The following rules refer only to divisibility by numbers relatively prime to 10. For
rules for division by all numbers, (which are sometimes harder), see this page.
General Rule
Any integer number can be represented as: 10a+b [1.1]
where b is non-negative integer less than 10: 0, 1, 2...9, and a is any
positive integer. [Actually, 10a+b can be a negative integer, where
0>b>-10, and a is a negative integer, but here we think of
non negatives for convenience in the knowledge that the negative version
is easily obtained].
Here we consider numbers relatively prime to 10.
If Equation 1.1 is divisible by a number relatively prime to 10, p, such as 2, 3, 5..., then: 10a+b=0 modp [1.2]
So: b=-10a modp [1.3]
The above is the same for both rules.
General Addition Rule
We are seeking an integer, m, such that a+mb=0 modp [1.4]
Substituting the value of b from 1.3 in 1.4: a+m(-10a)=0 modp [1.5]
Factorising: a(1-10m)=0 modp [1.6]
Now, 1.6 is divisible by p, if (1-10m) is.
(1-10m)=-(10m-1), so if (1-10m) is divisible by p, so is (10m-1), which is sometimes easier to use!
For instance, if p=7, then we need (10m-1) to be divisible by 7 (and we take the lowest value). If m=5, then: 1-10·5=49 (which is divisible by 7)
The
rule for 7, is therefore: add 5 times the unit digit to the rest of the
number, and if it is divisible by 7, then so is the number.
For instance, using the rule: 196 gives 5·6+19=49
So, 196 is divisible by 7.
General Subtraction Rule
We continue from above 10a+b=0 modp [1.2, repeated]
So: b=-10a modp [1.3, repeated]
We seek a rule: a-mb=0 mod p [1.2.1]
Substituting for b using Equation 1.3: a-(m(-10a))=0 mod p [1.2.2]
Simplifying: a+10ma=0 mod p [1.2.3]
Factorising: a(1+10m)=0 mod p [1.2.3]
For the left-hand side to be divisible by p, the (1+10m) must be
divisible by p. We search for the smallest m which makes (1+10m)
divisible by p.
For instance, if p=7, then m=2 (1+10·2=21, which is divisible by 7)
Therefore
the subtraction rule for 7 is: Double the last digit and add the rest
of the number. If this is divisible by 7, then so is the number.
Example:
315 is divisible by 7 because 31-2·5=21, which is divisible by 7 (or 21 gives 2-2·1=0, which is divisible by 7)
Reminder
When
seeking a rule, the addition rule uses (10m−1) and the subtraction rule
uses (10m+1). That is, minus for the addition rule, and plus for the
subtraction rule!
For instance, in seeking the addition rule for 3, using (10m-1), we need to find a number multiplied by 3 which ends in 9. We find 9, itself. 9+1=10, so m=1.
In seeking the subtraction rule, using (10m+1), we seek a number which when multiplied by 3 ends in 1. 3*7=21, so m=2;
In seeking an addition rule for 13, using (10m-1), we seek a number which when multiplied by 13 ends in 9. 3*13=39. So m=4.
The subtraction rule for 13, using (10m-1), requires a number ending in 1... 13*7=91. So m=9.
Explanation of the Method
We can write the following: 10a+b≡N [1.3.1] Where
N and a are non-negative integers, and b is a non-negative integer less
than 10 (or a and N are negative integers and b is a negative integer
greater than -10).
If N is divisible by an integer p, then so is the left-hand side. If we write Equation 1.3.1 modp, we get: 10a+b≡0 modp [1.3.2]
Dividing by 10, which we can do when p is prime (which is our assumption) so 10 and p are relatively prime: a+b/10≡0 modp [1.3.3]
So, if the original number is divisible by p, then [1.3.3] is true. We can write 1/10 as mp, where mp is the value of 1/10 for a given p. So: mp≡1/10 modp [1.3.4] And a+bmp≡0 modp [1.3.5] For example, if p≡7, m7≡1/10 (mod 7), so 10m7≡1 (mod 7), 3m7≡1, so by trial and error, m7≡5 (mod 7)
Addition and Subtraction Rules
We now have a rule: We multiply the units by 1/10 modp and add the rest of the number. If the result is 0 modp, then the number is divisible by p.
We can also have a subtraction rule where: np≡−1/10 (mod p)[1.3.6] And a-bnp≡0 (mod p) [1.3.7] For instance, if p=7
By adding [1.3.4] and [1.3.7], we get: mp+np≡0 mod p
We can write: mp+np≡p mod p because p mod p is 0. (we can write any multiple of p and get the same answer.)
Therefore,
the sum of the factors for the addition rule and the subtraction rule
is equal to p. This means that having discovered one of the
factors, we can easily compute the other.
For instance, with 7, the addition rule uses 5 and the subtraction rule uses 2, and 5+2=7!
Numbers for which the rule doesn't apply
We seek a number, m such that:
m=1/10 mod p
[1.3.8]
In other words: 10m=1 mod p
(Also m is called the multiplicative inverse in mod p)
If p=2 or p=5, then 10m mod p is zero. So these numbers cannot be used (because we require 10m mod p to be 1, not zero).
If p is an even number, then, because: 10m mod p=10m-qp [1.3.9]
where
q is an integer, then because 10m and p are even, the difference will
be an even number (and therefore cannot be equal to 1). For instance,
10 mod 8 is 2. There isn't a number, mod 8, which multiplied by 2 gives
1. (Actually there isn't a number at all, whatever the modulus!)
We
can use our formula to find rules for checking divisibility of numbers
which are not relatively prime to 10, that is, odd numbers
except those ending in 5 or 0.
Seeking the m's for the relatively-prime-to 10 numbers
It
is quite easy with practice to figure out these rules mentally using
the above, although here I will mention another method. Let us consider
13. So we seek an m: 10m=1 mod 13
Looking at an extract from the mod 13 times table for 10:
Table Mod 13
0
1
2
3
4
5
6
7
8
9
10
11
12
10
0
10
7
4
1
11
8
5
2
12
9
6
3
11
0
11
9
7
5
3
1
12
10
8
6
4
2
12
0
12
11
10
9
8
7
6
5
4
3
2
1
We note that 10·4=1 mod13.
So for the addition rule we multiply the units by 4 and add the rest of the number.
For the subtraction rule we want: 10m+1=0 mod 13
Or: 10m=-1 mod 13
-1 mod 13 =12 And: 12·9=1 mod13. So the subtraction rule is: subtract 9 times the units figure from the rest of the number.
Example for 13
Addition rule:
173 gives 17+4·3=29. 29 gives 2+4·9=38. 38 gives
3+4·8=35, which does not divide by 13 evenly so 173 doesn't have 13 as
a divisor.
Subtraction rule:
173 gives 17-9·3=-10. 10 (the divisors of 10 and -10
are numerically the same!) gives 1-9·0=1, so it still doesn't divide by
13.
Creating New Formula
This
is about creating new types formulae. Making up the formulae for
different numbers using this Method 2 has already been dealt with
above, and there is more below.
The
purpose here is to exercise our new knowledge and not to find
alternative formula in some practical way. The formula for Method 1
begin below.
We have used: 10a+b=N
Now, we might use: 100a+10b+c=N [1.4.1]
If N is divisible by p, then so is the left-hand side. So 100a+10b+c=0 mod p [1.4.2]
Dividing by 100: a+b/10+c/100=0 mod p [1.4.3]
We can write 1.4.3 as: a+bm1+cm2=0 mod p [1.4.4]
Where: m1=1/10 mod p, [1.4.5] and m2=1/100 mod p [1.4.6]
Seeking m's for 7
For instance, we can find these m's for 7. m1=1/10 mod 7, [from 1.4.5] and m2=1/100 mod 7 [from 1.4.6] 10 mod7=3 100 mod7=2
So the m's we seek are: m1=1/3 mod 7, m2=1/2 mod 7
From the table below, we find: m1=5 (3·5=1 mod7), and m2=4 (2·4=1 mod7)
Table Mod 7
0
1
2
3
4
5
6
0
0
0
0
0
0
0
0
1
0
1
2
3
4
5
6
2
0
2
4
6
1
3
5
3
0
3
6
2
5
1
4
4
0
4
1
5
2
6
3
5
0
5
3
1
6
4
2
6
0
6
5
4
3
2
1
Example
Rule: a+5b+4c=0 mod 7 Is 147 divisible by 7? 147 gives 1+5·4+4·7=49 49 gives 5·4+4·9=56. 56 gives 5·5+4·6=49 49 gives 5·4+4·9=56
Whilst our new rule works, it seems to circle after slowly reducing the number. We can rewrite it immediately as: a-(-5)b+4c=0 mod 7
And note that -5 mod 7=2. So a-2b+4c=0 mod 7
Trying our new new rule with 147: 147 gives 1-(2·4)+4·7=21 21 gives -(2·2)+4·1=0
The Division Rules
Division by 3
Addition Rule
For
the addition rule, we want (10m-1) to be divisible by 3, and when m=1,
this is so. So the rule is : Add the units digit to the rest of the
number, and if this is divisible by 3, then so is the number.
Example: 711 gives 1·1+71=72. 72 gives 2·1+7=9, which is divisible by 3, so 711 is divisible by 3.
Subtraction Rule
We need (1+10m) to be divisible by 3. The smallest m is 2. So the subtraction rule for 3 is Subtract twice the units digit from the rest of the number.
Example: 711 gives 71-2·1=69. 69 gives 6-2·9=12. 12 gives 1-2·2=-3, which is divisible by 3. In this case, these set of rules are more difficult than the Method 1 rule, so we would use that rule.
Division by 5
We
need (10m-1) or (10m+1) to be divisible by 5. There is no "m" for which
(10m-1) or (10m+1) is divisible by 5. (2 and 5 divide the base 10, and
neither has a rule in this system.) See the Method 1 method.
Division by 7
Addition Rule
We seek (10m-1) to be divisible by 7. When m=5, this is so.
Our rule is:
Add 5 times the units figure to the rest of the number, and if the
result is divisible by 5, then so is the number.
For instance: 861 gives 86+5·1=91. 91 gives 9+5·1-14. 14 gives 1+5·4=21. 21 gives 2+5·1=7, which is divisible by 7.
Subtraction Rule
We seek (10m+1) to be divisible by 7. When m=2, this is so.
Our rule is:
Subtract twice the units figure to the rest of the number, and if this
is divisible by 7, then so is the number.
For instance: 861 gives 86-2·1=84. 84 gives 8+2·8=24. 24 gives 2-2·4=7, which is divisible by 7.
Division by 13
Number
Plus Factor
Minus Factor
13
4
9
Addition Rule for 13
Add 4 times the unit figure to the rest of the number. If the result is divisible by 13, so is the number. Example 1092 gives 109+4·2=117. 117 gives 11+4·7=39 39 gives 3+4·9=39 So we cannot go further than 39, which is divisible by 13, so 1092 is divisible by 13.
Subtraction Rule for 13
Subtract 9 times the units figure from the rest of the number. If the result is divisible by 13, so is the number.
1092 gives 109-9·2=91. 91 gives 9-9·1=0, which is divisible by 13, so 1092 is divisible by 13
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