
#1
Dec2208, 09:06 AM

P: 3,177

What is the rule of division by 23?
Thanks in advance. 



#2
Dec2208, 11:57 AM

Sci Advisor
HW Helper
Thanks
P: 26,167





#3
Dec2208, 01:26 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

Nice link, tiny tim.




#4
Dec2208, 01:55 PM

P: 3,177

Rule of division by 23.
Well another question of mine is how do you prove these division rules, do we know for every prime number the rule of division, or not?
Obviously, you need some trial and error but afterwards you need to prove it. For example I am now rechecking the law of division by 3, and I know I should look at expansion in basis ten, i.e if [tex]a_m a_{m1}.....a_{1}=3n[/tex] where a_m,a_m1,....,a_1 are the digits of the number, [tex]\sum_{k=0}^{m}a_{k}10^k=3n[/tex], I need to prove that a_1+....+a_m is divisible by 3, any pointers? Thanks in advance. 



#5
Dec2208, 03:47 PM

PF Gold
P: 1,059

[tex]\sum_{k=0}^{m}a_{k}10^k=3n[/tex]
Since 10 ==1 Mod3, so are all its powers. 



#6
Dec2208, 06:36 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

I don't think there's a general rule … you just have to find a suitable multiple of the prime that ends in 9 or 1 … for 23, it's 69 (= 3 x 23), = 70  1 … so you add the unit times 70 to the tens and upward. (if 23  10a + b, then also 23  10a + 70b) 



#7
Dec2208, 06:41 PM

Sci Advisor
HW Helper
P: 3,680

http://front.math.ucdavis.edu/0001.5012 



#8
Dec2208, 08:53 PM

PF Gold
P: 1,059

GFGreathouse: Yes, but most are dumb like those for 7 and 23. Here's a link to some for the first thousand primes:
http://front.math.ucdavis.edu/0001.5012 However, they use the minus numbers, and for 23 that's 16. However, sometimes, it is easier to use the positive numbers that you add on, and that's 2316 = 7. (I don't see that they made that especially clear, though tinytim had already done that for 23.) 



#9
Dec2208, 10:57 PM

P: 3,177

What does the fact that the residue of 10 by 3 is 1 helps me here? 



#10
Dec2308, 02:58 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

so 10^{n} = 1 (mod3), and ∑ a_{n}10^{n} = ∑ a_{n }(mod3) 



#11
Dec2308, 07:01 AM

P: 3,177

Cheers. 



#12
Dec2308, 12:45 PM

Sci Advisor
HW Helper
P: 3,680

So on the subject of division rules:
It's clear that there are different sorts of divisibility rules here.
It's clear that any number of the form 2^m * 5^n has a test of the first sort. This is because base 10 factors as 2 * 5 so only the last max(m, n) digits need be considered. It's clear that numbers n appearing in the factorization of 9, 99, 999, ... have tests of the second sort, since 10^k = 1 (mod n), and so digits can be added in groups of k. So 11 has a divisibility test based on adding digits in pairs, for example. Any prime not in the factorization of 10 has a test of the third form. My question: Are there other kinds of tests than those listed here (or equivalent to these)? 



#13
Dec2308, 03:16 PM

Sci Advisor
PF Gold
P: 1,767

If you're looking at a last digit rule you need only consider:
If (p,10)=1 then [tex] n = 10m + r = ap \equiv m \mp kr =bp[/tex] where k comes from a two digit multiple of p which is one away from a multiple of 10. [tex]sp = 10k \pm 1[/tex] [tex] 10bp = 10m \mp 10k r = 10m+ r \mp (10k\pm 1)r = n \mp (10k\pm 1)r = n \mp sbp[/tex] Thus [tex] n = 10bp \pm sbp[/tex] You could also generate rules from multiples of p more than one away from a multiple of 10 but these would require you multiply the remaining digits of the number n by that number which is harder. You can work similarly with the last two digits working with 3 digit multiples of p and so on. Examples for multiples of 7: Example: 3x7 = 21 = 2x10 + 1. Double the last digit and subtract from remaining. Example: 7x7 = 49 = 5x10  1. Multiply last digit by 5 and add. 1792 > 179 + 10 = 189 > 18+45 = 63 > 6+21=28 > 2+40 = 42 > 4+10 = 14> 1+20 = 21 > 2+5=7. Higher order examples: For multiples of 11: 9x11 = 99=1x100  1. So multiply last two digits by 1 and add. For multiples of 211: 9x211 = 1899 = 19x1001. So multiply last 2 digits by 19 and add. 



#14
Dec2308, 05:04 PM

Sci Advisor
HW Helper
P: 3,680





#15
Dec2308, 05:26 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

Also numbers n appearing in the factorization of 11, 101, 1001, … for example 1001 = 7.11.13, and to check whether a large number is divisible by each of those numbers, partition it into groups of three digits, and add them. 



#16
Dec2308, 05:48 PM

Sci Advisor
PF Gold
P: 1,767

Well if it's simply a matter of doubling and you're good at doubling in your head then it isn't so bad. Except that since 2 divides 10 such a rule will have an equivalent half rule since you're choosing an even multiple of the prime. Let's say you're a wiz at tripling... Example: Divisible by 7? 1x7 = 7 = 1x103, so add the last digit to triple the remaining... 777 > 3x77 + 7 = 231+7=238> 3x23+8=69+8=77>3x7+7=28 > 6+8=14>3+4=7. Not too bad but as you say both less practical and there's an easier one. But that's the mechanics of how it works. Note also the add all digits for 3 and 9 come from the same type rule applied recursively. Divisible by 3 or 9? 3x3= 1x9 = 1x10  1 so add 1 times last digit to remainder... apply recursively and its the same as adding all digits. Note 9 is composite but these rules should work for composites provided they're mutually prime to 10. E.g. divisable by 33? 3x33 = 99 = 1001. Add last two digits to remaining... 50886> 508+86=594>5 + 94 = 99 check! Easier by stages to check for 3 and 11 separately but quicker to do both at once (maybe). 



#17
Dec2308, 06:36 PM

Sci Advisor
HW Helper
P: 3,680

For example, one borderlinepractical test for divisibility by 7 would be to add up a number by groups of 6: 111,222,333,444,555,666,777,888,999 is divisible by 7 iff 111+222333+444555+666777+888999 = 2222775 is, and 2,222,775 is divisible by 7 iff 2+222775 = 222,777 is. Of course to go further you need trial division (or the standard 'shorten by 1 digit' rule). But the analogue for 17 (summing digits in groups of 16) is seldom, if ever, worthwhile, and I can't imagine anyone working their way through that rule for 59. I'm more interested in seeing where the rules come from. So far they come from the factorization of * n^a, by looking at the last a digits * n^a  1, by summing in groups of a digits * n^a +/ n^b, by (alternate) summing 



#18
Jan2309, 12:14 AM

P: 3,177

Ok, Iv'e come to a bit difficulty here with 7.
I mean if I write down: [tex]a_0+10a_1+...+10^n a_n=7m[/tex] then the rule states that: [tex]7m2a_0+a_1=a_0+11a_1+...+10^n a_n[/tex] is divisble by 7, but I don't see how this is true, it means that a_12a_0 should be divisible by 7. for example 35=5*7 so the rule states that 355*2+3=28, and indeed 310=7, but here is where I am stuck, I don't see how do you prove it, anyone? 


Register to reply 
Related Discussions  
Synthetic division or Long division of polynomials?  General Math  6  
GPA of Upper Division & Lower Division courses and M.S degree  Academic Guidance  1  
Chain rule in division rule  Calculus & Beyond Homework  6  
Division with variables (I think I'd call this double division)  Precalculus Mathematics Homework  5  
Quotient rule for derivatives (algebraic division help?)  General Math  3 