Divisibility rules for prime numbers
Shyamal Das – Uttarpara, West Bengal, INDIA
The first prime number is 2. So, we will start with number 2.
Rule for 2: A number is divisible by 2, if its last figure is even, i.e,0,2,4,6,8.
Justification:
Rule for 3:
Let the number be 100a+ 10b+ c
Dividing by 2. We get, (100a+10b+c) / 2=(100a+10b)/2 + c/2
The 1st part = 100a+10b is divisible by 2, gives 50a+5b and
the 2nd part = c is divisible by 2, if the last figure = c is even.
A number is divisible by 3, if the sum of its digits is divisible by 3.
Justification: Let the number be 100a+ 10b+ c
Now , 100a+10b+c = (99a+9b) + (a+b+c)
The 1st part = 99a+9b is divisible by 3, gives 33a+3b and if,
the 2nd part = (a+b+c) is divisible by 3,
then, the given no. is divisible by 3
Rule for 5:
A number is divisible by 5, if its last figure is 0 or 5.
Justification:
Let the number be 100a+10b+c
Dividing by 5, we get, (100a+10b)/ 5 + c/5
the 1st part = (100a+10b) is divisible by 5, gives 20a+2b, and if
the 2nd part = c is 0 or 5
the given number is divisible by 5.
Rule for 7:
1) A number is divisible by 7, if after subtracting two times the last digit
from the rest, the remainder is divisible by 7.
2)A number is divisible by 7 (or11 or 13), if we observe the followings:
The figures of the given number is to be divided into groups containing
three digits each, as far as possible, counting from right to left. Then,
alternate groups are to be added. Now, smaller sum is to be subtracted
from the greater. If the remainder is zero or divisible by 7 (or11 or 13),
the given number is divisible by 7 (or 11 or 13)
Justification:
1) Let the number be 100a+10b+c
Here, the last digit = c and rest = 10a+b
as per the condition, let (10a+b-2c) be divisible by 7
Let 10a+b-2c = 7x , where, x=1,2,3….
Hence, 10a+b = 7x+2c
multiplying both sides by 10, we get, 100a+10b = 70x + 20c
Hence, 100a+10b+c = 70x+20c+c = 70x+21c = 7(10x+3c)
= divisible by 7
2) Let the number be 10000a+1000b+100c+10d+e
after dividing into two groups, we get 100c+10d+e and other is
10000a+1000b
as per the condition, let (100c+10d+e)- (10a+b) = 7x (or 11x or 13x, as the
case
may be)
or,100c+10d+e-10a-b = 7x
Hence, (10000a+1000b+100c+10d+e) – (100c+10d+e-10a-b) =
10010a+1001b
or,(10000a+1000b+100c+10d+e) – 7x = 1001(10a+b)
The given number = (10000a+1000b+100c+10d+e ) = 7*11*13(10a+b)+
7x (or 11x or13x)
Rule for 11:
1) A number is divisible by 11, if the difference between the sums of its
digits in the odd place and its digits in the even place is either zero,
or, divisible by 11.
2) A number is divisible by 11, if, after subtracting the last digit
from the rest, the remainder is divisible by 11.
Justification:
1) Let the number be 10000a+1000b+100c+10d+e
sum of the digits in the even place = a+c+e
sum of the digits in the odd place = b+d
let the difference = a+c+e-b-d be divisible by 11 =11x
now, (10000a+1000b+100c+10d+e) – (a+c+e-b-d)
=9999a+1001b+99c+11d = 11(909a+91b+9c+d)
Hence, (10000a+1000b+100c+10d+e) – 11x = 11(909a+91b+9c+d)
The given number = 10000a+1000b+100c+10d+e
=11(909a+91b+9c+d+x)
The given number is divisible by 11.
2) Let the number be 100a+10b+c
here, last digit =c and the rest = 10a+b
according to the condition, let 10a+b-c = 11x
Hence, 10a+b = 11x +c
Multiplying both sides by 10, we get, 100a+10b= 110x+10c
100a+10b+c = 110x+10c+c = 110x+11c =11(10x+c), divisible by11
Rule for 13:
A number is divisible by 13, if the sum of 4 times of last digit and
rest of the number is divisible by 13.
Justification:
Let the number be 100a+10b+c
Here, the last = c and the rest = 10a+b
According to the rule, let 10a+b+4c = 13x
Hence, 10a+b = 13x- 4 c
Multiplying both sides by 10, we get 100a+10b = 130x - 40c
100a+10b +c = 130x -40c+c = 130x -39c = 13(10x-3c)
The given number is divisible by 13.
Rule for 17:
A number is divisible by 17, if 5times of last digit is subtracted
from the rest, the remainder is divisible by
Justification:
Let the number be 100a+10b+c
Here, the last = c and the rest = 10a+b
According to the rule, let 10a+b-5c = 17x
Hence, 10a+b = 17x+ 5 c
Multiplying both sides by 10, we get 100a+10b = 170x +50c
100a+10b +c = 170x +50c+c = 170x +51c = 17(10x+3c)
The given number is divisible by 17.
Rule for 19:
Justification:
A number is divisible by 19, if the sum of twice of the last digit
and rest is divisible by 19.
Let the number be 100a+10b+c
Here, the last = c and the rest = 10a+b
According to the rule, let 10a+b+2c = 13x
Hence, 10a+b = 13x- 2 c
Multiplying both sides by 10, we get 100a+10b = 130x - 20c
100a+10b +c = 130x -20c+c = 130x -19c = 19(10x-c)
The given number is divisible by 19.
Deduction of a general formula to find out the divisibility rule for prime numbers.
Let the given number be 100a+10b+c
Here, the last digit = c and the rest = 10a+b
Let us suppose that by adding n times of c to the rest , 10a+b,
the sum is divisible by the Divisor i.e, by the Prime number .
Hence, 10a+b+nc = Dx, where D = Divisor and x = ±1,±2,±3…
Now, 10a+b = Dx – nc
Multiplying both sides by 10, we get, 100a+10b =10Dx – 10nc
100a+10b+c = 10Dx -10nc +c =10Dx +c(1- 10n)
Value of n Procedure Check for
Here, 1st part =
Divisor value of n Procedure
1
11
7
2
subtract
10Dx is divisible by Divisor,
2
7
11
1
subtract
and
2
+
19
13
4
add
if, the 2nd part=
3
+
29
17
5
subtract
c(1-10n) is divisible by Divisor,
3
31
19
2
add
4
+
13
then, the given
23
7
add
4
41
number is divisible by the
29
3
add
5
17
Divisor or the Prime number.
31
3
subtract
7
+
23
In other words, the
37
11
subtract
11
37
condition is 1 – 10n =Dy,
41
4
subtract
13
+
43
43
13
add
where, y =±1,±2,±3,…
14
47
47
14
subtract
Putting the value of y, it is
evident that all prime numbers consisting two digits or more have its last digit as 1,3,7 or9 only.
Given below, are two tables to find out the divisibility rules for the
Prime numbers up to 50.
TABLE-1
TABLE- 2
Example: Find out the divisors of 17017.
Solution:
st
1 step:
2nd step:
3rd step
4th step:
5th step:
Here, the last digit = 7 and rest =1701
LHS
1701
-7
--------1694 /11 = 154 *
-7
------1687 /7 = 241 *
-7
------1680 /31 = n.d
-7
------1673/ 41 =n.d
-7
------1666 /17 =98 *
6th step:
7th step:
Note: n.d means “not divisible”
Ans:
RHS
1701
+7
--------1708
+7
-------1715 /19 =n.d
+7
------1722 /29 =n.d
+7
------1729 /13 =133 *
+7
-----1736
+7
-----1743
+7
-----1750 /23 =n.d
17017 is divisible by7, 11, 13 and 17.
A format can be designed to find out the divisor/ divisors:
The given number = 54321
Last digit = 1
Rest =5432
n=1
Rest
- last digit
------/11
Rest
+last digit
-------
n=2
-------
------/7
/19
n= 3
-------
------/31
/29
n=4
-------
------/41
/13
n=5
-------
------/17
n=7
(multiply by 2)
-------
------/23
n = 11
(multiply by 4)
-------
------/37
n = 13
(multiply by 2)
-------
-------
/43
n = 14
--------
-----/47
An unique application of the divisibility rule
In case of the greater numbers i.e, the numbers consisting more than 5 digits or so, it can be
found out if a particular prime number is a divisor / factor of the same or not.
Example: To find out if 11 is a factor of 153967 ?
Soln:
For 11 , n = - 1 and last digit = 7, rest = 15396
15396
-7
-------15389
-9
-----1529
-9
----143
-3
---11
Ans:
11 is a factor of 153967.
Explanation: If 153967 is divisible by 11, then according to the rule, 15396 – 7 =15389 is
also divisible by 11. Similarlyly, 1538 – 9 = 1529 and 152 – 9 = 143 is
divisible by 11. This explains the procedure of the application.
Example:
Soln:
Ans:
Is 7 a divisor of 4748359 ?
For 7, n= - 2 and last digit =9, rest=474835
474835
-18
---------474817
-14
-------47467
-14
-----4732
-4
---469
-18
---28 which is divisible by 7.
yes, 7 is a divisor of 4748359.
Example: Show that 265943 is divisible by 19.
Soln:
For 19, n = +2 and last digit = 3 , rest=26594
26594
+6
-------26600
+0
-----2660
+0
----266
+12
----38 which is divisible by 19.
Ans:
265943 is divisible by 19 (proved)
The divisibility rule of other prime numbers starting from 53, can be found out
in the same way.