Divisibility rules for divisive primes


The study of the methods that can be used to determine whether a number is divisible by other numbers is an important topic in elementary number theory.

These are shortcuts for testing the factors of a number without resorting to division calculations.

The rules transform the divisibility of a given number by a divisor to the divisibility of a smaller number by the same divisor.

If the result is not obvious after applying it once, the rule should be applied again to the smaller number.

In children’s math textbooks, we will usually find the divisibility rules for 2, 3, 4, 5, 6, 8, 9, 11.

Even finding the divisibility rule for 7 in those books is a rarity.

In this article we present the divisibility rules for prime numbers in general and apply them to specific cases, for prime numbers below 50.

We present the rules with examples, in a simple way, to follow, understand and apply.

Divisibility rule for any prime divisor ‘p’:

Consider multiples of ‘p’ until (least multiple of ‘p’ + 1) is a multiple of 10, so that one-tenth of (least multiple of ‘p’ + 1) is a natural number.

Let’s say this natural number is ‘n’.

Thus, n = one tenth of (minimum multiple of ‘p’ + 1).

Find (p – n) too.

Example (me):

Let be the prime divisor 7.

The multiples of 7 are 1×7, 2×7, 3×7, 4×7, 5×7, 6×7,

7×7 (Got it. 7×7 = 49 and 49+1=50 is a multiple of 10).

So ‘n’ for 7 is one tenth of (least multiple of ‘p’ + 1) = (1/10)50 = 5

‘pn’ = 7 – 5 = 2.

Example (ii):

Let be the prime divisor 13.

The multiples of 13 are 1×13, 2×13,

3×13 (Got it. 3×13 = 39 and 39+1=40 is a multiple of 10.)

So ‘n’ for 13 is one tenth of (least multiple of ‘p’ + 1) = (1/10)40 = 4

‘pn’ = 13 – 4 = 9.

The values ​​of ‘n’ and ‘pn’ for other prime numbers below 50 are given below.

pnpn

7 5 2

13 4 9

17 12 5

19 2 17

23 7 16

29 3 26

31 28 3

37 26 11

41 37 4

43 13 30

47 33 14

After finding ‘n’ and ‘p-n’, the divisibility rule is as follows:

To find out if a number is divisible by ‘p’, take the last digit of the number, multiply it by ‘n’ and add it to the rest of the number.

or multiply it by ‘(p – n)’ and subtract it from the rest of the number.

If you get an answer that is divisible by ‘p’ (including zero), then the original number is divisible by ‘p’.

If you don’t know the divisibility of the new number, you can apply the rule again.

So, to form the rule, we have to choose ‘n’ or ‘p-n’.

We usually choose the lower of the two.

With this knowledge, let’s establish the divisibility rule for 7.

For 7, pn (= 2) is less than n (= 5).

Divisibility rule for 7:

To find out if a number is divisible by 7, take the last digit, multiply it by two, and subtract it from the rest of the number.

If you get an answer that is divisible by 7 (including zero), then the original number is divisible by 7.

If you don’t know the divisibility of the new number, you can apply the rule again.

Example 1:

Find if 49875 is divisible by 7 or not.

Solution :

To check if 49875 is divisible by 7:

Twice the last digit = 2 x 5 = 10; Rest of number = 4987

Subtracting, 4987 – 10 = 4977

To check if 4977 is divisible by 7:

Twice the last digit = 2 x 7 = 14; Rest of number = 497

Subtracting, 497 – 14 = 483

To check if 483 is divisible by 7:

Twice the last digit = 2 x 3 = 6; Rest of number = 48

By subtracting, 48 – 6 = 42 is divisible by 7. ( 42 = 6 x 7 )

So, 49875 is divisible by 7. Years.

Now, let’s establish the divisibility rule for 13.

For 13, n (= 4) is less than pn (= 9).

Divisibility rule for 13:

To find out if a number is divisible by 13, take the last digit, multiply it by 4, and add it to the rest of the number.

If you get an answer that is divisible by 13 (including zero), then the original number is divisible by 13.

If you don’t know the divisibility of the new number, you can apply the rule again.

Example 2:

Find if 46371 is divisible by 13 or not.

Solution :

To check if 46371 is divisible by 13:

4 x last digit = 4 x 1 = 4; Rest of number = 4637

Adding up, 4637 + 4 = 4641

To check if 4641 is divisible by 13:

4 x last digit = 4 x 1 = 4; Rest of number = 464

Adding, 464 + 4 = 468

To check if 468 is divisible by 13:

4 x last digit = 4 x 8 = 32; Rest of number = 46

Adding, 46 + 32 = 78 is divisible by 13. ( 78 = 6 x 13 )

(If you want, you can reapply the rule here. 4×8 + 7 = 39 = 3×13)

 

So, 46371 is divisible by 13. Years.

Now let’s state the divisibility rules for 19 and 31.

for 19, n = 2 is more convenient than (p – n) = 17.

So, the divisibility rule for 19 It is as follows.

To find out if a number is divisible by 19, take the last digit, multiply it by 2, and add it to the rest of the number.

If you get an answer that is divisible by 19 (including zero), then the original number is divisible by 19.

If you don’t know the divisibility of the new number, you can apply the rule again.

For 31, (p – n) = 3 is more convenient than n = 28.

So, the divisibility rule for 31 It is as follows.

To find out if a number is divisible by 31, take the last digit, multiply it by 3, and subtract it from the rest of the number.

If you get an answer that is divisible by 31 (including zero), then the original number is divisible by 31.

If you don’t know the divisibility of the new number, you can apply the rule again.

Thus, we can define the divisibility rule for any prime divisor.

The method for finding ‘n’ given above can also be extended to prime numbers above 50.

Before, we close the article, let’s see the proof of the divisibility rule for 7

Divisibility rule test for 7:

Let ‘D’ ( > 10 ) be the dividend.

Let D1 be the units digit and D2 the rest of the number of D.

that is, D = D1 + 10D2

we have to try

(i) if D2 – 2D1 is divisible by 7, then D is also divisible by 7

and (ii) if D is divisible by 7, then D2 – 2D1 is also divisible by 7.

Proof of (me):

D2 – 2D1 is divisible by 7.

So, D2 – 2D1 = 7k where k is any natural number.

Multiplying both sides by 10, we get

10D2 – 20D1 = 70k

Adding D1 to both sides, we get

(10D2 + D1) – 20D1 = 70k + D1

gold (10D2 + D1) = 70k + D1 + 20D1

or D = 70k + 21D1 = 7(10k + 3D1) = multiple of 7.

So, D is divisible by 7. (proved).

Proof of (ii):

D is divisible by 7

So, D1 + 10D2 is divisible by 7

D1 + 10D2 = 7k where k is any natural number.

Subtracting 21D1 from both sides, we get

10D2 – 20D1 = 7k – 21D1

or 10(D2 – 2D1) = 7(k – 3D1)

or 10(D2 – 2D1) is divisible by 7

Since 10 is not divisible by 7, (D2 – 2D1) is divisible by 7. (Prove).

Similarly, we can prove the divisibility rule for any prime divisor.