Download CSCI 190 Additional Final Practice Problems Solutions

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 2n1 (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 2n1 are 1, 2, 22 ,
2n1 (2n  1) are 1, 2, 22 ,
2n1 , p, 2 p, 2 p 2 ,
2n1 . Therefore, the divisors of
2 p n1 , 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   2n1 )  p(1  2  22   2n2 )  (2n  1)  p(2n1 1) 
(2n  1)  (2n 1)(2n1 1)  (2n 1)(1  2n1 1)  2n1 (2n 1) (2n  1)  (22n2  2n  1) 
9) Show that if p is prime then
Pf: Its proper divisors of
1  p  p2 
 p n1 
p n is not a perfect number all n  N ,.
p n are 1, p, p 2 ,
p n1 . 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.