قابلية القسمة

قابلية القسمة لعددين صحيحين طبيعين a و b. ما يكتب a=bc. حيث c عدد صحيح طبيعي. نقول أن b يقسم a.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

قابلية القسمة

هناك عدة قواعد لمعرفة قابلية القسمة فمثلا:

  • كل الأعداد التي رقم وحداتها(أحادها) يقسمه 2, يكون قابلا للقسمة على 2.
  • كل الأعداد التي رقم وحداتها(أحادها) 0 أو 5, يكون قابلا للقسمة على 5.
  • كل الأعداد التي مجموع أرقامها يقسمه 3, يكون قابلا للقسمة على 3.
  • كل الأعداد التي مجموع أرقام خاناتها الفردية يساوي مجموع أرقام خاناتها الزوجية,يكون قابلا القسمة على 11 ، مثل 6358 و 121 .


قواعد قابلية القسمة للأعداد 1–30

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. In some cases the process can be iterated until the divisibility is obvious; for others (such as examining the last n digits) the result must be examined by other means.

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.

Note: To test divisibility by any number that can be expressed as 2n or 5n, in which n is a positive integer, just examine the last n digits.

Note: To test divisibility by any number expressed as the product of prime factors , we can separately test for divisibility by each prime to its appropriate power. For example, testing divisibility by 24 (24 = 8*3 = 23*3) is equivalent to testing divisibility by 8 (23) and 3 simultaneously, thus we need only show divisibility by 8 and by 3 to prove divisibility by 24.

Divisor Divisibility condition Examples
1 No specific condition. Any integer is divisible by 1. 2 is divisible by 1.
2 The last digit is even (0, 2, 4, 6, or 8).[1][2] 1294: 4 is even.
3 Sum the digits. The result must be divisible by 3.[1][3][4] 405 → 4 + 0 + 5 = 9 and 636 → 6 + 3 + 6 = 15 which both are clearly divisible by 3.
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.
Subtract the quantity of the digits 2, 5, and 8 in the number from the quantity of the digits 1, 4, and 7 in the number. The result must be 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.
4 The last two digits form a number that is divisible by 4.[1][2] 40,832: 32 is divisible by 4.
If the tens digit is even, the ones digit must be 0, 4, or 8.
If the tens digit is odd, the ones digit must be 2 or 6.
40,832: 3 is odd, and the last digit is 2.
Double the tens digit, plus the ones digit is divisible by 4. 40832: 2 × 3 + 2 = 8, which is divisible by 4.
5 The last digit is 0 or 5.[1][2] 495: the last digit is 5.
6 It is divisible by 2 and by 3.[5] 1458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6.
7 Forming an alternating sum of blocks of three from right to left gives a multiple of 7[4][6] 1,369,851: 851 − 369 + 1 = 483 = 7 × 69
Adding 5 times the last digit to the rest gives a multiple of 7. (Works because 49 is divisible by 7.) 483: 48 + (3 × 5) = 63 = 7 × 9.
Subtracting 2 times the last digit from the rest gives a multiple of 7. (Works because 21 is divisible by 7.) 483: 48 − (3 × 2) = 42 = 7 × 6.
Subtracting 9 times the last digit from the rest gives a multiple of 7. (Works because 91 is divisible by 7.) 483: 48 − (3 × 9) = 21 = 7 × 3.
Adding 3 times the first digit to the next and then writing the rest gives a multiple of 7. (This works because 10a + b − 7a = 3a + b; the last number has the same remainder as 10a + b.) 483: 4×3 + 8 = 20,

203: 2×3 + 0 = 6, 63: 6×3 + 3 = 21.

Adding the last two digits to twice the rest gives a multiple of 7. (Works because 98 is divisible by 7.) 483,595: 95 + (2 × 4835) = 9765: 65 + (2 × 97) = 259: 59 + (2 × 2) = 63.
Multiply each digit (from right to left) by the digit in the corresponding position in this pattern (from left to right): 1, 3, 2, -1, -3, -2 (repeating for digits beyond the hundred-thousands place). Adding the results gives a multiple of 7. 483,595: (4 × (-2)) + (8 × (-3)) + (3 × (-1)) + (5 × 2) + (9 × 3) + (5 × 1) = 7.
Compute the remainder of each digit pair (from right to left) when divided by 7. Multiply the rightmost remainder by 1, the next to the left by 2 and the next by 4, repeating the pattern for digit pairs beyond the hundred-thousands place. Adding the results gives a multiple of 7. 194,536: 19|45|36 ; (5x4) + (3x2) + (1x1) = 27, so it is not divisible by 7

204,540: 20|45|40 ; (6x4) + (3x2) + (5x1) = 35, so it is divisible by 7

8 If the hundreds digit is even, the number formed by the last two digits must be divisible by 8. 624: 24.
If the hundreds digit is odd, the number obtained by the last two digits plus 4 must be divisible by 8. 352: 52 + 4 = 56.
Add the last digit to twice the rest. The result must be divisible by 8. 56: (5 × 2) + 6 = 16.
The last three digits are divisible by 8.[1][2] 34,152: Examine divisibility of just 152: 19 × 8
Add four times the hundreds digit to twice the tens digit to the ones digit. The result must be divisible by 8. 34,152: 4 × 1 + 5 × 2 + 2 = 16
9 Sum the digits. The result must be divisible by 9.[1][3][4] 2880: 2 + 8 + 8 + 0 = 18: 1 + 8 = 9.
10 The ones digit is 0.[2] 130: the ones digit is 0.
11 Form the alternating sum of the digits, or equivalently sum(odd) - sum(even). The result must be divisible by 11.[1][4] 918,082: 9 − 1 + 8 − 0 + 8 − 2 = 22 = 2 × 11.
Add the digits in blocks of two from right to left. The result must be divisible by 11.[1] 627: 6 + 27 = 33 = 3 × 11.
Subtract the last digit from the rest. The result must be divisible by 11. 627: 62 − 7 = 55 = 5 × 11.
Add the last digit to the hundreds place (add 10 times the last digit to the rest). The result must be divisible by 11. 627: 62 + 70 = 132: 13 + 20 = 33 = 3 × 11.
If the number of digits is even, add the first and subtract the last digit from the rest. The result must be divisible by 11. 918,082: the number of digits is even (6) → 1808 + 9 − 2 = 1815: 81 + 1 − 5 = 77 = 7 × 11
If the number of digits is odd, subtract the first and last digit from the rest. The result must be divisible by 11. 14,179: the number of digits is odd (5) → 417 − 1 − 9 = 407 = 37 × 11
12 It is divisible by 3 and by 4.[5] 324: it is divisible by 3 and by 4.
Subtract the last digit from twice the rest. The result must be divisible by 12. 324: 32 × 2 − 4 = 60 = 5 × 12.
13 Form the alternating sum of blocks of three from right to left. The result must be divisible by 13.[6] 2,911,272: 272 - 911 + 2 = -637
Add 4 times the last digit to the rest. The result must be divisible by 13. 637: 63 + 7 × 4 = 91, 9 + 1 × 4 = 13.
Subtract the last two digits from four times the rest. The result must be divisible by 13. 923: 9 × 4 - 23 = 13.
Subtract 9 times the last digit from the rest. The result must be divisible by 13. 637: 63 - 7 × 9 = 0.
14 It is divisible by 2 and by 7.[5] 224: it is divisible by 2 and by 7.
Add the last two digits to twice the rest. The result must be divisible by 14. 364: 3 × 2 + 64 = 70.
1764: 17 × 2 + 64 = 98.
15 It is divisible by 3 and by 5.[5] 390: it is divisible by 3 and by 5.
16 If the thousands digit is even, the number formed by the last three digits must be divisible by 16. 254,176: 176.
If the thousands digit is odd, the number formed by the last three digits plus 8 must be divisible by 16. 3408: 408 + 8 = 416.
Add the last two digits to four times the rest. The result must be divisible by 16. 176: 1 × 4 + 76 = 80.

1168: 11 × 4 + 68 = 112.

The last four digits must be divisible by 16.[1][2] 157,648: 7,648 = 478 × 16.
17 Subtract 5 times the last digit from the rest. (Works because 51 is divisible by 17.) 221: 22 − 1 × 5 = 17.
Subtract the last two digits from two times the rest. (Works because 102 is divisible by 17.) 4,675: 46 × 2 - 75 = 17.
Add 2 times the last digit to 3 times the rest. Drop trailing zeroes. (Works because (10a + b) × 2 − 17a = 3a + 2b; the last number has the same remainder as 10a + b.) 4,675: 467 × 3 + 5 × 2 = 1411; 238: 23 × 3 + 8 × 2 = 85.
18 It is divisible by 2 and by 9.[5] 342: it is divisible by 2 and by 9.
19 Add twice the last digit to the rest. (Works because (10a + b) × 2 − 19a = a + 2b; the last number has the same remainder as 10a + b.) 437: 43 + 7 × 2 = 57.
Add 4 times the last two digits to the rest. (Works because 399 is divisible by 19.) 6935: 69 + 35 × 4 = 209.
20 It is divisible by 10, and the tens digit is even. 360: is divisible by 10, and 6 is even.
The number formed by the last two digits is divisible by 20.[2] 480: 80 is divisible by 20.
It is divisible by 4 and 5. 480: it is divisible by 4 and 5.
21 Subtracting twice the last digit from the rest gives a multiple of 21. (Works because (10a + b) × 2 − 21a = −a + 2b; the last number has the same remainder as 10a + b.) 168: 16 − 8 × 2 = 0.
It is divisible by 3 and by 7.[5] 231: it is divisible by 3 and by 7.
22 It is divisible by 2 and by 11.[5] 352: it is divisible by 2 and by 11.
23 Add 7 times the last digit to the rest. (Works because 69 is divisible by 23.) 3128: 312 + 8 × 7 = 368. 36 + 8 × 7 = 92.
Add 3 times the last two digits to the rest. (Works because 299 is divisible by 23.) 1725: 17 + 25 × 3 = 92.
Subtract twice the last three digits from the rest. (Works because 2,001 is divisible by 23.) 2,068,965: 2,068 - 965 × 2 = 138.
24 It is divisible by 3 and by 8.[5] 552: it is divisible by 3 and by 8.
25 The last two digits are 00, 25, 50 or 75. 134,250: 50 is divisible by 25.
26 It is divisible by 2 and by 13.[5] 156: it is divisible by 2 and by 13.
Subtracting 5 times the last digit from 2 times the rest of the number gives a multiple of 26. (Works because 52 is divisible by 26.) 1248 : (124 ×2) - (8×5) =208=26×8
27 Sum the digits in blocks of three from right to left. (Works because 999 is divisible by 27.) 2,644,272: 2 + 644 + 272 = 918.
Subtract 8 times the last digit from the rest. (Works because 81 is divisible by 27.) 621: 62 − 1 × 8 = 54.
Subtract the last two digits from 8 times the rest. (Works because 108 is divisible by 27.) 6507: 65 × 8 - 7 = 520 - 7 = 513 = 27 × 19.
28 It is divisible by 4 and by 7.[5] 140: it is divisible by 4 and by 7.
29 Add three times the last digit to the rest. (Works because (10a + b) × 3 − 29a = a + 3b; the last number has the same remainder as 10a + b.) 348: 34 + 8 × 3 = 58.
Add 9 times the last two digits to the rest. (Works because 899 is divisible by 29.) 5510: 55 + 10 × 9 = 145 = 5 × 29.
Subtract twice the last three digits from the rest. (Works because 2,001 is divisible by 29.) 2,086,956: 2,086 - 956 × 2 = 174.
30 It is divisible by 3 and by 10.[5] 270: it is divisible by 3 and by 10.

بعد 30

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

القواسم المركبة

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 36, check divisibility by 4 and by 9.[5] Note that checking 3 and 12, or 2 and 18, 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 cannot 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 an inverse to 10 modulo the prime under consideration (does not work for 2 or 5) and use that as a multiplier to make the divisibility of the original number by that prime depend on the divisibility of the new (usually smaller) number by the same prime. Using 31 as an example, since 10 × (−3) = −30 = 1 mod 31, we get the rule for using y − 3x in the table above. Likewise, since 10 × (28) = 280 = 1 mod 31 also, we obtain a complementary rule y + 28x of the same kind - our choice of addition or subtraction being dictated by arithmetic convenience of the smaller value. In fact, this rule for prime divisors besides 2 and 5 is really a rule for divisibility by any integer relatively prime to 10 (including 33 and 39; see the table below). This is why the last divisibility condition in the tables above and below for any number relatively prime to 10 has the same kind of form (add or subtract some multiple of the last digit from the rest of the number).

أمثلة بارزة

The following table provides rules for some more notable divisors:

Divisor Divisibility condition Examples
31 Subtract three times the last digit from the rest. 837: 83 − 3×7 = 62
32 The number formed by the last five digits is divisible by 32.[1][2] 25,135,520: 35,520=1110×32
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.
Add the last two digits to 4 times the rest. 1312: (13×4) + 12 = 64.
33 Add 10 times the last digit to the rest. 627: 62 + 10×7 = 132,
13 + 10×2 = 33.
Add the digits in blocks of two from right to left. 2145: 21 + 45 = 66.
It is divisible by 3 and by 11. 627: 6-2+7 = 11 and 6 + 2 + 7 = 15 = 3 × 5
35 It is divisible by 7 and by 5. 595: 59 - (2×5) = 49 = 7×7. And the number ends in 5.
37 Take the digits in blocks of three from right to left and add each block. 2,651,272: 2 + 651 + 272 = 925. 925 = 37×25.
Subtract 11 times the last digit from the rest. 925: 92 − (5×11) = 37.
39 It is divisible by 3 and by 13. 351: 35 - 1 = 34 and 3 + 5 + 4 = 12 = 3 × 4
Add 4 times the last digit to the rest. 351: 35 + (1 × 4) = 39
41 Sum the digits in blocks of five from right to left. 72,841,536,727: 7 + 28,415 + 36,727 = 65,149 = 41×1,589.
Subtract 4 times the last digit from the rest. 738: 73 − 8 × 4 = 41.
43 Add 13 times the last digit to the rest. 36,249: 3624 + 9 × 13 = 3741,
374 + 1 × 13 = 387,
38 + 7 × 13 = 129,
12 + 9 × 13 = 129 = 43 × 3.
Subtract 3 times the last two digits from the rest. 36,249: 362 - 49 × 3 = 215 = 43 × 5.
45 It is divisible by 9 and by 5.[5] 2025: Ends in 5 and 2+0+2+5=9.
47 Subtract 14 times the last digit from the rest. 1,642,979: 164297 − 9 × 14 = 164171,
16417 − 14 = 16403,
1640 − 3 × 14 = 1598,
159 − 8 × 14 = 47.
Add the last two digits to 6 times the rest. 705: 7 × 6 + 5 = 47.
49 Add 5 times the last digit to the rest. 1,127: 112+(7×5)=147.
147: 14 + (7×5) = 49
Add the last two digits to 2 times the rest. 588: 5 × 2 + 88 = 98.
50 The last two digits are 00 or 50. 134,250: 50.
51 Number must be divisible by 3 and 17. 459: 4 × 2 - 59 = -51, and 4 + 5 + 9 = 18 = 3 × 6
Subtract 5 times the last digit from the rest. 204: 20-(4×5)=0
Subtract the last two digits from 2 times the rest. 459: 4 × 2 - 59 = -51.
53 Add 16 times the last digit to the rest. 3657: 365+(7×16)=477 = 9 × 53
Subtract the last two digits from 6 times the rest. 5777: 57 × 6 - 77 = 265.
55 Number must be divisible by 11 ending in 0 or 5.[5] 605: Ends in 5 and 60-5= 55 = 11×5.
57 Number must be divisible by 3 and 19. 3591: 359 + 1 × 2 = 361 = 19 × 19, and 3 + 5 + 9 + 1 = 15 = 3 × 5
Subtract 17 times the last digit from the rest. 3591: 359 − 17 = 342,
34 − 2 × 17 = 0.
59 Add 6 times the last digit to the rest. 295: 29 + 5×6= 59
61 Subtract 6 times the last digit from the rest. 732: 73-(2×6)=61
64 The number formed by the last six digits must be divisible by 64.[1][2] 2,640,000: 640,000 is divisible by 64.
65 Number must be divisible by 13 ending in 0 or 5.[5] 3,185: 318 + (5×4) = 338 = 13×26. And the number ends in 5.
67 Subtract twice the last two digits from the rest. 9112: 91 - 12×2= 67
Subtract 20 times the last digit from the rest. 4489: 448-9×20=448-180=268.
69 Number must be divisible by 3 and 23. 345: 3 + 4 + 5 = 12 = 3 × 4, and 34 + 5 × 9 = 69 = 3 × 23
Add 7 times the last digit to the rest. 345: 34 + 5×7 = 69
71 Subtract 7 times the last digit from the rest. 852: 85-(2×7)=71
73 Form the alternating sum of blocks of four from right to left. 220,241: 241 - 22 = 219.
Add 22 times the last digit from the rest. 5329: 532 + 22 × 9 = 730,
7 + 22 × 3 = 73.
75 Last two digits are 00, 25, 50 or 75, and the sum of all the digits must be divisible by 3.[5] 3675: 75 is at the end and 3 + 6 + 7 + 5 = 21 = 3×7.
77 Number is divisible by 7 and 11. 693: 69 - 3 = 66 = 11 × 6, and 69 - (6 × 2) = 63 = 7 × 9
Form the alternating sum of blocks of three from right to left. 76,923: 923 - 76 = 847.
79 Add 8 times the last digit to the rest. 711: 71 + 1×8= 79
81 Subtract 8 times the last digit from the rest. 162: 16-(2×8)=0
83 Add 25 times the last digit to the rest. 581: 58+(1×25)=83
Add the last three digits to four times the rest. 38,014: (4×38) + 14 = 166
85 Number must be divisible by 17 ending in 0 or 5. 30,855: 3085 - 25 = 3060 = 17×180. And the number ends in 5.
87 Number must be divisible by 29 with the sum of all its digits being divisible by 3. 2088: 208 + (8 × 3) = 232. 232 = 8 × 29

2 + 0 + 8 + 8 = 18 = 3 × 6

Subtract 26 times the last digit from the rest. 15138: 1513 − 8 × 26 = 1305,
130 − 5 × 26 = 0.
89 Add 9 times the last digit to the rest. 801: 80 + 1×9 = 89
Add the last two digits to eleven times the rest. 712: 12 + (7×11) = 89
91 Subtract 9 times the last digit from the rest. 182: 18 - (2×9) = 0
Form the alternating sum of blocks of three from right to left. 5,274,997: 5 - 274 + 997 = 728
Number is divisible by 7 and 13. 8281: 828+4 = 832. 83+8=91

828-2=826. 82-12=70.

95 Number must be divisible by 19 ending in 0 or 5. 51,585: 5158 + 10 = 5168,
516 + 16 = 532,
53 + 4 = 57 = 19×3. And the number ends in 5.
97 Subtract 29 times the last digit from the rest. 291: 29 - (1×29) = 0
Add the last two digits to 3 times the rest. 485: (3×4)+ 85 = 97
99 Number is divisible by 9 and 11. 891: 89 - 1 = 88.

8 + 9 + 1 = 18.

Add the digits in blocks of two from right to left. 144,837: 14 + 48 + 37 = 99.
100 Ends with at least two zeros. 14100: It has two zeros at the end.
101 Form the alternating sum of blocks of two from right to left. 40,299: 4 - 2 + 99 = 101.
103 Add 31 times the last digit to the rest. 585658: 58565 + (8×31) = 58813. 58813 : 103 = 571
Subtract the last two digits from 3 times the rest. 5356: (53×3) - 56 = 103
107 Subtract 32 times the last digit from the rest. 428: 42 - (8×32) = -214
Subtract the last two digits from 7 times the rest. 1712: 17 × 7 - 12 = 107
109 Add 11 times the last digit to the rest. 654: 65 + (11×4) = 109
111 Add the digits in blocks of three from right to left. 1,370,184: 1 + 370 + 184 = 555
113 Add 34 times the last digit from the rest. 3842: 384 + 34 × 2 = 452,
45 + 34 × 2 = 113.
121 Subtract 12 times the last digit from the rest. 847: 84 - 12 × 7 = 0
125 The number formed by the last three digits must be divisible by 125.[2] 2,125: 125 is divisible by 125.
127 Subtract 38 times the last digit from the rest. 4953: 495 - 38 × 3 = 381,
38 - 38 × 1 = 0.
128 The number formed by the last seven digits must be divisible by 128.[1][2]
131 Subtract 13 times the last digit from the rest. 1834: 183 - 13 × 4 = 131,
13 - 13 = 0.
137 Form the alternating sum of blocks of four from right to left. 340,171: 171 - 34 = 137.
139 Add 14 times the last digit from the rest. 1946: 194 + 14 × 6 = 278,
27 + 14 × 8 = 139.
143 Form the alternating sum of blocks of three from right to left. 1,774,487: 1 - 774 + 487 = -286
Add 43 times the last digit to the rest. 6149: 614 + 43 × 9 = 1001,
100 + 43 = 143.
The number must be divisible by 11 and 13. 2,431: 243 - 1 = 242. 242 = 11 × 22.
243 + 4 = 247. 247 = 13 × 19
149 Add 15 times the last digit from the rest. 2235: 223 + 15 × 5 = 298,
29 + 15 × 8 = 149.
151 Subtract 15 times the last digit from the rest. 66,893: 6689 - 15 × 3 = 6644 = 151×44.
157 Subtract 47 times the last digit from the rest. 7536: 753 - 47 × 6 = 471,
47 - 47 = 0.
163 Add 49 times the last digit to the rest. 26,569: 2656 + 441 = 3097 = 163×19.
167 Subtract 5 times the last two digits from the rest. 53,774: 537 - 5 × 74 = 167.
173 Add 52 times the last digit to the rest. 8996: 899 + 52 × 6 = 1211,
121 + 52 = 173.
179 Add 18 times the last digit to the rest. 3222: 322 + 18 × 2 = 358,
35 + 18 × 8 = 179.
181 Subtract 18 times the last digit from the rest. 3258: 325 - 18 × 8 = 181,
18 - 18 = 0.
191 Subtract 19 times the last digit from the rest. 3629: 362 - 19 × 9 = 191,
19 - 19 = 0.
193 Add 58 times the last digit to the rest. 11194: 1119 + 58 × 4 = 1351,
135 + 58 = 193.
197 Subtract 59 times the last digit from the rest. 11820: 118 - 59 × 2 = 0.
199 Add 20 times the last digit to the rest. 3980: 39 + 20 × 8 = 199.
200 Last two digits of the number are "00", and the third last digit is an even number. 34,400: The third last digit is 4, and the last two digits are zeroes.
211 Subtract 21 times the last digit from the rest. 44521: 4452 - 21 × 1 = 4431,
443 - 21 × 1 = 422,
42 - 21 × 2 = 0.
223 Add 67 times the last digit to the rest. 49729: 4972 + 67 × 9 = 5575,
557 + 67 × 5 = 892,
89 + 67 × 2 = 223.
225 Number must be divisible by 9 ending in "00", "25", "50", or "75". 15,075: 75 is at the end and 1 + 5 + 0 + 7 + 5 = 18 = 2×9.
227 Subtract 68 times the last digit from the rest. 51756: 5175 - 68 × 6 = 4767,
476 - 68 × 7 = 0.
229 Add 23 times the last digit to the rest. 52441: 5244 + 23 × 1 = 5267,
526 + 23 × 7 = 687,
68 + 23 × 7 = 229.
233 Add 70 times the last digit to the rest. 54289: 5428 + 70 × 9 = 6058,
605 + 70 × 8 = 1165,
116 + 70 × 5 = 466,
46 + 70 × 6 = 466 = 233 × 2.
239 Take the digits in blocks of seven from right to left and add each block. 1,560,000,083: 156 + 83 = 239.
Add 24 times the last digit to the rest. 57121: 5712 + 24 × 1 = 5736,
573 + 24 × 6 = 717,
71 + 24 × 7 = 239.
241 Subtract 24 times the last digit from the rest. 58081: 5808 - 24 × 1 = 5784,
578 - 24 × 4 = 482,
48 - 24 × 2 = 0.
250 The number formed by the last three digits must be divisible by 250.[1][2] 1,327,750: 750 is divisible by 250.
251 Subtract 25 times the last digit from the rest. 63001: 6300 - 25 × 1 = 6275,
627 - 25 × 5 = 502,
50 - 25 × 2 = 0.
256 The number formed by the last eight digits must be divisible by 256.[1][2]
257 Subtract 77 times the last digit from the rest. 66049: 6604 - 77 × 9 = 5911,
591 - 77 × 1 = 514 = 257 × 2.
263 Add 79 times the last digit to the rest. 69169: 6916 + 79 × 9 = 7627,
762 + 79 × 7 = 1315,
131 + 79 × 5 = 526,
52 + 79 × 6 = 526 = 263 × 2.
269 Add 27 times the last digit to the rest. 72361: 7236 + 27 × 1 = 7263,
726 + 27 × 3 = 807,
80 + 27 × 7 = 269.
271 Take the digits in blocks of five from right to left and add each block. 77,925,613,961: 7 + 79,256 + 13,961 = 93,224 = 271×344.
Subtract 27 times the last digit from the rest. 73441: 7344 - 27 × 1 = 7317,
731 - 27 × 7 = 542,
54 - 27 × 2 = 0.
277 Subtract 83 times the last digit from the rest. 76729: 7672 - 83 × 9 = 6925,
692 - 83 × 5 = 277.
281 Subtract 28 times the last digit from the rest. 78961: 7896 - 28 × 1 = 7868,
786 - 28 × 8 = 562,
56 - 28 × 2 = 0.
283 Add 85 times the last digit to the rest. 80089: 8008 + 85 × 9 = 8773,
877 + 85 × 3 = 1132,
113 + 85 × 2 = 283.
293 Add 88 times the last digit to the rest. 85849: 8584 + 88 × 9 = 9376,
937 + 88 × 6 = 1465,
146 + 88 × 5 = 586,
58 + 88 × 6 = 586 = 293 × 2.
300 Last two digits of the number are "00", and the result of sum the digits must be divisible by 3. 3,300: The result of sum the digits is 6, and the last two digits are zeroes.
329 Add 33 times the last digit to the rest. 9541:954+1×33=954+33=987. 987=3×329.
331 Subtract 33 times the last digit from the rest. 22177: 2217-231=1986. 1986=6×331.
333 Add the digits in blocks of three from right to left. 410,922: 410 + 922 = 1,332
369 Take the digits in blocks of five from right to left and add each block. 50243409: 43409+502=43911. 43911=369×119.
Add 37 times the last digit to the rest. 8487: 848+7×37=848+259=1107.
375 The number formed by the last three digits must be divisible by 125 and the sum of all digits is a multiple of 3. 140,625: 625 = 125×5 and 1 + 4 + 0 + 6 + 2 + 5 = 18 = 6×3.
499 Add the last three digits to two times the rest. 74,351: 74 × 2 + 351 = 499.
500 Ends with 000 or 500. 47,500 is divisible by 500.
512 The number formed by the last nine digits must be divisible by 512.[1][2]
625 Ends in 0000, 0625, 1250, 1875, 2500, 3125, 3750, 4375, 5000, 5625, 6250, 6875, 7500, 8125, 8750 or 9375.

Or, the number formed by the last four digits is divisible by 625.

567,886,875: 6875.
983 Add the last three digits to seventeen times the rest. 64878: 64×17+878=1966. 1966=2×983
987 Add the last three digits to thirteen times the rest. 30597: 30×13+597=987
Number must be divisible by 329 with the sum of all digits being divisible by 3. 547785: 5+4+7+7+8+5=36. 36=3×12

54778+5×33=54943. 5494+3×33=5593. 559+3×33=658. 658=2×329.

989 Add the last three digits to eleven times the rest. 21758: 21 × 11 = 231; 758 + 231 = 989
Number must be divisible by 23 and 43. 1978: 197+56=253. 253=11×23

197+104=301. 301=7×43.

993 Add the last three digits to seven times the rest. 986049: 49+6902=6951. 6951=7×993.
Number must be divisible by 331 with the sum of all digits being divisible by 3. 8937: 8+7=15. 15=3×5. (Note: 9 and 3 don't have to be in the sum, they are divisible by 3.)
893-231=662. 662=2×331.
997 Add the last three digits to three times the rest. 157,526: 157 × 3 + 526= 997
999 Add the digits in blocks of three from right to left. 235,764: 235 + 764 = 999
1000 Ends with at least three zeros. 2000 ends with 3 zeros


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

القاعدة المعممة لقابلية القسمة

To test for divisibility by D, where D ends in 1, 3, 7, or 9, the following method can be used.[7] Find any multiple of D ending in 9. (If D ends respectively in 1, 3, 7, or 9, then multiply by 9, 3, 7, or 1.) Then add 1 and divide by 10, denoting the result as m. Then a number N = 10t + q is divisible by D if and only if mq + t is divisible by D. If the number is too large, you can also break it down into several strings with e digits each, satisfying either 10e = 1 or 10e = -1 (mod D). The sum (or alternate sum) of the numbers have the same divisibility as the original one.

For example, to determine if 913 = 10×91 + 3 is divisible by 11, find that m = (11×9+1)÷10 = 10. Then mq+t = 10×3+91 = 121; this is divisible by 11 (with quotient 11), so 913 is also divisible by 11. As another example, to determine if 689 = 10×68 + 9 is divisible by 53, find that m = (53×3+1)÷10 = 16. Then mq+t = 16×9 + 68 = 212, which is divisible by 53 (with quotient 4); so 689 is also divisible by 53.

Alternatively, any number Q = 10c + d is divisible by n = 10a + b, such that gcd(n, 2, 5) = 1, if c + D(n)d = An for some integer A, where:

The first few terms of the sequence, generated by D(n) are 1, 1, 5, 1, 10, 4, 12, 2, ... (sequence A333448 in OEIS).

The piece wise form of D(n) and the sequence generated by it were first published by Bulgarian mathematician Ivan Stoykov in March 2020.[8]


انظر أيضاً

المراجع

  1. ^ أ ب ت ث ج ح خ د ذ ر ز س ش ص ض This follows from Pascal's criterion. See Kisačanin (1998), p. 100–101
  2. ^ أ ب ت ث ج ح خ د ذ ر ز س ش ص A number is divisible by 2m, 5m or 10m if and only if the number formed by the last m digits is divisible by that number. See Richmond & Richmond (2009), p. 105
  3. ^ أ ب Apostol (1976), p. 108
  4. ^ أ ب ت ث Richmond & Richmond (2009), Section 3.4 (Divisibility Tests), p. 102–108
  5. ^ أ ب ت ث ج ح خ د ذ ر ز س ش ص ض ط Richmond & Richmond (2009), Section 3.4 (Divisibility Tests), Theorem 3.4.3, p. 107
  6. ^ أ ب Kisačanin (1998), p. 101
  7. ^ Dunkels, Andrejs, "Comments on note 82.53—a generalized test for divisibility", Mathematical Gazette 84, March 2000, 79-81.
  8. ^ Stoykov, Ivan (March 2020). "OEIS A333448". Oeis A333448.

المصادر

وصلات خارجية