DIVISIBILITY BY 7 - RULE WITH PROOF, SOLVED EXAMPLES

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Please study
Multiplication Tables
and
Division before Divisibility By 7
They are prerequisites here.

Here we study methods that can be used to determine
whether a number is evenly divisible by Seven.

These are shortcuts for testing a number's factors
without resorting to division calculations.

The rules given below transform a given number's divisibility
by a divisor to a smaller number's divisibility by the same divisor.

If the result is not obvious after applying it once,
the rule should be applied again to the smaller number.

We present the rules with examples, in a simple way,
to follow, understand and apply. Great Deals on School & Homeschool Curriculum Books

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Rule for Divisibility By 7

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.


If you want Proof of Rule for Divisibility By 7 ,
you may see at the bottom of the page.

Example 1 of Rule for Divisibility By 7

Find whether 343 is divisible by 7 or not.

Solution :
Twice the last digit = 2 x 3 = 6; Rest of the number = 34
Subtracting, 34 - 6 = 28 is divisible by 7. ( 28 = 4 x 7)
∴ 343 is divisible by 7. Ans.

Example 2 of Rule for Divisibility By 7

Find whether 8965 is divisible by 7 or not.

Solution :
Twice the last digit = 2 x 5 = 10; Rest of the number = 896
Subtracting, 896 - 10 = 886
To check whether 886 is divisible by 7 :
Twice the last digit = 2 x 6 = 12; Rest of the number = 88
Subtracting, 88 - 12 = 66 is not divisible by 7. ( 66 = 9 x 7 + 3)
∴ 8965 is not divisible by 7. Ans.

Example 3 of Rule for Divisibility By 7

Find whether 49875 is divisible by 7 or not.

Solution :
To check whether 49875 is divisible by 7 :
Twice the last digit = 2 x 5 = 10; Rest of the number = 4987
Subtracting, 4987 - 10 = 4977

To check whether 4977 is divisible by 7 :
Twice the last digit = 2 x 7 = 14; Rest of the number = 497
Subtracting, 497 - 14 = 483

To check whether 483 is divisible by 7 :
Twice the last digit = 2 x 3 = 6; Rest of the number = 48
Subtracting, 48 - 6 = 42 is divisible by 7. ( 42 = 6 x 7 )

∴ 49875 is divisible by 7. Ans.

Example 4 of Rule for Divisibility By 7

Find whether 987651 is divisible by 7 or not.

Solution :
To check whether 987651 is divisible by 7 :
Twice the last digit = 2 x 1 = 2; Rest of the number = 98765
Subtracting, 98765 - 2 = 98763

To check whether 98763 is divisible by 7 :
Twice the last digit = 2 x 3 = 6; Rest of the number = 9876
Subtracting, 9876 - 6 = 9870

To check whether 987 is divisible by 7 :
Twice the last digit = 2 x 7 = 14; Rest of the number = 98
Subtracting, 98 - 14 = 84

To check whether 84 is divisible by 7 :
Twice the last digit = 2 x 4 = 8; Rest of the number = 8
Subtracting, 8 - 8 = 0 is divisible by 7.

∴ 987651 is divisible by 7. Ans.

Example 5 of Rule for Divisibility By 7

Find whether 986953 is divisible by 7 or not.

Solution :
To check whether 986953 is divisible by 7 :
Twice the last digit = 2 x 3 = 6; Rest of the number = 98695
Subtracting, 98695 - 6= 98689

To check Divisibility By 7 of 98689 :
Twice the last digit = 2 x 9 = 18; Rest of the number = 9868
Subtracting, 9868 - 18 = 9850

To check Divisibility By 7 of 985 :Twice the last digit = 2 x5 = 10; Rest of the number = 98
Subtracting, 98 - 10 = 88

To check Divisibility By 7 of 88 :Twice the last digit = 2 x 8 = 16; Rest of the number = 8
Subtracting, 8 - 16 = -8 is not divisible by 7.

∴ 986953 is not divisible by 7. Ans.

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Proof of Rule for Divisibility By 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.)

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

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