Best Way To Teach 3 Year Old To Do Math 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 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 into 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 these 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 that 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 of (minimum multiple of ‘p’ + 1).

Find (p – n) as well.

Example (i):

Let the prime divisor be 7.

Multiples of 7 are 1×7, 2×7, 3×7, 4×7, 5×7, 6×7,

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

So “n” for 7 is one tenth of (minimum multiple of “p” + 1) = (1/10)50 = 5

‘pn’ = 7 – 5 = 2.

Example (ii):

Let the prime divisor be 13.

The multiples of 13 are 1×13, 2×13,

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

So “n” for 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 prime numbers less than 50 are given below.

bp bp

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 “pn”, 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 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 the rule, we have to choose ‘n’ or ‘p-n’.

Usually, we choose the lower of the two.

With this knowledge, we will enunciate the rule of divisibility by 7.

For 7, pn (= 2) is less than an (= 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 remainder 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; The rest of the number = 4987

Remaining, 4987 – 10 = 4977

To check if 4977 is divisible by 7:

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

Remaining, 497 – 14 = 483

To check if 483 is divisible by 7:

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

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

Thus 49875 is divisible by 7. Ans.

Now, let’s state 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 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; The rest of the number = 4637

Sum, 4637 + 4 = 4641

To check if 4641 is divisible by 13:

4 x last digit = 4 x 1 = 4; The rest of the number = 464

Add up, 464 + 4 = 468

To check if 468 is divisible by 13:

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

Adding, 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)

 

Thus 46371 is divisible by 13. Ans.

Now we 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 is the following.

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 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 is the following.

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

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 greater than 50.

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

Divisibility test rule for 7:

Let ‘D’ (> 10) be the dividend.

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

that is, D = D1 + 10D2

We have to prove

(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 (i):

D2 – 2D1 is divisible by 7.

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

Therefore, 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. (proved.)

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

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