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## Divisibility Rules For Prime Divisors

Learning methods that can be used to determine whether a number is evenly divisible by other numbers is an important topic in early number theory.

These are shorthand methods for evaluating numerical problems without resorting to differential equations.

The rules convert a given number divisible by a divisor to a smaller number divisible by the same divisor.

If the effect is not visible after using it once, the principle should be used again in a small amount.

In children’s math books, we will find the rules for dividing by 2, 3, 4, 5, 6, 8, 9, 11.

Even finding the rule of division by 7, in those books is rare.

In this article, we show the rules for dividing prime numbers in general and use them in certain situations, for prime numbers, less than 50.

We show the rules with examples, in a way that is easy to follow, understand and apply.

**The division 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 (at least a multiple of ‘p’ + 1).

Find (p – n) again.

**Example (i) : **

Let the prime divisor be 7.

The maximum of 7 is 1×7, 2×7, 3×7, 4×7, 5×7, 6×7,

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

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

‘pn’ = 7 – 5 = 2.

**Example(s) : **

Let the prime divisor be 13.

The maximum of 13 is 1×13, 2×13,

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

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

‘pn’ = 13 – 4 = 9.

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

pn pn

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 division 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 other number.

or multiply it by ‘(p – n)’ and subtract from the whole 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 new partition number, you can use the rule again.

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

Usually, we choose the lower of the two.

With this information, let’s define the rule of division by 7.

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

**Rule 7 Breakdown:**

To find out, if a number is divisible by 7, take the last digit, multiply it by two, and subtract it from the other 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 new partition number, you can use the rule again.

**Example 1 :**

Find whether 49875 is divisible by 7 or not.

*Solution : *

To check if 49875 is divisible by 7:

Twice the last number = 2 x 5 = 10; Remaining number = 4987

Subtract, 4987 – 10 = 4977

To check if 4977 is divisible by 7:

Twice the last number = 2 x 7 = 14; Remaining number = 497

Subtract, 497 – 14 = 483

To check if 483 is divisible by 7:

Twice the last number = 2 x 3 = 6; Remaining number = 48

To subtract, 48 – 6 = 42 is divided by 7. (42 = 6 x 7)

So, 49875 is divisible by 7.

Now, let’s explain the division rule of 13.

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

**Rule 13 Division:**

To find out, if a number is divisible by 13, take the last digit, multiply it by 4, and add it to the other 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 new partition number, you can use the rule again.

**Example 2 :**

Find whether 46371 is divisible by 13 or not.

*Solution : *

To check if 46371 is divisible by 13:

4 x last digit = 4 x 1 = 4; Remaining number = 4637

In addition, 4637 + 4 = 4641

To check if 4641 is divisible by 13:

4 x last digit = 4 x 1 = 4; Remaining number = 464

In addition, 464 + 4 = 468

To check if 468 is divisible by 13:

4 x last digit = 4 x 8 = 32; Remaining number = 46

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

(if you want, you can use the rule again, here. 4×8 + 7 = 39 = 3 x 13)

So, 46371 is divisible by 13.

Now let’s explain the division rules of 19 and 31.

in 19, n = 2 is better than (p – n) = 17.

As for, **19 division rule **it is as follows.

To find out, if a number is divisible by 19, take the last number, multiply it by 2, and add it to the other 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 new partition number, you can use the rule again.

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

As for, **Rule of division 31 **it is as follows.

To find out, if a number is divisible by 31, take the last number, multiply it by 3, and subtract it from the other 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 new partition number, you can use the rule again.

In this way, we can define a division rule for any prime divisor.

The method of finding ‘n’ given above can be extended to higher numbers above 50 again.

Before, we close the article, let’s see the proof of the Law of Dividing by 7

**Proof of the Law of Separation in 7:**

Let ‘D’ ( > 10) be the reward.

Let D1 be the units digit and D2 be the remainder of D.

ie D = D1 + 10D2

We have to prove it

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

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

*Proof of (i) : *

D2 – 2D1 is divisible by 7.

Therefore, D2 – 2D1 = 7k where uk 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

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

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

Therefore, D is divisible by 7. (shown.)

*Evidence of (ii) : *

UD is divisible by 7

So, D1 + 10D2 is divisible by 7

D1 + 10D2 = 7k where uk 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) divided by 7

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

In a similar way, we can prove the division rule for any prime divisor.

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