I mean how to check whether a number is divisible by 7 or
not. So here I’m showing you the rule.

1.
Take the last digit of the given number.

2.
Double it.

3.
Subtract it from the rest of the number.

4.
If the answer is more than two digit number,
perform the above again of that new number.

5.
So if the result is 0 or is divisible by 7 then
the original number is also divisible by 7.

Now I’m showing you the above operation by taking two
examples.

Ex-1:

*322*
So by the above rule.

1.
Take the last digit of the given number.

**(Last digit in 322 is 2)**
2.
Double it.

*(when doubling the last digit 2 x2 =4)*
3.
Subtract it from the rest of the number.

*(Rest of the number is 32 so, 32-4=28)*
4.
If the answer is more than two digit number,
perform the above again.

*(This step is not needed in this as the subtracted value is only 2 digit number)*
5.
So if the result is 0 or is divisible by 7 then
the original number is also divisible by 7.

*(Yes the number 28 is divisible by 7 as we know. So the number 322 is also divisible by 7).*
Ex-2:

*2198*
Again from the above rule.

1.
Take the last digit of the given number.

*(Last digit in 2198 is 8)*
2.
Double it.

*(8x2=16)*
3.
Subtract it from the rest of the number.

*(219-16=203)***If the answer is more than two digit number, perform the above again.**

*4.*

*(Yes answer is more than two digit number i.e. 203, so we will again perform step 1, 2 and 3.)*

*So now the new number is 203, in 203 the last digit is 3.*

*3x2=6*

*20-6=14*
5.
So if the result is 0 or is divisible by 7 then
the original number is also divisible by 7.

*(So now we got two digit numbers i.e. 14 which we know is divisible by 7, since the number 2198 is also divisible by 7.)*
Now you can check the rules by
taking different numbers.

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