Difference between revisions of "Divisibility rules/Rule for 2 and powers of 2 proof"
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== Proof == | == Proof == | ||
− | + | ''An understanding of [[Introduction to modular arithmetic | basic modular arithmetic]] is necessary for this proof.'' | |
Let the [[base numbers | base-ten]] representation of <math>N</math> be <math>\underline{a_ka_{k-1}\cdots a_1a_0}</math> where the <math>a_i</math> are digits for each <math>i</math> and the underline is simply to note that this is a base-10 expression rather than a product. If <math>N</math> has no more than <math>n</math> digits, then the last <math>n</math> digits of <math>N</math> make up <math>N</math> itself, so the test is trivially true. If <math>N</math> has more than <math>n</math> digits, we note that: | Let the [[base numbers | base-ten]] representation of <math>N</math> be <math>\underline{a_ka_{k-1}\cdots a_1a_0}</math> where the <math>a_i</math> are digits for each <math>i</math> and the underline is simply to note that this is a base-10 expression rather than a product. If <math>N</math> has no more than <math>n</math> digits, then the last <math>n</math> digits of <math>N</math> make up <math>N</math> itself, so the test is trivially true. If <math>N</math> has more than <math>n</math> digits, we note that: |
Revision as of 12:31, 27 November 2008
A number is divisible by
if the last
digits of the number are divisible by
.
Proof
An understanding of basic modular arithmetic is necessary for this proof.
Let the base-ten representation of be
where the
are digits for each
and the underline is simply to note that this is a base-10 expression rather than a product. If
has no more than
digits, then the last
digits of
make up
itself, so the test is trivially true. If
has more than
digits, we note that:
![$N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0.$](http://latex.artofproblemsolving.com/2/9/8/298c376ff3a0f4ffb08ae3357165f9f92c517586.png)
Taking this we have
![]() |
![]() |
![]() |
because for ,
. Thus,
is divisible by
if and only if
![$10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10 a_1 + a_0 = \underline{a_{n-1}a_{n-2}\cdots a_1a_0}$](http://latex.artofproblemsolving.com/1/2/f/12fc0f5913011d282abb009ef802e2c608b9cfde.png)
is. But this says exactly what we claimed: the last digits of
are divisible by
if and only if
is divisible by
.