Question

we will say that a positive integer is cube-free if there is no integer such that . let be the set of all cube-free integers. we will say that a positive integer is modest if every prime factor of is less than 10 (i.e., 2, 3, 5 or 7). let be the set of all modest integers. what is (i.e., how many cube-free modest integers are there)? (hint: remember that every positive integer greater than 1 has a unique representation as a product of primes.)

Answer 1

There are 256 cube-free modest integers.

Step-by-step explanation:

We can find the number of cube-free modest integers by analyzing the intersection of their individual properties:

Modest Numbers: An integer is modest if all its prime factors are 2, 3, 5, or 7. This means we have 4 choices for each prime factor in the factorization.

Cube-free Numbers: An integer is cube-free if its prime factors have no exponent greater than 2. There are 3 possibilities for each prime factor exponent: 0, 1, or 2.

Intersection: Combining both conditions, we can choose any combination of these possibilities for each distinct prime factor in the factorization. Since each integer has a unique prime factorization, this essentially amounts to multiplying the number of choices for each factor.

Counting Combinations: Therefore, the number of cube-free modest integers is 4 raised to the power of the number of distinct prime factors that are 2, 3, 5, or 7.

Number of Prime Factors: Since there are 4 such primes (2, 3, 5, and 7), the exponent will be 4, resulting in:

|C ∩ M| = 4^4 = 256

Therefore, there are 256 cube-free modest integers.

Answer 2

The answer to
`\(|C \cap M|\)` (the number of cube-free modest integers) is
`\(4^4 = 256\)`.

To find the number of cube-free modest integers, let's consider the properties of cube-free and modest integers.

A positive integer n is modest if all its prime factors are 2, 3, 5, or 7. Additionally, n is cube-free if it's not divisible by any cube of a prime number.

Use these properties to find the intersection of C and M.

Firstly, consider the prime factorization of any positive integer greater than 1:

`$$n = p_1^(a_1) \cdot p_2^(a_2) \cdot \dots \cdot p_k^(a_k)$$`

Where
`\(p_1, p_2, \dots, p_k\)` are distinct primes, and
`\(a_1, a_2, \dots, a_k\)` are positive integers.

For an integer to be cube-free, it means that none of its prime factors have a cube in their power. In other words, each
`\(a_i\)` for
`\(p_i\)` (where
`\(p_i\)` is a prime factor of n should be less than 3.

For an integer to be modest, all of its prime factors should be 2, 3, 5, or 7.

So, let's calculate the number of integers that are both cube-free and modest:

• There are 3 possibilities for each prime factor: 2, 3, 5, or 7.
• Each integer can have at most one occurrence of each of these primes in its factorization.

Therefore, the number of cube-free modest integers will be the number of ways we can select primes 2, 3, 5, or 7 for each distinct prime power in the factorization of
`\(n\)`. This can be calculated as
`\(4^k\)` where
`\(k\)` is the number of distinct primes in the factorization of
`\(n\)`.

So, the number of cube-free modest integers will be
`\(4^k\)` where
`\(k\)` is the number of distinct primes that are 2, 3, 5, or 7.

Thus, The answer is
`\(4^4 = 256\)`.

The complete question is here:

We will say that a positive integer
`$n$` is cube-free if there is no integer
`$k > 1$` such that
`$k^3 \mid n$`. Let C be the set of all cube-free integers. We will say that a positive integer n is modest if every prime factor of n is less than 10 (i.e., 2, 3,5 or 7 ). Let M be the set of all modest integers.

What is
`$|C \cap M|$` (i.e., how many cube-free modest integers are there)?

(Hint: remember that every positive integer greater than 1 has a unique representation as a product of primes.)