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Difference between revisions of "Divisibility rules"
(→‎Divisibility Rule for 11)
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[[Divisibility rules/Rule for 11 proof | Proof]]
 
[[Divisibility rules/Rule for 11 proof | Proof]]
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===Proof===
 
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<math>10\equiv -1\pmod{11}</math>.  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:
 
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<math>3\cdot (-1)^3+8\cdot (-1)^2+0\cdot (-1)^1+6\cdot (-1)^0=-3+8-0+6\equiv 0\pmod{11}</math>.
 
  
 
== Divisibility Rule for 7 ==
 
== Divisibility Rule for 7 ==

Revision as of 08:17, 16 August 2006

These divisibility rules help determine when integers are divisible by particular other integers.


Divisibility Rule for 2 and Powers of 2

A number is divisible by $2^n$ if the last ${n}$ digits of the number are divisible by $2^n$.

Proof

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.

Proof

Divisibility Rule for 5 and Powers of 5

A number is divisible by $5^n$ if the last $n$ digits are divisible by that power of 5.

Proof


Divisibility Rule for 11

A number is divisible by 11 if the alternating sum of the digits is divisible by 11.

Proof

Divisibility Rule for 7

Rule 1: Partition $n$ into 3 digit numbers from the right ($d_3d_2d_1,d_6d_5d_4,\dots$). If the alternating sum ($d_3d_2d_1 - d_6d_5d_4 + d_9d_8d_7 - \dots$) is divisible by 7, then the number is divisible by 7.

Proof

Rule 2: Truncate the last digit of ${n}$, and double that digit, subtracting the rest of the number from the doubled last digit. If the absolute value of the result is a multiple of 7, then the number itself is.

Proof

Proof for Rule 2:

The divisibility rule would be $2n_0-k$, where $k=d_110^0+d_210^1+d_310^2+...$, where $d_{n-1}$ is the nth digit from the right (NOT the left) and we have $k-2n_0\equiv 2n_0+6k$ and since 2 is relatively prime to 7, $2n_0+6k\equiv n_0+3k\pmod{7}$. Then yet again $n_0+3k\equiv n_0+10k\pmod{7}$, and this is equivalent to our original number.

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 with the second divisibility rule for 7.

Proof

Proof

Let $n = d_0\cdot10^0 + d_1\cdot 10^1 +d_2\cdot 10^2 + \ldots$ be a positive integer with units digit $d_0$, tens digit $d_1$ and so on. Then $k=d_110^0+d_210^1+d_310^2+...$ is the result of truncating the last digit from $n$. Note that $n = 10k + d_0 \equiv d_0 - 3k \pmod 13$. Now $n \equiv 0 \pmod 13$ if and only if $4n \equiv 0 \pmod 13$, from which the rule follows ... (someone add the last 2 lines of pf.)

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 $p$, there exists some number $q$ such that an integer is divisible by $p$ if and only if truncating the last digit, multiplying it by $q$ and subtracting it from the remaining number gives us a result divisible by $p$. Divisibility rule 2 for 7 says that for $p = 7$, $q = 2$. The divisibility rule for 11 is equivalent to choosing $q = 1$. The divisibility rule for 3 is equivalent to choosing $q = -1$. These rules can also be found under the appropriate conditions in number bases other than 10.


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