Difference between revisions of "Divisibility rules"
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== Divisibility Rule for 13 == | == Divisibility Rule for 13 == | ||
− | + | Multiply the last digit by 4 and add it to the rest of the number. This process can be repeated for large numbers, as is with the divisibility rule for 7. | |
+ | ===Proof=== | ||
+ | Say <math>k=d_110^0+d_210^1+d_310^2+...</math>, where <math>d_{n+1}</math> is the nth digit from the right (NOT THE LEFT). The rule becomes <math>4d_0+k\equiv 4d_0-12k\pmod{13}</math>. Then <math>4d_0-12k\equiv d_0-3k\pmod{13}</math>, because you can divide by 4 in any instance mod 13. This goes further to <math>d_0+10k</math>, inspecting closely, this is our original number! | ||
==More general note == | ==More general note == |
Revision as of 10:52, 15 August 2006
These divisibility rules help determine when integers are divisible by particular other integers.
Contents
[hide]Divisibility Rule for 2 and Powers of 2
A number is divisible by if the last digits of the number are divisible by .
Divisibility Rule for 3 and 9
A number is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9, respectively. Note that this does not work for higher powers of 3. For instance, the sum of the digits of 1899 is divisible by 27, but 1899 is not itself divisible by 27.
Divisibility Rule for 5 and Powers of 5
A number is divisible by if the last n digits are divisible by that power of 5.
Proof
An understanding of basic modular arithmetic is necessary for this proof.
Consider, for example, the test for divisibility by :
Let be a positive integer. Then is divisible by if and only if the sum of the base-ten digits of is divisible by .
Arithmetic mod can be used to give an easy proof of this criterion:
Suppose that the base-ten representation of is
,
where is a digit for each . Then the numerical value of is given by
.
Now we know that, since , we have (mod ). So by the "exponentiation" property above, we have (mod ) for every .
Therefore, by repeated uses of the "addition" and "multiplication" properties, we can write
(mod ).
Therefore, we have
(mod ).
That is, differs from the sum of its digits by a multiple of . It follows, then, that is a multiple of if and only if the sum of its digits is a multiple of .
Since we also have , the same proof also gives the divisibility rule for 3. The proof fails for 27 because .
Divisibility Rule for 11
A number is divisible by 11 if the alternating sum of the digits is divisible by 11.
Proof
. Since the number is in base 10, as we assume, the digits would be of powers of 10, or in mod 11, powers of -1. The number 3806 thus becomes:
.
Divisibility Rule for 7
Rule 1: Partition into 3 digit numbers from the right (). If the alternating sum () is divisible by 7, then the number is divisible by 7.
Rule 2: Truncate the last digit of , and subtract twice that digit from the remaining number. If the result is divisible by 7, then the number is divisible by 7. This process can be repeated for large numbers.
Divisibility Rule for 13
Multiply the last digit by 4 and add it to the rest of the number. This process can be repeated for large numbers, as is with the divisibility rule for 7.
Proof
Say , where is the nth digit from the right (NOT THE LEFT). The rule becomes . Then , because you can divide by 4 in any instance mod 13. This goes further to , inspecting closely, this is our original number!
More general note
For every prime number other than 2 and 5, there exists a rule similar to rule 2 for divisibility by 7. For a general prime , there exists some number such that an integer is divisible by if and only if truncating the last digit, multiplying it by and subtracting it from the remaining number gives us a result divisible by . Divisibility rule 2 for 7 says that for , . The divisibility rule for 11 is equivalent to choosing . The divisibility rule for 3 is equivalent to choosing . These rules can also be found under the appropriate conditions in number bases other than 10.
Example Problems
Resources
Books
- The AoPS Introduction to Number Theory by Mathew Crawford.
Classes