CSCE 222 CSCE222 Pham-Gia-hw7 – Texas A&M

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CSCE 222 CSCE222 Pham-Gia-hw7 – Texas A&M

Resources.Discrete Mathematics and Its Applications (textbook)

Problem 1.

(16 points) Section 6.1, Exercise 24, page 396

Solution.

(a)The integers between 1000 and 9999 are: 9999 – 1000 + 1 = 9000The number of integers divisible by 9 among these are:b9000/9c= 1000.

(b)The number of integers are even among 1000 and 9999 are: 9000/2 = 4500.(half even, half odd)

(c)Since the domain is between 1000 to 9999, all numbers in domain have 4 digits.The number of integers have distinct digits are:9*9*8*7 = 4536.(The first digit has 9 choice from 1 to 9 because it must be different from 0.The second digit has 10-1 choice.The third digit has 10-1-1 choice.The last digit has 10-1-1-1 choice.Note that we eleminate the cases of chosen digit.)

(d)The number of integers not divisible by 3 between 1000 and 9999 are:9000 –b9000/3c= 6000.

(e)integers divisible by 5 or 7:[9999/5+9999/7–9999/(5*7)]–[999/5+999/7–999/(5*7)] = 3142–313 = 2829

(f)integers not divisible by either 5 or 7:9000 – 2829 = 6171

(g)integers are divisible by 5 but not by 7:[9999/5–9999/(5*7)]–[999/5–999/(5*7)] = (1999–285)–(199–28) =1714–171 = 1543

(h)

CSCE 222 CSCE222 Pham-Gia-hw7 – Texas A&M

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