Tag Archives: factoring

Defactorization

One hobby I enjoy is counting things. As I play with things or count things, I discover new things. Today, I’m going over factoring and what I have found while playing with factors. I’ve been doing this for a long time. It’s a high time I explain my theory.

Counting the Factors

Defactorization is the process where you count out the factors and then use the factor count as the number’s answer. There are two ways to count out the factors. First way is the long method. The other way is the short method.

The Long Method:

The long method is where you try to plug in a number to see if it fits into the number. One way to try is by using the divisibility test. Here are the rules:

  1. All natural numbers are divisible by 1.
  2. A number is divisible by 2 if the number is an even number. All even numbers have a 2, 4, 6, 8, or 0 in the one’s digit.
  3. A number is divisible by 3 if the sum of the digits is divisible by 3. Most smaller numbers (like 3 or less digits) that fit this rule have a sum of 3, 6, 9, 12, 15, or 18 if using its digits.
  4. A number is divisible by 4 if the form of the last two digits is divisible by 4. To tell if it’s divisible by 4, it has to be an even number to be divisible by 4. The ten’s digit must be even if the one’s digit is 0, 4, or 8. The ten’s digit must be odd if the one’s digit is 2 or 6.
  5. A number is divisible by 5 if the number ends with a 5 or 0.
  6. A number is divisible by 6 if the sum of the digits is divisible by 3 AND is an even number.
  7. A number is divisible by 7 if the difference between all but the last digit and double the last digit is divisible by 7. Basically, you have to double the last digit and subtract it from the rest of the number. If the result is divisible by 7, then the number is divisible by 7.
  8. A number is divisible by 8 if the form of the last three digits is divisible by 8. In order to divide evenly by 8, it has to be divisible by 4. The hundred’s digit must be even if the last two digits are 00, 08, 16, 24, 32, 40, 48. 56. 64. 72, 80, 88, or 96. The hundred’s digit must be odd if the last two digits are 04, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, or 92.
  9. A number is divisible by 9 if the sum of the digits is divisible by 9.
  10. A number is divisible by 10 if the number ends with a 0.

I can’t go over higher numbers, but if you divide all of these numbers out of the big picture and the result isn’t a prime number, you still have a lot to go.

Let’s try the number 2,520 and see if it passes the divisibility test.

  1. 2,520 is divisible by 1 since it’s a natural number.
  2. 2,520 is divisible by 2 since it’s an even number.
  3. 2,520 is divisible by 3 since the sum of the digits is 9, which is divisible by 3.
  4. 2,520 is divisible by 4 since the form of the last two digits is 20, which is divisible by 4.
  5. 2,520 is divisible by 5 since it ends with a 5 or 0.
  6. 2,520 is divisible by 6 since it’s an even number divisible by 3.
  7. 2,520 is divisible by 7. Without the 0, the number is 252. The difference between the first two numbers and double the last number is 21, which is divisible by 7.
  8. 2,520 is divisible by 8 since the form of the last three digits is 520, which is divisible by 8.
  9. 2,520 is divisible by 9 since the sum of the digits is 9, which is obviously divisible by 9.
  10. 2,520 is divisible by 10 since it ends with a 0.

Fact: 2,520 is the least common multiple of the first 10 numbers.

Now let’s try 541. Remember, if the number isn’t divisible by 2, then it can’t be divisible by other even numbers, including 4, 6, and 8. If the number isn’t divisible by 3, it can’t be divisible by 6, 9 or other multiples of 3. So let’s start.

  1. 541 is divisible by 1 since it’s a natural number.
  2. 541 isn’t divisible by 2 since it’s an odd number.
  3. 541 isn’t divisible by 3 since the sum of the digits is 10, which is not divisible by 3.
  4. 541 isn’t divisible by 4 since it isn’t divisible by 2.
  5. 541 isn’t divisible by 5 since it ends with a 1, which is neither 5 or 0.
  6. 541 isn’t divisible by 6 since it fails both the 2 and 3 divisibility tests.
  7. 541 isn’t divisible by 7 since it the difference between the first two numbers and double the last digit is 52, which isn’t divisible by 7.
  8. 541 isn’t divisible by 8 since it isn’t divisible by 4.
  9. 541 isn’t divisible by 9 since the sum of the digits is 10, which is not divisible by 9.
  10. 541 isn’t divisible by 10 since it doesn’t end with a 0.

So 541 fails the divisibility test of all but one of them. Try plugging in the other prime numbers. It fails on them too. In fact, the closest number after plugging in the other numbers to the midpoint, 23, is not a factor. Therefore, 541 is a prime number.

Our number to factor is 24. To factor out a number, you need to see if it passes the divisibility test. If if passes for that number, then the number is one of the factors. Then you divide the number by that factor, and you put in both the divisor and the quotient in the factor list.

  • 24 is divisible by 1. A rule of thumb is that all numbers have one and itself as two of the factors. So 1 and 24 are factors.
  • 24 is divisible by 2 since it’s an even number. 24 divided by 2 is 12, making 2 and 12 factors.
  • 24 is divisible by 3 since the sum of the digits is 6, which is divisible by 3. 24 divided by 3 is 8, making 3 and 8 factors.
  • 24 is divisible by 4 since the ten’s digit is even while the number ends with a 4. 24 divided by 4 is 6, making both 4 and 6 factors.
  • 24 is not divisible by 5 since the number ends with a 4, which is neither 5 or 0. Therefore, 5 is not a factor.

When we keep factoring until we get to the point where both numbers are at their closest, then we can do no more. So our factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

Now let’s count how many factors there are. All I see are 8 factors. So the factor count of 24 is 8.

The Short Method:

The short method is better off for larger numbers since it takes longer to use the long method. The short method is tree factoring. To do this, you divide the lowest divisible number from the big number to branch off. Once one divisor was stripped off from the other, we get two branches, which are the divisor and the quotient. For every iteration you make, the first branch goes down as the same number as the other number breaks down. The whole process continues until all what remains are prime numbers.

Once again, we will be using the number 24. Here is a diagram of the process:

Tree Factoring

To word it out, 24 is our main number. To branch off from the number, I once again use the divisibility test. We can’t use 1 since that doesn’t really do anything. According to the divisibility test, 24 is divisible by 2, so we can “branch off” 2 from the number. 24 divided by 2 gets us 12, so our numbers are 2 and 12. We already taken care of 2, so we should go to 12 now. Once again, we have another number divisible by 2, so we take 2 down again, as we get 2 and 6. The 2 we broke off earlier can move down as 2 again as we got a new branch. 6 is divisible by 2, so we “branch off” one more time, and our results are 2 and 3. Since all we have left are prime numbers, we count how many of each prime number we see. There are three 2’s, but one 3. For every repeat of a number, we give it a superscript that we call “exponents” as the count is the exponent’s number. If there are one of each, the exponent is 1. If there are two of it, the exponent is 2. If there are three of the same number, the exponent is three. Our prime factorization is 2^3*3.

How can we count factors by using prime factorization? To do this, we strip off the exponents from the number. Notice that a number without an exponent really has an exponent. The only reason why it doesn’t have an exponent is because the exponent is 1. 2^3*3 is the same as 2^3*3^1. The exponents in this prime factorization are 3 and 1. The next step is to add 1 to each exponent, which results in 4 and 2. Then we multiply what remains. Our result is 8. Anything familiar? 8 was the factor count of the long method, and we got the same from the short method.

So to count how many factors by using the short method, here is the following:

  1. Put the number into prime factorization. You can do this through tree factoring like in the diagram.
  2. Strip the exponents from the bases. We only want to focus on the exponents. Remember that a number without an exponent has an exponent of 1.
  3. Add 1 to every exponent.
  4. Multiply what remains.

And here we are, the factor count. For every new prime number introduced, we have a new number to multiply to the number. Here are the most common re-occurrences:

  • A simple prime number with an exponent of 1 has 2 factors.
  • A prime number with an exponent of 2 has 3 factors.
  • A prime number with an exponent of 3 has 4 factors.
  • A prime number with an exponent of 4 has 5 factors.
  • Two prime numbers with both exponents being 1 has 4 factors.
  • When one has an exponent of 2, but the other has 1, there are 6 factors.
  • When one has an exponent of 3, but the other has 1, there are 8 factors.
  • When both have an exponent of 2, there are 9 factors.
  • When one has an exponent of 3, but the other is 2, there are 12 factors.
  • When both numbers have an exponent of 3, there are 16 factors.
  • When there are three prime numbers with an exponent of 1, there are 8 factors.
  • When one prime number has an exponent of 2, but the other two have 1, there are 12 factors.
  • When one prime number has an exponent of 3, but the other two have 1, there are 16 factors.
  • When two prime numbers have an exponent of 2, but the other has 1, there are 18 factors.
  • When all three prime numbers have an exponent of 2, there are 27 factors.
  • When there are four prime factors with an exponent of 1, there are 16 factors.

Defactorization

Defactorization is when we use the factor count as the number’s secondary number. The primary number is the case in point. The secondary number is the factor count. And then, we continue defactorizing until we get to the point when we have no new results. This number is 2. Some numbers have more iterations than others. The level of defactorization is the number of iterations it takes to get to 2. Here is a chart of the first 100 numbers color-coded by their levels of defactorization:

Defactorization Chart

I will now explain what the colors mean.

  • Black – there is only one black number on this chart, which is 1. Since 1 is neither prime nor composite, there is no way we can get to 2. The level of defactorization is 0.
  • Gray – prime numbers. A number is considered a prime number if the factors are 1 and itself. Since it takes only one iteration to get to 2 (including 2 itself), the level of defactorization is 1.
  • Red – numbers where the secondary number (or factor count) is a prime number. Notice that all prime numbers except for 2 are odd-numbered. All numbers with an odd-number of factors are perfect squares, but not all perfect squares have a prime number of factors. The only ones with a possibility are the ones with only one prime factor. Because of this, these numbers are the rarest type of numbers. Since the secondary number is a prime number other than 2, the level of defactorization of these numbers is 2.
  • Yellow – numbers where the tertiary number is a prime number. As we continue piling up, we start getting numbers where the secondary number is a composite number. At this point, the prime number (odd-numbered) in the hierarchy is the tertiary number. These numbers are one of the most common numbers. On the chart seen above, there are 34 of these numbers. In order to be this kind of number, the secondary number must have a prime number as its factor count (besides 2). They can have 4 factors, 9 factors, 16 factors, or 25 factors. However, the majority of these numbers are 4-factored numbers. A number has 4 factors either because they are cubes of prime numbers (like 8) or have no composite factors except for itself while there are two prime factors. Next in line are the 16-factored numbers. They are mostly numbers with four prime factors with an exponent of 1 or have three prime factors when one has an exponent of three as the other to have exponents of one. The level of defactorization of these numbers is 3.
  • Green – numbers where the quaternary number is a prime number. As we continue piling up, this is when we begin having quaternary numbers in the hierarchy (besides 2). These numbers take 4 iterations to get to the number 2. There are no numbers of this type from 10 and under, and only four of these numbers under 25. As we start diving into larger numbers, these get increasingly common. We start seeing them when we have 6-factored numbers. Then we see 8-factored numbers, and 10-factored numbers, 14-factored numbers, 15-factored numbers, and so on. But the most common numbers of this type have 6 or 8 factors. A number has 6 factors when there are two prime factors, but one has a maximum exponent of 2 while the other is 1. A number has 8 factors when there are three prime factors with an exponent of 1, two prime factors when one has an exponent of 3 (like 24), or when there is only one prime factor, but has an exponent of 7. The level of defactorization of these numbers is 4.
  • Cyan – numbers where the quaternary number is a composite number (but the quaternary number’s factor count is prime). These are the golden numbers of defactorization. They take 5 iterations (that’s right, 5 iterations) to get to 2. As numbers keep getting bigger, they have more factors. Eventually, we may get to a factor count where the level of defactorization of that number is 4. The smallest number of this kind is 60. Most numbers of this type have 12 factors, either because there are three prime factors as only one has a maximum exponent of 2 while the others are 1, or because there are two prime factors where one has an exponent of 3 while the other’s is 2. The first few numbers of this type have 12 factors. The first number of this type to not have 12 factors is 180. All numbers of this type under 300 are even numbers, and all numbers of this type under 100 are divisible by 3. Notice that there are only five two-digit numbers of this type, which are 60, 72, 84, 90, and 96. They are rare at first, but they begin getting more common when there are three digits or more. The first odd number of this type is 315. The level of defactorization of these numbers is 5.

Coding:

Aside to the color-coding of these numbers, there is another type of coding. The level of defactorization can be simply expressed by having a number next to the letter L. For instance, L3 numbers are numbers where the level of defactorization is 3, also the numbers where the prime number is the tertiary number. Here are the list of levels and their stats:

  • L0: Level of defactorization is 0. 1 is the only number of this level.
  • L1: Level of defactorization is 1. The primary number is a prime number. The first two numbers of this level are 2 and 3.
  • L2: Level of defactorization is 2. The secondary number is a prime number. The first number of this level is 4.
  • L3: Level of defactorization is 3. The tertiary number is a prime number. The first number of this level is 6.
  • L4: Level of defactorization is 4. The quaternary number is a prime number. The first number of this level is 12.
  • L5: Level of defactorization is 5. The quaternary number is an L2 composite number. The first number of this level is 60.

L5 isn’t the highest level, but it is the highest level from numbers 1 to 1,000. There’s only one number under 10,000 to be at an even higher level.

Proof of Defactorization:

If you want to know what I mean by iterations, I mean by factor counting each step until I can do this no more. The reason why 2 is the stopping point is because if I continue doing this, I’m only going to get the same result. It’s all about getting unique results. I’ll begin with this process.

  • L1 numbers: 3 is a good example. The factors are 1 and 3. The factor count is 2. That’s one iteration.
  • L2 numbers: 4 has 3 factors. The factor count is 3. The second factor count is 2. That’s two iterations.
  • L3 numbers: 36 has 9 factors. The factor count is 9. The second factor count is 3. The third factor count is 2. That’s three iterations.
  • L4 numbers: 48 has 10 factors. The factor count is 10. The second factor count is 4. The third factor count is 3. The fourth factor count is 2. That’s four iterations.
  • L5 numbers: 72 has 12 factors. The factor count is 12. The second factor count is 6. The third factor count is 4. The fourth factor count is 3. The fifth factor count is 2. That’s five iterations.

I was able to do this with the first 100 numbers and first 1,000 numbers, but not the first 10,000 numbers. As the factor count is a higher level, the primary number adds one more to its level.

And that’s it about defactorization. You now know a new theory.

Advertisements