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COT 3100 Fall 2002
Homework #5 Solutions
1)
a) Use Euclid's Algorithm to find the greatest common divisor of 962 and 629.
962 = 1*629 + 333
629 = 1*333 + 296
333 = 1*296 + 37
296 = 8*37
GCD(962, 629) = 37
b) Use the Extended Euclidean Algorithm to find integers x and y such that
962x+629y = gcd(962, 629).
333  296 = 37
629  333 = 296
962  629 = 333
⇒
(962  629)  (629  333) = 37
962 + 333  2*629 = 37
962 + (962  629)  2*629 = 37
2*962  3*629 = 37
x = 2, y=3 is a solution.
2)
Let x and y be integers. If 13  (2x+5y), prove there are no integer solutions to 3x+y =
2003.
Since 13  (2x+5y), let 2x+5y = 13a, where a
∈
Z.
3x + y = 13(x+2y)  5(2x+5y)
= 13(x+2y)  5(13a)
= 13(x + 2y  5a)
Thus, we can conclude, since x,y and a are integers, that 13  (3x + y).
But, we see that 13  2003 is false. Thus, it is impossible for any integer
solutions to exist to the given equation.
3)
Given the prime factorization of n is p
a
q
b
, where p and q are prime and a and b are
positive integers, determine the number of factors of n. (Hint: Note that each factor of n
MUST only contain p and q in its prime factorization, and can not contain more than a p's
or b q's.) As an example, find the total number of factors of 108 and list each of these.
This is a counting question.
..)
Each factor of n must be of the form p
x
q
y
, where x and y are nonnegative
integers with x
≤
a, and y
≤
b. Imagine making a table of all factors. Each
row could be labelled by a value of x, and each column by a value of y.
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View Full Document There are a+1 possible values of x (0, 1, 2, .
.., a) and b+1 possible values
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This document was uploaded on 07/14/2011.
 Spring '09

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