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COT 3100 Homework # 5 Solutions
Fall 2000
1)
Use Euclid’s Algorithm to find the greatest common divisor(GCD) of 245 and 455.
455 = 1x245 + 210
245 = 1x210 + 35
210 = 6x35
Thus, the gcd(455, 245) = 35.
2)
Show that if 11  (6x + 5y), then there are no integer solutions to the equation
9x + 13y = 20000
Since 11  (6x + 5y), we must have that 6x+5y = 11A, for some integer A.
9x + 13y = 33(x+y) – 4(6x+5y)
(You can verify this by multiplying out the right
hand side.)
= 11*3(x+y) – 4*11A
= 11(3(x+y) – 4A)
Thus, we can conclude that 11  (9x + 13y).
However, we can do long division to verify that 11  20000 is false. Thus, since one
side of the equation is divisible by 11 and the other is not, the equation can never be
true, thus having no integer solutions in x and y.
3)
Prove that if 3  2x + 7y, then 60  (20x – 80y).
Since 3  2x + 7y, 2x+7y = 3A for some integer A.
We must prove that 20x – 80y = 60c, for some integer c.
20x – 80y = 20(x – 4y)
= 20(3(x+y) – (x – 4y)), once again you can verify this by multiplying out
what is inside the parentheses.
= 20(3(x+y) – 3A)
=(20)(3)(x+y – A)
=60(x + y – A)
Thus we can conclude that 60  (20x – 80y).
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View Full Document 4)
Using mod rules, find the remainder when you divide 5
192
by 7.
5
192
≡
(5
2
)
96
(mod 7)
≡
(25)
96
(mod 7)
≡
(4)
96
(mod 7)
≡
(4
2
)
48
(mod 7)
≡
(16)
48
(mod 7)
≡
(2)
48
(mod 7)
≡
(2
3
)
16
(mod 7)
≡
(8)
16
(mod 7)
≡
(1)
16
(mod 7)
≡
1 (mod 7)
Thus, the desired remainder is 1.
5)
Let a and b be two positive integer. If a is odd, show that 9ab + 5b
2
is even.
First we will show that the product of two odd numbers is odd.
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This document was uploaded on 06/09/2011.
 Fall '09

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