How to find divisibility test for a prime number?
Divide 100 , 1000 , 10000 by the prime number. Note quotients and remainders.
Select that set of where the remainder and quotients are both small numbers.It should be preferably single digit.
Suppose we want t find divisibility test for 13. (1) 100/13 = 7; remainder 9. 13 x 7 + 9 = 100.
(2) 1000/13=76; remainder 12 or 13 x 76 +12 = 1000 (3) 10000/13=769; remainder 3
Best selection will be 7 & 9. i.e. for 100.
any natural number can be expressed as: 1000a+100b+10c+d
Now 910a+90a+91b+9b+10c+d.
(1) Since we selected the pair of quotient and remainder for 100; leave aside last 2 digits(= no. of zeros to the right of 100).
(2) 90a+9b=9(10a+b). So multiply the number left after leaving aside last 2 digits, by 9 (remainder)
(3) Divide the sum by 13. Note Quotient. If remainder is zero the number (dividend) under consideration is divisible by 13.
(4) We know 910a+91b will be divisible by 13.and answer of division will be (a) Quotient= 7x(10a+b) i.e. 7 x the left number after leaving aside last 2 digits; (b) remiander zero since 91=13 x 7. 7 is fixed part of test.
(5) Add the quotients in step (3) & (4). This is the quotient when the dividend is divided by 13.
(6) Note the test can be used as another method for division. In this method only advantage is that dividend has 2 digits less than the original dividend.
Divide 100 , 1000 , 10000 by the prime number. Note quotients and remainders.
Select that set of where the remainder and quotients are both small numbers.It should be preferably single digit.
Suppose we want t find divisibility test for 13. (1) 100/13 = 7; remainder 9. 13 x 7 + 9 = 100.
(2) 1000/13=76; remainder 12 or 13 x 76 +12 = 1000 (3) 10000/13=769; remainder 3
Best selection will be 7 & 9. i.e. for 100.
any natural number can be expressed as: 1000a+100b+10c+d
Now 910a+90a+91b+9b+10c+d.
(1) Since we selected the pair of quotient and remainder for 100; leave aside last 2 digits(= no. of zeros to the right of 100).
(2) 90a+9b=9(10a+b). So multiply the number left after leaving aside last 2 digits, by 9 (remainder)
(3) Divide the sum by 13. Note Quotient. If remainder is zero the number (dividend) under consideration is divisible by 13.
(4) We know 910a+91b will be divisible by 13.and answer of division will be (a) Quotient= 7x(10a+b) i.e. 7 x the left number after leaving aside last 2 digits; (b) remiander zero since 91=13 x 7. 7 is fixed part of test.
(5) Add the quotients in step (3) & (4). This is the quotient when the dividend is divided by 13.
(6) Note the test can be used as another method for division. In this method only advantage is that dividend has 2 digits less than the original dividend.
Option 2:
You also can find divisibility test for any odd prime number as follows:
Step 1: Find at which multiple of the unit place digit of that prime number, you will get 1 at the right-most place.
Ex. 37.
7x3 = 21.
Multiply 37 by 3
37 x 3 = 111. 11 | 1.
Step2: So the test for 37 is multiply the unit place digit of the given number by 11 and find its difference with rest of the number.
Step 3: If this number is difficult to divide, continue the step 2 till it is easy to divide.
Ex. 123
12 | 3
so , 3 x 11 = 33
33 - 12 = 21.
21 is not divisible by 37 so 123 is not divisible by 37.
