Difference between revisions of "Divisibility rules/Rule for 11 proof"
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Let <math>N = a_na_{n-1}\cdots a_1a_0</math> where the <math>a_i</math> are [[base numbers | base-ten]] numbers. Then <math>N = 10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10 a_1 + a_0.</math> | Let <math>N = a_na_{n-1}\cdots a_1a_0</math> where the <math>a_i</math> are [[base numbers | base-ten]] numbers. Then <math>N = 10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10 a_1 + a_0.</math> | ||
| − | Note that <math>10\equiv -1\pmod{11}</math>. Thus <math> 10^n a_n + 10^{n-1} a_{n-1} + \cdots + 10a_1 + a_0 \equiv (-1)^n a_n + (-1)^{n-1} a_{n-1} + \cdots -a_1 + a_0. </math> | + | Note that <math>10\equiv -1\pmod{11}</math>. Thus <math> 10^n a_n\! +\! 10^{n-1} a_{n-1} + \cdots + 10a_1 + a_0 \equiv (-1)^n a_n + (-1)^{n-1} a_{n-1} + \cdots -a_1 + a_0. </math> |
This is the alternating sum of the digits of <math>N</math>, which is what we wanted. | This is the alternating sum of the digits of <math>N</math>, which is what we wanted. | ||
Revision as of 08:19, 16 August 2006
A number 
is divisible by 11 if the alternating sum of the digits is divisible by 11.
Proof
An understanding of basic modular arithmetic is necessary for this proof.
Let 
where the 
are base-ten numbers. Then 
Note that 
. Thus 
This is the alternating sum of the digits of , which is what we wanted.
