In Medieval
times and before, there was a great use of tests to check calculations.
Arithmetic calculations by hand (and sometimes by machine)
often go
awry, and checking is important. Nowadays, these checks are less
common, due to faith in the electronic computer and calculators. When,
however,
we do checks, we tend to repeat our calculations, which might
leads us to repeat the error; consequently, repeated checks might
continue to produce the same error. The divisibility tests enable us to
check our calculations in a different way, therefore avoiding repeating
our mistakes. Of course, divisibility tests can also be used to find
factors, etc...
On this page, I write
about what I call the traditional or conventional method (for want of a name!) of divisibility checks, and on
another page a method called Method 2.
Intuitive Approach
In
the days when a computer meant "one who computes" and referred to a
person who did arithmetic, there was a demand for tests to determine
the accuracy of the arithmetic (pencil and paper arithmetic is
notoriously subject to error!). Many of these tests have a long history, going back to Medieval times and even
hundreds or thousands of years before.
To
take a simpler
example. Suppose we have multiplied a number by 7 and we have 378. To
check the arithmetic, we can begin by subtracting 7 from the number (if
we subtract a multiple of 7 from the number, what remains must also be
divisible by 7, if the number is so divisible).
We keep the remainder (371) and
look at more of the number. Clearly 70 is divisible by 7, so we can simplify further by subtracting 70, leaving 301
In
the remainder so far, we have 301, and 3·7=21, so we can subtract 21,
getting 280. Discarding the 0, we have 28 remaining, and as this is
divisible by 7, so is the number 378, confirming our calculations.
Alternatively, with 378, we could note that 4x7=28, and we can
subtract this from 378, leaving 350. Dropping the 0, we note that
7x5=35, so the number, 378, is divisible by 7.
General Principle
We can represent integers (up to 99999), where a, b, c, d, and e are 0, 1, 2, ... 9 and N is a nonnegative integer, as follows [1.1] (Naturally, larger numbers can be represented in a similar manner) We can write out abcde as follows: [1.2]
If
N is divisible by a whole number n, then [1.2] is divisible by n. That is, the
right-hand side, mod n, is zero, if N is divisible by n.
Taking the example of dividing by 7, [1.2] (with some terms dropped for clarity) can be written: [1.3]
If
the number, N, is divisible by 7, the multiples of 7 can be dropped
(because they are 0, mod 7, and because they are divisible by 7, so
they need not be taken into account). The remainder is: [1.4]
That
is, we can say that if N is divisible by 7, then [1.4] must be
divisible by 7. Also, if [1.4] is divisible by 7, then N is divisible
by 7 because [1.4] is the remainder on dividing [1.3] by 7.
We can write a longer version of [1.3]: [1.5] Here the coefficients are 1, 3, 2, 6, 4, 5....
The same number can be expressed as follows (with the coefficients negative smaller numbers: [1.6]
In this case, the formula becomes easier to remember and simpler to use, with the coefficients 1, 3, 2, -1, -2, -3: [1.7]
Naturally, the negative numbers are the complements of the positive ones they replace.
Divisibility by 2
Seeking the rule
We first form a table, beginning with the units and continuing to the tens, hundreds, etc.., as follows:
Powers of 10
Mod 2
1
1
10
0
100
0
1000
0
From the table we can make a formula by substituting in 1.2
the Mod 2 values for the various powers of 10. In this case all the
coefficients (powers of 10 mod 2) are zero, except the last number: [1.3]
So a number is divisible by 2, if the last digit is. In other words, it is divisible by 2 if it is an even number.
Divisibility by 3
In order to find the coefficients of a formula, we make a table for 3:
Powers of 10
Mod 3
1
1
10
1
100
1
1000
1
In this case, when we substitute the values of Mod 3 from the table into 1.2, all the coefficients of our formula for 3 are 1: [1]
Which means that if the sum of its digits are divisible by 3, then so is the number.
For instance, we know 762 is divisible by 3 because 7+6+2=15, is divisible by 3 (or 15 gives 1+5=6, which is divisible by 3).
The rule is: add the units digit to twice the
tens digit, and if the result is divisible by 4, then so is the number (ignore any other digits).
For example: 756 gives 2·5+6=16, which is divisible by 4 (Ignore the 7).
The common-sense rule (the rule often stated, but based on different principles
from this one) is that a number is divisible by 4 If the last two digits are divisible by 4
This is because 100 and higher numbers are divisible by 4 if the last 2 digits are so divisible.
After
the units, the coefficients are zero, so if the units figure is
divisible by 5, then so is the number. In other words, A number is divisible by 5 if it ends in 5 or 0
The coefficient for the units figure is 1, and all the rest are 4.
The
rule is therefore, add the units figure to 4 times the rest of the
number, and if this is divisible by 6, then so is the number. The alternative rule is Subtract from the units figure twice the rest of the number
Example:
Is 384 divisible by 6? First Rule: 384 gives 4+4·38=76 76 gives 6+4·6=30 30 gives 0+4·3=12 12 gives 2+4·1=6 Second Rule: 384 gives 4-(2·38)=72 72 gives 2-(2·7)=12 12 gives 2-(2·1)=0 It seems that it is often easier to check whether the number is even and whether it divides by 3!
divisibility by 7
The table of coefficients follows:
Power of 10
Mod 7
Mod 7, alternative
1
1
1
10
3
3
100
2
2
1000
6
-1
10000
4
-3
100000
5
-2
1000000
1
1
10000000
3
3
100000000
2
2
1000000000
6
-1
10000000000
4
-3
In
the first formula, we make a sum by multiplying the units, tens, hundreds.,
digits by respectively 1, 3, 2. That is, substituting the Mod 7 values for the powers of 10 in Equation 1.2. We then multiply the thousands, ten-thousands and hundred-thousands digits by 6, 4, 5, respectively and repeat these
sequences, as necessary [3.1]
The
second formula uses the "alternative" numbers, and is easier to
remember: the sequences being 1, 3, 2, followed by -1, -3, -2 and the
sequences repeat at the millions digit. [3.2]
The
patterns repeat. The pattern for Equation 3.2 is particularly easy, being 1, 3,
2 followed by the negatives: -1, -3, -2, and this pattern repeats.
Example 2: 861861 gives 1+3·6+2·8-1-(3·6)-(2·8)=0, which is divisible by 7! Example 3 1771938 gives 8+9+18-1-21-14+1=0, hence divisible by 7.
This formula has received a bad press, and in the using, it is probably, in general, superior to the corresponding one from Method 2.
For example, using 147: Method 2: 147 gives 14-2·7=0, so it divides by 7 Method 1: 147 gives 7+3·4+2·1=21. 21 gives 1+3·2=7 The Method 2 seems easier, but... Method 2: 117649 gives 11764-2·9=11746. 11746 gives 1174-2·6=1162. 1162 gives 116-2·2=112. 112 gives 11-2·2=7 Method 1: 117649 gives 9+(3·4)+(2·6)-(7·3)-(1·2)+1=21. 21 gives 1+3·2=7 With even larger numbers, the Method 1 seems even more superior to the apparently easy-looking Method 2. For instance, is 311973482284542371301330321821976049 divisible
by 7? While the number is hard to check, and too big for a calculator
or normal computer range, using Method 1 as a check is much more
efficient.
divisibility by 8
Consulting the table below, we note that the rule for 8 is: Add the units figure to twice the tens figure and add four times the 100's figure.
Note that the rule is a continuation of the rule for divisibility by 4 (add the units digit to twice the tens digit).
Power of 10
Mod 8
1
1
10
2
100
4
1000
0
10000
0
Example
Is 176 divisible by 8? Using the rule: 176 gives 4·1+2·7+6=24 24 gives 2·2+4=8 176 is therefore divisible by 8
Is 590295810358705651712 divisible by 8? Using the rule: 590295810358705651712 gives 4·7+2·1+2=32 32 gives 2·3+2=8 590295810358705651712 is therefore divisible by 8
divisibility by 9
Part
of the table of the coefficients for the formula for 9 is below. It
tells us all the coefficients are 1. This means the rule for
divisibility by 9 is If the sum of the digits is divisible by 9, then so is the number
Power of 10
Mod 9
1
1
10
1
100
1
1000
1
divisibility by 10
Part of the table of coefficients for 10 follows:
Power of 10
Mod 10
1
1
10
0
100
0
Except
for the units coefficient, all the coefficients are zero. So, if the
units figure is divisible by 10, then so is the number. The only digit:
0, 1, ...9 that is divisible by 10 is 0. So: A number is divisible by 10 if it ends in 0
Divisibility by 11
As usual, the 11's coefficients:
Power of 10
Mod 11
Alternative Mods
1
1
1
10
10
-1
100
1
1
1000
10
-1
10000
1
1
Using the alternative mods: A number is divisible by 11, if the sum of the differences between adjacent pairs of digits is divisible by 11 That is, abcde is divisible by 11 if: (a-b)+(c-d)+e is divisible by 11.
Alternatively, if we say the units digit is digit number 1 and the tens is digit number 2, etc., then: A
number is divisible by 11 if the sum of the odd placed number equals
that of the even placed, or the difference is divisible by 11.
Or A number is divisible by 11 if the difference between
Examples
55 gives 5-5=0, which is divisible by 11, so 55 is divisible
132 gives 1-3+2=0, which is divisible by 11, so 132 is divisible
8679 gives 8-6 + 7-9=0, so 8679 is divisible by 11.
34716 gives (3-4)+(7-1)+6=11, which is divisible by 11, so 34716 is divisible.
Divisibility by 12
Below is the table of coefficients for 12.
Power of 10
Mod 12
Alternative Mods
1
1
1
10
10
-2
100
4
4
1000
4
4
The rule for 12 is: Add
the units number, ten times the tens number and 4 times the rest of the
number, and if the sum is divisible by 12, so is the number.
Example
3624 gives 4+10·2+4·36=168 168 gives 8+10·6+4·1=72 72 gives 2+10·7=72, which is divisible by 12
The alternative rule (using the alternative mods) is: Subtract
twice the tens digit from the units digit and add 4 times the rest of
the number. If it is divisible by 12, then so is the number
Example
3624 gives 4-(2·2)+4·36=144 144 gives 4-2·4+4·1=0, which is divisible by 12
Divisibility by 13
From the table below, the pattern is 1, -3, -4, -1, 3, 4, 1....
Power of 10
Mod 13
Alternative Mods
1
1
1
10
10
-3
100
9
-4
1000
12
-1
10000
3
3
100000
4
4
1000000
1
1
10000000
10
-3
100000000
9
-4
1000000000
12
-1
10000000000
3
3
The pattern goes 1, -3, -4 and the negatives of these numbers follow: -1, 3, 4. The whole pattern repeats.
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