Some divisibility rules for positive integers:
1) A positive integer is divisible by 2 iff its one’s digit is even.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as 100a  10b  c ,
but 100a  10b  c  2  50a  5b   c , so if the one’s digit, c is even(divisible by 2) then the
integer abc will also be divisible by 2. If the number is divisible by 2, then c must be divisible
by 2(even).
2) A positive integer is divisible by 3 iff the sum of its digits is divisible by 3. This process may
be repeated.
Examples: 243 is divisible by 3, but 271 is not.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as 100a  10b  c ,
but 100a  10b  c  99a  9b  a  b  c  3  33a  3b    a  b  c  , so if the sum of the digits,
the sum of the digits
a  b  c , is divisible by 3 then the integer abc will also be divisible by 3. If the number is
divisible by 3, then the sum of the digits must be divisible by 3.
3) A positive integer is divisible by 4 iff the ten’s and one’s digits form a two-digit integer
divisible by 4.
Examples: 724 is divisible by 4, but 726 is not.
Here’s why:
Suppose we have the four-digit integer abcd, then its value can be expressed as
1000a  100b  10c  d , but
1000a  100b  10c  d  1000a  100b   10c  d   4  250a  25b   10c  d 
, so if the
the two-digit integer cd
ten’s and one’s two-digit integer, cd, is divisible by 4 then the integer abcd will also be divisible
by 4. If the number is divisible by 4, then the ten’s and one’s two-digit integer, cd, is divisible
by 4. (Note: It still works if c is zero.)
4) A positive integer is divisible by 5 iff its one’s digit is either a 5 or a 0.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as 100a  10b  c ,
but 100a  10b  c  5 20a  2b  c , so if the one’s digit, c is divisible by 5, then the integer
abc will also be divisible by 5, but the only digits divisible by 5 are 0 and 5. If the number is
divisible by 5, then the one’s digit is divisible by 5(5 or 0).
5) A positive integer is divisible by 6 iff it’s both divisible by 2 and divisible by 3.
6) A positive integer is divisible by 7 iff when you remove the one’s digit from the integer and
then subtract twice the one’s digit from the new integer, you get an integer divisible by 7.
This process may be repeated.
Examples: 714 is divisible by 7 since 71 – 8 = 63, but 423 is not since 42 – 6 = 36.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as 100a  10b  c ,
but
100a  10b  c  90a  9b  3c  10a  b  2c 
new integer minus twice one's digit
 9 10a  b  2c   21c  10a  b  2c 
 10 
10a  b  2c 
 7   3c 
new integer minus twice one's digit
, so if the new integer minus twice the one’s digit, 10a  b  2c , is divisible by 7 then so is the
original integer abc. If the number is divisible by 7, then so is the new integer minus twice the
one’s digit.
Or
A positive integer is divisible by 7 iff when you remove the one’s digit from the integer and
then subtract nine times the one’s digit from the new integer, you get an integer divisible by 7.
This process may be repeated.
Examples: 714 is divisible by 7 since 71 – 36 = 35, but 423 is not since 42 – 27 = 15.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as 100a  10b  c ,
but
100a  10b  c  90a  9b  10c 
10a  b  9c 
new integer minus 9 times one's digit
 9 10a  b  9c   91c  10a  b  9c 
 10 
10a  b  9c 
 7  13c 
new integer minus 9 times one's digit
, so if the new integer minus nine times the one’s digit, 10a  b  9c , is divisible by 7 then so is
the original integer abc. If the number is divisible by 7, then so is the new integer minus nine
times the one’s digit.
Or
A positive integer with more than three digits is divisible by 7 iff when you split the digits into
groups of three starting from the right and alternately add and subtract these three digit numbers
you get a result which is divisible by 7.
Examples: 1412236 is divisible by 7 since 1 – 412 + 236 = -175, but 130747591 is not since
130 – 747 + 591 = -26.
Here’s why:
Suppose we have the five-digit integer abcde, then its value can be expressed as
10000a  1000b  100c  10d  e , but
10000a  1000b  100c  10d  e  10010a  1001b   10a  b   100c  10d  e 
integer ab
2 100c  10d  e   2 10a  b 
integer cde
, so if the


 7  143 10a  b   10a  b   100c  10d  e  


integer cde
 integer ab

integer ab minus the integer cde is divisible by 7 then the integer abcde will also be divisible by
7. If the number is divisible by 7, then so is the integer ab minus the integer cde.
7) A positive integer is divisible by 8 iff the hundred’s, ten’s, and one’s digits form a three-digit
integer divisible by 8.
Examples: 1240 is divisible by 8, since 240 is, 3238 is not, since 238 is not even divisible by 4.
Here’s why:
Suppose we have the five-digit integer abcde, then its value can be expressed as
10000a  1000b  100c  10d  e , but
10000a  1000b  100c  10d  e  10000a  1000b   100c  10d  e 
 8 1250a  125b   100c  10d  e  , so if the hundred’s,
the three-digit integer cde
ten’s, and one’s three-digit integer, cde, is divisible by 8 then the integer abcde will also be
divisible by 8. If the number is divisible by 8, then the hundred’s, ten’s, and one’s three-digit
integer is divisible by 8. (Note: It still works if c or d is zero.)
8) A positive integer is divisible by 9 iff the sum of its digits is divisible by 9. This process may
be repeated.
Examples: 243 is divisible by 9, but 9996 is not.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as 100a  10b  c ,
but 100a  10b  c  99a  9b  a  b  c  9 11a  b    a  b  c  , so if the sum of the digits,
the sum of the digits
a  b  c , is divisible by 9 then the integer abc will also be divisible by 9. If the number is
divisible by 9, then the sum of the digits is divisible by 9.
9) A positive integer is divisible by 11 iff when you remove the one’s digit from the integer
and then subtract the one’s digit from the new integer, you get an integer divisible by 11.
This process may be repeated.
Examples: 1001 is divisible by 11 since 100 – 1 = 99, but 423 is not since 42 – 3 = 38.
Here’s why:
Suppose we have the three-digit integer abc, then its value can be expressed as 100a  10b  c ,
but
100a  10b  c  90a  9b  2c  10a  b  c 
new integer minus one's digit
 9 10a  b  c   11c  10a  b  c 
 10 
10a  b  c 
 11c
new integer minus one's digit
, so if the new integer minus the one’s digit, 10a  b  c , is divisible by 11 then so is the original
integer abc. If the number is divisible by 11, then the new integer minus the one’s digit is
divisible by 11.
Or
A positive integer is divisible by 11 iff when you subtract the sum of the ten’s digit and every
other digit to the left from the sum of the one’s digit and every other digit to the left you get a
number divisible by 11. This process may be repeated.
Examples: 9031 is divisible by 11 since
 4  3   2   5 .
1  0   3  9   11,
but 423 is not since
Here’s why:
Suppose we have the five-digit integer abcde, then its value can be expressed as
10,000a  1,000b  100c  10d  e , but
10,000a  1,000b  100c  10d  e   9,999  1 a  1,001  1 b   99  1 c  11  1 d  e
 9,999a  1,001b  99c  11d   a  c  e    b  d 



 11   909a  91b  9c  d    a  c  e    b  d  


 alternating from one's alternating from ten's 
, so if the sum of the alternating digits from the one’s digit minus the sum of the alternating
digits from the ten’s digit,  a  c  e    b  d  , is divisible by 11 then so is the original integer
abcde. If the number is divisible by 11, then the difference of the alternating sums is divisible
by 11.
Or
A positive integer with more than three digits is divisible by 11 iff when you split the digits into
groups of three starting from the right and alternately add and subtract these three digit numbers
you get a result which is divisible by 11.
Examples: 1412290 is divisible by 11 since 1 – 412 + 290 = -121, but 130747591 is not since
130 – 747 + 591 = -26.
Here’s why:
Suppose we have the five-digit integer abcde, then its value can be expressed as
10000a  1000b  100c  10d  e , but
10000a  1000b  100c  10d  e  10010a  1001b   10a  b   100c  10d  e 
integer ab
2 100c  10d  e   2 10a  b 
integer cde
, so if the



 11  9110a  b   10a  b   100c  10d  e  


integer cde
 integer ab

integer ab minus the integer cde is divisible by 11 then the integer abcde will also be divisible
by 11. If the number is divisible by 11, then so is the integer ab minus the integer cde.