Casting Out Nines - Numberphile
video description
The guest said if you wanted to get your -numerological number- or something like that you add all the letters in you name as numbers and keep adding those digits until you get a single digit. The host asked if it mattered if you did it separately for first name and last name or if you had to do all the letters at the same time and the numerologist said it didn't matter.
I started playing around with that and noticed that no mater how you do a sum, this reduction always gives you the same digit.
Say, if you wanted to do 15+245+33 you can reduce them by doing all the digits at the same time:
1+5+2+4+5+3+3 -> 23 -> 5
or by reducing each number separately:
15->6, 245->11->2, 33->6 and then 6+2+6 = 14 -> 5
Either way, that reduction is the same as the reduction of the initial sum, 293 -> 14 -> 5.
I guess I learned something useful from a pseudoscience: D
Date: 2022-04-08
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Comments and reviews: 9
La
Just remember not to cast out 9s if it would leave you with no digits in the number (45, 634572234, 999, etc, since no number other than 0 has a digital root of 0.
However, this doesn't make it harder to figure out the digital root of, for instance, 634572234. In fact, it makes it easier. If you have a number where casting out 9s would eliminate every digit, that number will always have a digital root of 9.
This is because any number evenly divisible by 9 has a digital root of 9 (in fact, according to Martin Gardner, the fastest known way to manually test if a large number is divisible by 9 is to check its digital root.
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Just remember not to cast out 9s if it would leave you with no digits in the number (45, 634572234, 999, etc, since no number other than 0 has a digital root of 0.
However, this doesn't make it harder to figure out the digital root of, for instance, 634572234. In fact, it makes it easier. If you have a number where casting out 9s would eliminate every digit, that number will always have a digital root of 9.
This is because any number evenly divisible by 9 has a digital root of 9 (in fact, according to Martin Gardner, the fastest known way to manually test if a large number is divisible by 9 is to check its digital root.
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Akanksha
I have a doubt!
As I can see that (2039 + 172 + 1218 + 3091 = 6520), and we can evaluate this equation by adding all digits and matching them (4 in this video. Which concludes that our answer is correct.
But What If, By Mistake, I calculated the same equation result as 6250 (2039 + 172 + 1218 + 3091 = 6250 or 6160, still we are able to conclude that our answer is correct, but which is actually not.
So this shows that, we can only conclude that our answer is definitely wrong, but can never conclude that our answer is definitely right. Am I correct?
Can you explain if I misunderstood something.
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I have a doubt!
As I can see that (2039 + 172 + 1218 + 3091 = 6520), and we can evaluate this equation by adding all digits and matching them (4 in this video. Which concludes that our answer is correct.
But What If, By Mistake, I calculated the same equation result as 6250 (2039 + 172 + 1218 + 3091 = 6250 or 6160, still we are able to conclude that our answer is correct, but which is actually not.
So this shows that, we can only conclude that our answer is definitely wrong, but can never conclude that our answer is definitely right. Am I correct?
Can you explain if I misunderstood something.
reply
Aerxis
Just be careful when you define -digital root-, for the number zero. For example, the digital root of zero is zero, according to the video. If you add 9, the digital root changes to 9. If a (natural) number is greater than zero, the digital root remains unchanged when you add 9, because at least the first digit should be non-zero. So you should either define the digital root of zero to be 9, or you change the invariance theorem in order to accommodate this irregularity. Negative numbers (for subtraction) can be considered by adding nine to any non-positive digital root.
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Just be careful when you define -digital root-, for the number zero. For example, the digital root of zero is zero, according to the video. If you add 9, the digital root changes to 9. If a (natural) number is greater than zero, the digital root remains unchanged when you add 9, because at least the first digit should be non-zero. So you should either define the digital root of zero to be 9, or you change the invariance theorem in order to accommodate this irregularity. Negative numbers (for subtraction) can be considered by adding nine to any non-positive digital root.
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Charles
Some may not realize, in a long list of numbers to add, you can also cast out nines between the rows.
For example, adding
647 +
3521 = will quickly give you a 1 as the root. 6+3=9, 4+5=9, 7+2=9.
I learned this back in the early 80's when the cashiers at Radio Shack were writing prices on paper pads and adding them by calculator to tell me how much to pay for my latest project. I could easily verify their numbers while viewing their receipt pad upside down across the counter.
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Some may not realize, in a long list of numbers to add, you can also cast out nines between the rows.
For example, adding
647 +
3521 = will quickly give you a 1 as the root. 6+3=9, 4+5=9, 7+2=9.
I learned this back in the early 80's when the cashiers at Radio Shack were writing prices on paper pads and adding them by calculator to tell me how much to pay for my latest project. I could easily verify their numbers while viewing their receipt pad upside down across the counter.
reply
Math
It's also very easy to check that everything is correct modulo 2. The sum between several numbers will be even if and only if an even number of inputs are odd. Same goes for differences because subtraction is just adding the negative. And the product between multiple numbers will be even if and only if at least one input is even.
With this easy extra precaution, you can be sure if you messed up your arithmetic, you have to be off by a multiple of 18.
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It's also very easy to check that everything is correct modulo 2. The sum between several numbers will be even if and only if an even number of inputs are odd. Same goes for differences because subtraction is just adding the negative. And the product between multiple numbers will be even if and only if at least one input is even.
With this easy extra precaution, you can be sure if you messed up your arithmetic, you have to be off by a multiple of 18.
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Jawsome
Sometimes I liked to play around with numbers and add up digits and such, and for the longest time I knew in base 10 that if you were to add up the digits of a number and 9 was in the number, then you could just forget about the nine and nothing would change. I never knew about the application, however, and it's cool that this can be applied to everyday life.
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Sometimes I liked to play around with numbers and add up digits and such, and for the longest time I knew in base 10 that if you were to add up the digits of a number and 9 was in the number, then you could just forget about the nine and nothing would change. I never knew about the application, however, and it's cool that this can be applied to everyday life.
reply
Nifty
9=99=999=9999 and so on, mod9. Therefore every power of 10 is congruent (or equal in shorthand) to 1.
1234 = 1000-1 + 100-2 + 10-3 + 1-4 = 1-1 + 1-2 + 1-3 + 1-4 = 1+2+3+4. Thus the number equals its digital root (mod9.
So if you add two numbers, that's the same as adding their digital roots in Z_9. Their sum in Z must be equal to their sum in Z_9.
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9=99=999=9999 and so on, mod9. Therefore every power of 10 is congruent (or equal in shorthand) to 1.
1234 = 1000-1 + 100-2 + 10-3 + 1-4 = 1-1 + 1-2 + 1-3 + 1-4 = 1+2+3+4. Thus the number equals its digital root (mod9.
So if you add two numbers, that's the same as adding their digital roots in Z_9. Their sum in Z must be equal to their sum in Z_9.
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---
Let's see 10 as 9+1 and 100 as 99+1 and so on.
Let's see the number 201. We can write it as 2(99+1) + 1.
We know 99 is devided by 9 and so we get 2+1 as the remaining.
When we know the remainings of the numbers we're calculating and the remaining of the answer and those are the same, it's quite accurate to conclude that we calculated correctly.
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Let's see 10 as 9+1 and 100 as 99+1 and so on.
Let's see the number 201. We can write it as 2(99+1) + 1.
We know 99 is devided by 9 and so we get 2+1 as the remaining.
When we know the remainings of the numbers we're calculating and the remaining of the answer and those are the same, it's quite accurate to conclude that we calculated correctly.
reply
Z-Statistic
I refer to these things as -cycles-. I always try to use cycles in my arithmetic whenever I can either to do it or check it because position in cycles is much easier to manage than multiple digits. Well, we all already do this; digits are just cycles of length 10. This is a cycle of length 9. It's always kind of fun to find a cycle in the numbers.
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I refer to these things as -cycles-. I always try to use cycles in my arithmetic whenever I can either to do it or check it because position in cycles is much easier to manage than multiple digits. Well, we all already do this; digits are just cycles of length 10. This is a cycle of length 9. It's always kind of fun to find a cycle in the numbers.
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