Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 year ago

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

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  • 1 year ago
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    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.

    • andell1 year agoReport

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

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  • Anonymous
    1 year ago

    n

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