Divisibility rule

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A divisibility rule is a method that can be used to determine whether a number divides other numbers. Divisibility rules are useful when it is necessary to determine a number's factors and a calculator is not present, or if the number is too high for the calculator to compute. Although divisibility rules can be defined for any base, only rules for decimal are given here.

2 through 20

The rules given below transform a given number into a generally smaller number while preserving divisibility by the divisor of interest. Therefore unless otherwise noted the resulting number should be evaluated for divisibility by the same divisor.

For divisors with multiple rules, the rules are generally ordered first for those appropriate for numbers with many digits, then those useful for numbers with fewer digits.

If the result is not obvious by examination, the same or another rule should be applied to the result.

Divisor	Divisibility Condition	Examples
2	The last digit is even (0, 2, 4, 6, or 8).	1,294: 4 is even.
3	Sum the digits.	405: 4 + 0 + 5 = 9.
4	The number formed by the last two digits is divisible by 4, as follows:
	If the tens digit is even, the last digit is divisible by 4 (0, 4, 8).	184: 8 is even, and 4 is divisible by 4.
	If the tens digit is odd, the last digit plus 2 is divisible by 4 (2, 6).	5,496: 9 is odd, and 6+2 is divisible by 4.
5	The last digit is 0 or 5.	490: the last digit is 0.
6	It is divisible by 2 and by 3.	24: it is divisible by 2 and by 3.
7	Sum the digits in alternate blocks of three from right to left, then difference the two sums.	2,911,272: − (2 + 272) + 911 = 637
	Sum the number with the last two digits removed, doubled, plus the last two digits.	364: (3x2) + 64 = 70.
	Sum the number with the last digit removed with 5 times the last digit.	364: 36 + (5×4) = 56.
	Difference the number with the last digit removed with 2 times the last digit.	364: 36 − (2×4) = 28.
8	The number formed by the last three digits is divisible by 8, as follows:
	If the hundreds digit is even, examine the number formed by the last two digits.	624: 24.
	If the hundreds digit is odd, examine the number formed by the last two digits plus 4.	352: 52+4 = 56.
	Sum the number with the last digit removed, doubled, plus the last digit.	56: (5x2) + 6 = 16.
9	Sum the digits.	2,880: 2 + 8 + 8 + 0 = 18.
10	The last digit is 0.	130: the last digit is 0.
11	Sum the digits in blocks of two from right to left.	627: 6 + 27 = 33.
	Difference the number with the last digit removed with the last digit.	627: 62 - 7 = 55.
	Sum alternate digits, then difference the two sums.	182,919: − (1 + 2 + 1) + (8 + 9 + 9) = 22.
12	It is divisible by 3 and by 4.	324: it is divisible by 3 and by 4.
	Difference the number with the last digit removed, doubled, with the last digit.	324: (32x2) − 4 = 60.
13	Sum the digits in alternate blocks of three from right to left, then difference the two sums.	2,911,272: − (2 + 272) + 911 = 637
	Sum the number with the last digit removed with 4 times the last digit.	338: 33 + (8×4) = 65.
	Difference the number with the last digit removed with 9 times the last digit.	637: 63 − (7×9) = 0.
14	It is divisible by 2 and by 7.	224: it is divisible by 2 and by 7.
	Sum the number with the last two digits removed, doubled, plus the last two digits.	364: (3x2) + 64 = 70.
15	It is divisible by 3 and by 5.	390: it is divisible by 3 and by 5.
16	The number formed by the last four digits is divisible by 16, as follows:
	If the thousands digit is even, examine the number formed by the last three digits.	254,176: 176.
	If the thousands digit is odd, examine the number formed by the last three digits plus 8.	3,408: 408+8 = 416.
	Sum the number with the last two digits removed, times 4, plus the last two digits.	176: (1x4) + 76 = 80.
17	Difference the number with the last two digits removed, doubled, with the last digit.	187: − (1x2) + 87 = 85.
	Difference the number with the last digit removed with 5 times the last digit.	85: − 8 + (5×5) = 17.
18	It is divisible by 2 and by 9.	342: it is divisible by 2 and by 9.
19	Sum the number with the last digit removed with twice the last digit.	437: 43 + (7x2) = 57.
20	It is divisible by 10, and the tens digit is even.	360: is divisible by 10, and 6 is even.

Beyond 20

Divisibility properties can be determined in two ways, depending on the type of the divisor.

Composite divisors
A number is divisible by a given divisor, if it is divisible by the highest power of each of its prime factors. For example, to determine divisibility by 24, one would check divisibility by 8 and by 3. Note that checking 4 and 6, or 2 and 12, would not be sufficient. A table of prime factors may be useful.

A composite divisor may also have a rule formed using the same procedure as for a prime divisor, given below, with the caveat that the manipulations involved may not introduce any factor which is present in the divisor. For instance, one can not make a rule for 14 that involves multiplying the equation by 7. This is not an issue for prime divisors because they have no smaller factors.

Prime divisors
The goal is to find a multiple of the divisor which is close to a round number, and then manipulate the representation of the dividend to produce that number, as shown in the proofs section below. Using 17 as an example, we might find 3x17 = 51, and 6x17 = 102. Both these numbers produce rules in the table above, one using 50y + 5z, and the other using 100y + z.

Notable examples
The following table provides rules for a few more notable divisors:

Divisor	Divisibility Condition	Examples
25	The number formed by the last two digits is divisible by 25.	134,250: 50 is divisible by 25.
27	Sum the digits in blocks of three from right to left.	2,644,272: 2 + 644 + 272 = 918.
	Difference the number with the last digit removed with 8 times the last digit.	621: 62 − (1×8) = 54.
32	The number formed by the last five digits is divisible by 32, as follows:
	If the ten thousands digit is even, examine the number formed by the last four digits.	41,312: 1312.
	If the ten thousands digit is odd, examine the number formed by the last four digits plus 16.	254,176: 4176+16 = 4192.
	Sum the number with the last two digits removed, times 4, plus the last two digits.	1,312: (13x4) + 12 = 64.
33	Sum the digits in blocks of two from right to left.	627: 6 + 27 = 33.
37	Sum the digits in blocks of three from right to left.	2,651,272: 2 + 651 + 272 = 925.
	Difference the number with the last digit removed with 11 times the last digit.	925: 92 − (5x11) = 37.
49	Sum the number with the last digit removed with 5 times the last digit.	1,127: 112 + (7×5) = 147.

Proofs

Proof using basic algebra

Many of the simpler rules can be produced using only algebraic manipulation, creating binomials and rearranging them. By writing a number as the sum of each digit times a power of 10 each digit's power can be manipulated individually.

Case where all digits are summed

Using 3 as an example, 3 divides [10^n - 1] (9, 99, 999, etc.) for all [n]. So, a number such as

[100\cdot a + 10\cdot b + c]

can be rewritten as

[(99 + 1) \cdot a + (9 + 1) \cdot b + c]

and then

[99 \cdot a + 9 \cdot b + 1 \cdot a + 1 \cdot b + c].

Then, since we are only interested in whether the number [abc] is divisible by 3, not the quotient, we can eliminate all parts that we know are a mulitple of 3, i.e., the multiples of 9 and 99. The remaining result is

[a + b + c \,\!]

which is exactly the sum of the digits.

This is also true for numbers such as 4 and 25, except that once a high enough power of 10 is reached, each of those terms can be eliminated directly.

Case where only the last digit(s) are removed

Most numbers do not divide 9 or 10 evenly, but do divide a higher power of [10^n] or [10^n - 1]. In this case the number is still written in powers of 10, but not fully expanded.

For example, 7 does not divide 9 or 10, but does divide 98, which is close to 100. Thus, proceed from

[100 \cdot a + b]

where in this case a is any integer, and b can range from 0 to 99. Next,

[(98+2) \cdot a + b]

and again expanding

[98 \cdot a + 2 \cdot a + b],

and after eliminating the known multiple of 7, the result is

[2 \cdot a + b],

which is the rule "double the number formed by all but the last two digits, then add the last two digits".

Case where the last digit(s) must be multiplied by a factor

The representation of the number may also be multiplied by any other factor not present in the divisor without changing its divisibility. After observing that 7 divides 21, we can perform the following:

[10 \cdot a + b],

after multiplying by 2, becomes

[20 \cdot a + 2 \cdot b],

and then

[(21 - 1) \cdot a + 2 \cdot b].

Eliminating the 21 gives

[ -1 \cdot a + 2 \cdot b],

and multiplying by −1 gives

[ a - 2 \cdot b].

Either of the last two rules may be used, depending on which is easier to perform. They correspond to the rule "difference the number with the last digit removed, with twice the last digit".

Proof using modular arithmetic

This section will illustrate the basic method; all the rules can be derived following the same procedure. The following requires a basic grounding in modular arithmetic; the proofs rest on the basic fact that a divides b if and only if b ≡ 0 (mod a).

For 2ⁿ or 5ⁿ:

Only the last n digits need to be checked.

[10^n = 2^n \cdot 5^n \equiv 0 \pmod 5^n}]

Representing x as [10^n \cdot y + z],

[x = 10^n \cdot y + z \equiv z \pmod 5^n}]

and the divisibility of x is the same as that of z.

For 7:

Since 1000 ≡ −1 (mod 7) we can do the following:

Representing x as [1000 \cdot y + z],

[x = 1000 \cdot y + z \equiv -y + z \pmod]

Or using 20 ≡ −1 (mod 7) with x as [10 \cdot y + z], we can multiply both sides of [10 \cdot y + z \equiv 0] by two and then

[20 \cdot y + 2 \cdot z \equiv -y + 2 \cdot z]

which must also be 0 if 7 divides x.

See also

• Divisor
• Table of divisors — A table of prime and non-prime divisors for 1–1000
• Table of prime factors — For determining prime factors
• Integer factorization — for more sophisticated techniques

External links

• [Divisibility Criteria] at cut-the-knot
• [Divisibility by 9 and 11] at cut-the-knot
• [Divisibility by 7] at cut-the-knot
• [Divisibility by 81] at cut-the-knot

 

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