Fall2011+Week+3

Chapter 3, problems 1 - 31 are the source of our challenges for this week. I did numbers 4, 6 and 11 in class. From this vast set of 3 examples, spend a happy 1.5 hours posting some problems or questions! Where are Kayla and Megan? My apologies- I had no access to a computer for the weekend. user:kml003

3a) gcd(12, 236)=4  4 divides 28, so from 12x≡28 (mod 236) can be written as the new congruence 3x≡7 (mod 59).  Finding s, t such that 3s+59t=1  59=19(3)+2  3=1(2)+1  1=3-2  1=3-(59-19(3))  1=3(20)-59  s=20 and t=-1  Finding x:  x≡7*20 (mod 59)  x≡140 (mod 59)≡22 (mod 59)  All solutions are of the form x≡22+59h, where h is an integer.

b) gcd(12, 236)=4  4 does not divide 30, so 12x≡30 (mod 236) has no solution.  user:meganternes

5) a. gcd(4883, 4369) 4883=4369(1)+514  4369=514(8)+257  514=257(2)+0  So gcd(4883, 4369)=257  user:kml003

9] x ≡ 2 (mod 7) and x ≡ 3 (mod 10). What is x congruent to mod 70? First, we have to check that 7 and 10 a relatively prime. They are, as gcd(7, 10) = 1. Then we can use the following: x = b+nk (mod m*n) and b + nk ≡ a (mod m) [From page 77]. In this case b = 2, n = 7, a = 3, and m = 10 2 + 7*k≡ 3 (mod 10) -> 7*k≡ 1 (mod 10) ≡ 21 (mod 10), so k = 3. Plugging k back in, we get x = 2 + 7*3 ≡ 23 (mod 70). user:ian.hart

10] From the story problem, we can write it as follows: x ≡ 1 (mod 3) [Rows of 3, 1 left over] x ≡ 2 (mod 4) [Rows of 4, 2 left over] x ≡ 3 (mod 5) [Rows of 5, 3 left over]

to find x = min number of people, find what x is congruent to mod 60 (60 = 3*4*5) knowing that these are all relatively prime, or their respective gcds all = 1, we can do the same as we did for #9, just twice. Starting with n = 3, m = 4, mn = 12: x = 1 + 3k 1 + 3*k ≡ 2 (mod 4) -> 3*k ≡ 1 (mod 4) ≡ 9 (mod 4), so k = 3. x ≡ 1 + 3*3 = 10 (mod 12). Now let n = 12, m = 5, mn = 60 x = 10 + 12j 10 + 12*j ≡ 3 (mod 5) -> 12*j ≡ -7 (mod 5) ≡ 48 (mod 5), so j = 4. x ≡ 10 + 12*4 = 58 (mod 60) So the smallest number of people who could be in the parade is 58. The next smallest number to fit the criteria would be 58 + 60 = 118 user:ian.hart

12. 2^10203 = ? mod 101 2^10203 = 2^(100)(102)+3 = 2^(100)(102) * 2^3 2^(100)(102)=1 mod 101 according to FLT 1*2^3 = 8 mod 101 user:bea001

13) Finding the last two digits is like working mod 100. gcd(123, 100)=1. Using Euler's φ-function,  φ(100) = 100(1 - 1/2)(1 - 1/5) = 40   123^(562)=123^(40*14+2)=123^(40*14)*123^(2) ≡123^(2)=15129 (mod 100)  Therefore, the last 2 digits of 123^562=29  user:kml003

14a) 7^2≡49≡1 (mod 4)  7^3≡(7^2)*7≡1*7≡3 (mod 4)  7^4≡3*7≡1  7^5≡1*7≡3  7^6≡3*7≡1  7^7≡1*7≡3 (mod 4)  or  7^2≡49≡1 (mod 4)  7^3≡1*7≡3  7^4≡3*7≡1  7=3+4 so 7^7≡3*1≡3 (mod 4)

b) From a we have 7^7≡3, so 7^(7^7)≡7^3≡3  The last digit of 7^(7^7) is 3.  user:meganternes

15. a) ϕ (d) for all divisors of 10 - (1, 2, 5, 10)  ϕ (1)=0, ϕ (2)=1 {1}, ϕ (5)=4 {1, 2, 3,4}, ϕ (10)=4 {1, 3, 7, 9}  sum = 0+1+4+4=9  b)   ϕ (d) for all divisors of 12 - (1, 2, 3, 4, 6, 12) ϕ (1)=0, ϕ(2)=1, ϕ(3)=2 {1,2} ϕ(4)=2 {1, 3} ϕ(6)=2 {1, 5} ϕ(12)=4 {1, 5, 7, 11} sum = 0+1+2+2+2+4=11 c) let n>= 1, **Σ** ϕ (d) for all divisors of n= n-1 user:bea001

18a] Using the inverse 2x2 formula, I got the following matrix: __ 1 __ * [1 -6 ] (mod 26) -5 [-1 1 ]  Next, I found the multiplicative inverse of 1/-5. -5 * x ≡ 1 (mod 26) gives x = 5. So multiplying by -1/5 is the same as multiplying by 5. The matrix with that multiplication is as follows: [5 -30 ] (mod 26) = [ 5 22] (mod 26) [-5 5 ] = [ 21 5] user:ian.hart

18b] The matrix A (mod 26) is invertable as long as |A| is relatively prime to 26. So 1 - b = x where {x | gcd(x, 26) = 1}, which means that A is invertable so long as b = 1 - x where {x | gcd(x, 26) = 1}. Assuming we just want to look integers less than 26 (since we're working mod 26) without any factors of 2 or 13 (26 = 13*2). Therefore, x can equal 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, and 25 which means b can equal -2 (24), -4 (22), -6 (20), -8 (18), -10 (16), -14 (12), -16 (10), -18 (8), -20 (6), -22 (4), or -24 (2). user:ian.hart

19) The matrix [3 5 is invertible mod p if and only if its determinant and p are relatively prime.  7 3]  determinant = -26  Of the prime numbers less than 26, 2 and 13 are the only ones not relatively prime to the determinant; 2 and 13 are both factors of -26 (or 26). For the primes p=3, 5, 7, 11, 17, 19, and 23, there is some integer x such that -26x≡1 (mod p). The primes larger than 26 must all be relatively prime to the determinant, otherwise they would not be prime. (I think that last sentence is right, otherwise I'm just making stuff up.)  user:meganternes