We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.

A few months ago, I heard a simple and elegant way to find multiples of $7$:

Cut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.



Cut digits into pairs:

$2, 15, 53$

Multiply $53$ by $1, 15$ by $2, 2$ by $4$:

$8, 30, 53$



As $91$ is a multiple of $7$ ($13 \cdot 7$), then $21553$ is too.

This works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 → 05 = 7 - 2, 112 → 12 = 14 - 2, \cdots)$

I figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.

Exemple with $91$:


$9, 1$

$9\cdot3, 1\cdot1$

$27, 1$


My question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$?

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    $\begingroup$ You'll want to see this KCd blurb. $\endgroup$ Dec 8, 2011 at 15:49
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    $\begingroup$ I've seen the following method for 7: Take the last digit, double it, and subtract it from the first digits. So $21553$ leads to $2155-3*2=2149$. Next step is $214-2*9=196$. Next $19-2*6=7$. $\endgroup$ Dec 8, 2011 at 16:04
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    $\begingroup$ How can we find divisors of 23? Very easily: they are just $\pm 1$ and $\pm 23$. Presumably, you want to recognize multiples of 23? $\endgroup$ Dec 8, 2011 at 16:11
  • $\begingroup$ For $23$, here is something fairly simple, for numbers that have $3$ or $4$ digits, say $abcd$. Take the number $ab$, add to it $3$ times the number $cd$. Call the result $t$. Then $abcd$ is a multiple of $23$ iff $t$ is. One could probably tweak this to be even simpler to execute. $\endgroup$ Dec 8, 2011 at 16:16
  • $\begingroup$ @ArturoMagidin Yes, I meant multiples. It's corrected. $\endgroup$
    – SteeveDroz
    Dec 9, 2011 at 8:46

7 Answers 7


One needn't memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \ $ is $\rm\: (d_2\cdot 10 + d_1)\ 10 + d_0\ \equiv\ (d_2\cdot 3 + d_1)\ 3 + d_0\ (mod\ 7)\ $ since $\rm\ 10\equiv 3\ (mod\ 7)\:.\:$ So we compute the remainder $\rm\ (mod\ 7)\ $ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\rm\:(mod\ 7)\:.\:$

For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\ d_{n-1}\ $ by $\rm\ d_n\cdot 3 + d_{n-1}\:\ (mod\ 7),\:$ namely

$\rm\qquad\phantom{\equiv} \color{red}{4\ 3}\ 2\ 1\ 1$

$\rm\qquad\equiv\phantom{4} \color{green}{1\ 2}\ 1\ 1\quad $ by $\rm\quad \color{red}4\cdot 3 + \color{red}3\ \equiv\ \color{green}1 $

$\rm\qquad\equiv\phantom{4\ 3} \color{royalblue}{5\ 1}\ 1\quad $ by $\rm\quad \color{green}1\cdot 3 + \color{green}2\ \equiv\ \color{royalblue}5 $

$\rm\qquad\equiv\phantom{4\ 3\ 5} \color{brown}{2\ 1}\quad $ by $\rm\quad \color{royalblue}5\cdot 3 + \color{royalblue}1\ \equiv\ \color{brown}2 $

$\rm\qquad\equiv\phantom{4\ 3\ 5\ 2} 0\quad $ by $\rm\quad \color{brown}2\cdot 3 + \color{brown}1\ \equiv\ 0 $

Hence $\rm\ 43211\equiv 0\:\ (mod\ 7)\:,\:$ indeed $\rm\ 43211 = 7\cdot 6173\:.\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\rm\: \pm\{0,1,2,3\}\ \:(mod\ 7)\:.$ Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9\:$).

  • $\begingroup$ Cool trick! Thanks $\endgroup$ Dec 3, 2020 at 22:02

One approach is to find some higher multiple that makes it easy. For your example of $23$, note that $3*23+1=70$, so $a(70-1)$ will be a multiple of $23$. Now you have a single digit multiply followed by a subtraction. If you pick $a=301, a(70-1)=301(70-1)=301*70-301=2107-301=1806$, which is a multiple of $23$

  • $\begingroup$ I generally find this method much easier than the standard modular divisibility test. It does require that you memorize a starting list of multiples, but it's not so bad because you can just multiply them by powers of $10$. $\endgroup$ Dec 8, 2011 at 18:28
  • $\begingroup$ @QiaochuYuan: It seemed the OP's request was to find multiples, not do a divisibility test. But this generalizes. In the spirit of the test for $7$, to test divisibility by $23$, you can take off the last digit, multiply the rest by $7$ and add the last digit. Keep going until you have only two digits left. You just have to find a multiple that ends in $1$ or $9$ to do this. $\endgroup$ Dec 8, 2011 at 18:40

In general, if you are doing things by dividing the number into groups of $k$ digits, you can think of it as looking at the number in base $10^k.$

More generally, if $B$ is a base, and $n$ is a number with no common factors with $B$, then you can always find an $m$ such that $Bm\equiv 1\pmod n$. Then $X\equiv 0\pmod n$ if and only if $Xm\equiv 0\pmod n$. But if $X=Bu+v$, then $Xm\equiv u+mv\pmod n$. So if we take the last digit, base B, multiply it by $m$ and add it to the other digits, the result is divisible by $n$ if and only if the original number was divisible by $n$.

In the case of $n=23$ base $B=10$, you get $m=7$, so you can take the last digit, multiply it by seven, and add it to the rest.

Or, if you use $B=100$, you get $m=3$, and you can take the lsat two digits, multiply by 3, and add to the other digits.

In general, you can always find a $k$ such that $10^k=1\pmod n$ if $n$ is not even or divisible by 5. Then if you take base $B=10^k$, you get $m=1$, and you can separate $m$ into groups of $k$ digits and add them. That's hardly useful when $k$ is large. For example, the smallest $k$ for $n=23$ is $k=22$, so this part only helps if your starting number was more than 23 digits long.

Usually, you want to find a relatively small pair $(m,k)$ so that $m10^k\equiv \pm 1\pmod n$. Then you take the last $k$ digits, multiplied by $m$, and add to or subtract from the other digits, depending on whether $+1$ or $-1$.


The following is a simple method to check divisibility by $7$ or $13$:

Cut the digits in pairs of 3 and calculate their alternating sum. If this is a multiple of 7 or 13, the original number is.

For example

$12345631241$ leads to

$$241-639+345-12=-65 \,.$$

Thus our number is a multiple of $13$, but not a multiple of $7$.

This works because $1001=7*11*13$ meaning that any number of the form $abcabc$ is a multiple of $7, 11$ and $13$...The trick also works for 11, but there is another simple trick.

Method 2 If the number is relatively prime to 10 (if it is not, you can make it), look for a multiple on $n$ which ends in 1.

Let say that this multiple is $a_1..a_k1$.

Then you simply pick the large number and subtract $a_1...a_k*$last digit from the remaining digits.

The trick works because $a_1...ak1$*last digit is always a multiple of $n$, and subtracting this from the original number you get a multiple of 10..Since $n$ is relatively prime to 10, you can erase the 0 at the end...

A simple such example, for $7$ the smalest such desires multiple is ... 21, which leads to the criteria posted by Thomas Andrews .

Also, for small numbers the following


I will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$ z = a_0 + a_1 \times 10 + a_2 \times 10^2 + a_3 \times10^3 + \cdots + a_n \times 10^n$$

Lets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have,

$$ 10 \equiv 3, 10^2 \equiv 2, 10^3 \equiv -1, 10^4 \equiv -3, 10^5 \equiv -2, 10^6 \equiv 1,$$

The successive remainder then repeat. Thus our integer $z$ is divisible by $7$ iff if the remainder expression $$ r= a_0 + 3a_1 +2a_2 -a_3-3a_4-2a_5+a_6+3a_7+\cdots$$ is divisible by $7$

To understand why the divisibility of $r$ indicate the divisibility of $z$, find $z-t$ which is given by :$$z-t = a_1 \times (10-3) + a_2 \times (10^2-2) + a_3 \times (10^3+1) + \cdots + a_6 \times (10^6-1) + \cdots $$ Since all this numbers $ (10-3),(10^2-2),(10^3+1),\cdots$ are congruent to 0 modulo $7,z-t$ is also, and therefore $z$ leaves the same remainder on division by $7$ as $r$ does.

Using this approach we can derive divisibility of any integer.


This can be explained using Horner's method as below. z=a0 + a1 ×10 + a2 ×102 + a3 × 10^3+⋯+ an × 10^n

i.e. Z = ((((...(an × 10 + an-1)*10 + an-2 )*10 + an-3).....)*10 + a0

                           ( Shift-add representation of number)

Z % 7 =  ((((...(an × 3 + an-1)*3 + an-2 )*3 + an-3).....)*3 + a0  (Mod 7)

Now evaluate from left to right using modulo algebra.

To know more about shift- add representation of number and its application in interbase conversion and divisibility refer tinyurl.com/mlxk8pw .

                            Vitthal Jadhav

I taught myself that if all the numbers in a number added up equal a multiple of 3 then that original number is a multiple of 3.. I'm sure most of you who are good with numbers know this. But I found it this out for myself and im not very good with numbers.. So lets say 801 that is a multiple of 3 because 8+0+1=9 and since 9 is a multiple of 8 than so is 801.. This blew my mind. Please dont make fun of me. I'm sure moat of you already know this. I just think ita cool that I found this out for myself


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