Difference between revisions of "Divisibility rules/Rule for 2 and powers of 2 proof"
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== Proof == | == Proof == | ||
+ | === Basic Idea === | ||
+ | Let the number <math>N</math> be <math>(10^n)k + p</math> where k and p are integers and <math>p<(10^n)</math>. Since <math>\frac{10^n}{2^n}</math> is <math>5^n</math>, <math>(10^n)</math> is a multiple of <math>2^n</math>, meaning <math>(10^n)k</math> is also a multiple of <math>2^n</math>. As long as p is a multiple of <math>2^n</math>, then <math>N</math> is a multiple of <math>2^n</math>. Since <math>(10^n)k</math> has <math>n</math> trailing 0's, <math>p</math> is the last <math>n</math> digits of the number <math>n</math>. | ||
+ | |||
+ | === Concise === | ||
''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 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: |
− | + | <center><math> N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0. </math></center> | |
− | Taking <math> | + | Taking this <math>\mod 2^n</math> we have |
− | {| class="wikitable" style="margin: 1em auto 1em auto | + | {| class="wikitable" style="margin: 1em auto 1em auto" |
− | | <math>N</math> || <math>= 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 | + | | <math>N</math> || <math>= 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0 </math> |
− | |||
− | |||
|- | |- | ||
− | | || <math>\equiv a_{n-1}a_{n-2}\cdots | + | | || <math>\equiv 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10 a_1 + a_0 \pmod{2^n}</math> |
|} | |} | ||
− | Thus, if the last <math>n</math> digits of <math>N</math> are divisible by <math>2^n</math> | + | because for <math>i \geq n</math>, <math>10^i \equiv 0 \pmod{2^n}</math>. Thus, <math>N</math> is divisible by <math>2^n</math> if and only if |
+ | |||
+ | <center><math>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}</math></center> | ||
+ | |||
+ | is. But this says exactly what we claimed: the last <math>n</math> digits of <math>N</math> are divisible by <math>2^n</math> if and only if <math>N</math> is divisible by <math>2^n</math>. | ||
== See also == | == See also == | ||
* [[Divisibility rules | Back to divisibility rules]] | * [[Divisibility rules | Back to divisibility rules]] | ||
+ | [[Category:Divisibility Rules]] |
Latest revision as of 16:46, 3 May 2020
A number is divisible by
if the last
digits of the number are divisible by
.
Contents
Proof
Basic Idea
Let the number be
where k and p are integers and
. Since
is
,
is a multiple of
, meaning
is also a multiple of
. As long as p is a multiple of
, then
is a multiple of
. Since
has
trailing 0's,
is the last
digits of the number
.
Concise
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
.