Discrete Mathematics Assignment: Detailed Solutions and Calculations

This document presents a comprehensive solution to a discrete mathematics assignment, addressing three key questions. The first question involves calculating binomial coefficients, greatest common divisors (GCD), and least common multiples (LCM). The second question focuses on solving a linear equation graphically and identifying a solution with natural number coordinates. The third question delves into number theory, including the Euclidean algorithm to find the GCD of two large numbers, solving Diophantine equations, and finding general solutions based on given parameters. The solution includes detailed calculations, explanations, and references to relevant mathematical concepts and techniques. The assignment covers a range of topics within discrete mathematics, providing a valuable resource for students studying this subject.

Running head: DISCRETE MATHEMATICS 0
DISCRETE MATHEMATICS
Name of Student
Institution Affiliation

DISCRETE MATHEMATICS 2
Question one
We can write as:
=
=
We can write binomial coefficient (Marie, 2012).
as
=
Greatest common divisor(HCF) is:
= 28
Least common multiple(LCM) is:
Multiplication of numbers = HCF * LCM
Question two
The model question is shown as below<
Through using the graph to obtain the x, y-intercepts and the graph
is as below

DISCRETE MATHEMATICS 3
And the only coordinate with both as natural number in the line
graph is at ( 27,13). (Sandor, 2014).
x= 27 while y= 13
31.70× 27+116.90× 13 = 855.90 +1519.7
= $ 2375.6
Therefore
The number of items that cost $ 31.70 each = 27
The number of items which cost $116.90 each = 13
Question three
(a) a = 2637110736
b = 2961110016
2961110016 = 2637110736 + 323999280
2637110736 = 8*323999280 + 45116496
323999280 = 7*45116496 + 8183808
45116496 = 5*8183808 + 4197456
8183808 = 4197456 + 3986352
4197456 = 3986352 + 211104
3986352 = 18*211104 + 186480

DISCRETE MATHEMATICS 4
211104 = 186480 + 24624
186480 = 7* 24624 + 14112
24624 = 14112 + 10512
14112 = 10512 + 3600
10512 = 2*3600 + 3312
3600 = 3312 + 288
3312 = 11*288 + 144
288 = 2*144 + 0
The gcd is g = 144.
(b) Because the techniques used in lectures are not given, it is not possible
to know which method to use. Nevertheless answer is below. This will be
verified in (d) (Sandor, 2014).
x = 9860829 and y = -8781875.
(c) 6768 = 144 * 47.
Thus 6768 is a multiple of g.
527760 = 144 * 3665.
We have from (b), x = 9860829 and y = -8781875.
s = 3665 * 9860829 and t = 3665 * -8781875
=> s = 36139938285 and t = -32185571875.
General solution according to Diophantine is
s = 36139938285 + k * 2961110016 / 144 and t = -32185571875 - k *
2637110736
=> s = 36139938285 - 20563264k and t = -32185571875 - 18313269k.
(d) We have from (c)
s = 36139938285 and t = -32185571875.

DISCRETE MATHEMATICS 5
=> as + bt = 2637110736 * 36139938285 + 2961110016 * -32185571875
= 95305019249750927760 - 95305019249750400000
= 527760.
Some of the problems of using this are that it requires a higher level of
accuracy which can be limited by the use of a scientific calculator.

DISCRETE MATHEMATICS 6
Reference
marie, D. M. (2012). Figurate Numbers. London: World Scientific.
Sandor, J. (2014). Handbook of Number Theory II, Volume 2. Chicago:
Springer Science & Business Media.