Does this have a name? (recreational math)

[freeztar]

1191=1+1+9+1=12=1+2=3

 

or

 

38976=3+8+9+7+6=33=3+3=6

 

Is there any use for such a "math game" or is it just that, a game?

 

Also, is there some rule that defines what single digit integer you end up with?

Perhaps an equation that explains it?

[Pyrotex]

I believe some forms of numerology (voodoo with numbers)

are based on such calculations.

 

There are a couple of math tricks that do, too. ;)

 

Add up all the digits in a number of any size.

If the sum contains more than one digit, repeat the operation.

Continue until you have a single digit.

 

If the single digit is exactly divisible by 3, then so is the original number.

If the single digit is exactly divisible by 9, then so is the original number.

[CraigD]

I’ve usually heard it referred to by the rather unimaginative name “digit summing”, and usually is written as the function [math]s()[/math] Doing it repeatedly until [math]s(n)=n[/math] doesn’t have a common name, AFAIK, so I’d just call it, in keeping with unimaginative tradition, “repeated digit summing”

 

It is, obviously, dependent on your numeration system. For ordinary Arabic numeration, the base, for weirder systems (eg: Roman numeral), weird and complicated factors I’ve not much contemplated.

 

Sticking with ordinary numeration, this means that the function is usually written with a notation of the base system, [math]s_b()[/math]. Our usual digit summing of everyday decimal numbers, then, could be written something like [math]s_{10}(s_{10}(s_{10}( \ldots s_{10}(n) \ldots )))[/math]

 

As an example of what I mean, [math]s_10(125)= 1+2+5 = 8[/math], while [math]s_2(125)= 1+1+1+1+1+0+1 = 6[/math] and [math]s_5(125)= 1+0+0+0 = 1[/MATH]. So when we speak of a number’s digit sum, we have to specify another integer, its base. Digit summing isn’t a single-argument function, nor a number’s digit sum an attribute of it as a “pure number”.

 

One of the basic theorems about digit summing is one of modular congruance [math]s_b(n) \,\mbox{mod}\, (b-1) = n \,\mbox{mod}\, (b-1)[/math]. For example, notice that the remainder (modulo) of 125 divided by 9 is 8, the same as [MATH]s_{10}(125)[/MATH], and 125 mod 4 = 1, the same as its digit sum. Since all numbers modulo 1 have the same value of zero, this isn’t very interesting for base 2 digit sums. Because this theorem makes it so easy to find the “repeated digit sum” of a number – for base 10, its just n mod 9, writing “9” if the result is “0” – nobody much, AFAIK, bothers searching for nifty ways to do it.

 

Lots more can be found at places like Wolfram’s page on the subject.

[freeztar]

Thanks for the replies guys!

 

Pyro, thanks for reminding me about the 3 and 6 tricks, I had forgotten about that. It can be helpful in quickly (if not entirely accurately) checking your answers in some cases.

 

Craig, I read your post about 3 times and have come to the conclusion that I need to do some background reading to understand this better. I checked out the link you provided, but my lack of ability in calculus precludes me from gaining much understanding from that page. :(

 

Once I wrap my head around the base system, this should all make perfect sense. Thanks!

 

Oh yeah, one quick question:

 

125 mod 4 = 1, the same as its digit sum

 

I understand the (125 mod 4 = 1) part, but wouldn't the digit sum be of 125 still be 8, which is obviously [math]\ne[/math] to 1. :confused:

[CraigD]

I understand the (125 mod 4 = 1) part, but wouldn't the digit sum be of 125 still be 8, which is obviously [math]ne[/math] to 1.

[math]125_{10} = 1\cdot5^3 + 0\cdot5^2 + 0\cdot5^1+ 0\cdot5^0 = 1000_{5}[/math], so [math]s_5(125_{10}) = s_5(1000_{5}) = 1 + 0 + 0 + 0 = 1[/math]

 

A quick refresher (if you’re about my age, and went to a US public school, you probably had this in grades 1-6 – love that old “new math” :)) on converting from base 10 to any base system:

Divide repeatedly by the base, taking the remainder as each digit, right to left, of the new numeral.

For example, converting 1234 (base 10) to base 7, we get:

1234 ÷ 7 = 176 r 2

176 ÷ 7 = 25 r 1

25 ÷ 7 = 3 r 4

3 ÷ 7 = 0 r 3

So [math]1234_{10} = 3412_7[/math]

 

Pyro’s observation

If the single digit is exactly divisible by 9, then so is the original number.

Is a special case of [math]s_b(n) \,\mbox{mod}\, (b-1) = n \,\mbox{mod}\, (b-1)[/math] where b=10 and [math]n \,\mbox{mod}\, (b-1) = 0[/math]. If you substitute “has the same remainder when divided” for “is exactly divisible”, the observation applies to any number.

 

The key to the basic math of these digit sums, and to most “digit” math, is modular arithmetic. Though typically not much taught except in “advanced” math classes, modular arithmetic is old (though not as old as you might guess), very intuitive, and not very hard to gain a basic proficiency in. The wikipedia link above and its links are not bad introductions. Visualizing it as “clock arithmetic” is a pretty good intuitive approach. Set theorists love modular arithmetic, as it provides an easy way to have small number sets that are closed under addition and subtraction. It comes in handy in computer engineering and low-to-mid level programming, as the registers on computer arithmetic units are usually modulo some power of 2 (eg: 8, 16, 32, 64 bits).

[freeztar]

[math]125_{10} = 1cdot5^3 + 0cdot5^2 + 0cdot5^1+ 0cdot5^0 = 1000_{5}[/math], so [math]s_5(125_{10}) = s_5(1000_{5}) = 1 + 0 + 0 + 0 = 1[/math]

 

Thanks for your patient explanation Craig. I'm rusty, which is obvious from...

 

Pyro’s observation is a special case of [math]s_b(n) ,mbox{mod}, (b-1) = n ,mbox{mod}, (b-1)[/math] where b=10 and [math]n ,mbox{mod}, (b-1) = 0[/math]. If you substitute “has the same remainder when divided” for “is exactly divisible”, the observation applies to any number.

...That's what threw me off. :doh:

 

The key to the basic math of these digit sums, and to most “digit” math, is modular arithmetic. Though typically not much taught except in “advanced” math classes, modular arithmetic is old (though not as old as you might guess), very intuitive, and not very hard to gain a basic proficiency in. The wikipedia link above and its links are not bad introductions. Visualizing it as “clock arithmetic” is a pretty good intuitive approach. Set theorists love modular arithmetic, as it provides an easy way to have small number sets that are closed under addition and subtraction. It comes in handy in computer engineering and low-to-mid level programming, as the registers on computer arithmetic units are usually modulo some power of 2 (eg: 8, 16, 32, 64 bits).

 

That's the ironic part for me. I'm familiar with hex, but I've never made any connection from that to base 10, only base 2. Taak for opening my "third (mathematical) eye" on this. ;)

I'll study up on modular arithmetic per the link provided. :)

[sanctus]

And freez a trick to remember it when you check for divisibility you can drop all the nines (in base 10, or all the fives in base 6) eg

937--> 9+3+7=19-->1+9=10-->1+0=1

 

and

 

37-->3+7=10-->1+0=1

[C1ay]

Also see Digit Sum -- from Wolfram MathWorld for some more interesting trivia on the topic...