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
- 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