Concept used:

Check the by divisibility rule of 7.

To check if a number is evenly divisible by 7: Take the last digit of the number, double it Then subtract the result from the rest of the number If the resulting number is evenly divisible by 7

Calculation:

1) 519 

If we multiply last digit by 2 we get, 18

Then subtract 18 from 51, we get 33

Which is not divisible by 7

∴  519 is not divisible by 7.

2) 983

If we multiply last digit by 2 we get, 6

Then subtract 6 from 98, we get 92

Which is not divisible by 7

∴  983 is not divisible by 7.

3) 568

If we multiply last digit by 2 we get, 16

Then subtract 16 from 56, we get 40

Which is not divisible by 7

∴  568 is not divisible by 7.

4) 364

If we multiply last digit by 2 we get, 8

Then subtract 8 from 36, we get 28

Which is divisible by 7

∴  364 is divisible by 7.

Divisibility Test for 7

To Find out if a number is divisible by 7 or not, follow these steps:
1. Separate the last digit from the rest of the number. Let us call the rest of the number the truncated number. The truncated number has one less digit than the original number or the previous truncated number.
2. Double the last digit and subtract it from the truncated number.
3. Check if this result is sufficiently small so that you can immediately say if this is divisible by 7. If it is divisible by 7, then so was the original number. If it is not divisible by 7, then neither was the original number.
4. If the number is still too large to visually check if it is divisible, apply this rule over and over again as necessary.

E.g.Check 6132.

The last digit is 2 and the truncated number is 613.

Twice of 2 is 4. So subtract 4 from the truncated number 613 i.e. 613 – 4 = 609.
Again, the last digit is now 9, and the truncated number is 60.

Twice of 9 is 18. Subtract it from the truncated number 60, i.e. 60 – 18 = 42.

Now 42 is small enough to check visually.

We know that 42 is divisible by 7, so we can tell that 6132 is divisible by 7 also.

NOTE:

Explaining this step is long. But actually using it is a very short and time-saving method. This method is especially useful in Geometry and Mensuration problems where the value of π plays an important role.