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Problem:

If we line up the students in this class in 2 rows, there is one student left. If we line up them in 3 rows, there are two left. If we line up them in 5 rows, there are 3 left. How many number of students in this class?

Definition. Given a, b are integers and n is a positive integer, a[pic]b (mod n) , or a and b are congruent modulo n, if n|(a-b).

Example. 3 [pic]0 (mod 3), 4[pic] 1 (mod 3), 5 [pic]2 (mod 3), 6[pic] 0 (mod 3), -1[pic] 2 (mod 3),

-2[pic] 1 (mod 3), …

Now, we can collect all integers that are congruent in the same set, called the congruent class, as defined in the following:

Definition. The set of integers congruent to r (mod n) is called the congruence class of r (mod n), is the set

[r]={ r + kn | k[pic] Z}

The collection of [0], [1], .., [n-1] is denoted by Zn.

Definition. The Integers modulo n, or Zn , is defined as

Zn ={[0], [1], [2], …, [n-1]}

Theorem. The following are equivalent: i) a[pic]b (mod n) ii) [a]=[b] iii) There are exactly n congruence classes (mod n), i.e., [0], [1], [2], …, [n-1]

The modulo arithmetic + and * do not depend on the representation, as in the following:

Theorem. If a, b, c, and d are integers, a[pic]b (mod n) and c[pic]d (mod n) i) a +c[pic]b +d (mod n) ii) ac[pic]bd (mod n)

Definition. If [r] and [s] are congruence classes (mod n), then

[r]+[s]=[r+s]

[r][s]=[rs]

Example. In Z4 ={[0], [1], [2], [3]}

[1]+[1]=[2], [1]+[2]=[3], [1]+[3]=[4]=[0], [2]+[2]=[0], [2]+[3]=[1], [3]+[3]=[2]

[2][2]=[4]=[0], [2][3]=[6]=[2], [3][3]=[9]=[1]

Definition. A non-zero element [r] [pic]Zn is called a zero-divisor if there exists a non-zero element [s] [pic]Zn such that [r][s]=[0].

Definition. A non-zero element [r] [pic]Zn is called a unit if there exists a non-zero element [s] [pic]Zn such that [r][s]=[1]. [s] is called a multiplicative inverse of [r], or [r]-1.

Example. In Z4,

[2] is a zero-divisor and [3] is a unit. The multiplicative inverse of [3] is itself.

Example.

What is [15]-1 (mod 2)?

Solution:

Since 15*1 (mod 2)= 1 (mod 2), [15]-1 =[1].

Theorem. For a congruence class [r] [pic]Zn, the following are equivalent: i) gcd (m, n)=1 ii) [m] has a multiplicative inverse. iii) [m] is not a zero-divisor in Zn.

Example.

In Z4, n=4, m=3, gcd(3,4)=1. Hence [3] has a multiplicative inverse, which is itself.

Corollary.

Every non-zero [r] in Zn has a multiplicative inverse if and only if n is a prime,.

Example.

In Z5, n=5, [1]-1 = [1], [2]-1 = [3], [3]-1 = [2], [4]-1 = [4].

Definition. A function f has a period of r (positive integer) if f(x+r)=f(x), for every x.

Example.

In Z4, calculate the Fibonacci sequence.

[0], [1], [1], [2], [3], [1], [0], [1], …

The period is 6.

Theorem. (The Chinese Remainder Theorem)

Assume that a1, a2, …, am are integers, and n1, n2, …, nm are positive integers such that gcd (ni, nj)=1 if i [pic] j (ni, nj are called relatively prime).

Then the system equation X[pic] a1 (mod n1) X[pic] a2 (mod n2) … X[pic] am (mod nm)

Has a unique solution X modulo N=n1 *n2 *…*nm

Proof:

Let Ni = [pic][pic] = [pic]. Obviously, Ni [pic] 0 (mod nj), where i=1, 2, …, m and i[pic]j

Then gcd (Ni, ni)=1. Hence there is a multiplicative inverse Ri (=Ni-1) of Ni (mod ni).

Let X= [pic]. Since X= [pic][pic] ai (mod ni), X is the unique solution.

Example.

If we line up the students in this class in 2 rows, there is one student left. If we line up them in 3 rows, there are two left. If we line up them in 5 rows, there are 3 left. How many number of students in this class?

That is, to solve the system equation X[pic] 1 (mod 2) X[pic] 2 (mod 3) X[pic] 3 (mod 5)

Solution: m=3, a1=1, a2=2, a3=3, n1 =2, n2=3, n3=5. N= n1*n2* n3 =30

N1 = n2* n3 = 15, R1 =1 (mod 2)

N2 = n1* n3 = 10, R2 =1 (mod 3)

N3 = n1* n2 = 6, R3 =1 (mod 5)

X= [pic]=15*1*1+10*1*2+6*1*3=53 (mod 30)=23 (mod 30)

Ans: 23, 53, 83, 113, …

Applications.

A. DNA Model (Z4) The genetic code in DNA of organism is in the form of double helix, each consisting of a sequence of nucleotides: T(Thymine), A(Adenine), C(Cytosine) and G(Guanine). The double helix is governed by Chargaff’s Rules: T pairs with A and G pairs with C. If we set T=0, A=2, G=1 and C=3, then every helix can interchange to another strand of helix by using modulo arithmetic (adding 2) in Z4. For example, one strand of the human TSH- [pic]gene has the genetic code as GGTCACCACAGCATCTGCTCACCAATGCAAAGTAAG This can be represented in by Z4 as 1103233232132030130323322013222102221 After adding 2 (mod 4) respectively, the other helix code is 3321011010310212312101100231000320003 The genetic code is CCAGTGGTGTCGTAGACGAGTGGTTACGTTTCATTTC B. Public Key Encryption/Decryption (RSA Algorithm) The public key encryption is used by a sender to transmit a secret key. The sender first encrypts the key (called a plain text T, e.g., a positive integer from the key’s ASCII code) using a public key (displayed from a web site) and transmit it (called a cipher text C) to the other person. The receiver will use the corresponding private key to decode the message and know the secret key. Then the receiver will use the secret key to communicate with the sender. The idea is that even people know the public key, it is supposed very hard to find the corresponding private key to decode it. One such algorithm is called RSA algorithm (developed in 1976 by Rivest, Shamir and Adleman at MIT) is described as follows: (PQ, E) is the public key) 1) Choose 2 different prime numbers P and Q. (In practice, each has 200 digits). 2) Choose an odd number E such that gcd (E, (P-1)(Q-1)) = 1 3) By the theorem above, find a multiplicative inverse of E (mod (P-1)(Q-1)), called D (the private key). That is ED [pic]1 (mod (P-1)(Q-1)) Assume that ED=1+m(P-1)(Q-1), where m is an integer 4) The cipher text C is created by C=TE (mod PQ). 5) To decipher C, the receiver performs

CD (mod PQ)= TED (mod PQ)= (T Tm(P-1)(Q-1)) (mod PQ) If gcd(T,P)=1, then by Fermat theorem, Tm(P-1) [pic] 1 (mod P), or Tm(P-1)(Q-1) [pic] 1 (mod P) If gcd(T,Q)=1, then by Fermat theorem, Tm(Q-1) [pic] 1 (mod Q), or Tm(P-1)(Q-1) [pic] 1 (mod Q) Therefore, if gcd(T,P)=1 and gcd(T,Q)=1, then by Chinese Reminder Theorem that Tm(P-1)(Q-1) [pic] 1 (mod PQ), or CD (mod PQ)=T (mod PQ)

The receiver can recover the original plain text T.

Note: 1) Choose 2 large primes P and Q:

i) The Fermat’s Little Theorem (Fermat’s primes p) If p is prime and a is an integer not divisible by p, then ap-1 [pic]1 (mod p). Example. Let a=2, p=5. 24 = 16 [pic]1 (mod 5)

However, this does not guarantee that if an-1 [pic]1 (mod n), then n is a prime. This composite number n is called a Carmichael number (or a pseudo-prime) Example. Let a=2, n=561=3*11*17 Then 2560=(22) 280 [pic]1280 (mod 3) [pic]1 (mod 3) [pic]1 (mod 561). n=561 is not a prime.

ii) The Euclid’s primes Let Nn = p1 * p2 * …* pn +1, where p1, p2, …, pn are the first n primes in order). Then N1, N2, N3, N4, N5, N11, N75, N171, and N172 are primes. No other Nn are primes for 1 [pic]n[pic] 200.

iii) The Mersenne primes If p is a prime, then Mp =2p -1 is called a Mersenne number. Lucas Theorem. Mp is prime if and only if Mp divides S, where S0 = 4, S1 = 42 -2=14, S2 = 142 -2, …, Sk = Sk-12 -2.

iv) The Gaussian primes 2k If a prime has the form 2 +1, it is a Gaussian prime. Only when k=0, 1, 2, 3, 4, it is prime. 2) Choose an odd integer E that is not divisible by (P-1)(Q-1) 3) D can be found using the Euclidean algorithm and Extended Euclidean algorithm as follows: Procedure moduloMultiplicativeInverse (given a, b are positive integers and a [pic]b, modulo n) // stores all quotient in array q[0], q[1], … and stores as count the number of steps used in Euclidean //Algorithm // The following is the Euclidean Algorithm x=a y=b i=0 While y [pic] 0 q[i]=x div y r=x mod y x=y y=r i++

count=i

// the following is the extended Euclidean algorithm xprev=0 x=1 for i=0 to count-2 xnew=( xprev-x*q[i]) mod n xprev=x x=xnew // make x non-negative while x

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...Green Computing Research Project – Part 1 CIS 517 – IT Project Management Green Computing Research Project Computer science educators are uniquely positioned to promote greater awareness of Green Computing, using the academic setting to encourage environmentally conscious use of technology. This paper reports on practical techniques that can engage faculty and students, enabling Green Computing to be integrated into the classroom and research laboratory. Analysis and empirical evaluation of each reported technique is given, comparing the efficacy of each in terms of energy, environmental and financial cost savings. These results are provided as technological and economic evidence for the benefits of “Going Green,” and to promote education in Green Computing in the classroom, department and research lab (Shen, W, 2005). We Are Big, Inc. is an international firm with over 100,000 employees in different countries. The goal of the project is to find environment friendly energy resources to reduce cost and increase energy efficiency. Green Computing Research Project has been authorized on January 24, 2013 as the date of authorization. Project Name: Green Computing Research Project Date of Authorization: January 24, 2013 Project Manager: Hatem Numan 1-320-761-8787 Hatem@wearebig.com Project Start date: January 24, 2013 Project End: July 24, 2013 Project budget: $500,000 Project Objectives: The objectiveof the project is to preparea detailed...

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