Why do the divisibility "rules" work?
The Power of Modulo Arithmetic
Modulo arithmetic is a powerful tool that can be used to test for
divisibility by any number. The great disadvantage of using modulo
arithmetic to test for divisibility is the fact that it is usually a slow
method. Under some circumstances, however, the application of modulo
arithmetic leads to divisibility rules that can be used. These rules are
important to mathematics because they save us lots of time and effort.
In fact, many of the divisibility rules that are commonly used (rules for 3,
9, 11) have their roots in modulo arithmetic. This page will show you how
some of these common divisibility rules are connected to modulo arithmetic.
Some Background Information
Consider any number (call the number N) written in base 10. Let n be the
number of digits in N, and Di be the "i"th digit (reading right to left) in
the number, i = 1 to n. This number can be expressed by the following
summation:
N = [Dn * 10^(n1)] + [Dn1 * 10^(n2)] + ... + [D2 * 10^1] + [D1 * 10^0].
Consider the number N = 6788. This number contains 4 digits, so n = 4. The
first digit is D1 = 8, the second digit is D2 = 8, the third digit is D3 =
7, and the fourth is D4 = 6, so
6788 = [6 * 10^3] + [7 * 10^2] + [8 * 10^1] + [8 * 10^0].
Applying the Rules of Modulo Arithmetic
The rules of modulo arithmetic state that the number N is divisible by some number (P) if the above expression is also divisible by P after the base (10) is replaced by the remainder of 10 divided by P. In compact notation, this remainder is denoted by (10 mod P). Thus, N is divisible by P if the following expression is also divisible by P:
[Dn * (10 mod P)^(n1)] + ... + [D2 * (10 mod P)^1] + [D1 * (10 mod P)^0].
Origin of Divisibility Rules for 3, 9, and 11
We can now see how the divisibility rules for P = 3, 9, and 11 are rooted in modulo arithmetic.
Consider the case where P = 3.
Because 10 mod 3 is equal to 1, any number is divisible by 3 if the following expression is also divisible by 3:
[Dn * (1)^n1] + ... + [D2 * (1)^1] + [D1* (1)^0].
Because 1 raised to any power is equal to 1, the above expression can be simplified as:
Dn + ... + D2 + D1 + D0,
which is equal to the sum of digits in the number. Thus modulo arithmetic allows us to state that any number is divisible by 3 if the sum of its digits is also divisible by 3.
Now consider the case where P = 9.
Because 10 mod 9 is also 1, any number is divisible by 9 is the sum of its
digits are also divisible by 9. Thus, the divisibility rules for 3 and 9
come directly from modulo arithmetic.
Finally, consider the case where P = 11.
Because 10 mod 11 is equal to 1, any number is divisible by 11 if the following expression is also divisible by 11:
[Dn * (1)^n1] + ... + [D2 * (1)^1] + [D1* (1)^0].
Because 1 raised to any even power is equal to 1 and 1 raised to any odd power is equal to 1, the above expression becomes a summation of the digits involving alternated signs:
Dn ...  D2 + D1 (for even n)
and
Dn + ...  D2 + D1 (for odd n).
This "alternating" summation rule for P = 11 is well known and has two versions. The first version states a number is divisible by 11 if the alternating sum of the digits is also divisilbe by 11 (i.e., the alternating sum is a muliple of 11). The second version states that a number is divisible by 11 if the sum of every other digit starting with the rightmost digit is equal to the sum of every other digit starting with the second digit from the right (i.e., the alternating sum is 0). Both of these versions come directly from the rules of modulo arithmetic.
Bases Other Than 10
Up to this point, we have only considered numbers written in base 10. A
number can, however, be written in any base (B). Such a number can be
expressed by the following summation:
N = [Dn * B^n1] + [Dn1 * B^(n2)] + ... + [D2 * B^1] + [D1* B^0].
The rules of modulo arithmetic apply no matter what base is used. These
rules tell us that N is divisible by P if the following expression is also
divisible by P:
[Dn * (B mod P)^n1] + ... + [D2 * (B mod P)^1] + [D1* (B mod P)^0].
Most bases other than 10 are difficult to use, but bases that are powers
of 10 can easily be used to construct divisibility rules using modulo
arithmetic.
Consider writing the base10 number N = 1233457 in base100. This can be
done by starting with the rightmost digit and grouping the digits in pairs
of two. Each grouping of two digits is considered a single "digit" when the
number is written in base100. Written in base100, the number is 1 23 34 57
where 57, 34, 23, and 1 are considered single "digits." In a similar
fashion, this number written in base1000 is 1 233 457 where 457, 233, and 1
are considered single "digits." Once the base10 digits are grouped to form
the "digits," the above expression can be used to test for divisibility.
Consider P = 7.
Use the base, B = 1000. Because 1000 mod 7 = 1, the alternating summation rule (used for P = 11 in base10) can be applied. Before applying this rule, the base10 digits must be properly grouped to form base1000 "digits."
Consider P = 11.
Use the base, B = 100. Because 100 mod 11 = 1, a number written in base100
is divisible by 11 if the sum of its "digits" is divisible by 11. Before
applying this rule, the base10 digits must be properly grouped to form
base100 "digits."
Consider P = 13.
Use the base, B = 1000. Because 1000 mod 13 = 1, the alternating summation
rule can be applied. Before applying this rule, the base10 digits must be
properly grouped to form base1000 "digits."
Back to Rules for 3,9,11,7,13,17...
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