Discrete mathematics

Order these steps to produce a proof of the uniqueness of the prime factorization of a positive integer.

1. We use a proof by contradicition 2. We have n=p1 p2.....ps and n=q1q2....qt 3. Set the two factorizations equal and remove common primes 4. Because p1 divides the left-hand side, it also divides the right-hand side 5. But this is a contradiction since no prime can divide another prime 6. Hence, there can be at most one factorization

Which of the following statements are true?

The ratio of the number of primes not exceeding x and x/ln x approaches 1 as x grows without bound. We can use the prime number theorem to estimate the odds that a randomly chosen number is prime. Factoring and primality testing have become important in the applications of number theory to cryptography.

Which of the following conjectures have not been settled?

There are infinitely many Mersene primes. There are infinitely many primes of the form n2 + 1. The twin prime conjecture, which says that there are infinitely many pairs of primes that differ by 2 Goldbach's conjecture, which says every even integer greater than two is the sum of two primes

Match each expression involving modular arithmetic on the left with its value on the right.

(16^2 mod 20) mod 7 2 (16^2 mod 20)^2 mod 7 4 (5^2 mod 7)^3 mod 9 1 (5 mod 7)^2 mod 11 3

When adding the binary numbers a = (10110)2 and b = (10011)2, for which of these indices iis the carry ci equal to 1.

1 2 4

Which of these are true, where m is a positive integer and a, b, c, and d are integers?

1. (a + b) mod m = ((a mod m) + (b mod m)) mod m 2. If a ≡ b (mod m) and c ≡ d (mod m), then ac ≡ bd (mod m). 3. If a ≡ b (mod m) and c ≡ d (mod m), then a + c ≡ b + d (mod m).

Arrange these steps to convert 23 to binary.

1. 23= 2*11 +1 2. 11= 2*5+1 3. 5= 2*2 +1 4. 2 = 2*1+1 5. 1 =2*0 +1

Arrange the following steps into a proof that if a, b, and c are positive integers such that gcd(a, b) = 1 and a | bc, then a | c.

1. Assume that gcd(alb)=1 2. It follows that there are integers s and t such that 3. The equation sa +tb=1 4. We note that a 5. a | sac+tbc by part I of theorem a of section 4.1 6. Because sac+tbc=c

Arrange the following steps in order to add two binary numbers a and b.

1. Begin by adding the right most bits 2. This gives a +b =c*2 +s 3. Then add the next pair of bits along with carry, 4. Continue this process, adding the corresponding bits and the carry to compute the next bit 5. At the last stage, add 6. The leading bit of the sum is

Order the steps to produce a solution by back substitution of the congruences x ≡ 3 (mod 6), x ≡ 2 (mod 5), and x ≡ 4 (mod 7).

1. Form the equation x=6t + 3 form the first congruence 2. Solve this congruence to obtain t=4 ( mod 5) 3. Substitute the equation for t into the first equation to obtain 4. Substitute the equation for x in terms of u into the final congruence to yield 5. Translate the last congruence into u=7v +6 6. Convert to a congruence to obtain the solution

If a, b, and c are integers and if a ≠ 0, which of these must be true?

1. If a | b and a | c, then a | 7b + 3c. 2. If a | b and a | c, then a | b - c. 3. If a | b and a | c, then a | -5b + 4c.

Arrange the steps below to produce a demonstration by trial division that 151 is prime.

1. Note that sq151 = 12.28... 2. The prime numbers less then 12.28 are 2, 3, 5, 7, and 11 3. So, we only need to check whether 151 is dividable by 2, 3, 5, 7, and 11 4, We find that there is a nonzero remainder who we divide 151 by 2, 3, 5, 7, and 11 5. Hence, 151 is prime

Order the steps below to produce the pseudo code of Algorithm 5 for finding ba mod m efficiently, were a is an n-digit number.

1. Procedure modular exponentiation 2. x: =1 power: =b mod m 3. for I:=0 to k-1 4. if a=1 then x:=(x*power(mod) m 5. power:=(power * power) mod m 6. return x

Order the following steps to produce the algorithm for computing q = a div d and r = a mod d from integers a and d > 0.

1. Set the variable q equal to 1 and the variable r equal to the absolute value of a. 2. As long as r is greater then or equal to d, set r equal to r-d and q to q+1 3. If a is negative and r is greater then 0, then set r to d-r and q to -(q+1) 4. Return q and r

Which of the following statements are true?

1. The element 1 is the identity element for multiplication modulo m. 2. If a ≠ 0 belongs to Zm, then m - a is an additive inverse of a modulo m. 3. If a and b belong to Zm, then a +m b = b +m a.

Order the following steps of the procedure for finding Bézout coefficients of two integers a and b, by working backward through the Euclidean algorithm.

1. Use the Euclidean algorithm to find the gcd 2. We now have a sequence of equations in which the next t the last 3. Rewrite this next to the last equation as gcd 4. The equation prior to the once used above expresses 5. Substituting this expression 6 Work backward, expressing

Put the steps in order to prove that if m1, m2, ..., mn are pairwise relatively prime positive integers greater than one and a1, a2, ..., an are arbitrary integers, then the system x ≡ a1 (mod m1), x ≡ a2 (mod m2), ..., x ≡ an (mod mn) has a solution.

1. WE show that a solution exists by constructing 2. Let mk=m/mk for k=1,2,...,n 3. Because mj and mk have no common factors greater then 1 4. By the theorem stating the conditions for the existence 5. We form x= 6. Because Mk*Yk=1(modMk),

Put these statements in order to produce a proof that the integers a and b are congruent modulo m if and only if there is an integer k such that a = b + km.

1. We Perform a direct proof in both directions, first proving sufficiency and then necessity 2. if a= b (mod m) then m | (a-b) 3. Then there is an integer k such that km=a-b, so that a=b+km 4. Now, we perform a direct proof of the other direction, beginning with the assumption that there is an integer k such that a=b+km 5. If there is an integer k such that a=b+km, then km=a-b 6. Hence, m divides a-b, so that a=b(mod m).

Put these statements in order to produce a proof that the integers a and b are congruent modulo m if and only if there is an integer k such that a = b + km.

1. We perform a direct proof in both directions, first proving sufficiency and then necessity. 2. if a= b (mod m) then m | (a-b) 3. Then there is an integer k such that km=a-b, so that a=b+km 4. Now, we perform a direct proof of the other direction, beginning with the assumption that there is an integer k such that a=b+km 5. If there is an integer k such that a=b+km, then km=a-b 6. Hence, m divides a-b, so that a=b(mod m).

Put the steps in order to produce a proof that there are infinitely many primes.

1. We use a proof by contradiction and begin by assuming that there are finitely many primes 2. We can enumerate the primes as p1, p2,.... pn. Then we form 3. Either Q is prime or it is composite and can be written as a product of two or more primes 4. None of the primes p in the original list divide Q, since if p divides Q, then 5. Hence, there is a prime not in the list 6. Since the original assumption that we can enumerate

Arrange the following steps into a proof that if a | b and a | c, then a | (b + c).

1. We will use a direct proof 2. Suppose that a | b and a | c 3. Then it follows that there must exist integers s and t with b=a-s and c=a*t 4. Hence, b+c=a*s+a*t=a(s+t) 5. Therefore, a | b+c

Put these steps in order to form a proof that if a and m are relatively prime integers and m > 1, then an inverse of a modulo m exists.

1. assume that a and m are relatively prime 2. By bezout's theorem and the fact that gcd 3. WE see that sa+tm=1(mod m) 4. It follows that sa=a(mod ,m) 5. Hence, s is an inverse of a modulo m

Which of these are true? (Select all that are correct.) 0 mod d = 0 for all d 0 div d = 0 for all d ≠ 0 a = d(a div d) + a mod d d | a if and only if a mod d = 0

1. d | a if and only if a mod d = 0 2. a = d(a div d) + a mod d 3. 0 div d = 0 for all d ≠ 0 4. 0 mod d = 0 for all d 0 mod d = 0 for all d 0 div d = 0 for all d ≠ 0 a = d(a div d) + a mod d d | a if and only if a mod d = 0

Put the following steps in order to produce the algorithm for multiplying two n-bit (binary) integers a = (an-1an-2⋅⋅⋅a1a0) and b = (bn-1bn-2⋅⋅⋅b1b0).

1. if b=1, let the partial product c equal a 2. for each index position I>1, if b=1 let the partial product c equal a shifted I places 3. Otherwise, if b=0, let the partial product c equal 0 4. set p equal to the sum of the partial products 5. return p

Put in order these steps of the pseudocode of the Euclidean algorithm for finding the greatest common divisor of a and b.

1. procedure gcd(a, b: positive integers) 2. x:=a y:=b 3. While y does not = 0 4. r: = x mod y 5. x:=y y:=r 6 return x

Match the pair of integers a and d > 0 with the unique integers q and r with 0 ≤ r < d such that a = d⋅q + r. 1.a = 80, d = 9 2. a = -40, d = 7 3. a = 99, d = 11 4. a = 0, d = 13 5. a = 51, d = 6

1. q=8, r=8 2. q=-6 , r-2 3. q=9 , r=0 4. q=0 , r=0 5. q=8 , r=3

Arrange the steps in order to produce a proof that if n is a composite integer, then n has a prime divisor less than or equal to √n.

1. suppose that n is composite and therefore is has a factor a 2. Hence, n=a*b, where b is a positive integer greater the n1 3. We will prove bu contradiction that a<=√n 4. If a> √n and b >√n then a*b> √n√n=n 5 Hence a <=√n or b <=√n because both a and b are divisors of n, we see that n has a positive divisor not exceeding √n 6. This divisor is either prime or has a prime divisor less then or equal

Find π(x)/(x/ln x) for x = 20 to three decimal places, where π(x) is the number of primes not exceeding x

1.198

Which of these numbers appear as remainders in one of the steps used to find the GCD (using the Euclidean algorithm) of 160 and 90.

10 20 70

Which of these equations arise when working backward through the Euclidean algorithm to find the Bézout coefficients of 160 and 90?

10 = 4⋅160 - 7⋅90 10 = 70 - 3⋅20 10 = 4⋅70 - 3⋅90

Which of the following pairs of numbers are relatively prime?

10 and 27 3 and 5 13 and 22

Which of these statements are true? (Select all that are correct.)

10 mod 6 = 4 18 div 3 = 6

Match the integer on the left with the ordered pairs on the right where the first component is the remainder upon division by 11 and the second is the remainder upon division by 13.

100 matches Choice, (1,9) 77 matches Choice, (0, 12) 63 matches Choice, (8, 11) 84 matches Choice, (7, 6) 133 matches Choice, (1, 3)

Given 5x ≡ 2 (mod m), match modulus m on the left with the solution x on the right.

11 matches Choice, 7 13 matches Choice, 3 16 matches Choice, 10 19 matches Choice, 8

Match the numbers on the left with the description on the right.

11 matches Choice, Prime Prime 12 matches Choice, Composite and not a power of a prime Composite and not a power of a prime 1 matches Choice, Neither prime nor composite Neither prime nor composite 9 matches Choice, A power of a prime A power of a prime

Which of these is true?

12 +15 6 = 3 in Z15 1 +7 2 = 3 in Z7 8 ⋅14 9 = 2 in Z14

Which of these is true

12^100 ≡ 1 (mod 101) 12^130 ≡ 1 (mod 131)

Which of these statements are true? (Select all that are true.)

15 ≡ 5 (mod 2) 17 ≡ 5 (mod 6)

Match each integer with one that divides it. Instructions

18 -----------------------------------9 -14-----------------------------------7 11------------------------------------11 65----------------------------------13

Match the pairs of numbers on the left with their greatest common divisor on the right

18 and 22 matches Choice, 2 17 and 34 matches Choice, 17 45 and 50 matches Choice, 5 36 and 48 matches Choice, 12 11 and 17 matches Choice, 1

Match the pairs of numbers on the left with their greatest common divisor on the right. You should use the Euclidean algorithm to compute the GCDs.

1836 and 3549 matches Choice, 3 693 and 763 matches Choice, 7 421 and 461 matches Choice, 1 470 and 2020 matches Choice, 10

Use trial division of 33,339,600 to find its prime factorization. Then match the primes in the left hand column with the corresponding exponents in the right hand column.

2-------------- 4 3---------------5 5--------------- 2 7----------------3

Which integers are pairwise relatively prime?

21, 22, and 23 7, 10, and 13

What does (10110)2 equal in decimal notation?

22

Which of these are prime?

23 29 31 17

What does (5017)8 equal in decimal notation?

2575

Find the least common multiple for the pair 25,725 = 3^1⋅5^2⋅7^3 and 2,613,600 = 2^5⋅3^3⋅5^2⋅11^2.

2^5⋅3^3⋅5^2⋅7^3⋅11^2.

Which of the following is the correct octal expansion of 1776?

3360

Match the octal number on the left with its binary expansion on the right.

4534 matches Choice, 100 101 011 100 100 101 011 100 1111 matches Choice, 001 001 001 001 001 001 001 001 2375 matches Choice, 010 011 111 101 010 011 111 101 2017 matches Choice, 010 000 001 111 010 000 001 111 7532 matches Choice, 111 101 011 010 111 101 011 010

Match the pairs of integers on the left with the Bézout coefficients on the right

50 and 35 matches Choice, -2 and 3 100 and 73 matches Choice, -27 and 37 21 and 77 matches Choice, 4 and -1 45 and 150 matches Choice, -3 and 1

Which of these is true?

561 is a Carmichael number. There are infinitely many Carmichael numbers.

Match each linear congruence on the left with its solutions on the right.

5x ≡ 2 (mod 17) matches Choice, x ≡ 14 (mod 17) 11x ≡ 13 (mod 17) matches Choice, x ≡ 12 (mod 17) 9x ≡ 14 (mod 17) matches Choice, x ≡ 11 (mod 17) 2x ≡ 10 (mod 17) matches Choice, x ≡ 5 (mod 17) 8x ≡ 15 (mod 17) matches Choice, x ≡ 4 (mod 17)

Which of these are primitive roots modulo 13?

6 7 2 11

Match the numbers (on the left) to their greatest common divisors (on the right).

617,400 = 23⋅32⋅52⋅73 and 1,890 = 21⋅33⋅51⋅71. matches Choice, 2^1⋅ 3^2⋅ 5^1 ⋅ 7^1 21,600 = 25⋅33⋅52 and 504 = 23⋅32⋅71. matches Choice, 2^3 ⋅ 3^2⋅ 5^0⋅ 7^0 525 = 31⋅52⋅71 and 4,410 = 21⋅32⋅51⋅72. matches Choice, 2^0⋅ 3^1⋅ 5^1⋅7^1 875 = 53⋅71 and 14,700 = 22⋅31⋅52⋅72 matches Choice, 2^0⋅ 3^0⋅ 5^2⋅7^1

Which of these integers are pseudoprime to the base 2?

645 561 341

Match the integer on the left with its inverse modulo 99. You can find the inverses by working backward through the steps of the Euclidean algorithm.

7 matches Choice, 85 13 matches Choice, 61 17 matches Choice, 35 19 matches Choice, 73 25 matches Choice, 4

Match the left-hand side with the right-hand side. Use Fermat's little theorem to help do this problem.

7111 mod 13 matches Choice, 5 6313 mod 13 matches Choice, 6 5227 mod 13 matches Choice, 8 4666 mod 13 matches Choice, 1

Given that 8 ≡ 3 (mod 5) and 9 ≡ 4 (mod 5), which of the following are true?

8 + 9 ≡ 3 + 4 (mod 5) 8 ⋅ 9 ≡ 3 ⋅ 4 (mod 5)

Which of the following statements are true?

A positive integer greater than 1 is prime if it has exactly two positive divisors. The integer 2 is prime. A positive integer greater than 1 that is not composite is prime. There are four primes less than 10.

Match the hexadecimal number on the left with its binary expansion on the right.

ABCD ----- 1010 1011 1100 1101 1010 1011 1100 1101 789A------- 111 1000 1001 1010 111 1000 1001 1010 1234 --------- 1 0010 0011 0100 1 0010 0011 0100 F2A5 -------- 1111 0010 1010 0101 1111 0010 1010 0101

Which of the following statements are true?

Binary expansions are used by computers to do arithmetic with integers. (347)8 = 231

The representation of n given by n = akbk + ak-1bk-1 + ⋅⋅⋅ + a1b + a0 is called the (BLANK 1) b expansion of n. Setting b = 10 gives the (BLANK 2) expansion. Setting b = 2 gives the (BLANK 3) expansion. Setting b = 8 gives the (BLANK 4)expansion.

Blank 1: base Blank 2: decimal Blank 3: binary Blank 4: octal

A (BLANK 1) integer n that satisfies the congruence bn-1 ≡ 1 (mod n) for all positive integers b with gcd(b, n) = 1 is called a (BLANK 2) number.

Blank 1: composite Blank 2: Carmichael

Let b be a positive integer. If n is a composite positive integer, and bn-1 ≡ 1 (mod n), then n is called a (BLANK 1) to the (BLANK 2) b.

Blank 1: pseudoprime Blank 2: base

Match the property name on the left with the definition on the right.

Closure If a and b belong to Zm, then a+mb and a ⋅mb belong to Zm. If a and b belong to Zm, then a +m b and a ⋅mbbelong to Zm. Associativity If a, b and c belong to Zm, then (a+mb) +mc=a+m(b+mc) and (a⋅mb) ⋅mc=a⋅m(b⋅mc). If a, b and c belong to Zm, then (a +m b) +m c = a +m (b +m c) and (a ⋅m b) ⋅m c = a ⋅m (b ⋅m c). Commutativity If a and b belong to Zm, then a+mb=b+maand a ⋅mb=b⋅ma. If a and b belong to Zm, then a +m b = b +m a and a ⋅mb = b ⋅m a. Identity elements If a belongs to Zm, then a+m0=0 +ma=a and a⋅m1=1 ⋅ma=a. If a belongs to Zm, then a +m 0 = 0 +m a = a and a ⋅m 1 = 1 ⋅m a = a. Additive Inverses If a≠ 0 belongs to Zm, then a+m(m-a)=0 and 0 +m0 = 0. If a ≠ 0 belongs to Zm, then a +m (m - a) = 0 and 0 +m 0 = 0. Distributivity If a, b and c belong to Zm, then a⋅m(b+mc)=(a⋅mb) +m(a⋅mc) and (a+mb) ⋅mc=(a⋅mc) +m(b⋅mc). If a, b and c belong to Zm, then a ⋅m (b +m c )= (a ⋅m b) +m (a ⋅m c) and (a +m b) ⋅m c = (a ⋅m c) +m (b ⋅m c).

Which of the following are true?

Every integer greater than 1 can be written uniquely as a prime or as the product of primes in nondecreasing order. Every integer greater than 1 can be written uniquely as a product of prime powers where the primes are listed in increasing order.

True or false: f(n) = n2 - n + 41 is prime for all positive integers n.

FALSE

True or false: If a | b + 1 and a | c + 1, then a | b + c + 1 for all integers a, b, and c, where a ≠0.

False

If a and b are integers and m is a positive integer, then a is ________________to b ___________________________m if m divides a - b. We call the relationship a ≡ b (mod m) a ______________where m is the ___________________.

If a and b are integers and m is a positive integer, then a is CONGRUENT to b MODULO m if m divides a - b. We call the relationship a ≡ b (mod m) a CONGRUENCE where m is the MODULUS OR MODULI

Which of these are correct

If a and m > 1 are relatively prime, then the inverse of a modulo m exists. We can find the inverse of a modulo m when gcd(a, m) = 1 by reversing the steps in Euclidean algorithm. We can quickly find the inverse of a modulo m when gcd(a, m) = 1 by knowing the Bézout coefficients.

Which of these is true where m is a positive integer and a and b are integers?

If a mod m = b mod m, then a ≡ b (mod m). If a ≡ b (mod m), then a mod m = b mod m.

If a, b, and c are integers and if a ≠ 0, which of these must be true?

If a | b and b | c, then a | c. If a | b, then a | b⋅c for all integers c. If a | b and a | c, then a | (b + c).

Which of the following statements are true?

The complexity of addition of two n-bit integers is O(n) Multiplication (following the conventional algorithm) of two n-bit integers uses O(n2) bit additions.

Which of these statements are true?

The four smallest Mersenne primes are 3, 7, 31, and 127. Mersenne primes are of the form 2p - 1, where p is prime.

Which of the following statements are true?

The integers 5, 11, 17, 23, and 29 form an arithmetic progression of prime numbers. There is an arithmetic progression with 10 primes. There are infinitely many primes in the arithmetic progression 4k + 3 as k = 0, 1, 2, ....

Which of these statements about the sieve of Eratosthenes is true?

The last step involves eliminating all remaining numbers divisible by the greatest prime not exceeding n‾√. In the first step of the sieve, all even numbers other than 2 are eliminated. Given a positive integer n, the sieve of Eratosthenes can be used to find all the primes less than or equal to n.

Which of the following statements are true?

To convert binary to octal, starting from the right, group the bits into three-digit binary numbers and convert each into an octal digit. Each hexadecimal digit corresponds to four binary digits.

Which of these are true?

We can solve the linear congruence ax ≡ b (mod m) where a and m are relatively prime integers and m > 1 by multiplying by the inverse of a modulo m. The linear congruence ax ≡ b (mod m) where a and m are relatively prime integers and m > 1 always has a unique solution modulo m.

Match the systems of linear congruences on the left to their solutions on the right.

x ≡ 3 (mod 6) x ≡ 2 (mod 5) x ≡ 4 (mod 7) matches Choice, x ≡ 207 (mod 210) x ≡ 2 (mod 3) x ≡ 4 (mod 10) x ≡ 6 (mod 7) matches Choice, x ≡ 104 (mod 210) x ≡ 1 (mod 3) x ≡ 4 (mod 5) x ≡ 6 (mod 14) matches Choice, x ≡ 34 (mod 210) x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ 3 (mod 5) x ≡ 4 (mod 7) matches Choice, x ≡ 53 (mod 210)