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I would like to get a quote on getting this assignment completed. If possible, I would also like whoever did my second assignment to do this one. They did it ahead of schedule and without any plagiarism. Plagiarism has been a huge problem on 2 out of the 3 assignments I have give you, with no legitimate response back. Sincerely, Chris Ivy.

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Individual Assignment, Week 4 for Chris Ivy SECTION 6.1 Exercise #2) Show that 12!+ 1 is divisible by 13, by grouping together pairs of inverses modulo 13 that occur in 12!. Exercise #4) What is the remainder when 5!25! is divided by 31? Exercise #6) What is the remainder when 7 × 8 × 9 × 15 × 16 × 17× 23 × 24 × 25 × 43 is divided by 11? Exercise #10) What is the remainder when 62000 is divided by 11? Exercise #14) Using Fermat’s little theorem, find the last digit of the base 7 expansion of 3100. Exercise #22) Show that 30 | (n9 - n) for all positive integers n. Exercise #42) If p is prime and k is a positive integer less than p, then the binomial coefficient pk is divisible by p. Use this fact and the binomial theorem to show that if a and b are integers, then (a+b)p = ap + bp (mod p). SECTION 7.1 Exercise #2) Find the value of the Euler phi-function at each of these integers. a) 100 b) 256 c) 1001 d) 2×3×5×7×11×13 e) 10! f ) 20! Exercise #4) Find all positive integers n such that Ø(n) has each of these values. Be sure to prove that you have found all solutions. a) 1 b) 2 c) 3 d) 4 Exercise #8) Show that there is no positive integer n such that Ø(n) = 14. Exercise #12) For which positive integers n is Ø(n) divisible by 4? Exercise # 16) Show that if n is a positive integer having k distinct odd prime divisors, then Ø(n) is divisible by 2k. Exercise #32) Show that if m and n are positive integers with m | n, then Ø(m) | Ø(n). SECTION 7.2 Exercise #2) a) 36 b) 99 c) 144 d) 2 × 3 × 5 × 7 × 11 × 13 × 17 × 19 e) 2 × 32 × 53 × 74× 115 × 134 × 175 × 195 f) 20! Exercise #4) For which positive integers n is the sum of divisors of n odd? Exercise #8) Which positive integers have exactly two positive divisors? Exercise #12) Show that the equation s(n) = k has at most a finite number of solutions when k is a positive integer. SECTION 13.1 Exercise #2) Show that if (x, y, z) is a primitive Pythagorean triple, then either x or y...

005_o7tkapo-vp1w31uv.docx

Robert · Accepted Answer

SECTION 6.1
Sol: (2)
        
     
Note that  12! 1 (1)(2 7)(3 9)(4 10) 5 8 6 11 12 1 1 1 1 1
1 1 1 1 0 mod13 .
Therefore 13 divides 12! + 1.
        
  
Sol: (4)
     
 
   
5
5
We can write 5! 25! as,
                5! 25! 1 2 3 4 5 25!
                          30 29 28 27 26 25!
                          -1 30!
                          -1 -1
                          1 mod
     
     


   31 ,  by Wilson's theorem,
So finally we can see that the remainder is 1.

Sol: (6)
First we find
       7 8 9 15 16 17 23 24 25 43 7 8 9 4 5 6 1 2 3 10 10! 1 (mod 11). 
So we can see that the remainder is 10.
                     
Sol: (10)
   
10
200
2000 10 200
From Fermat's little theorem, we know that
                                        6 1 (mod 11). 
Then                            6     6 1 1 mod 11
So we can see that the remainder is 1.

  

Sol: (14)
100
100
6
100
We aim to find the last digit of 3  in its base 7 expansion. Equivalently, we would 
like to determine 3  mod 7. 
By Fermat's little theorem 
                        3  1 mod 7; 
hence          3  

6 14 4 4
100
= ( 3 ) ·3  3  81  4 mod 7.
So we can see that from the last step, the last digit of the base 7 expansion 3  is 4.
  

Sol: (22)
   
 
 
3
9 3
9 5 4 5
9 9
We have 
                                                0 mod3 ,
                                                0 mod5 ,
Since  and  have the same parity,  0 mod 2 ,
 By the
n n n n
n n n n n n n
n n n n
   
     
 
9
9
9
 Chinese remainder theorem, since both  and 0 are solutions to the 
system 0 (mod 2), 0 (mod 3), and 0 (mod 5),
we have 0  (mod 2 3 5). 
Therefore 30 divides .
n n
x x x
n n
n n

  
   


Sol: (42)
 
 
0 0
0
We have 
      0 0 ...
            Since 0 mod  
When   1 1. 
p
p k p k p p p p
k
p
a b a b a b a b b a
k
p
p
k
k p


 
         
 
 
 
 
  


SECTION 7.1
Sol: (2) (a)  
                       2 2 2 2100 2 5 2 5 4 2 25 5 40            
Sol: (2) (b) 
                    8 8 7256 2 2 2 128       
Sol: (2) (c) 
                              1001 7 11 13 7 11 13 7 1 11 1 13 1 7200                
Sol: (2) (d)
             
      
2 3 5 7 11 13 2 3 5 7 11 13
                               2 1 3 1 5 1 7 1 11 1 13 1
                               1 2 4 6 10 12
                               5760
                
      
     

 
Sol: (2) (e)
   
    
8 4 2
8 7 4 3 2
10! 2 3 5 7
2 2 3 3 5 5 7 1
829440
    
    

 
Sol: (2) (f)
   
        
18 8 5 2
18 17 8 7 4 3 2
20! 2 3 5 7 11 13 17 19
2 2 3 3 5 5 7 7 11 1 13 1 17 1 19 1
416,084,687,585,280,000
        
        


Sol: (4) (a)
     
     
 
1 2
1 1
1 1
1 2
1 11
1 1
1 1
1 1
If  1,  let 2 ...  be the prime factorization of . If 0 then 
                         2 ...  
and 
   if 0 then ... .
If    1,  then
r
r r
r r
a a ak
r
a a a ak
r r
a a a a
r r
n n p p p n k
n p p p p
k n p p p p
n



 
 
  
  
   
  either   1;     1    2.n or k and n  
 
Sol: (4) (b)  
                       1 1
1 1 1 1
1
1 1
1
1 1 1 1
Using the notation developed in part (a),
If    2,  then either  2 and   4;  or  1 and  -  2,
so  3 and 6;  or 0 and  -  2,  so  3 and 3.
a a
a a a a
n k n k p p
p n k p p p n
 

    
     

Sol: (4) (c) 
                           1 1 11 1
Using the notation developed in part (a),
If    3,  then  -  3,  
which is impossible, so there are no solutions..
a a
n p p    
 Sol: (4) (d)                       
   1Using the notation developed in part (a), if  , then 1   = 4. Therefore, 
no odd prime can appear in the factorization of  to a power higher than 1. Further, 1 
must be a divisor of 4, so  
t tp n p p n
n p
p
 

 k 1
must be one of 2; 3, or 5. Say 2 3 5 ,  where  and  are 
0 or 1. Note that 2 2 which must divide 4, so  is either 0, 1, 2, or 3. If   3, 
then     0, and so one solution is  8. If   2,
k a b
k
n a b
k k
a b n k
 

 
     
 
2
k
 then 2 = 2 which forces 
  1 and   0,  so a second solution is   12. If   0  1,  then 2 =1. This 
forces   0 and   1. This gives us two more solutions   5 and   10.
Having exhausted all
a b n k or
a b n n

   
   
 possibilities, we have the complete set of solutions: 5, 8, 10, and 12.

Sol: (8) 
  1 1 11 1 1
1 1 1 1
1 1 1
If 14,  then 7  for some odd prime . Since the only factors of 14 are 
2 and 7, either  = 7 and 1 and hence 1 = 6 14  which is false, or 7 1, 
but 1 is even, so 1 =14 or =
a a
n p p p
p a p p
p p p
  
  
   15 which is not prime. 
Therefore there are no solutions.

Sol:

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