14,443
edits
(→Divisibility rules for numbers 1–20: add "id" attributes to table for easier linking) 

Note: To test divisibility by any number that can be expressed as the product of prime factors <math>p_1^n p_2^m p_3^q</math>, we can separately test for divisibility by each prime to its appropriate power. For example, testing divisibility by 18 (18 = 9*2 = 3<sup>2</sup>*2) is equivalent to testing divisibility by 9 (3<sup>2</sup>) and 2 simultaneously, thus we need only show divisibility by 9 and by 2 to prove divisibility by 18.
<! Note: the "id" attributes are there to allow direct linking to this table as e.g. [[Divisibility rule#7]]. >
{ class="wikitable"
! Divisor
! Examples

id=1 '''[[1 (number)1]]'''
 No special condition. Any integer is divisible by 1.
 2 is divisible by 1.

id=2 '''[[2 (number)2]]'''
 The last digit is even (0, 2, 4, 6, or 8).<ref name="Pascal'scriterion">This follows from Pascal's criterion. See Kisačanin (1998), [{{Google booksplainurl=yid=BFtOuh5xGOwCpage=101text=A number is divisible by}} p. 100–101]</ref><ref name="lastmdigits">A number is divisible by 2<sup>''m''</sup>, 5<sup>''m''</sup> or 10<sup>''m''</sup> if and only if the number formed by the last ''m'' digits is divisible by that number. See Richmond & Richmond (2009), [{{Google booksplainurl=yid=HucyKYx0_WwCpage=105text=formed by the last}} p. 105]</ref>
 1,294: 4 is even.

id=3 rowspan=2 '''[[3 (number)3]]'''
 Sum the digits. If the result is divisible by 3, then the original number is divisible by 3.<ref name="Pascal'scriterion"/><ref name="apostol1976">Apostol (1976), [{{Google booksplainurl=yid=Il64dZELHEICpage=108text=sum of its digits}} p. 108]</ref><ref name="RichmondRichmond2009">Richmond & Richmond (2009), [{{Google booksplainurl=yid=HucyKYx0_WwCpage=102text=divisible by}} Section 3.4 (Divisibility Tests), p. 102–108]</ref>
 405 → 4 + 0 + 5 = 9 and 636 → 6 + 3 + 6 = 15 which both are clearly divisible by 3.<br>16,499,205,854,376 → 1+6+4+9+9+2+0+5+8+5+4+3+7+6 sums to 69 → 6 + 9 = 15 → 1 + 5 = 6, which is clearly divisible by 3.
 Using the example above: 16,499,205,854,376 has '''four''' of the digits 1, 4 and 7 and '''four''' of the digits 2, 5 and 8; ∴ Since 4 − 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3.

id=4 rowspan=3 '''[[4 (number)4]]'''
 Examine the last two digits.<ref name="Pascal'scriterion"/><ref name="lastmdigits"/>
 40832: 32 is divisible by 4.
 40832: 2 × 3 + 2 = 8, which is divisible by 4.

id=5 '''[[5 (number)5]]'''
 The last digit is 0 or 5.<ref name="Pascal'scriterion"/><ref name="lastmdigits"/>
 495: the last digit is 5.

id=6 '''[[6 (number)6]]'''
 It is divisible by 2 and by 3.<ref name="productofcoprimes">Richmond & Richmond (2009), [{{Google booksplainurl=yid=HucyKYx0_WwCpage=102text=divisible by the product}} Section 3.4 (Divisibility Tests), Theorem 3.4.3, p. 107]</ref>
 1,458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6.

id=7 rowspan=5 '''[[7 (number)7]]'''
 Form the [[alternating sum]] of blocks of three from right to left.<ref name="RichmondRichmond2009"/><ref name="alternatingsumofblocksofthree">Kisačanin (1998), [{{Google booksplainurl=yid=BFtOuh5xGOwCpage=101text=third criterion for 11}} p. 101]</ref>
 1,369,851: 851 − 369 + 1 = 483 = 7 × 69
 483595: (4 × (2)) + (8 × (3)) + (3 × (1)) + (5 × 2) + (9 × 3) + (5 × 1) = 7.

id=8 rowspan=5 '''[[8 (number)8]]'''
style="borderbottom: hidden;" If the hundreds digit is even, examine the number formed by the last two digits.
style="borderbottom: hidden;" 624: 24.
 34152: 4 × 1 + 5 × 2 + 2 = 16

id=9 '''[[9 (number)9]]'''
 Sum the digits. If the result is divisible by 9, then the original number is divisible by 9.<ref name="Pascal'scriterion"/><ref name="apostol1976"/><ref name="RichmondRichmond2009"/>
 2,880: 2 + 8 + 8 + 0 = 18: 1 + 8 = 9.

id=10 '''[[10 (number)10]]'''
 The last digit is 0.<ref name="lastmdigits"/>
 130: the last digit is 0.

id=11 rowspan=6 '''[[11 (number)11]]'''
 Form the alternating sum of the digits.<ref name="Pascal'scriterion"/><ref name="RichmondRichmond2009"/>
 918,082: 9 − 1 + 8 − 0 + 8 − 2 = 22.
 14,179: the number of digits is odd (5) → 417 − 1 − 9 = 407 = 37 × 11

id=12 rowspan=2 '''[[12 (number)12]]'''
 It is divisible by 3 and by 4.<ref name="productofcoprimes"/>
 324: it is divisible by 3 and by 4.
 324: 32 × 2 − 4 = 60.

id=13 rowspan=3 '''[[13 (number)13]]'''
 Form the [[alternating sum]] of blocks of three from right to left.<ref name="alternatingsumofblocksofthree"/>
 2,911,272: −2 + 911 − 272 = 637
 637: 63  63 = 0.

id=14 rowspan=2 '''[[14 (number)14]]'''
 It is divisible by 2 and by 7.<ref name="productofcoprimes"/>
 224: it is divisible by 2 and by 7.
 364: 3 × 2 + 64 = 70.<br />1764: 17 × 2 + 64 = 98.

id=15 '''[[15 (number)15]]'''
 It is divisible by 3 and by 5.<ref name="productofcoprimes"/>
 390: it is divisible by 3 and by 5.

id=16 rowspan=4 '''[[16 (number)16]]'''
style="borderbottom: hidden;" If the thousands digit is even, examine the number formed by the last three digits.
style="borderbottom: hidden;" 254,176: 176.
 157,648: 7,648 = 478 × 16.

id=17 '''[[17 (number)17]]'''
 Subtract 5 times the last digit from the rest.
 221: 22 − 1 × 5 = 17.

id=18 '''[[18 (number)18]]'''
 It is divisible by 2 and by 9.<ref name="productofcoprimes"/>
 342: it is divisible by 2 and by 9.

id=19 '''[[19 (number)19]]'''
 Add twice the last digit to the rest.
 437: 43 + 7 × 2 = 57.

id=20 rowspan=2 '''[[20 (number)20]]'''
 It is divisible by 10, and the tens digit is even.
 360: is divisible by 10, and 6 is even.
