# Rule of division by 23.

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What is the rule of division by 23?

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HallsofIvy
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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 $$a_m a_{m-1}.....a_{1}=3n$$ where a_m,a_m-1,....,a_1 are the digits of the number, $$\sum_{k=0}^{m}a_{k}10^k=3n$$, I need to prove that a_1+....+a_m is divisible by 3, any pointers?

$$\sum_{k=0}^{m}a_{k}10^k=3n$$

Since 10 ==1 Mod3, so are all its powers.

tiny-tim
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69 = 23 x 3

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?
Hi loop quantum gravity !

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)

CRGreathouse
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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?
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

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 23-16 = 7. (I don't see that they made that especially clear, though tiny-tim had already done that for 23.)

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$$\sum_{k=0}^{m}a_{k}10^k=3n$$

Since 10 ==1 Mod3, so are all its powers.
I'm not sure I understand, I need to prove that it's divisible iff $$a_0+...+a_m$$ is a multiple of 3.
What does the fact that the residue of 10 by 3 is 1 helps me here?

tiny-tim
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I'm not sure I understand, I need to prove that it's divisible iff $$a_0+...+a_m$$ is a multiple of 3.
What does the fact that the residue of 10 by 3 is 1 helps me here?
Because if a = b = 1 (mod3), then ab = 1 (mod 3) …

so 10n = 1 (mod3),

and ∑ an10n = ∑ an (mod3)

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Because if a = b = 1 (mod3), then ab = 1 (mod 3) …

so 10n = 1 (mod3),

and ∑ an10n = ∑ an (mod3)
I figured it out eventually for both 3 and 5, the next task of mine is to prove it for 7, wish me luck. (-:

Cheers.

CRGreathouse
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So on the subject of division rules:
It's clear that there are different sorts of divisibility rules here.
• Some, like those for 2 and 5, allow testing a huge number in one step -- they just look at a fixed number of digits, one in these cases.
• Others, like those for 3 and 9, reduce testing the number to testing a number a fraction of its size. In particular, each application of the rule reduces the number logarithmically: 123456789 is divisible by 3 iff 1+2+3+4+5+6+7+8+9 = 45 is.
• Finally, some rules reduce the size of the number by only a fixed number of digits, like the tests for 7 and 23.

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)?

jambaugh
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If you're looking at a last digit rule you need only consider:

If (p,10)=1 then
$$n = 10m + r = ap \equiv m \mp kr =bp$$
where k comes from a two digit multiple of p which is one away from a multiple of 10.
$$sp = 10k \pm 1$$

$$10bp = 10m \mp 10k r = 10m+ r \mp (10k\pm 1)r = n \mp (10k\pm 1)r = n \mp sbp$$
Thus
$$n = 10bp \pm sbp$$

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 = 19x100-1. So multiply last 2 digits by 19 and add.

CRGreathouse
Homework Helper
If you're looking at a last digit rule you need only consider:

If (p,10)=1 then
$$n = 10m + r = ap \equiv m \mp kr =bp$$
where k comes from a two digit multiple of p which is one away from a multiple of 10.
$$sp = 10k \pm 1$$

$$10bp = 10m \mp 10k r = 10m+ r \mp (10k\pm 1)r = n \mp (10k\pm 1)r = n \mp sbp$$
Thus
$$n = 10bp \pm sbp$$
OK, that's the third kind I mentioned: a rule to shorten the number by one.

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.
Good point. I'm not sure that those sorts of rules would be practical compared to long division,though.

tiny-tim
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1001

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.

My question: Are there other kinds of tests than those listed here (or equivalent to these)?
Hi CRGreathouse!

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.

jambaugh
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[...]
Good point. I'm not sure that those sorts of rules would be practical compared to long division,though.
Oh! You want practical!

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 = 1x10-3, 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 = 100-1. 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).

CRGreathouse
Homework Helper
Oh! You want practical!
Actually, not really. I just thought the lack of practicality was worth mentioning.

For example, one borderline-practical 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

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

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Sorry I had a mistake it should be:
$$7m-2a_0+a_n=-a_0+....+(1+10^n)a_n$$
I still don't see how a_n-2a_0 is divisble by 7, do you?

loop quantum gravity: I still don't see how a_n-2a_0 is divisble by 7, do you?

All you have to do is write the original number as 10a+b, where b is from 0 to 9, as base 10 goes, and 10a is the remaining number ending in 0.

Then look at a-2b, if this is congruent to 0 Mod 7, we have $$a \equiv 2b \bmod 7$$

So the original equation gives $$10a +b \equiv 20 b + b \equiv 21b \equiv 0 \bmod 7$$

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Thanks, robert.