Aptitude - Numbers - Discussion

There is a way to check if a number is divisible by 7 or not.
Take 555681.

Step i ; 2x last number. i.e.: 2(1).
Step ii ; now subtract (step i ) from left out number ..ie : 55568 - 2(1) = 55566.

Check the new number for divisibility by 7.

If you are not sure, then repeat the steps for 55566.

5556- 2(6) = 5556-12= 5544 not sue, then repeat.
554- 2(4) = 554-8 = 546 again.
54-2(6) = 54- 12 = 42 which is obviously divisible by 7 if you are not yet satisfied.
4-2(4) = 4-4 = 0.

If the number we were checking didn't end up t zero going through this process, then it's not divisible by 7.

The LCM OF 987 is 3, 7, 47.
So, it means that the number 555681 will be divisible by 3, 7, 47.

So according to the divisibility rule of 3.
If the sum of all the digits in a number is divisible by 3 then the number will divide by 3.

So, let's see 5 + 5 + 5 + 6 + 8 + 1 = 30.
So, 30/3=10 which means the number 555681 will divide by 3.

And there is no divisibility rule for 7 and 47 so you have to divide the number by 7 to get the result.

Let's check 555681/7=79383.
And for 555681/47=11823.

Here is the another method. From the answer you could say last two digits are correct so you can assume that the number would be three digits and last two digits are 6 and 3. By iterations you can calculate 100th place digit also the number would be 563.

The sum of the digits in answer results in 30.

How is it divisible by 7? Option D also results in same sum 30. Then option D does not answer. Please explain. I'm unable to get it.

We can solve this question by dividing each option by 987.

As we can see only option C is dividing by 987.

i.e) 555681/987 = 563.

So option C is the correct answer.

The option C and D are both divisible by 3, then how if the only option C will be right?

5+5+5+6+8+1 = 30 then how can it divisible by 7 and 47?

Can anyone explain me please?

@Nidhi.

There are the Factors of the 987 by prime factorization.

Please explain me "the formula for the divisible of 47 & 7.

I'am not getting. Can anyone please explain me briefly?