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Difference between revisions of "Divisibility rules"

m (→‎Type 2 blocks -- 111 and 11111)
m (→‎Bigger blocks -- 10001 and 10000001)
 
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Rule 3:  "Tail-End divisibility."  Note.  This only tells you if it is divisible and NOT the remainder.  Take a number say 12345.  Look at the last digit and add or subtract a multiple of 7 to make it zero.  In this case we get 12380 or 12310 (both are acceptable; I am using the former).  Lop off the ending 0's and repeat.  1238 - 28 ==> 1210 ==> 121 - 21 ==> 100 ==>  1  NOPE.  Works in general with numbers that are relatively prime to the base (and works GREAT in binary).  Here's one that works.  12348 - 28 ==> 12320 ==> 1232 +28 ==> 1260 ==> 126 + 14 ==> 14 YAY!
 
Rule 3:  "Tail-End divisibility."  Note.  This only tells you if it is divisible and NOT the remainder.  Take a number say 12345.  Look at the last digit and add or subtract a multiple of 7 to make it zero.  In this case we get 12380 or 12310 (both are acceptable; I am using the former).  Lop off the ending 0's and repeat.  1238 - 28 ==> 1210 ==> 121 - 21 ==> 100 ==>  1  NOPE.  Works in general with numbers that are relatively prime to the base (and works GREAT in binary).  Here's one that works.  12348 - 28 ==> 12320 ==> 1232 +28 ==> 1260 ==> 126 + 14 ==> 14 YAY!
 +
 +
Tiny extension to tell you the remainder:
 +
 +
Count how many zeroes you lop off and mod 6.
 +
 +
Multiply mod 7 with the corresponding number
 +
 +
Zeroes (mod 6)  Number to multiply by
 +
 +
      0                    1
 +
      1                    3
 +
      2                    2
 +
      3                    6
 +
      4                    4
 +
      5                    5
 +
 +
And that's the remainder.
  
 
=== Divisibility Rule for 10 and Powers of 10===
 
=== Divisibility Rule for 10 and Powers of 10===
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=== General Rule for Primes ===
 
=== General Rule for Primes ===
 
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 <math>p</math>, there exists some number <math>q</math> such that an integer is divisible by <math>p</math> if and only if truncating the last digit, multiplying it by <math>q</math> and subtracting it from the remaining number gives us a result divisible by <math>p</math>.  Divisibility rule 2 for 7 says that for <math>p = 7</math>, <math>q = 2</math>.  The divisibility rule for 11 is equivalent to choosing <math>q = 1</math>.  The divisibility rule for 3 is equivalent to choosing <math>q = -1</math>.  These rules can also be found under the appropriate conditions in [[number base]]s other than 10.  Also note that these rules exist in two forms: if <math>q</math> is replaced by <math>p - q</math> then subtraction may be replaced with addition.  We see one instance of this in the divisibility rule for 13: we could multiply by 9 and subtract rather than multiplying by 4 and adding.
 
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 <math>p</math>, there exists some number <math>q</math> such that an integer is divisible by <math>p</math> if and only if truncating the last digit, multiplying it by <math>q</math> and subtracting it from the remaining number gives us a result divisible by <math>p</math>.  Divisibility rule 2 for 7 says that for <math>p = 7</math>, <math>q = 2</math>.  The divisibility rule for 11 is equivalent to choosing <math>q = 1</math>.  The divisibility rule for 3 is equivalent to choosing <math>q = -1</math>.  These rules can also be found under the appropriate conditions in [[number base]]s other than 10.  Also note that these rules exist in two forms: if <math>q</math> is replaced by <math>p - q</math> then subtraction may be replaced with addition.  We see one instance of this in the divisibility rule for 13: we could multiply by 9 and subtract rather than multiplying by 4 and adding.
 +
 +
<math>q</math> is any number so that <math>p|(10q+1)</math>
  
 
=== Divisibility Rule for 13 ===
 
=== Divisibility Rule for 13 ===
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Why 49?  For taking pesky <math>7^2</math> out of a root.
 
Why 49?  For taking pesky <math>7^2</math> out of a root.
  
−
Useful until 2300.  Round up to the nearest 50, call it <math>A</math>, and subtract the original number, call this <math>D</math>.  If <math>A = 50 \cdot D</math> , it is divisible by 49.
+
Useful below 4900.  Round up to a multiple of 50, call it <math>A</math>, and subtract the original number, call this <math>D</math>.  If <math>A = 50 \cdot D</math> , it is divisible by 49.
  
 
Examples:
 
Examples:
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1470.  Round up: <math>A = 1500</math>. Difference: <math>D = 30</math>.  <math>A = 50 \cdot D</math>? Yes!
 
1470.  Round up: <math>A = 1500</math>. Difference: <math>D = 30</math>.  <math>A = 50 \cdot D</math>? Yes!
 +
 +
Extension to work for all numbers.
 +
Floor divide by 4950, multiply by 50, and add to <math>A</math> before calculating <math>D</math>
  
 
[[Divisibility rules/Rule for 49 proof | Proof]]
 
[[Divisibility rules/Rule for 49 proof | Proof]]
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====Bigger blocks -- 10001 and 10000001====
 
====Bigger blocks -- 10001 and 10000001====
 
10001 has block size length 4, and factors nicely into 73*137.
 
10001 has block size length 4, and factors nicely into 73*137.
 +
 
1000001 has block size 6, and factors into 17*5882353. 5882353 isn't much use, but 17 is, when we're testing a large number.
 
1000001 has block size 6, and factors into 17*5882353. 5882353 isn't much use, but 17 is, when we're testing a large number.
 +
 
====Type 2 blocks -- 111 and 11111====
 
====Type 2 blocks -- 111 and 11111====
 
A different type of test can be yielded from adding all the blocks, but again starting with the end.
 
A different type of test can be yielded from adding all the blocks, but again starting with the end.

Latest revision as of 21:11, 22 September 2024

These divisibility rules help determine when positive integers are divisible by particular other integers. All of these rules apply for base-10 only -- other bases have their own, different versions of these rules.

Divisibility Videos

https://youtu.be/bIipw2XSMgU https://youtu.be/6xNkyDgIhEE?t=1699

Basics

Divisibility Rule for 2 and Powers of 2

A number is divisible by $2^n$ if and only if the last ${n}$ digits of the number are divisible by $2^n$. Thus, in particular, a number is divisible by 2 if and only if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8.

Proof

Divisibility Rule for 3 and 9

A number is divisible by 3 or 9 if and only 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 and only if the last $n$ digits are divisible by that power of 5.

Proof

Divisibility Rule for 7

Rule 1: Partition $N$ into 3 digit numbers from the right ($d_3d_2d_1,d_6d_5d_4,\dots$). The alternating sum ($d_3d_2d_1 - d_6d_5d_4 + d_9d_8d_7 - \dots$) is divisible by 7 if and only if $N$ is divisible by 7.

Proof

Rule 2: Truncate the last digit of $N$, double that digit, and subtract it from the rest of the number (or vice-versa). $N$ is divisible by 7 if and only if the result is divisible by 7.

Proof

Rule 3: "Tail-End divisibility." Note. This only tells you if it is divisible and NOT the remainder. Take a number say 12345. Look at the last digit and add or subtract a multiple of 7 to make it zero. In this case we get 12380 or 12310 (both are acceptable; I am using the former). Lop off the ending 0's and repeat. 1238 - 28 ==> 1210 ==> 121 - 21 ==> 100 ==> 1 NOPE. Works in general with numbers that are relatively prime to the base (and works GREAT in binary). Here's one that works. 12348 - 28 ==> 12320 ==> 1232 +28 ==> 1260 ==> 126 + 14 ==> 14 YAY!

Tiny extension to tell you the remainder:

Count how many zeroes you lop off and mod 6.

Multiply mod 7 with the corresponding number 

Zeroes (mod 6)   Number to multiply by

     0                    1
     1                    3
     2                    2
     3                    6
     4                    4
     5                    5

And that's the remainder.

Divisibility Rule for 10 and Powers of 10

If a number is power of 10, define it as a power of 10. The exponent is the number of zeros that should be at the end of a number for it to be divisible by that power of 10.

Example: A number needs to have 6 zeroes at the end of it to be divisible by 1,000,000 because $1,000,000=10^6$.

Divisibility Rule for 11

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

Proof

General Rule for Composites

A number is divisible by $N$, where the prime factorization of $N$ is $p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n}$, if the number is divisible by each of $p_1^{e_1}, p_2^{e_2},\ldots, p_n^{e_n}$.

Example

For the example, we will check if 55682168544 is divisible by 36.

The prime factorization of 36 to be $2^2\cdot 3^2$. Thus we must check for divisibility by 4 and 9 to see if it's divisible by 36.

  • Since the last two digits, 44, of the number is divisible by 4, so is the entire number.
  • To check for divisibility by 9, we look to see if the sum of the digits is divisible by 9. The sum of the digits is 54 which is divisible by 9.

Thus, the number is divisible by both 4 and 9 and must be divisible by 36.

Advanced

General Rule for Primes

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. Also note that these rules exist in two forms: if $q$ is replaced by $p - q$ then subtraction may be replaced with addition. We see one instance of this in the divisibility rule for 13: we could multiply by 9 and subtract rather than multiplying by 4 and adding.

$q$ is any number so that $p|(10q+1)$

Divisibility Rule for 13

Rule 1: Truncate the last digit, multiply it by 4 and add it to the rest of the number. The result is divisible by 13 if and only if the original number was divisble by 13. This process can be repeated for large numbers, as with the second divisibility rule for 7.

Proof

Rule 2: Partition $N$ into 3 digit numbers from the right ($d_3d_2d_1,d_6d_5d_4,\dots$). The alternating sum ($d_3d_2d_1 - d_6d_5d_4 + d_9d_8d_7 - \dots$) is divisible by 13 if and only if $N$ is divisible by 13.

Proof

Rule 3: Works for $1 \leq N \leq 1000$. Let $K = 3N$. If $K$ is odd add 39 to $K$. Round $K$ up to the nearest multiple of 80, call the result $T$. Find $R = (T - K)/2$. Check: Is $T = R \cdot 80$.

Proof

Divisibility Rule for 17

Truncate the last digit, multiply it by 5 and subtract from the remaining leading number. The number is divisible if and only if the result is divisible. The process can be repeated for any number.

Proof

Divisibility Rule for 19

Truncate the last digit, multiply it by 2 and add to the remaining leading number. The number is divisible if and only if the result is divisible. This can also be repeated for large numbers.

Proof

Divisibility Rule for 29

Truncate the last digit, multiply it by 3 and add to the remaining leading number. The number is divisible if and only if the result is divisible. This can also be repeated for large numbers.

Proof

Divisibility Rule for 49

Why 49? For taking pesky $7^2$ out of a root.

Useful below 4900. Round up to a multiple of 50, call it $A$, and subtract the original number, call this $D$. If $A = 50 \cdot D$ , it is divisible by 49.

Examples:

49. Round up: $A = 50$. Difference: $D = 1$. $A = 50 \cdot D$? Yes!

1501. Round up: $A = 1550$. Difference: $D = 49$. $A = 50 \cdot D$? No!

1470. Round up: $A = 1500$. Difference: $D = 30$. $A = 50 \cdot D$? Yes!

Extension to work for all numbers. Floor divide by 4950, multiply by 50, and add to $A$ before calculating $D$

Proof

Special

Mod-preserving tests

These tests allow you take the modulo operation easily.

Mod-preserving for 7

Multiply the first digit by 3 and add it to the rest.

Mod-preserving for 13

Multiply the first digit by 3 and subtract it from the rest

Block tests

As a bonus, these are also mod-preserving

Small blocks -- 101 and 1001

The divisibility for 101 test is simple: Alternate adding and subtracting blocks of two digits starting from the end two, which are added.

Ex. 1102314 by 101

- 01 + 10 - 23 + 14 ← last block is always two digits and positive =0 so 1102314 is divisible by 101

The divisibility for 1001 is the same, but with blocks of three. (Starting with the end **three**, this time)

The 1001 test also works for all it's divisors. The most useful are 7, 11, and 13.

Bigger blocks -- 10001 and 10000001

10001 has block size length 4, and factors nicely into 73*137.

1000001 has block size 6, and factors into 17*5882353. 5882353 isn't much use, but 17 is, when we're testing a large number.

Type 2 blocks -- 111 and 11111

A different type of test can be yielded from adding all the blocks, but again starting with the end.

111 has a block length of three, and factors into 37 and 3.

11111 has a length of five, and factors to 41 and 271.

1111111, with a length of seven, can provide a test for 239 and 4649, if you ever need it.

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