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Difference between revisions of "Divisibility rules"

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These '''divisibility rules''' help determine when [[positive integer]]s are [[divisibility | divisible]] by particular other [[integer]]s.  All of these rules apply for [[Base number| base-10]] ''only'' -- other bases have their own, different versions of these rules.
 
These '''divisibility rules''' help determine when [[positive integer]]s are [[divisibility | divisible]] by particular other [[integer]]s.  All of these rules apply for [[Base number| 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
  
−
== Divisibility Rule for 2 and Powers of 2 ==
+
==Basics==
 +
 
 +
=== Divisibility Rule for 2 and Powers of 2 ===
 
A number is divisible by <math>2^n</math> if and only if the last <math>{n}</math> digits of the number are divisible by <math>2^n</math>.  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.
 
A number is divisible by <math>2^n</math> if and only if the last <math>{n}</math> digits of the number are divisible by <math>2^n</math>.  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.
  
 
[[Divisibility rules/Rule for 2 and powers of 2 proof | Proof]]
 
[[Divisibility rules/Rule for 2 and powers of 2 proof | Proof]]
  
−
== Divisibility Rule for 3 and 9==
+
=== 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.
 
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.
  
 
[[Divisibility rules/Rule for 3 and 9 proof | Proof]]
 
[[Divisibility rules/Rule for 3 and 9 proof | Proof]]
  
−
== Divisibility Rule for 5 and Powers of 5 ==
+
=== Divisibility Rule for 5 and Powers of 5 ===
 
A number is divisible by <math>5^n</math> if and only if the last <math>n</math> digits are divisible by that power of 5.
 
A number is divisible by <math>5^n</math> if and only if the last <math>n</math> digits are divisible by that power of 5.
  
 
[[Divisibility rules/Rule for 5 and powers of 5 proof | Proof]]
 
[[Divisibility rules/Rule for 5 and powers of 5 proof | Proof]]
  
−
== Divisibility Rule for 7 ==
+
=== Divisibility Rule for 7 ===
 
Rule 1:  Partition <math>N</math> into 3 digit numbers from the right (<math>d_3d_2d_1,d_6d_5d_4,\dots</math>).  The alternating sum (<math>d_3d_2d_1 - d_6d_5d_4 + d_9d_8d_7 - \dots</math>) is divisible by 7 if and only if <math>N</math> is divisible by 7.
 
Rule 1:  Partition <math>N</math> into 3 digit numbers from the right (<math>d_3d_2d_1,d_6d_5d_4,\dots</math>).  The alternating sum (<math>d_3d_2d_1 - d_6d_5d_4 + d_9d_8d_7 - \dots</math>) is divisible by 7 if and only if <math>N</math> is divisible by 7.
  
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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!
  
−
== Divisibility Rule for 10 and Powers of 10==
+
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.
 
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.
  
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A number needs to have 6 zeroes at the end of it to be divisible by 1,000,000 because <math>1,000,000=10^6</math>.
 
A number needs to have 6 zeroes at the end of it to be divisible by 1,000,000 because <math>1,000,000=10^6</math>.
  
−
== Divisibility Rule for 11 ==
+
=== Divisibility Rule for 11 ===
−
A number is divisible by 11 if and only if the [[alternating sum]] of the digits is divisible by 11.
+
A number is divisible by 11 if the [[alternating sum]] of the digits is divisible by 11.
  
 
[[Divisibility rules/Rule for 11 proof | Proof]]
 
[[Divisibility rules/Rule for 11 proof | Proof]]
  
−
== Divisibility Rule for 13 ==
+
=== General Rule for Composites ===
 +
A number is divisible by <math>N</math>, where the [[prime factorization]] of <math>N</math> is <math>p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n}</math>, if the number is divisible by each of <math>p_1^{e_1}, p_2^{e_2},\ldots, p_n^{e_n}</math>.
 +
 
 +
==== Example ====
 +
 
 +
For the example, we will check if 55682168544 is divisible by 36.
 +
 
 +
The prime factorization of 36 to be <math>2^2\cdot 3^2</math>.  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 <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 ===
 
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.   
 
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.   
  
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[[Divisibility rules/Rule 2 for 13 proof | Proof]]
 
[[Divisibility rules/Rule 2 for 13 proof | Proof]]
  
−
==Divisibility Rule for 17==
+
Rule 3:  Works for <math>1 \leq N \leq 1000</math>.  Let <math>K = 3N</math>.  If <math>K</math> is odd add 39 to <math>K</math>.  Round <math>K</math> up to the nearest multiple of 80, call the result <math>T</math>.  Find <math>R = (T - K)/2</math>.  Check: Is <math>T = R \cdot 80</math>.
 +
 
 +
[[Divisibility rules/Rule 3 for 13 proof | 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.
 
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.
  
 
[[Divisibility rules/Rule for 17 proof | Proof]]
 
[[Divisibility rules/Rule for 17 proof | Proof]]
  
−
==Divisibility Rule for 19==
+
===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.
 
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.
  
 
[[Divisibility rules/Rule for 19 proof | Proof]]
 
[[Divisibility rules/Rule for 19 proof | Proof]]
  
−
==More general note for primes ==
+
===Divisibility Rule for 29===
−
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.
+
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.
 +
 
 +
[[Divisibility rules/Rule for 29 proof | Proof]]
 +
 
 +
===Divisibility Rule for 49===
 +
Why 49?  For taking pesky <math>7^2</math> out of a root.
 +
 
 +
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:
 +
 
 +
49.  Round up: <math>A = 50</math>. Difference: <math>D = 1</math>.  <math>A = 50 \cdot D</math>? Yes!
 +
 
 +
1501.  Round up: <math>A = 1550</math>. Difference: <math>D = 49</math>.  <math>A = 50 \cdot D</math>? No!
 +
 
 +
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]]
 +
 
 +
==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.
  
−
== More general note for composites ==
+
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.
−
A number is divisible by <math>N</math>, where the [[prime factorization]] of <math>N</math> is <math>p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n}</math>, if the number is divisible by each of <math>p_1^{e_1}, p_2^{e_2},\ldots, p_n^{e_n}</math>.
 
  
−
=== Example ===
+
====Type 2 blocks -- 111 and 11111====
−
Is 55682168544 divisible by 36?
+
A different type of test can be yielded from adding all the blocks, but again starting with the end.
  
−
==== Solution ====
+
111 has a block length of three, and factors into 37 and 3.
−
First, we find the prime factorization of 36 to be <math>2^2\cdot 3^2</math>.  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.
+
11111 has a length of five, and factors to 41 and 271.
  
−
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.
+
1111111, with a length of seven, can provide a test for 239 and 4649, if you ever need it.
  
−
Thus, the number is divisible by both 4 and 9 and must be divisible by 36.
+
== Problems ==
  
−
== Example Problems ==
+
* Practice Problems on [https://artofproblemsolving.com/alcumus/ Alcumus]
−
* [[2006_AMC_10B_Problems/Problem_25 | 2006 AMC 10B Problem 25]]
+
** Divisibility (Prealgebra)
 +
* [[2000 AMC 8 Problems/Problem 11]]
 +
* [[2006 AMC 10B Problems/Problem 25]]
  
  
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* [[Math books]]
 
* [[Math books]]
 
* [[Mathematics competitions]]
 
* [[Mathematics competitions]]
 +
 +
[[Category:Divisibility Rules]]

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.

Problems


Resources

Books

Classes


See also

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