Something I’ve wanted to work on for a LONG time. Any thoughts would be greatly appreciated, either on my brain dump here or on another system that has worked better than my idea.
I’m thinking about using my PAO system as mnemonics for prime factor multiplication. Each character has a certain associated object, but also all the objects that would be factors of this character’s number.
For example, 42, character “Rainy”, has a horn (42), but also a ball of snow (02), a ham (03), and a sack (07).
Also, 48, character “Riff”, has a roof (48), but also four balls of snow (02) and a ham (03).
So whenever I glance at a number, I can quickly recall what all the prime factors are, and by extension what all the factors are because some of those prime factors can be multiplied together.
Fun application: whenever I see the number 35 and the number 14 together, I can easily multiply them by reducing them to prime factors.
35 = 75
14 = 72
(7 * 7) * (2 * 5) = 490 easily.
I’m fairly new to speed math, but I thought I’d go ahead and throw this out there. Feel free to steer me in a better direction if there’s an easier system for prime-factor-based mnemonics.
It is great to know the factors of numbers.
It can indeed help with multiplication.
Also, instead of memorizing, it is great practice to actually calculate them. Think about it, how would you factor 78 into 6 and 13?
Do you immediately see that 78 is divisible by 3?
Randomly factor all two digit numbers, then try your hand at three digit numbers. This helps in getting the nessesary number sense.
It is not widely known that elaborate algorithms exists to factor numbers. With one of these for example 1,118,117 can be factored into 7, 11, 13 and 1117. You will be surprised to see how easy this is btw.
Let me know if you like to hear more about this, please.
If you want to know the factors of a number, one option is to check all primes below the number and see if it is a factor.
You can do this up until the square root of the number and no further.
Not everybody understands this.
If I want to see if 97 is prime or that it has factors, I can check all integers from 1-10, 10 being the next integer after the square root of 97.
The reason for this is that if there was an factor bigger than 10 the other factor needs to be smaller than 10.
2 numbers bigger than 10 multiplied is certainly bigger than 97.
If numbers are big, then we need tools to determine factors quickly. Fortunately they exists.
For 2, 3 and 5 it is easy to see if a number can be evenly divided by this.
For 7, 11 and 13 there is a great way of certifying that one of these is a factor or not.
This is based on the fact that 7 X 11 X 13 = 1001. And this is a great number to work with.
If I multiply let’s say 123 by 1001, my answer is 123,123.
If I add let’s say 13 to this, my answer is 123,136.
If I subtract 123 X 1001 from 123,136 then the result is 13.
Look at the numbers. A quick way to do this is subtracting the first 3 digits (123) from digit 4-6 or 136-123 = 13.
If I start with 123,136 and want to check for 7, 11, or 13 I can subtract a number of 1001s from this number.
Because assuming 123,136 is divisible 7, 11 or 13, then 123,136 - (123 X 1001) is also.
If this is unclear let it sink in for a minute. This is essentially what I am saying:
If X is divisible by 7, then x-7 is also and x - 2X7, and x-3X7, etc.
If 35 is divisible by 7, then 28 is also, and 21, and14, and 7.
Procedure:
From 123,136. Split into 2 groups of 3 digits: 123 and 136. 136-123 = 13.
If 123,136 is divisible by 7, 1, or 13, then 13 is also.
13 is divisible by 13 only, so now we know that 123,136 is divisible by 13. And not by 7 and not by 11.
Another example:
345,455.
Split into 2 groups of 3 digits: 345 and 455. 455-345 = 110.
110 = 11 X 10.
So 345,455 is divisible by 11.
What this does is turning a 6 digit problem into a 3 digit problem.
Knowing the above I hope you see that 1,118,117 can be split into 1,117,000 and 1,117.
1,118,117 = 1,117 X 1,001.
If you don’t then just follow the procedure:
The first 3 digits are 111. 111 X 1001 = 111,111 In order to get rid of the first 3 digits of 1,118,117 we can subtract 111,111 ten times. So we subtract 1,111,110 from 1,118,117 = 7,007. Now again we can subtract 7 X 1,001 = 7,007.
Remainder is zero.
Since 1,118,117 divides evenly into 1,001 it shows that 7 as well as 11 as well as 13 are factors.
In other words, if 1001 is a factor of 1,118,117 then 7 and 11 and 13 are as well.
