# Help with bonus question?

I reuploaded this question because the formula was really hard to understand I hope this is better. And it's due tomorrow please help!!

p= [10^(2n)-1]/ [3(10^(n)+1)]

-n is a positive integer

-the sum of the digits of the number p is 567

Find n

### 2 Answers

- az_lenderLv 71 year agoFavorite Answer
Good, much better typing !! That in itself is a good lesson.

p = [10^(2n) - 1] / [3(10^n + 1)].

Let's see, if n = 1, you get

99/33 = 3. Obviously sum of digits in not 567!

If n=2 you get

9999/303 = 33. Again the sum of the digits is not 567.

Hmm, as I detect a pattern, I may try ALGEBRA...

Note that [10^(2n) - 1]/[10^n + 1] = 10^n - 1, that's "Algebra 2."

SO, apparently p = (10^n - 1)/3.

Whose digits would all be 3's.

How many 3's are in 567? I guess 189 threes.

So that's 10^190 - 1.

n = 190.

Great, glad you typed it correctly this time.

- Anonymous1 year ago
n

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This is a pretty good explanation, except the answer is 189 not 190