Answer to Question #251540 in Discrete Mathematics for Ben

Question

M is a subset of the set of natural numbers. 10 elements of the set are prime numbers, and the rest are divisible by either 2, or 3, or 5. Determine the cardinality of the set if it contains: 70 numbers that are divisible by 2; 60 numbers divisible by 3; 80 divisible by 5; 98 multiples by 2 or by 3; 95 multiples by 2 or by 5; 102 multiples by 3 or by 5; 20 numbers divisible by 30

Answer 1

Solution:

(i) 10 elements of the set are prime numbers. 70 numbers that are divisible by 2.

The set will have (10+70) numbers.

So, the cardinality = 80

(ii) 10 elements of the set are prime numbers. 60 numbers divisible by 3.

The set will have (10+60) numbers.

So, the cardinality = 70

(iii) 10 elements of the set are prime numbers. 80 divisible by 5.

The set will have (10+80) numbers.

So, the cardinality = 90

(iv) 10 elements of the set are prime numbers. 98 multiples by 2 or by 3.

The set will have (10+98) numbers.

So, the cardinality = 108

(v) 10 elements of the set are prime numbers. 95 multiples by 2 or by 5.

The set will have (10+95) numbers.

So, the cardinality = 105

(vi) 10 elements of the set are prime numbers. 102 multiples by 3 or by 5.

The set will have (10+102) numbers.

So, the cardinality = 112

(vii) 10 elements of the set are prime numbers. 20 numbers divisible by 30.

The set will have (10+20) numbers.

So, the cardinality = 30