{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hmk4_key - COT 3100 Homework 4 Spring 2001 Assigned Due...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
COT 3100 Homework # 4 Spring 2001 Assigned: 3/26/01 Due: 4/9,10/01 (in class) 1) Use Euclid’s Algorithm to find the greatest common divisor(GCD) of 851 and 444. 851 = 1x444 + 407 444 = 1x407 + 37 407 = 11x37, so gcd(851, 444) = 37. 2) Show that if 17 | (3x + 8y), then there are no integer solutions to the equation 22x + 2y = 1602 22x + 2y = 34(x + y) – 4(3x + 8y) = 17(2x+2y) – 4(17)(c), for some integer c, using given information. = 17(2x + 2y – 4c) Thus, if x and y are integers, then 17 | (22x + 2y). However, 1602 is NOT divisible by 17, thus the two sides of the given equation can not be equal since there are not equivalent mod 17. 3) Using mod rules, find the remainder when you divide 7 214 by 13. 7 214 (7 2 ) 107 (mod 13) (49) 107 (mod 13) (-3) 107 (mod 13) ((-3) 4 ) 26 (-3) 3 (mod 13) (81) 26 (-3) 3 (mod 13) (3) 26 (-27)(mod 13) (3) 26 (3)(-9)(mod 13) (3) 27 (-9) (mod 13) (3 3 ) 9 (-9) (mod 13) (27) 9 (4) (mod 13) 1 9 (4) (mod 13) 4 (mod 13), so the remainder desired is 4. 4) Let a and b be two positive integer. Show that 3ab + 7ab 2 is even. 3ab + 7ab 2 = ab(3 + 7b)
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
If either a or b is even, the quantity above is automatically even since a and b are both divisors of the quantity. Thus, if the quantity is to be odd, then both a and b must be odd. But in this case, 7b would be an odd number, making 3+7b an even number. Thus, in all possible cases, at least one of the three factors a, b, (3+7b) is even, making 3ab + 7ab 2 even for all positive integers a and b.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern