# Is 4 9 Short For A 13 Year Old Boy Divisibility Rules For Prime Divisors

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## Divisibility Rules For Prime Divisors

The study of the methods that can be used to determine whether a number is equally 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 fraction into the divisibility of a smaller number by the same divisor.

If the result is not obvious after a single application, the rule should be applied again to a smaller number.

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

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

In this article, we present general rules of divisibility of prime numbers and apply them to special 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 one tenth of (least multiple of ‘p’ + 1) is a natural number.

Let’s say this natural number is ‘n’.

So n = one tenth (the smallest multiple of ‘p’ + 1).

Find (p – n) also.

Examples):

Let the prime divisor be 7.

More than 7 are 1×7, 2×7, 3×7, 4×7, 5×7, 6×7,

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

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

‘pn’ = 7 – 5 = 2.

Example (ii):

Let the prime divisor be 13.

Multiples of 13 are 1×13, 2×13,

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

So ‘n’ for 13 is one tenth (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.

mon mon

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:

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

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

If you get an answer 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 a rule, we have to choose either ‘n’ or ‘p-n’.

We usually choose the lower of the two.

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

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

Divisibility rule by 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 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 whether 49875 is divisible by 7 or not.

Solution:

To check if 49875 is divisible by 7:

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

Subtracting, 4987 – 10 = 4977

To check if 4977 is divisible by 7:

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

Subtracting, 497 – 14 = 483

To check if 483 is divisible by 7:

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

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

So 49875 is divisible by 7. Ans.

Now, state the divisibility rule for 13.

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

Divisibility rule by 13:

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

If you get an answer 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 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; Remainder of the number = 4637

Adding, 4637 + 4 = 4641

To check if 4641 is divisible by 13:

4 x last digit = 4 x 1 = 4; Remainder of the number = 464

Adding, 464 + 4 = 468

To check if 468 is divisible by 13:

4 x last digit = 4 x 8 = 32; Remainder of the number = 46

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

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

So 46371 is divisible by 13. Ans.

Now state the divisibility rules for 19 and 31.

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

so, divisibility rule for 19 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 remainder of the number.

If you get an answer 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, divisibility rule for 31 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 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.

This way, we can define a divisibility rule for any prime divisor.

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

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

Divisibility Proof Rule for 7:

Let ‘D’ ( > 10 ) be the dividend.

Let D1 be the unit 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 also divisible by 7

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

Evidence (i):

D2 – 2D1 is divisible by 7.

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

We multiply 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. (proved.)

Evidence (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. (Proved.)

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

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