Difference between revisions of "Divisibility rules/Rule for 11 proof"
Erictu2003 (talk | contribs) m |
|||
| (9 intermediate revisions by 4 users not shown) | |||
| Line 4: | Line 4: | ||
''An understanding of [[Introduction to modular arithmetic | basic modular arithmetic]] is necessary for this proof.'' | ''An understanding of [[Introduction to modular arithmetic | basic modular arithmetic]] is necessary for this proof.'' | ||
| − | Let <math>N = | + | Let <math>N = a_ka_{k-1}\cdots a_1a_0</math> where the <math>a_i</math> are [[base numbers | base-ten]] numbers. Then <math>N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0.</math> |
| − | Note that <math>10\equiv -1\pmod{11}</math>. Thus | + | Note that <math>10\equiv -1\pmod{11}</math>. Thus <math> 10^k a_k\!\! +\!\!10^{k-1} a_{k-1}\! +\! \cdots + 10a_1 + a_0 \equiv (-1)^k a_k + (-1)^{k-1} a_{k-1} + \cdots -a_1 + a_0 \pmod {11}. </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. | ||
| + | |||
| + | *** | ||
| + | Here is another way that doesn't require knowledge of modular arithmetic. Suppose we have a 3-digit number that is expressed in the form: | ||
| + | <math>100a+10b+c</math><br /> | ||
| + | we then can transpose this into: | ||
| + | <math>99a+a+11b-b+c</math><br /> | ||
| + | and that equals: | ||
| + | <math>(99a+11b)+(a-b+c)</math><br /> | ||
| + | which equals | ||
| + | <math>11(9a+b)+(a-b+c)</math><br /> | ||
| + | Since the first addend, <math>11(9a+b)</math> will always be divisible by 11, we just need to make sure that <math>(a-b+c)</math> is divisible by 11. | ||
| + | |||
| + | You can use this for any number. Here it is again, with an even-numbered digit number: | ||
| + | |||
| + | <math>1000a+100b+10c+d</math><br /> | ||
| + | <math>1001a-a+99b+b+11c-c+d</math><br /> | ||
| + | <math>(1001a+99b+11c)-(a-b+c-d)</math><br /> | ||
| + | <math>11(91a+9b+c)-(a-b+c-d)</math><br /> | ||
| + | So you just need to check <math>(a-b+c-d)</math> for divisibility with 11. | ||
== See also == | == See also == | ||
* [[Divisibility rules | Back to divisibility rules]] | * [[Divisibility rules | Back to divisibility rules]] | ||
| + | [[Category:Divisibility Rules]] | ||
Revision as of 11:12, 27 August 2017
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.
Here is another way that doesn't require knowledge of modular arithmetic. Suppose we have a 3-digit number that is expressed in the form:
we then can transpose this into:
and that equals:
which equals
Since the first addend, 
will always be divisible by 11, we just need to make sure that 
is divisible by 11.
You can use this for any number. Here it is again, with an even-numbered digit number:
So you just need to check 
for divisibility with 11.
