What is the divisibility rule for seven?

Answer 1

To find out, if a number is divisible by 7, take the last digit, double it, 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 new number's divisibility, you can apply the rule again.

Eg: Take the number 343

The units digit is 3.

Doubling this 3,we get 6.

Subtracting 3 from 34(that is the remaining 2 digits in the number), we get 34-6=28.

As 28 is divisible by 7, The whole number, i.e., 343 is divisible by 7.

Proof of Divisibilty Rule for 7

Let 'D' ( > 10 ) be the dividend.

Let D1 be the units' digit

and D2 be the rest of the number of D.

i.e. 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 ⇒ 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

⇒ (10D2 + D1) = 70k + D1 + 20D1

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

⇒ D is divisible by 7. (proved.)

Proof of (ii) :

D is divisible by 7 ⇒ 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

⇒ 10(D2 - 2D1) = 7(k - 3D1)

⇒ 10(D2 - 2D1) is divisible by 7

Since 10 is not divisible by 7,

(D2 - 2D1) is divisible by 7. (proved.)