Difference between revisions of "Divisibility rules/Rule for 2 and powers of 2 proof"
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== Proof == | == Proof == | ||
=== Basic Idea === | === 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> | + | 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 === | === Concise === |
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:
Taking this we have
because for , . Thus, is divisible by if and only if
is. But this says exactly what we claimed: the last digits of are divisible by if and only if is divisible by .