A simple calculation will reveal that 72 has 12 factors including 1 and 72. How many natural numbers less than 100 have 12 such factors
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To find the natural numbers less than 100 that have 12 factors, we can first list down the prime factorization of 12.
The prime factorization of 12 is 2^2 * 3^1.
Now, we need to consider the factors in terms of their exponent values.
For a factor to have 12 factors, it can have:
- Exponent of 2 as 11 and exponent of 3 as 1, or
- Exponent of 2 as 2 and exponent of 3 as 2.
Now, let's find the natural numbers that satisfy these conditions.
1. Exponent of 2 as 11 and exponent of 3 as 1:
We need to find numbers of the form 2^11 * 3. But since we want the natural numbers to be less than 100, we need to find how many times 2^11 * 3 is less than 100.
2^11 * 3 = 2048 * 3 = 6144.
The number 6144 is the only natural number less than 100 that has 12 factors with the exponent of 2 as 11 and exponent of 3 as 1.
2. Exponent of 2 as 2 and exponent of 3 as 2:
We need to find numbers of the form 2^2 * 3^2. But since we want the natural numbers to be less than 100, we need to find how many times 2^2 * 3^2 is less than 100.
2^2 * 3^2 = 4 * 9 = 36.
The numbers 36, 72, and 90 are the natural numbers less than 100 that have 12 factors with the exponent of 2 as 2 and exponent of 3 as 2.
Therefore, there are a total of 4 natural numbers less than 100 that have 12 factors. These numbers are 36, 72, 90, and 6144.
To find the natural numbers less than 100 that have 12 factors, we need to find the numbers whose factorization results in exactly 12 factors.
Let's analyze the factors of 72 to understand the pattern. The prime factorization of 72 is 2^3 * 3^2. We can calculate the number of factors using the formula (a + 1) * (b + 1), where a and b are the exponents of the prime factors. In this case, (3 + 1) * (2 + 1) = 4 * 3 = 12.
The prime factorization of a number determines the number of factors it has. So, to find the numbers less than 100 with 12 factors, we can look for numbers whose prime factorization follows the form p^2 * q^3, where p and q are distinct prime numbers.
We can write down all the prime numbers less than 100:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Now, let's find pairs of distinct prime numbers (p, q) such that p^2 * q^3 is less than 100 and has 12 factors:
For p = 2:
2^2 * q^3 = 4 * q^3
In this case, the possible values of q can be found by taking the cube root of integers less than or equal to 100/4 = 25. Cube roots of 1, 2, 3, 4, 5 will give us distinct prime numbers. If we substitute the values of q, we will obtain numbers whose factorization results in 12 factors.
For p = 3:
3^2 * q^3 = 9 * q^3
In this case, the possible values of q can be found by taking the cube root of integers less than or equal to 100/9 = 11. Cube roots of 1, 2, 3 will give us distinct prime numbers. If we substitute the values of q, we will obtain numbers whose factorization results in 12 factors.
Following these steps for p = 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97, we can find numbers whose factorization results in 12 factors.
By listing the numbers we find, we can count the total number of natural numbers less than 100 that have 12 factors.