𝑏) + 186 ⟹ 11𝑏 − 2𝑎 = 186
𝑎 − 5𝑏 = 2(11𝑏 − 2𝑎) + 138 ⟹ 5𝑎 − 27𝑏 = 138

Surname 3

11𝑏 − 2𝑎 = 5𝑎 − 27𝑏 + 48 ⟹ 38𝑏 − 7𝑎 = 48
5𝑎 − 27𝑏 = 2(38𝑏 − 7𝑎) + 42 ⟹ 19𝑎 − 103𝑏 = 42
19𝑎 − 103𝑏 = 7.6 𝑡ℎ𝑢𝑠 6 =
Hence 6 =

1
7

1
(19𝑎 − 103𝑏)
7

(19.513 − 103.187)

2.) We know that if 𝑎│𝑏 then 𝑎𝑘 = 𝑏 for some integer 𝑘 and If 𝑎 │𝑐 then 𝑎𝑚 = 𝑐 for some
integer 𝑚
Thus 𝑏𝑐 = (𝑎𝑘)(𝑎𝑚) = 𝑘𝑚𝑎2 = 𝑛 𝑎2 for some 𝑘𝑚 = 𝑛 ∈ 𝑧. Hence 𝑎2 │𝑏𝑐
3.) We find the gcd(21,15) and then find the numbers between 0 and 10 with the found 𝑔𝑐𝑑
Thus (21,15) = 3 and thus 0, 1, 3, 6 and ...


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