Divisibility Rule for 7:
- Take the last (rightmost) digit,
- double it,
- Subtract it from the rest (or vice versa),
- check whether the result is divisible by 7,
- if yes, the original is divisible by 7. (This may have to be done a few rounds)
Why it works? The concept is explained below.
Let the number be "ab". It can be re-written as 10a + b.
We doubled the last digit"b" ==> 2b. For what?
Let's deviate a bit.
If a number is divisible by 7, doubling it will also be divisible by 7.
Example: 21 /7 = 3, doubling 21 ==> 42 / 7 = 6 (still divisible after 21 is doubled)
Therefore 10a + b ==> doubling it ==> becomes 20a + 2b.
By the rule stated at the top, we need to subtract "2b" from "a" (the rest of the digit besides the first digit), that is, a - 2b, for divisibility checking.
This means the 20a + 2b need to be negated and to be added to 21a to achieve the targetted a - 2b.
Result become, -(20a + 2b) + 21a ==> -20a -2b + 21a ==> a -2b.
Note now, the added 21a is definitely divisible by 7, therefore, leaving -(20a + 2b) as the deciding factor for divisibility consideration.
If -(20a + 2b) is divisible by 7, then it implies that the a - 2b is also divisible by 7!
Why go through such hassle of doing a - 2b when we can just double the original number, that is, (20a + 2b) of above and get the answer?
The reason is that by doubling the original number, we will get an even larger number that the calculator cannot handle or display. Therefore the rule using a - 2b has to be used since the number of digits is reduced from original. Clear?
For Concept on Divisibility by:
5 and 6 Click here
3 and 4 Click here
2 Click here
11 Click here