For The Quadratic Formula Does A Have To Be 1 Divisibility Rules For Prime Divisors

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Divisibility Rules For Prime Divisors

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

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

The rules change the divisibility of a given number by the divisor into the divisibility of a smaller number by the same divisor.

If there is no result after applying once, the rule should be applied again in a small number.

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

It is also rare to find the divisibility rule for 7 in those books.

In this article, we introduce divisibility rules for prime numbers in general and apply them to special cases for prime numbers less than 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’ (least multiple of ‘p’ + 1) to be a multiple of 10, so that one tenth (least multiple of ‘p’ + 1) is a natural number.

Let it be a natural number ‘n’.

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

Also find (p – n).

Example (i):

Let the prime divisor be 7.

Multiply 7 by 1×7, 2×7, 3×7, 4×7, 5×7, 6×7,

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

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

‘pn’ = 7 – 5 = 2.

Example (ii):

Let the prime divisor be 13.

13 is a product of 1×13, 2×13,

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

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

‘pn’ = 13 – 4 = 9.

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

pn pn

7 5 2

1349

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 ‘pn’, the divisibility rule is as follows:

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

Or multiply it by ‘(p – n)’ and subtract it from the remaining number.

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

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

So to make a rule, we have to choose ‘n’ or ‘pn’.

Usually, we choose the lower of the two.

With this knowledge, let us state the division rule for 7.

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

Divisibility Rule of 7:

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

If you get an answer that is divisible by 7 (including zero), 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 whether 49875 is divisible by 7:

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

Subtracting, 4987 – 10 = 4977

To check if 4977 is divisible by 7:

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

Subtracting, 497 – 14 = 483

To check if 483 is divisible by 7:

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

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

So, 49875 is divisible by 7. Answer.

Now, state the division rule for 13.

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

Divisibility rule for 13:

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

If you get an answer that is divisible by 13 (including zero), 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; Remaining number = 4637

Adding, 4637 + 4 = 4641

To check if 4641 is divisible by 13:

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

Adding, 464 + 4 = 468

To check if 468 is divisible by 13:

4 x last digit = 4 x 8 = 32; Remaining 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 x 13)

 

So, 46371 is divisible by 13. Answer.

Now state the divisibility rules of 19 and 31.

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

but, Divisibility rule for 19 is as follows.

To find out, divide a number by 19, take the last digit, multiply it by 2, and add it to the remainder.

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

but, Division rule for 31 is as follows.

To find out, divide a number by 31, take the last digit, multiply it by 3, and subtract it from the remainder.

If you get an answer that is divisible by 31 (including zero), 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 above method of finding ‘n’ can be extended to prime numbers above 50 as well.

Before closing the article, let’s look at the proof of the divisibility rule of 7

Proof of the divisibility rule for 7:

Consider ‘D’ (> 10) as dividend.

Let D1 be the unit digit and let D2 be the remainder of D.

That is 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.

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

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

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

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

(ii) Evidence of:

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. (Attested.)

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

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