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 20202020 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 20202020.

2020=22×5×1012020 = 2^2 \times 5 \times 101

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 NN has the prime factorization:

N=p1e1×p2e2××pkekN = p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}

Then, the total number of factors of NN is given by:

(e1+1)(e2+1)(ek+1)(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)

For 2020=22×51×10112020 = 2^2 \times 5^1 \times 101^1, the total number of factors is:

(2+1)(1+1)(1+1)=3×2×2=12(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12

So, 2020 has 12 factors in total.

Step 3: Listing All Factors of 2020

The factors of 20202020 are:

1,2,4,5,10,20,101,202,404,505,1010,20201, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020

Step 4: Identifying Factors with More Than 3 Factors

Now we need to identify which of these factors themselves have more than 3 divisors.

  • 11: has 1 factor (itself).
  • 22: has 2 factors (1,21, 2).
  • 44: has 3 factors (1,2,41, 2, 4).
  • 55: has 2 factors (1,51, 5).
  • 1010: has 4 factors (1,2,5,101, 2, 5, 10) → More than 3 factors.
  • 2020: has 6 factors (1,2,4,5,10,201, 2, 4, 5, 10, 20) → More than 3 factors.
  • 101101: has 2 factors (1,1011, 101).
  • 202202: has 4 factors (1,2,101,2021, 2, 101, 202) → More than 3 factors.
  • 404404: has 6 factors (1,2,4,101,202,4041, 2, 4, 101, 202, 404) → More than 3 factors.
  • 505505: has 4 factors (1,5,101,5051, 5, 101, 505) → More than 3 factors.
  • 10101010: has 8 factors (1,2,5,10,101,202,505,10101, 2, 5, 10, 101, 202, 505, 1010) → More than 3 factors.
  • 20202020: has 12 factors → More than 3 factors.

Step 5: Conclusion

The factors of 20202020 that have more than 3 factors are:

10,20,202,404,505,1010,202010, 20, 202, 404, 505, 1010, 2020

There are 7 factors of 20202020 that have more than 3 factors.

Would you like further clarification or a more detailed breakdown of any step?

Related Questions:

  1. How do you find the prime factorization of large numbers?
  2. What is the general method for determining the number of divisors of a number?
  3. How can you determine if a number has exactly 3 factors?
  4. What are the properties of numbers with a small number of divisors?
  5. 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