Math Problem Statement
How many positive integer factors of 2020 have more than 3 factors?
Solution
To solve this problem, we need to find the number of positive integer factors of that themselves have more than 3 factors. Here's how we can approach the solution step by step.
Step 1: Prime Factorization of 2020
We start by finding the prime factorization of .
Step 2: Total Number of Factors of 2020
To calculate the total number of factors of a number, we use its prime factorization. If a number has the prime factorization:
Then, the total number of factors of is given by:
For , the total number of factors is:
So, 2020 has 12 factors in total.
Step 3: Listing All Factors of 2020
The factors of are:
Step 4: Identifying Factors with More Than 3 Factors
Now we need to identify which of these factors themselves have more than 3 divisors.
- : has 1 factor (itself).
- : has 2 factors ().
- : has 3 factors ().
- : has 2 factors ().
- : has 4 factors () → More than 3 factors.
- : has 6 factors () → More than 3 factors.
- : has 2 factors ().
- : has 4 factors () → More than 3 factors.
- : has 6 factors () → More than 3 factors.
- : has 4 factors () → More than 3 factors.
- : has 8 factors () → More than 3 factors.
- : has 12 factors → More than 3 factors.
Step 5: Conclusion
The factors of that have more than 3 factors are:
There are 7 factors of that have more than 3 factors.
Would you like further clarification or a more detailed breakdown of any step?
Related Questions:
- How do you find the prime factorization of large numbers?
- What is the general method for determining the number of divisors of a number?
- How can you determine if a number has exactly 3 factors?
- What are the properties of numbers with a small number of divisors?
- How can we use the prime factorization to solve problems about divisibility?
Tip:
To quickly determine if a factor has more than 3 divisors, use its prime factorization to count its divisors directly.
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Math Problem Analysis
Mathematical Concepts
Number Theory
Prime Factorization
Divisors
Formulas
Total number of divisors of N = (e1 + 1)(e2 + 1)...(ek + 1), where N = p1^e1 × p2^e2 × ... × pk^ek
Theorems
Divisor Counting Theorem
Suitable Grade Level
Grades 8-10