# How to determine if a large number (say in quintillion, 19 digits) is perfectly divisible by 999 without using the process of division?

### 5 Answers

- 4 weeks agoFavorite Answer
999999 is divisible by 999

999999999 is divisible by 999

In general

10^(3t) - 1 is divisible by 999

So let's say you have a number like 1,378,492,265,446,978,132. Is it divisible by 999? Well, we know that 999,999,999,999,999,999 is divisible by 999, so we can start there

1,378,492,265,446,978,132 - 999,999,999,999,999,999 =>

378,492,265,446,978,133

999,999,999,999,999 * 378 = 378,000,000,000,000 - 378

378,492,265,446,978,133 - (378,000,000,000,000 - 378) =>

492,265,446,978,133 + 378 =>

492,265,446,978,511

999,999,999,999 * 492 = 492,000,000,000,000 - 492

265,446,978,511 + 492 = 265,446,979,003

Do you see a pattern emerging?

1 , 378 , 492 , 265 , 446 , 978 , 132

Break it up and add each 3 digit section

1 + 378 + 492 + 265 + 446 + 978 + 132 =>

379 + 757 + 1424 + 132 =>

2692

2 , 692

692 + 2 = 694

694 is not divisible by 999, so 1,378,492,265,446,978,132 is not divisible by 999.

Let's use a number that is divisible by 999 and see if it works:

1 , 488 , 038 , 328 , 957 , 537 , 648

1 + 488 + 038 + 328 + 957 + 537 + 648 = 2 , 997

2 + 997 = 999

Looks like it works.

- geezerLv 74 weeks ago
Add up the digits .. if that is divisable by 3 then the number is divisable by 999

- ?Lv 74 weeks ago
If the sum of the digits are divisible by 3, then the large number is divisible by 999.

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- smallLv 74 weeks ago
I have a simple process which uses only the addition process to get the answer and it would also determine the remnant if the large number is not perfectly divisible by 999.... and even the process of addition needs to be applied not more than twice.

so you are answering your own question?