Divisibility Rules

Can divided by	If	Example
1	No specific condition.	$\text{Any integer is divisible by } 1$
2	The last digit is even (0, 2, 4, 6, or 8)	$1,294 \to 4 \text{ is even.}$
3	The sum of the digits must be divisible by 3.	$609 → 6 + 0 + 9 = 15 \text{ which is clearly divisible by 3}$
	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.	$16,499,205,854,376 \text{has four of the digits 1, 4 and 7} \\ \text{and four of the digits 2, 5 and 8; Since } 4 − 4 = 0 \\ \text{is a multiple of 3, the number 16,499,205,854,376 is divisible by 3}$
	Subtracting 2 times the last digit from the rest gives a multiple of 3.	$405 \to 40 - 5 \times 2 = 40 - 10 = 30 = 3 \times 10$
4	The last two digits form a number that is divisible by 4.	$40,832 \to 32 \text{ is divisible by 4}$
	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.	$40,832 \to 3 \text{ is odd, and the last digit is } 2$
	The sum of the ones digit and double the tens digit is divisible by 4.	$40,832 \to 2 × 3 + 2 = 8 \text{ which is divisible by} 4$
5	The last digit is 0 or 5.	$495 \to \text{ the last digit is 5}$
6	It is divisible by 2 and by 3.	$1,458 \to 1 + 4 + 5 + 8 = 18 \text{so it is divisible by 3 and the last digit is even,} \\ \text{hence the number is divisible by 6.}$
6	Sum the ones digit, 4 times the 10s digit, 4 times the 100s digit, 4 times the 1000s digit, etc. If the result is divisible by 6, so is the original number. (Works because $10^{n}=4 \pmod 6 \text{ for } 𝑛 > 1$	$1,458 \to (4 × 1) + (4 × 4) + (4 × 5) + 8 = 4 + 16 + 20 + 8 = 48$
7	Forming an alternating sum of blocks of three from right to left gives a multiple of 7	$1,369,851 \to 851 − 369 + 1 = 483 = 7 × 69$
	Adding 5 times the last digit to the rest gives a multiple of 7. (Works because 49 is divisible by 7.)	$483 \to 48 + (3 × 5) = 63 = 7 × 9$
	Subtracting 2 times the last digit from the rest gives a multiple of 7. (Works because 21 is divisible by 7.)	$483 \to 48 − (3 × 2) = 42 = 7 × 6$
	Subtracting 9 times the last digit from the rest gives a multiple of 7. (Works because 91 is divisible by 7.)	$483 \to 48 − (3 × 2) = 42 = 7 × 6$
	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.)	$483 \to 4 × 3 + 8 = 20 \\ 203 \to 2 × 3 + 0 = 6 \\ 63 \to 6 × 3 + 3 = 21$
	Adding the last two digits to twice the rest gives a multiple of 7. (Works because 98 is divisible by 7.)	$483,595 \to 95 + (2 × 4835) = 9765 \\ 65 + (2 × 97) = 259 \to 59 + (2 × 2) = 63$
	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.	$483,595 \to (4 × (−2)) + (8 × (−3)) + (3 × (−1)) \\ + (5 × 2) + (9 × 3) + (5 × 1) = 7$
	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.	$194,536 \to 19\|45\|36 \to (5 \times 4) + (3 \times 2) + (1 \times 1) = 27 \\ \text{ so it is not divisible by 7 } \\ 204,540 \to 20\|45\|40 \to (6 \times 4) + (3 \times 2) + (5 \times 1) = 35 \\ \text{ so it is divisible by 7.}$
8	If the hundreds digit is even, the number formed by the last two digits must be divisible by 8.	$624 \to 24$
	If the hundreds digit is odd, the number obtained by the last two digits must be 4 times an odd number.	$352 \to 52 = 4 \times 13$
	Add the last digit to twice the rest. The result must be divisible by 8.	$56 \to (5 × 2) + 6 = 16$
	The last three digits are divisible by 8.	$34,152 \to \text{Examine divisibility of just 152} \to 19 × 8$
	The sum of the ones digit, double the tens digit, and four times the hundreds digit is divisible by 8.	$34,152 \to 4 × 1 + 5 × 2 + 2 = 16$
9	The sum of the digits must be divisible by 9	$2,880 \to 2 + 8 + 8 + 0 = 18: 1 + 8 = 9$
9	Subtracting 8 times the last digit from the rest gives a multiple of 9. (Works because 81 is divisible by 9)	$2,880 \to 288 - 0 \times 8 = 288 - 0 = 288 = 9 \times 32$
10	The last digit is 0.[3]	$130 \to \text{ the ones digit is 0.}$
10	It is divisible by 2 and by 5	$130 \to \text{ it is divisible by 2 and by 5}$
11	Form the alternating sum of the digits, or equivalently sum(odd) - sum(even). The result must be divisible by 11.	$918,082 \to 9 − 1 + 8 − 0 + 8 − 2 = 22 = 2 \times 11.$
	Add the digits in blocks of two from right to left. The result must be divisible by 11.	$627 \to 6 + 27 = 33 = 3 \times 11$
	Subtract the last digit from the rest. The result must be divisible by 11.	$627 \to 62 − 7 = 55 = 5 \times 11.$
	Add 10 times the last digit to the rest. The result must be divisible by 11. (Works because 99 is divisible by 11).	$627 \to 62 + 70 = 132 \to 13 + 20 = 33 = 3 \times 11$
	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.	$918,082 \to \text{the number of digits is even} \\ 1808 + 9 − 2 = 1815 \to 81 + 1 − 5 = 77 = 7 \times 11$
	If the number of digits is odd, subtract the first and last digit from the rest. The result must be divisible by 11.	$14,179 \to \text{the number of digits is odd (5)} \\ 417 − 1 − 9 = 407 = 37 \times 11$
12	It is divisible by 3 and by 4.	$324 \to \text{it is divisible by 3 and by 4.}$
12	Subtract the last digit from twice the rest. The result must be divisible by 12.	$324 \to 32 \times 2 − 4 = 60 = 5 \times 12.$
13	Form the alternating sum of blocks of three from right to left. The result must be divisible by 13.	$2,911,272 \to 272 − 911 + 2 = −637$
	Add 4 times the last digit to the rest. The result must be divisible by 13. (Works because 39 is divisible by 13).	$637 \to 63 + 7 × 4 = 91 \\ 9 + 1 × 4 = 13.$
	Subtract the last two digits from four times the rest. The result must be divisible by 13.	$923 \to 9 \times 4 − 23 = 13.$
	Subtract 9 times the last digit from the rest. The result must be divisible by 13. (Works because 91 is divisible by 13).	$637 \to 63 - 7 \times 9 = 0.$
14	It is divisible by 2 and by 7.	$224 \to \text{it is divisible by 2 and by 7.}$
14	Add the last two digits to twice the rest. The result must be divisible by 14.	$364 \to 3 \times 2 + 64 = 70 \\ 1,764 \to 17 \times 2 + 64 = 98.$
15	It is divisible by 3 and by 5.	$390 \to \text{it is divisible by 3 and by 5.}$
16	If the thousands digit is even, the number formed by the last three digits must be divisible by 16.	$254,176 \to 176$
	If the thousands digit is odd, the number formed by the last three digits must be 8 times an odd number.	$3408 \to 408 = 8 \times 51$
	Add the last two digits to four times the rest. The result must be divisible by 16.	$176 \to 1 \times 4 + 76 = 80 \\ 1,168 \to 11 \times 4 + 68 = 112$
	The last four digits must be divisible by 16.	$157,648 \to 7,648 = 478 \times 16$
	The sum of the ones digit, double the tens digit, four times the hundreds digit, and eight times the thousands digit is divisible by 16.	$157,648 \to 7 \times 8 + 6 \times 4 + 4 \times 2 + 8 = 96$
17	Subtract 5 times the last digit from the rest. (Works because 51 is divisible by 17.)	$221 \to 22 − 1 \times 5 = 17$
	Add 12 times the last digit to the rest. (Works because 119 is divisible by 17).	$221 \to 22 + 1 \times 12 = 22 + 12 = 34 = 17 \times 2$
	Subtract the last two digits from two times the rest. (Works because 102 is divisible by 17.)	$4,675 \to 46 \times 2 − 75 = 17$
	Add 2 times the last digit to 3 times the rest. Drop trailing zeroes. (Works because (10a + b) × 2 − 17a = 3a + 2b; since 17 is a prime and 2 is coprime with 17, 3a + 2b is divisible by 17 if and only if 10a + b is.)	$4,675 \to 467 \times 3 + 5 \times 2 = 1,411 \\ 238 \to 23 \times 3 + 8 \times 2 = 85$
18	It is divisible by 2 and by 9.[6]	$342 \to \text{it is divisible by 2 and by 9.}$
19	Add twice the last digit to the rest. (Works because (10a + b) × 2 − 19a = a + 2b; since 19 is a prime and 2 is coprime with 19, a + 2b is divisible by 19 if and only if 10a + b is.)	$437\ to 43 + 7 \times 2 = 57.$
19	Add 4 times the last two digits to the rest. (Works because 399 is divisible by 19.)	$6,935 \to 69 + 35 \times 4 = 209$
20	It is divisible by 10, and the tens digit is even.	$360 \to \text{is divisible by 10, and 6 is even.}$
	The last two digits are 00, 20, 40, 60 or 80.[3]	$480 \to 80$
	It is divisible by 4 and 5.	$480\to \text{it is divisible by 4 and 5.}$
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.)	$168 \to 16 − 8 \times 2 = 0$
	Suming 19 times the last digit to the rest gives a multiple of 21. (Works because 189 is divisible by 21).	$441 \to 44 + 1 \times 19 = 44 + 19 = 63 = 21 \times 3$
	It is divisible by 3 and by 7.[6]	$231 \to \text{it is divisible by 3 and by 7.}$
22	It is divisible by 2 and by 11.[6]	$352 \to \text{it is divisible by 2 and by 11}$
23	Add 7 times the last digit to the rest. (Works because 69 is divisible by 23.)	$3,128 \to 312 + 8 \to 7 = 368. 36 + 8 \times 7 = 92$
	Add 3 times the last two digits to the rest. (Works because 299 is divisible by 23.)	$1,725 \to 17 + 25 \times 3 = 92$
	Subtract 16 times the last digit from the rest. (Works because 161 is divisible by 23).	$1,012 \to 101 - 2 \times 16 = 101 - 32 = 69 = 23 \times 3$
	Subtract twice the last three digits from the rest. (Works because 2,001 is divisible by 23.)	$2,068,965 \to 2,068 − 965 \times 2 = 138$
24	It is divisible by 3 and by 8.[6]	$552 \to \text{it is divisible by 3 and by 8.}$
25	The last two digits are 00, 25, 50 or 75.	$134,250 \to 50 \text{is divisible by 25.}$
26	It is divisible by 2 and by 13.[6]	$156 \to \text{it is divisible by 2 and by 13.}$
26	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.)	$1,248 \to (124 \times 2) - (8 \times 5) = 208 = 26 \times 8$
27	Sum the digits in blocks of three from right to left. (Works because 999 is divisible by 27.)	$2,644,272 \to 2 + 644 + 272 = 918$
	Subtract 8 times the last digit from the rest. (Works because 81 is divisible by 27.)	$621 \to 62 − 1 \times 8 = 54$
	Sum 19 times the last digit from the rest. (Works because 189 is divisible by 27).	$1,026 \to 102 + 6 \times 19 = 102 + 114 = 216 = 27 \times 8$
	Subtract the last two digits from 8 times the rest. (Works because 108 is divisible by 27.)	$6,507: 65 × 8 − 7 = 520 − 7 = 513 = 27 × 19$
28	It is divisible by 4 and by 7.[6]	$140 \to \text{it is divisible by 4 and by 7.}$
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.)	$348\to 34 + 8 \times 3 = 58$
	Add 9 times the last two digits to the rest. (Works because 899 is divisible by 29.)	$5,510 \to 55 + 10 \times 9 = 145 = 5 \times 29$
	Subtract 26 times the last digit from the rest. (Works because 261 is divisible by 29).	$1,015 \to 101 - 5 \times 26 = 101 - 130 = -29 = 29 \times -1$
	Subtract twice the last three digits from the rest. (Works because 2,001 is divisible by 29.)	$2,086,956 \to 2,086 − 956 \times 2 = 174$
30	It is divisible by 3 and by 10.[6]	$270 \to \text{it is divisible by 3 and by 10}$
	It is divisible by 2, by 3 and by 5	$270 \to \text{it is divisible by 2, by 3 and by 5}$
	It is divisible by 2 and by 15	$270 \to \text{it is divisible by 2 and by 15}$
	It is divisible by 5 and by 6	$270 \to \text{it is divisible by 5 and by 6}$

UCH Viki'den alıntılanmıştır. https://wiki.ulascemh.com/doku.php?id=en:math:algebra:divisibility