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Difference between revisions of "Divisibility rules/Rule for 2 and powers of 2 proof"

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== Proof ==
 
== Proof ==
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=== Basic Idea ===
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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>.
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=== 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.''
  
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Let <math>N = a_ka_{k-1}\cdots a_1a_0</math> be the [[base numbers | base-ten]] expression for <math>N</math>, where the <math>a_i</math> are [[digit]]s.
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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:
  
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Thus <center><math> N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0. </math></center>
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<center><math> N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0. </math></center>
  
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Taking <math>N</math> mod <math>2^n</math> gives
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Taking this <math>\mod 2^n</math> we have
  
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{| class="wikitable" style="margin: 1em auto 1em auto;height:100px"
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{| class="wikitable" style="margin: 1em auto 1em auto"
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| <math>N</math> || <math>= 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0</math>
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| <math>N</math> || <math>= 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0 </math>
 
|-
 
|-
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| || <math>\displaystyle \equiv 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + a_1 + a_0 \mod 2^n </math>
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| || <math>\equiv 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10 a_1 + a_0 \pmod{2^n}</math>
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|-
 
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| || <math>= a_{n-1}a_{n-2}\cdots a_1a_0</math>
 
 
|}
 
|}
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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
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<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>
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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 $N$ is divisible by $2^n$ if the last ${n}$ digits of the number are divisible by $2^n$.

Proof

Basic Idea

Let the number $N$ be $(10^n)k + p$ where k and p are integers and $p<(10^n)$. Since $\frac{10^n}{2^n}$ is $5^n$, $(10^n)$ is a multiple of $2^n$, meaning $(10^n)k$ is also a multiple of $2^n$. As long as p is a multiple of $2^n$, then $N$ is a multiple of $2^n$. Since $(10^n)k$ has $n$ trailing 0's, $p$ is the last $n$ digits of the number $n$.

Concise

An understanding of basic modular arithmetic is necessary for this proof.

Let the base-ten representation of $N$ be $\underline{a_ka_{k-1}\cdots a_1a_0}$ where the $a_i$ are digits for each $i$ and the underline is simply to note that this is a base-10 expression rather than a product. If $N$ has no more than $n$ digits, then the last $n$ digits of $N$ make up $N$ itself, so the test is trivially true. If $N$ has more than $n$ digits, we note that:

$N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0.$

Taking this $\mod 2^n$ we have

$N$ $= 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0$
$\equiv 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10 a_1 + a_0 \pmod{2^n}$

because for $i \geq n$, $10^i \equiv 0 \pmod{2^n}$. Thus, $N$ is divisible by $2^n$ if and only if

$10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10 a_1 + a_0 = \underline{a_{n-1}a_{n-2}\cdots a_1a_0}$

is. But this says exactly what we claimed: the last $n$ digits of $N$ are divisible by $2^n$ if and only if $N$ is divisible by $2^n$.

See also

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