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To solve the problem where \( N^3 \) has 70 factors, and we need to find the minimum value of \( N \), let’s proceed step by step.

### Step 1: Prime Factorization of \( N \)

Assume \( N \) can be expressed as a product of prime factors:
\[
N = p_1^{e_1} \cdot p_2^{e_2} \cdot \dots \cdot p_k^{e_k}
\]
where \( p_1, p_2, \dots, p_k \) are distinct primes and \( e_1, e_2, \dots, e_k \) are their respective powers.

### Step 2: Factors of \( N^3 \)

The number of factors of any number is determined by its prime factorization. If \( N = p_1^{e_1} \cdot p_2^{e_2} \cdot \dots \cdot p_k^{e_k} \), then:
\[
N^3 = (p_1^{e_1})^3 \cdot (p_2^{e_2})^3 \cdot \dots \cdot (p_k^{e_k})^3 = p_1^{3e_1} \cdot p_2^{3e_2} \cdot \dots \cdot p_k^{3e_k}
\]

The number of divisors of \( N^3 \) is given by:
\[
(3e_1 + 1)(3e_2 + 1) \cdot \dots \cdot (3e_k + 1)
\]
We know from the problem statement that the number of factors is 70. Thus, we need to solve:
\[
(3e_1 + 1)(3e_2 + 1) \cdot \dots \cdot (3e_k + 1) = 70
\]

### Step 3: Factorization of 70

The prime factorization of 70 is:
\[
70 = 2 \cdot 5 \cdot 7
\]

This gives us potential values for the factors \( (3e_1 + 1), (3e_2 + 1), \dots \).

### Step 4: Assigning Values

Let's try to assign these values to \( 3e_1 + 1 \), \( 3e_2 + 1 \), etc.

1. Set \( 3e_1 + 1 = 7 \), which gives \( e_1 = 2 \).
2. Set \( 3e_2 + 1 = 5 \), which gives \( e_2 = 1 \).
3. Set \( 3e_3 + 1 = 2 \), which gives \( e_3 = 0 \).

Thus, the prime factorization of \( N \) is:
\[
N = p_1^2 \cdot p_2^1
\]
where \( p_1 \) and \( p_2 \) are distinct primes.

### Step 5: Minimum Value of \( N \)

To minimize \( N \), we choose the smallest primes for \( p_1 \) and \( p_2 \). Let \( p_1 = 2 \) and \( p_2 = 3 \). Therefore:
\[
N = 2^2 \cdot 3 = 4 \cdot 3 = 12
\]

### Final Answer:
The minimum value of \( N \) is **12**.

Would you like more details or have any further questions?

### Related Questions:
1. How do you calculate the number of divisors of a number?
2. What is the significance of prime factorization in divisor problems?
3. Can you explain why we used the factorization of 70 in this problem?
4. How would the solution change if the number of divisors was different?
5. What are some similar problems involving the number of divisors?

### Tip:
When solving divisor problems involving powers, remember that the formula for divisors depends on the exponents in the prime factorization of the number.

If N³ has 70 factors, then find the minimum value of N.

Learn how to determine the minimum value of N when N³ has 70 factors, using prime factorization and divisor counting techniques. Suitable for students in Grades 9-12.

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