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CSCI 190 Additional Final Practice Problems Successive squaring, Solving Power modulo, Powers modulo m, perfect numbers 1) Use successive squaring to compute 5120 (mod 47) Power 5 power (mod 47) 1 2 51 5 52 25(mod 47) 22 4 54 (52 )2 (25)2 625 14(mod 47) 23 8 58 (54 )2 142 196 8(mod 47) 516 (58 )2 82 64 17(mod 47) 2 4 16 25 32 532 (516 )2 (17)2 7(mod 47) 26 64 564 (532 )2 (7)2 2(mod 47) 120=2(60)+0 60=2(30)+0 30-2(15)+0 15=2(7)+1 7=2(3)+1 3=2(1)+1 1=2(0)+1 Thus 120=64+32+16+8 And 5120 564 532 516 58 2 7 17 8 1904 27(mod 47) 2) Use successive squaring to compute Power 3 power (mod13) 1 31 3 381 (mod13) 2 32 9(mod13) 22 4 34 (32 )2 (9)2 81 3(mod13) 23 8 38 (34 )2 32 9(mod13) 316 (38 )2 92 81 3(mod13) 2 4 16 25 32 332 (316 )2 (3)2 9(mod13) 26 64 364 (332 )2 (9)2 3(mod13) 81=2(40)+1 40=2(20)+0 20-2(10)+0 10=2(5)+0 5=2(2)+1 2=2(1)+0 1=2(0)+1 Thus 81=64+16+1 And 581 364 316 31 3 3 3 27 1(mod13) 3) Solve x7 7(mod 41) Solutions: To solve for x, you need to make the exponent of x equal 1. In regular algebra, you would raise both sides by 1/7. So you need to find the inverse of 7. By the Euler’s Theorem, a ( m) 1(mod m) if (a,m)=1. So first find the inverse of 7 mod (41) 40 . Use the Euclidean algorithm. 40=7(5)+5 7=5(1)+2 5=2(2)+1 Then 1=5-2(2)=5-2(7-5(1))=5(3)+7(--2)=(40-7(5))(3)+7(-2)=40(3)+7(-17) Thus the inverse of 7 is -17 mod 40. Adding 40, -17+40=23 is the inverse of 7 mod 40. x7 7(mod 41) ( x7 )23 (7)23 (mod 41) x 723 (mod 41) Power 7 power (mod 41) 1 2 71 7 72 49 8(mod 41) 22 4 74 (72 )2 (8)2 23(mod 41) 23 8 78 (74 )2 232 529 (4)(mod 41) 716 (78 )2 (4)2 16(mod 41) 2 4 16 We have 723 7 72 74 716 7 8 23 (16)(mod 41) 17 6 26(mod 41) Therefore, x 26(mod 41) 4) Solve x21 7(mod 23) Solution: To solve for x, the exponent of x need to become 1. In regular algebra, you would raise both sides by 1/21. So you need to find the inverse of 21. By the Euler’s Theorem, Since a ( m) 1(mod m) if (a,m)=1. So first find the inverse of 21 mod (23) 22 . 21 1(mod 22) and (1)(1) 1(mod 22) , (21)(21) 1(mod 22) . Therefore, the inverse of 21 is 21 mod 22. Therefore, x 21 7(mod 23) ( x 21 )21 (7)21 (mod 23) x 721 (mod 23) Using successive squaring, Power 7 power (mod 23) 1 2 71 7 72 49 3(mod 23) 22 4 74 (72 )2 (3)2 9(mod 23) 23 8 78 (74 )2 92 81 12(mod 23) 2 4 16 716 (78 )2 122 144 6(mod 23) 721 7 74 716 7 9 6(mod 23) 17 6 10(mod 23) We have Therefore, x 10(mod 23) 5) Use Euler’s Theorem to simplify Solution: By the Euler’s Theorem, (10) 4,34 1(mod10) . 3310 (mod10) a ( m) 1(mod m) if (a,m)=1. Since . Thus all the exponents that are multiple of 4 becomes 1 3310 3108 32 1 9 9(mod10) mod 10. 6) Use Fermat’s Little Theorem to simplify Solution: Since 53 is prime, and (5, 53)=1, 5105 (mod 53) 552 1(mod 53) by FLT. Thus all the exponents that are multiple of 52 become 1 mod 53. 105 5 5 104 Therefore, 5 1 5 5(mod 53) n N , if 2n 1 is prime, then n is odd or n is 2. (recall that n prime numbers of the form 2 1 are called Mersenne primes) 7) Prove that for all Pf: We know from Math 50 that (a k 1 a k 2 where a k 1, k 1 factors as (a 1)(a k 1 a k 2 a 1) , a 1) contains k terms. (it it easy to see that a=1 is a zero of the polynomial. Apply synthetic using 1 to get the remainder equals 0.) Proof by contrapositive. First observe that the negation of “n is odd or n is 2” is “ n is even and n is not 2”. That is, suppose n is even greater than 2. This implies n=2k for some k>1. Then using Math 50 Clearly 2n 1 22k 1 (22 )k 1 (22 1)((22 )k 1 (22 )k 2 (22 1) 1 . In addition, ((22 )k 1 (22 )k 2 1) . 1) 1 since k 1 means this expression contains more than one term. Therefore, 2n 1 is a composite number. 8) A number is a perfect number if the sum of its proper divisors the number itself. For example, 6 is a perfect number since its proper divisors are 1,2,3 and 1+2+3=6. It can be verified easily that 28 is also a perfect number. Prove that if Pf: Let 2n 1 is prime for some n N , then 2n1 (2n 1) is a perfect number. p 2n 1 . Since p 2n 1 is prime by assumption, 1 and p are the only divisors of 2n 1 . The divisors of 2n1 are 1, 2, 22 , 2n1 (2n 1) are 1, 2, 22 , 2n1 , p, 2 p, 2 p 2 , 2n1 . Therefore, the divisors of 2 p n1 , From the list, remove 2n 1 p since we are to sum its proper divisors. Therefore, the sum of its proper divisors is (1 2 22 2n1 ) p(1 2 22 2n2 ) (2n 1) p(2n1 1) (2n 1) (2n 1)(2n1 1) (2n 1)(1 2n1 1) 2n1 (2n 1) (2n 1) (22n2 2n 1) 9) Show that if p is prime then Pf: Its proper divisors of 1 p p2 p n1 p n is not a perfect number all n N ,. p n are 1, p, p 2 , p n1 . Their sum is pn 1 pn 1 . Since p n , p n is not a perfect number. p 1 p 1 Proofs to study (there wiil be one questions on the final) 1. Prove that every nonleaf in a tree is a cut vertex 2. Let G be a tree. Then a new graph G’ obtained from G by adding one edge has exactly one cycle. 3. There are infinitely many prime numbers. 4. \ 2 is an irrational number.